qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
141347 | Finding [imath]\int x^xdx[/imath]
I'm trying to find [imath]\int x^x \, dx[/imath], but the only thing I know how to do is this: Let [imath]u=x^x[/imath]. [imath]\begin{align} \int x^x \, dx&=\int u \, du\\[6pt] &=\frac{u^2}{2}\\[6pt] &=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &=\frac{x^{2x}}{2} \end{align}[/imath] But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong. | 407069 | Integrating [imath]x^x[/imath] and getting a graph
I've heard many times of functions that cannot be integrated. For example, [imath]x^x[/imath], which is the most common. But what I don't know is how could you, even if the graph has no equation, plot this integral. If anyone could give me a graph, I would be extremely pleased. Or at least, could you tell me how to graph it? Maybe in Wolfram Mathematica? |
119904 | Units and Nilpotents
If [imath]ua = au[/imath], where [imath]u[/imath] is a unit and [imath]a[/imath] is a nilpotent, show that [imath]u+a[/imath] is a unit. I've been working on this problem for an hour that I tried to construct an element [imath]x \in R[/imath] such that [imath]x(u+a) = 1 = (u+a)x[/imath]. After tried several elements and manipulated [imath]ua = au[/imath], I still couldn't find any clue. Can anybody give me a hint? | 325318 | Prove that if matrix [imath]A[/imath] is nilpotent, then [imath]I+A[/imath] is invertible.
So my friend and I are working on this and here is what we have so far. We want to show that [imath]\exists \, B[/imath] s.t. [imath](I+A)B = I[/imath]. We considered the fact that [imath]I - A^k = I[/imath] for some positive [imath]k[/imath]. Now, if [imath]B = (I-A+A^2-A^3+ \cdots -A^{k-1})[/imath], then [imath](I+A)B = I-A^k = I[/imath]. My question is: in matrix [imath]B[/imath], why is the sign for [imath]A^{k-1}[/imath] negative? Couldn't it be positive, in which case we'd get [imath](I+A)B = I + A^k[/imath]? Thank you. |
55679 | Weak-to-weak continuous operator which is not norm-continuous
Can one give a "relatively easy" example of a linear mapping [imath]T\colon X\to X[/imath] ([imath]X[/imath] a Banach space) which is a) weak-to-weak continuous b) weak*-to-weak* continuous ([imath]X=Y^*[/imath]) but not norm-to-norm continuous (not bounded). This needs some choice I guess. | 2480904 | Endomorphisms in the weak-* topology
Let [imath](X,\|\cdot\|)[/imath] be a Banach space and [imath](X^*,\|\cdot\|_{op})[/imath] its continuous dual Banach space. Let [imath]\text{End}_{op}(X^*)[/imath] be the continuous linear endomorphisms of [imath]X^*[/imath] with respect to [imath]\|\cdot\|_{op}[/imath] and [imath]\text{End}_*(X^*)[/imath] the continuous linear endomorphisms with respect to the weak-* topology of [imath]X^*[/imath]. Does [imath]\text{End}_{op}(X^*)=\text{End}_*(X^*)[/imath]? If not when are they equal and how do they compare in general? Also references and proofs are welcome. |
143222 | What does [imath]dx[/imath] mean?
[imath]dx[/imath] appears in differential equations, such us derivatives and integrals. For example, a function [imath]f(x)[/imath] its first derivative is [imath]\dfrac{d}{dx}f(x)[/imath] and its integral [imath]\displaystyle\int f(x)dx[/imath]. But I don't really understand what [imath]dx[/imath] is. | 433172 | What's [imath]dy[/imath] and [imath]dx[/imath], in differential equations?
I can't understand why we multiply by [imath]dx[/imath] in differential equations to integrate? People say so that we know what are we integrating with respect to, but this is obviously not true right? What does [imath]\;x\,dx = (y-y^2)\,dy\;[/imath], for example, mean? If [imath]dx[/imath] or [imath]dy[/imath] are very very small with a limit that goes to zero how can we multiply by zero? What's a differential? This is not an exact duplicate. I still don't know what a differential is. |
242779 | Limit of [imath]L^p[/imath] norm
Could someone help me prove that given a finite measure space [imath](X, \mathcal{M}, \sigma)[/imath] and a measurable function [imath]f:X\to\mathbb{R}[/imath] in [imath]L^\infty[/imath] and some [imath]L^q[/imath], [imath]\displaystyle\lim_{p\to\infty}\|f\|_p=\|f\|_\infty[/imath]? I don't know where to start. | 388227 | Finding [imath]\lim\limits_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}[/imath] for continuous [imath]f:[0,1]\to[0,\infty)[/imath]
Find [imath]\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}[/imath]if [imath]f:[0,1]\rightarrow(0,\infty)[/imath] is a continuous function. My attempt: Say [imath]f(x)[/imath] has a max. value [imath]M[/imath]. Then [imath]\left(\int_0^1(f(x))^ndx\right)^\frac{1}{n}\leq\left(\int_0^1M^ndx\right)^\frac{1}{n} =M[/imath] I cannot figure out what to do next. |
28905 | Expected time to roll all 1 through 6 on a die
What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die? The order in which the numbers appear does not matter. I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation. We were given the answer [imath](1-(\frac56)^n)^6 = .5[/imath] or [imath]n = 12.152[/imath] Can someone please explain this to me, possibly with a link to a general topic? | 459374 | How many times must you roll a die until each side has appeared?
Let [imath]X[/imath] be the random variable which denotes the number of times a die has been rolled till each side has appeared. The order does not matter. We are trying to find [imath]E[X][/imath]. Let [imath]X_i[/imath] be a random variable which denotes how many times a die has to be rolled till side i has appeared. So, [imath]E[X]= E[X1+X2+X3+X4+X5+X6] = E[X1]+E[X2]+E[X3]+E[X4]+E[X5]+E[X6][/imath] [imath]E[X1]=E[X2]=E[X3]=E[X4]=E[X5]=E[X6]=6[/imath] [imath]E[X]=36[/imath]? Why is this solution wrong? |
155 | How can you prove that a function has no closed form integral?
I've come across statements in the past along the lines of "function [imath]f(x)[/imath] has no closed form integral", which I assume means that there is no combination of the operations: addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions , which when differentiated gives the function [imath]f(x)[/imath]. I've heard this said about the function [imath]f(x) = x^x[/imath], for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called? Merged with "How to prove that some functions don't have a primitive" by Ismael: Sometimes we are told that some functions like [imath]$\dfrac{\sin(x)}{x}$[/imath] don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion? | 659777 | What's wrong with [imath]cos(t) /t[/imath] and [imath]e^{-t^2}[/imath]?
Usually whenever these functions pop up in the computation of a proper integral, one is stucked finding their antiderivative. Has it been proved that their antiderivative has no closed form in terms of a polynomial in [imath]sin, cos, tan, exp, ln,... [/imath]? EDIT: the question is indeed a duplicate. For French speakers, this issue is completely assessed here 1995 Competitive exam |
20969 | Prove [imath]0! = 1[/imath] from first principles
How can I prove from first principles that [imath]0![/imath] is equal to [imath]1[/imath]? | 226449 | Why [imath]0![/imath] is equal to [imath]1[/imath]?
Many counting formulas involving factorials can make sense for the case [imath]n= 0[/imath] if we define [imath]0!=1 [/imath]; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question: If [imath]A[/imath] is an associative and commutative ring, then we can define an unary operation on the set of all the finite subsets of our ring, denoted by [imath]+ \left(A\right) [/imath] and [imath]\times \left(A\right)[/imath]. While it is intuitive to define [imath]+ \left( \emptyset \right) =0[/imath], why should the product of zero number of elements be [imath]1[/imath]? Does the fact that [imath]0! =1[/imath] have anything to do with 1 being the multiplication unity of integers? |
239566 | Subset of a finite set is finite
We define [imath]A[/imath] to be a finite set if there is a bijection between [imath]A[/imath] and a set of the form [imath]\{0,\ldots,n-1\}[/imath] for some [imath]n\in\mathbb N[/imath]. How can we prove that a subset of a finite set is finite? It is of course sufficient to show that for a subset of [imath]\{0,\ldots,n-1\}[/imath]. But how do I do that? | 732374 | How do i prove that a subset of a finite subset is finite?
Let [imath]n\in\omega[/imath] and [imath]F\subset n[/imath]. How do i prove that [imath]F[/imath] is finite? I know that this can be proven by constructing a function in a following way: take [imath]f(0)[/imath] as the least element of [imath]F[/imath] and then take [imath]f(1)[/imath] as the least element of [imath]F\setminus \{f(0)\}[/imath] and continue this process. Since [imath]F\subset n[/imath], [imath]F[/imath] has a maximal element. Thus this process must be done in a finite step. However, i don't know how to prove this precisely. How do i write down "this process must be done in a finite step" in first-order logic and so on. Please help :) |
146477 | How can I determine the number of unique hands of size H for a given deck of cards?
I'm working on a card game, which uses a non-standard deck of cards. Since I'm still tweaking the layout of the deck, I've been using variables as follows: Hand size: [imath]H[/imath] Number of suits: [imath]S[/imath] Number of ranks: [imath]R[/imath] Number of copies of each card: [imath]C[/imath] Thus, the total number of unique cards in the deck is [imath]S*R[/imath], and the absolute total number of cards in the deck is [imath]S*R*C[/imath]. Since there are duplicates for each card, I'm trying to find the number of unique hands of size [imath]H[/imath] that are possible for given [imath]S[/imath], [imath]R[/imath], [imath]C[/imath], and [imath]H[/imath]. If I'm remembering correctly, [imath]{S*R*C\choose H}[/imath] would over count the number of unique hands in this case. How would I handle this calculation with duplicate cards? EDIT: As an aside, how would calculations of unique number of winning hands be different with duplicate cards? I'm thinking of having general types of winning hands, such as [imath]N[/imath]-flushes, [imath]N[/imath]-straights, [imath]N[/imath]-of-a-kinds, etc (restricted to [imath]1 \le N \le H[/imath], naturally). | 394848 | Partition integer into n parts, with constraint on each part
[imath]x_1,x_2,...,x_n[/imath] are integer numbers in the range [0,B-1]. Count the number of solution for [imath]x_1+x_2+...+x_n=k[/imath]. I know this problem is similar to the one here Number of ways of partitioning a sum into ordered non-negative summands But now there is constraint on the range of [imath]x_i[/imath]. In the problem text it gives a hint to use principle of inclusion and exclusion. Does anyone have a clue? |
19536 | If [imath]|f(z)|\lt a|q(z)|[/imath] for some [imath]a\gt 0[/imath], then [imath]f=bq[/imath] for some [imath]b\in \mathbb C[/imath]
If [imath]q\colon\mathbb{C}\to\mathbb{C}[/imath] is a polynomial, [imath]f\colon\mathbb{C}\to\mathbb{C}[/imath] is analytic on all of [imath]\mathbb{C}[/imath], and if there exists [imath]a\gt 0[/imath] such that [imath]|f(z)| \lt a|q(z)|[/imath] for every [imath]z\in \mathbb{C}[/imath], then [imath]f = bq[/imath] for some [imath]b\in \mathbb{C}[/imath]. Can an arbitrary analytic function (on all of [imath]\mathbb{C}[/imath]) replace [imath]q[/imath]? | 52121 | Property of Entire Functions
Suppose [imath]f[/imath] and [imath]g[/imath] are entire functions with [imath]|f(z)|\leq|g(z)|[/imath] for all [imath]z[/imath]. How can we show that [imath]f=cg[/imath] for some complex constant [imath]c[/imath]? Thanks for any help :) |
155166 | Finite Sum of Power?
Can someone tell me how to get a closed form for [imath]\sum_{k=1}^n k^p[/imath] For [imath]p = 1[/imath], it's just the classic [imath]\frac{n(n+1)}2[/imath]. What is it for [imath]p > 1[/imath]? | 326663 | How to calculate [imath] 1^k+2^k+3^k+\cdots+N^k [/imath] with given values of [imath]N[/imath] and [imath]k[/imath]?
Here [imath] 1<N<10^9[/imath] and [imath]0<k<50[/imath] So we have to calculate it in order of [imath]O(\log N)[/imath]. |
30040 | Limits: How to evaluate [imath]\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x[/imath]
This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates. What methods can be used to evaluate the limit [imath]\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.[/imath] In other words, if I am given a polynomial [imath]P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0[/imath], how would I find [imath]\lim_{x\rightarrow\infty} P(x)^{1/n}-x.[/imath] For example, how would I evaluate limits such as [imath]\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x[/imath] or [imath]\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.[/imath] | 359441 | How do I find this limit: [imath]\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2[/imath]
[imath] \lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2 [/imath] The answer is [imath] \frac{-3}{2} [/imath] according to Wolfram alpha. |
73550 | The limit of truncated sums of harmonic series, [imath]\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}[/imath]
What is the sum of the 'second half' of the harmonic series? [imath]$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$[/imath] More precisely, what is the limit of the above sequence of partial sums? | 326545 | Limit of definite sum equals [imath]\ln(2)[/imath]
I have to show the following equality: [imath]\lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=\log(2)[/imath] I've been playing with it for almost an hour, mainly with the taylor expansion of [imath]\ln(2)[/imath]. It looks very similar to what I need, but it has an alternating sign which sits in my way. Can anyone point me in the right direction? |
2746 | Is there an elementary proof that [imath]\sum \limits_{k=1}^n \frac1k[/imath] is never an integer?
If [imath]n>1[/imath] is an integer, then [imath]\sum \limits_{k=1}^n \frac1k[/imath] is not an integer. If you know Bertrand's Postulate, then you know there must be a prime [imath]p[/imath] between [imath]n/2[/imath] and [imath]n[/imath], so [imath]\frac 1p[/imath] appears in the sum, but [imath]\frac{1}{2p}[/imath] does not. Aside from [imath]\frac 1p[/imath], every other term [imath]\frac 1k[/imath] has [imath]k[/imath] divisible only by primes smaller than [imath]p[/imath]. We can combine all those terms to get [imath]\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab[/imath], where [imath]b[/imath] is not divisible by [imath]p[/imath]. If this were an integer, then (multiplying by [imath]b[/imath]) [imath]\frac bp +a[/imath] would also be an integer, which it isn't since [imath]b[/imath] isn't divisible by [imath]p[/imath]. Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. | 339267 | Sum [imath]\sum \frac{1}{n}\not \in N[/imath]
If [imath]S_n[/imath] denote sum of [imath]n[/imath] terms of H.P. [imath]\frac{1}{2},\frac{1}{3},\frac{1}{4}[/imath] ..... , Then prove using summation of series that [imath]S_n\not\in N[/imath] [imath]\forall \ n \in N[/imath]; |
109167 | If [imath]A[/imath] is compact and [imath]B[/imath] is closed, show [imath]d(A,B)[/imath] is achieved
Let [imath]A, B[/imath] be subsets of a metric space [imath]X[/imath]. If [imath]A[/imath] is compact and [imath]B[/imath] is closed, show that the distance between [imath]A[/imath] and [imath]B[/imath] is achieved. Attempt at a proof: Let [imath]A[/imath] be compact and [imath]B[/imath] be closed. Let [imath]m=d(A,B)=\inf_{b\in B} d(A,B)[/imath]. Then, there are two possibilities: (a) [imath]\exists b\in B[/imath], [imath]d(a,b)=m[/imath]. If this is the case, we're done. (b) [imath]\forall b\in B[/imath], [imath]d(a,b)>m[/imath]. In this case, there exists a sequence [imath]\{b_n\}\subseteq B:[/imath] [imath]d(a,b_n)\rightarrow m[/imath] as [imath]n\rightarrow\infty[/imath] by definition of infinum. Then there exists a subsequence [imath]\{b_{n_k}\}[/imath]: [imath]d(b_{n_1},a)>d(b_{n_2},a)>...[/imath] which is monotonic decreasing. Then note that [imath]d(b_{n_k},a)<d(b_{n_1},a)<\infty[/imath]. So it's a bounded sequence. Now I want to show that it has a convergent subsequence that converges to [imath]b\in B[/imath] and then I want to do the same for [imath]A[/imath]. And then finally to show that [imath]d(a,b)=m[/imath] in fact. Any clues to how to get there? | 1777619 | Can the distance between two subsets be reached?
Given a metric space [imath](X,d)[/imath]. Let [imath]A[/imath] and [imath]B[/imath] be two nonempty subsets of [imath]X[/imath]. If [imath]A[/imath] is compact and [imath]B[/imath] is closed, then there exist [imath]x_0\in A[/imath] and [imath]y_0\in B[/imath], such that [imath]d(A,B)=d(x_0,y_0)[/imath]? I know that, if [imath]A[/imath] and [imath]B[/imath] are both compact, then the conclusion is right. But if [imath]B[/imath] is just a closed subset? Thank you for any help. |
51292 | Relation of this antisymmetric matrix [imath]r = \left(\begin{smallmatrix}0 &1\\-1&0\end{smallmatrix}\right)[/imath] to [imath]i[/imath]
I was reviewing some matrices and found this interesting if [imath]r = \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}[/imath] then [imath]rr=-I[/imath], also [imath]\exp{(\theta r)} = \cos\theta I + \sin\theta r[/imath] No wonder, the matrix [imath]R(\theta) = e^{\theta r}[/imath] is the 2d rotation matrix, just like [imath]e^{i\theta}[/imath] rotates a vector in the Argand plane. I have a very cursory knowledge of complex analysis, so I would like to know where I can find the details, i.e what is the unifying theme and in which literature can it be found. | 801204 | Matrices and Complex Numbers
Given this set: [imath] S=\left\{\begin{bmatrix}a&-b\\b&a\end{bmatrix}\middle|\,a,b\in\Bbb R\right\} [/imath] Part I: Why is this set equivalent to the set of all complex numbers a+bi (when both are under multiplication?) There is one matrix that corresponds to a specific complex number. Can this example be found and how can it be demonstrated to give equivalent answers? Part II: What is a formula for the multiplicative inverse of the matrix shown in the set, using knowledge on inverses of complex numbers? |
169706 | Which of the following metric spaces are complete?
[NBHM_2006_PhD Screening test_Topology] Which of the following metric spaces are complete? [imath]X_1=(0,1), d(x,y)=|\tan x-\tan y|[/imath] [imath]X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}[/imath] [imath]X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y[/imath] [imath]X_4=\mathbb{R}, d(x,y)=|e^x-e^y|[/imath] [imath]2[/imath] is complete as closed subset of a complete metric space is complete and the metric is also equivalent to our usual metric. [imath]3[/imath] is also complete as every Cauchy sequence is constant ultimately hence convergent. [imath]4[/imath] is not complete I am sure but not able to find out a counter example, not sure about 1.thank you for help. | 639219 | Metric space [imath]X=(0,1)[/imath] with [imath]d(x,y):= \vert \tan x-\tan y \vert[/imath] complete or not?
Let [imath]X=(0,1)[/imath] be a metric space with metric defined as [imath]d(x,y):= \vert \tan x-\tan y \vert[/imath] for all [imath]x,y \in (0,1)[/imath]. The question is whether [imath]X[/imath] is complete with respect to the given metric or not. The answer is given to be "Yes", but I think the sequence [imath]x_{n}=\frac{1}{n}[/imath] serves as a counterexample. Can anyone tell me whether I am correct or not? |
297315 | Limit point of a set (Real Analysis)
To learn the definition of a closed set - a set which contains all its limit points, I want to fully understand the concept of a limit point. What exactly is a limit point? I think that a limit point is a point 'at the end' of (each subset) of the set. For example [imath]a[/imath] and [imath]b[/imath] if the set is [imath][a,b][/imath], or the points [imath]x^2+y^2 = 1[/imath] for the set which is the (filled) unit circle. In my book, the definition is as follows: A point [imath]x[/imath] is a limit point of a set [imath]A[/imath] is every [imath]\epsilon[/imath]-neighborhood [imath]V_{\epsilon}(x)[/imath] of [imath]x[/imath] intersects the set [imath]A[/imath] in some point other then [imath]x[/imath] What is so special about a limit point? The only thing I read is that if [imath]x[/imath] is a limit point of [imath]A[/imath], there must be another [imath]y \in A[/imath], such that if [imath]\epsilon\gt0[/imath], then [imath]|x-y|<\epsilon[/imath] But why is that not true for 'non'-limit points? How can I distinguish these points from point that are not limit points in the set ? | 104489 | Limit points and interior points
I am reading Rudin's book on real analysis and am stuck on a few definitions. First, here is the definition of a limit/interior point (not word to word from Rudin) but these definitions are worded from me (an undergrad student) so please correct me if they are not rigorous. The context here is basic topology and these are metric sets with the distance function as the metric. A point [imath]p[/imath] of a set [imath]E[/imath] is a limit point if every neighborhood of [imath]p[/imath] contains a point [imath]q \neq p[/imath] such that [imath] q \in E[/imath] Also, an interior point is defined as A point [imath]p[/imath] of a set [imath]E[/imath] is an interior point if there is a neighborhood [imath]N_r\{p\}[/imath] that is contained in [imath]E[/imath] (ie, is a subset of E). I understand interior points. Ofcourse given a point [imath]p[/imath] you can have any radius [imath]r[/imath] that makes this neighborhood fit into the set. Thats how I see it, thats how I picture it. I can't understand limit points. It seems trivial to me that lets say you have a point [imath]p[/imath]. Then one of its neighborhood is exactly the set in which it is contained, right? ie, you can pick a radius big enough that the neighborhood fits in the set. Ofcourse I know this is false. Our professor gave us an example of a subset being the integers. He said this subset has no limit points, but I can't see how. |
27096 | The cardinality of the set of all finite subsets of an infinite set
Let [imath]X[/imath] be an infinite set of cardinality [imath]|X|[/imath], and let [imath]S[/imath] be the set of all finite subests of [imath]X[/imath]. How can we show that Card([imath]S[/imath])[imath]=|X|[/imath]? Can anyone help, please? | 620977 | Proving that [imath]|\{B\subseteq S: |B|<\infty \}|=|S|[/imath]
I've some elementary set theory problem that I came across with: Let [imath]S\subseteq\mathbb{R}[/imath] be infinite set, and let [imath]A=\{B\subseteq S: |B|<\infty \}[/imath]. I'm interested in showing that cardinality of set [imath]A[/imath] is equals to cardinality of [imath]S[/imath], i.e [imath]|A|=|S|[/imath]. I thought using Cantor–Bernstein–Schroeder theorem in some way. |
67364 | Sequence sum question: [imath]\sum_{n=0}^{\infty}nk^n[/imath]
I am very confused about how to compute [imath]\sum_{n=0}^{\infty}nk^n.[/imath] Can anybody help me? | 1306830 | How to compute [imath]\sum_{x=0}^{\infty}xa^x[/imath]?
Need a hint to compute [imath]\displaystyle \sum_{x=0}^\infty xa^x[/imath] and [imath]\displaystyle \sum_{x=0}^\infty x^2a^x[/imath], where [imath]a \in (0,1)[/imath]. |
299104 | Showing that [imath]f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1[/imath] for functions [imath]f[/imath], [imath]g[/imath], and [imath]h[/imath] defined by certain power series
I'm having trouble with this question, I have found the interval of convergence of [imath]h(x)[/imath] to be [imath](-\infty, \infty)[/imath], but I don't know how to use that for the question as well as the hint. Any help would be appreciated. Thanks! If [imath]f(x) =\sum_{n = 0}^\infty \frac{x^{3n}}{(3n)!},\qquad g(x) =\sum_{n = 0}^\infty \frac{x^{3n+1}}{(3n+1)!},\qquad h(x) = \sum_{n = 0}^\infty \frac{x^{3n+2}}{(3n+2)!}[/imath] show that [imath]f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1.[/imath] Hint: show that [imath]h'(x) = g(x)[/imath]. | 247053 | What's the background of this exercise?
I found this interesting exercise on a calculus book (Stewart) Let [imath] u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdots [/imath] [imath] v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\cdots [/imath] [imath] w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots [/imath] Show that [imath]u^3+v^3+w^3-3uvw=1[/imath] It turns out to be an interesting application of the 3rd root of unity, which greatly simplified the (could be) tedious calculation. I wonder if it has any deeper interpretation. (At least I don't see how to easily generalize it.) Can anybody explain this? (I don't need help on solving this problem.) |
82140 | Show that [imath]f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1[/imath] when [imath]n[/imath] is a positive integer
Letting [imath]f_n[/imath] be the Fibonacci numbers, show that [imath]f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1[/imath] when [imath]n[/imath] is a positive integer. Just some homework help. Need to prove. Thank you in advance. | 1711271 | Let [imath]F_n[/imath] be the Fibonacci sequence
Let [imath]F_n[/imath] be the Fibonacci sequence with [imath]F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2}, n\geq2\\ Use\;the\;mathematical\; Induction\; show\;that\;\sum_{i=0}^{n}(-1)^iF_i= (-1)^n F_{n-1}-1, For\;every\;positive\;integers[/imath] |
6273 | What does E mean in 9.0122222900391E-5?
I am not a mathematician(IANAM), however I wish I could be. My question: I often find this at the bottom of pages. Page generated in 0.00013899803161621 Sometimes, I come across Page generated in 9.0122222900391E-5 What does that time mean? I tried searching Wikipedia for E and maths but found the e mathematical constant with a scary looking graph. My guess is E stand for Exponential or something and -5 is the power it is raised to. And the time is a really small number. But that doesn't make sense when compared to the other time in the question is 0.00013899803161621 bigger than 9.0122222900391E-5? If it means x times [imath]10^{-5}[/imath], then 9.0122222900391E-5 will be 0.000090122222900391 which is smaller than 0.00013899803161621 | 2964486 | What does [imath]2.5437065e\!−\!5[/imath] mean? [scientific notation]
[imath]\frac{0.314^5}{120}=2.5437065e\!−\!5[/imath] I did this calculation by my android phone's calculator. I can't understand the result I got (The RHS of calculation). Please help. What is this? |
3215 | Convergence of [imath]\sqrt{n}x_{n}[/imath] where [imath]x_{n+1} = \sin(x_{n})[/imath]
Consider the sequence defined as [imath]x_1 = 1[/imath] [imath]x_{n+1} = \sin x_n[/imath] I think I was able to show that the sequence [imath]\sqrt{n} x_{n}[/imath] converges to [imath]\sqrt{3}[/imath] by a tedious elementary method which I wasn't too happy about. (I think I did this by showing that [imath]\sqrt{\frac{3}{n+1}} < x_{n} < \sqrt{\frac{3}{n}}[/imath], don't remember exactly) This looks like it should be a standard problem. Does anyone know a simple (and preferably elementary) proof for the fact that the sequence [imath]\sqrt{n}x_{n}[/imath] converges to [imath]\sqrt{3}[/imath]? | 425464 | How fast does [imath]f_n = \sin f_{n-1}[/imath] approach zero?
The sequence [imath]f(n) = \sin(\sin(\sin(......(1)......)))[/imath] approaches zero like [imath]\sqrt{3/n}[/imath], as has been asked and answered here a few times. So [imath]f(n)[/imath] would get below [imath]1/n[/imath] after [imath]3n^2[/imath] steps, but it seems to get there about [imath]\ln n[/imath] steps earlier: 0.1 after 295, 0.01 after 29992, 0.001 after 2999989 steps. Is that numerical roundoff error or not? |
12906 | The staircase paradox, or why [imath]\pi\ne4[/imath]
What is wrong with this proof? Is [imath]\pi=4?[/imath] | 436271 | How does [imath]a^2 + b^2 = c^2[/imath] work with ‘steps’?
We all know that [imath]a^2+b^2=c^2[/imath] in a right-angled triangle, and therefore, that [imath]c<a+b[/imath], so that walking along the red line would be shorter than using the two black lines to get from top left to bottom right in the following graphic: Now, let's assume that the direct way using the red line is blocked, but instead, we can use the green way in the following picture: Obviously, the green way isn't any shorter than the black one, it's just [imath]a/2+b/2+a/2+b/2 = a+b[/imath]. Now, we can divide the green path again, just like the black path, and get to the purple path. Dividing this one in two halfs again, we get the yellow path: Now obviously, the yellow path is still as long as the black path from the beginning, it's just [imath]8*a/8+8*b/8=a+b[/imath]. But if we do this segmentation again and again, we approximate the red line - without making the way any shorter. Why is this so? |
69386 | Inequality involving [imath]\limsup[/imath] and [imath]\liminf[/imath]: [imath] \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)[/imath]
This may have been asked before, however I was unable to find any duplicate. This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14: If [imath](a_n)[/imath] is a sequence in [imath]\mathbb R[/imath] and [imath]a_n > 0[/imath] for every [imath]n[/imath]. Then show: [imath] \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)[/imath] The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities. Thanks. | 321198 | Proof of limit inequality
Prove that for any sequence [imath]\{x_n\}[/imath] of positive real numbers [imath]\lim\text{sup}\sqrt[n]{x_n}\leq \lim\text{sup}\frac{x_{n+1}}{x_n}.[/imath] My attempt: Let [imath]A = \lim\text{sup}\frac{x_{n+1}}{x_n}[/imath]. Suppose [imath]A<\infty[/imath] and choose [imath]\epsilon >0[/imath], then [imath]\exists[/imath] an integer [imath]N[/imath] so that [imath]N\le n \implies \frac{x_{n+1}}{x_n}\le \epsilon[/imath]. But I do not know how to proceed in finishing the proof? |
67068 | Combinatorial proof for two identities
Does exist a combinatorial proof for the following two identities ? [imath]\sum_{k = 0}^{n} \binom{x+k}{k} = \binom{x+n+1}{n}[/imath] [imath]\sum_{k = 0}^{n} k\binom{n}{k} = n2^{n-1}[/imath] I know how to derive the identites from [imath](1+x)^n[/imath] , but I am searching for a combinatorial proof ? | 357063 | Proving that [imath]\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }[/imath]
I have to prove that: [imath]\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }[/imath] I tried to open up the right side with Pascal's definition that: [imath] { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}[/imath] Here is what I came up with, and I am sure it is wrong because it does not equal the left side: [imath] {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 } [/imath] Which, again, probably is wrong because it is not equal [imath]\sum_{k=0}^{n} { m+k \choose m}[/imath]. Any help is appreciated |
32199 | What does the math notation [imath]\sum[/imath] mean?
I have come across this symbol a few times, and I am not sure what it "does" or what it means: [imath]\Large\sum[/imath] | 2428172 | Calculating the length (norm) of a vector
How do you interpret this equation? I've seen far simpler equations for solving the magnitude, like the square root of all of the elements of a vector squared added to one another. What exactly is going on with the Greek symbol in that equation? Is there something I'm missing? [vector magnitude calculation][1] 2.6 The lenght (norm) of a vector The lenght ( or norm ) of a vector a is: [imath]\left\|{a}\right\|=\sqrt{aa}=\sqrt{\sum_{i=1}^n a_i^2}[/imath] |
1064 | Perfect set without rationals
Give an example of a perfect set in [imath]\mathbb R^n[/imath] that does not contain any of the rationals. (Or prove that it does not exist). | 925330 | Is there a non-empty subset of [imath]\mathbb{R} [/imath] like [imath]A[/imath] such that the set of accumulation points of [imath]A[/imath] is itself and [imath]A\cap\mathbb{Q} = \emptyset [/imath]
Is there a non-empty subset of [imath]\mathbb{R} [/imath] like [imath]A[/imath] such that the set of accumulation points of [imath]A[/imath] is [imath]A[/imath] and [imath]A\cap\mathbb{Q}=\emptyset\,[/imath]? |
291071 | Let [imath]G[/imath] be a finite group and [imath]H\triangleleft G[/imath] a normal subgroup. Prove that [imath]|G/H| =|G|[/imath] if, and only if, [imath]H = \{e\}[/imath].
The group [imath]G[/imath] is a finite group, a group with finite number of elements, and [imath]H\triangleleft G [/imath]a normal subgroup. How can we prove that the index [imath]|G/H|=|G|[/imath] iff [imath]H=\{e\}[/imath], the identity element? | 296694 | [imath]|G/H|= |G|\Leftrightarrow H=\{e\}[/imath]
Please help me with the following problem. Prove that if [imath]|H|=\{e\}[/imath] then [imath]|G/H|= |G|[/imath]. Then show that if [imath]|G/H|=|G|[/imath] then [imath]H={e}[/imath]. |
11601 | Proof that a Combination is an integer
From its definition a combination [imath]\binom{n}{k}[/imath], is the number of distinct subsets of size [imath]k[/imath] from a set of [imath]n[/imath] elements. This is clearly an integer, however I was curious as to why the expression [imath]\frac{n!}{k!(n-k)!}[/imath] always evaluates to an integer. So far I figured: [imath]n![/imath], is clearly divisible by [imath]k![/imath], and [imath](n-k)![/imath], individually, but I could not seem to make the jump to proof that that [imath]n![/imath] is divisible by their product. | 442568 | Binomial coefficient formula: Why the dividend is a multiple of divisor
Looking at the formula for binomial co-effient, [imath] \binom{n}{r}= \frac {n(n-1)...(n-r+1)}{ 1(2)(3)...(r)} [/imath] I am wondering why [imath] n(n-1)...(n-r+1) [/imath] is a multiple of [imath] 1(2)(3)...(r) [/imath] . I understand from the applications of this formula that only integer values makes sense as in [imath] n C r [/imath]. But, why are these two products be related in such a way, purely thinking of them as product of numbers and from algebra, how can we prove that they will always be related that way. |
46822 | Density of irrationals
I came across the following problem: Show that if [imath]x[/imath] and [imath]y[/imath] are real numbers with [imath]x <y[/imath], then there exists an irrational number [imath]t[/imath] such that [imath]x < t < y[/imath]. We know that [imath]y-x>0[/imath]. By the Archimedean property, there exists a positive integer [imath]n[/imath] such that [imath]n(y-x)>1[/imath] or [imath]1/n < y-x[/imath]. There exists an integer [imath]m[/imath] such that [imath]m \leq nx < m+1[/imath] or [imath]\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y[/imath]. This is essentially the proof for the denseness of the rationals. Instead of [imath]\large \frac{m+1}{n}[/imath] I need something of the form [imath]\large\frac{\text{irrational}}{n}[/imath]. How would I get the numerator? | 2295648 | Proof there is an irrational number [imath]r[/imath] in every intervall [imath]a < r < b[/imath]
Proof that for [imath]a,b \in \mathbb{R}[/imath] there is an irrational number [imath]r[/imath] so that [imath]a < r < b[/imath]. Basically, proof, that between any two irrationals, there is another irrational r. I'm sure there are already many ways out there how to do it, however I have troubles proving it in the following way: (1) For every [imath]x,y \in \mathbb{R}[/imath] there is a bijective function between [imath][0,1][/imath] and [imath][x,y][/imath] (already proven) (2) [imath]\frac{\sqrt{2}}{2} \in ]0,1[[/imath] (3) Now when mapping [imath][0,1][/imath] onto [imath][x,y][/imath] [imath]\frac{\sqrt{2}}{2}[/imath] will also be mapped into the new intervall, therefore there has to be an irrational number in [imath][x,y][/imath] Now the problem I see is, that for example [imath]\frac{\sqrt{2}}{2}[/imath] could be mapped onto a rational number and therefore I'd have to proof, that there is a different irrational in [imath][x,y][/imath]. It'd be nice if you could help me complete the proof. |
60578 | What is the term for a factorial type operation, but with summation instead of products?
(Pardon if this seems a bit beginner, this is my first post in math - trying to improve my knowledge while tackling Project Euler problems) I'm aware of Sigma notation, but is there a function/name for e.g. [imath] 4 + 3 + 2 + 1 \longrightarrow 10 ,[/imath] similar to [imath]4! = 4 \cdot 3 \cdot 2 \cdot 1 ,[/imath] which uses multiplication? Edit: I found what I was looking for, but is there a name for this type of summation? | 593318 | Factorial, but with addition
Is there a notation for addition form of factorial? [imath]5! = 5\times4\times3\times2\times1[/imath] That's pretty obvious. But I'm wondering what I'd need to use to describe [imath]5+4+3+2+1[/imath] like the factorial [imath]5![/imath] way. EDIT: I know about the formula. I want to know if there's a short notation. |
141196 | Highest power of a prime [imath]p[/imath] dividing [imath]N![/imath]
How does one find the highest power of a prime [imath]p[/imath] that divides [imath]N![/imath] and other related products? Related question: How many zeros are there at the end of [imath]N![/imath]? This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details. | 1143699 | Why does [imath]\sum\limits_{k=1}^\infty \lfloor m/(n^k)\rfloor[/imath] give you the number of times that [imath]n[/imath] divides [imath]m![/imath]?
If [imath]n[/imath] is a prime less than [imath]m[/imath], with [imath]n,m \in \mathbb N[/imath], why does [imath]\sum_{k=1}^\infty \left\lfloor \frac{m}{n^k}\right\rfloor[/imath] give you the number of times that [imath]n[/imath] divides [imath]m![/imath]? Examples: [imath]n=13[/imath] [imath]m=321[/imath] [imath]\sum\limits_{k=1}^\infty \lfloor 321/(13^k)\rfloor=25[/imath] [imath]n=5[/imath] [imath]m=321[/imath] [imath]\sum\limits_{k=1}^\infty \lfloor 321/(5^k)\rfloor=78[/imath] In fact, FactorInteger[321!]={{2, 318}, {3, 157}, {5, 78}, {7, 51}, {11, 31}, {13, 25}, {17, 19}, {19, 16}, {23, 13}, {29, 11}, {31, 10}, {37, 8}, {41, 7}, {43, 7}, {47, 6}, {53, 6}, {59, 5}, {61, 5}, {67, 4}, {71, 4}, {73, 4}, {79, 4}, {83, 3}, {89, 3}, {97, 3}, {101, 3}, {103, 3}, {107, 3}, {109, 2}, {113, 2}, {127, 2}, {131, 2}, {137, 2}, {139, 2}, {149, 2}, {151, 2}, {157, 2}, {163, 1}, {167, 1}, {173, 1}, {179, 1}, {181, 1}, {191, 1}, {193, 1}, {197, 1}, {199, 1}, {211, 1}, {223, 1}, {227, 1}, {229, 1}, {233, 1}, {239, 1}, {241, 1}, {251, 1}, {257, 1}, {263, 1}, {269, 1}, {271, 1}, {277, 1}, {281, 1}, {283, 1}, {293, 1}, {307, 1}, {311, 1}, {313, 1}, {317, 1}} and so on, for every [imath]n,m ∈ N[/imath], [imath]n[/imath] prime and [imath]<m[/imath]...? Can someone explain? |
83850 | Two Limits Equal - Proof that [imath]\lim_{n\to\infty }a_n=L[/imath] implies [imath]\lim_{n\to\infty }\frac{\sum_1^na_k}n=L[/imath]
Problem: Given that [imath]\lim_{n\to\infty }a_n=L[/imath] and [imath]m_n=\frac{\sum_{1}^{n}a_k}{n}[/imath]. Prove that [imath]\lim m_n=L[/imath] Proof: We have [imath]\sum_{1}^{n}a_k=na_k[/imath], so [imath]m_n=\frac{\sum_{1}^{n}a_k}{n}=\frac{na_k}{n}=a_k[/imath] and [imath]\lim a_k=L.\square[/imath] Is this a correct proof? I am confused with the subscripts [imath]m[/imath] and [imath]k[/imath] and not sure if I am using the right one in the right spot, in the proof. Thanks. | 403972 | How to prove [imath]\lim_{n \to \infty}\frac{a_1+a_2+\dots+a_n}{n}=l.[/imath]
When Let [imath]{\{a_n}\}[/imath] be a convergent sequence, [imath]\lim_{n \to \infty}a_n=l.[/imath] Prove that [imath]\lim_{n \to \infty}\dfrac{a_1+a_2+\dots+a_n}{n}=l.[/imath] Hint : Prove that the sequence [imath]{\{a_n}\}[/imath] is bounded first. Please provide me hint or full solution. Thanks. |
63950 | Proof that an integral domain that is a finite-dimensional [imath]F[/imath]-vector space is in fact a field
I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof? Lemma 2.2.3. Let [imath]F[/imath] be a field and [imath]R[/imath] an integral domain that is a finite-dimensional [imath]F[/imath]-vector space. Then [imath]R[/imath] is a field. Proof. We need to show that any nonzero [imath]r \in R[/imath] has an inverse. Consider [imath]\{1, r, r^2, \cdots\}[/imath]. This is an infinite set of elements of [imath]R[/imath], and by hypothesis [imath]R[/imath] is finite dimensional as an [imath]F[/imath]-vector space, so this set is linearly dependent. Hence [imath]\sum_{i=0}^n{c_i r^i} = 0[/imath] for some [imath]n[/imath] and some [imath]c_i \in F[/imath] not all zero. It then goes on to show, given the above, that we can derive an inverse for [imath]r[/imath]. However, if I consider examples like [imath]r = 2 \in Q[\sqrt{2}][/imath], [imath]r = \sqrt{2} \in Q[\sqrt{2}][/imath] or [imath]r = 2 \in Q[X]/{<X^2>}[/imath], the set [imath]\{1, r, r^2, ...\}[/imath] doesn't look linearly dependent to me. I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something? [Edit] Well yes, I am. Somehow I had managed to discount the possibility of any [imath]c_i[/imath] being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding. | 390126 | Proving an Integral domain is a field.
Let [imath]R[/imath] be an integral domain containing a field [imath]F[/imath] as a subring. Show that if [imath]R[/imath] is a finite dimensional vector space over [imath]F[/imath], then [imath]R[/imath] is a field. This is a Ph.D. entrance question, I recently appeared. Somehow I need to prove that every element in [imath]R[/imath] has an inverse, but can't figure out how. |
149655 | To show that Fermat number [imath]F_{5}[/imath] is divisible by [imath]641[/imath].
How can I show that Fermat number [imath]F_{5}=2^{2^5}+1[/imath] is divisible by [imath]641[/imath]. | 876481 | Proving the divisibility of large numbers without making large calculations
How would you you show that [imath]2^{32}+1[/imath] is divisible by [imath]641[/imath] without making large calculations? |
136922 | [imath]X[/imath] is Hausdorff if and only if the diagonal of [imath]X\times X[/imath] is closed
Let [imath]X[/imath] be a topological space. The diagonal of [imath]X \times X[/imath] is the subset [imath]D = \{(x,x)\in X\times X\mid x \in X\}.[/imath] Show that [imath]X[/imath] is Hausdorff if and only if [imath]D[/imath] is closed in [imath]X \times X[/imath]. First, I tried to show that [imath]X \times X \setminus D[/imath] is open using the fact that [imath]X \times X[/imath] is Hausdorff (because [imath]X[/imath] is Hausdorff), but I couldn't find an open set that contains a point outside [imath]D[/imath] and is disjoint to it... | 712125 | Prove that a topological space [imath](X, \tau)[/imath] is [imath]T_2[/imath] if and only if the diagonal [imath]D=\{(x,x):x \in X\}[/imath] is closed subset of [imath]X\times X[/imath]
Prove that a topological space [imath](X, \tau)[/imath] is [imath]T_2[/imath] if and only if the diagonal [imath]D=\{(x,x):x \in X\}[/imath] is closed subset of the product space [imath]X\times X[/imath] => assume that [imath](X, \tau)[/imath] is [imath]T_2[/imath], I know that [imath]\forall x,y \in X[/imath], there exists [imath]U,V \subset X[/imath] are open and disjoint sets such that [imath]x\in U[/imath] and [imath]y \in V[/imath]. I think that [imath]U \times U= \{(x,x):x\in U\}[/imath] but I'm not sure this is correct, I don't know how to continue from here for the converse, I can't see how the hypothesis help me prove [imath](X, \tau)[/imath] is [imath]T_2[/imath] |
36502 | Isometries of [imath]\mathbb{R}^n[/imath]
Let [imath]f:\mathbb{R}^n\rightarrow \mathbb{R}^n[/imath] be such that [imath]\left\| f(x)-f(y)\right\| =\left\| x-y\right\|[/imath]. Is [imath]f[/imath] necessarily surjective? If this is so, you can prove (Mazur-Ulam Theorem) that [imath]f[/imath] is affine, and hence you could classify all isometries of [imath]\mathbb{R}^n[/imath]. However, at the moment, I can't think of any good ideas to prove that [imath]f[/imath] is surjective. For that matter, is it even the case that [imath]f[/imath] must be surjective? Any ideas would be most welcomed. Thanks much! | 516599 | Compact metric space
Let [imath](X,d)[/imath] be a compact metric space and [imath]f : X \to X[/imath] be isometric, i.e. for every [imath]x,y \in X[/imath] : [imath] d(f(x),f(y)) = d(x,y) [/imath]. How I can show the following? [imath]f(X) = X[/imath] My first thought was to show the surjection [imath] x \in X \setminus f(x) [/imath]. Is this the right approach? Thanks in advance |
136626 | [imath]\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}[/imath] is infinite
How do I prove that [imath] \displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}[/imath] is infinite? | 411289 | Is [imath]\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} = 0[/imath]?
When I was trying to solve a problem to find the radius of convergence of the power series [imath]\sum \frac{2^nz^n}{n!}[/imath] I fully understand that the ratio test works well in this one and the radius of convergence is [imath]\infty[/imath]. However, knowing that the root test gives a better span of the ratio test and it is necessary to prove it, I wanted to be able to find the radius of convergence using the ratio test. Thus obtaining the following [imath]\begin{align} \lim_{n \to \infty} \sqrt[n]{\frac{2^nz^n}{n!}} & = |2z|\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} \\ \\ & = 0\\ \\ & \lt 1 \\ \\ & \Rightarrow R = \infty \end{align}[/imath] must be true. So, I was thinking that [imath]\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}}[/imath] must be equal to 0. Is there a direct proof of this ? I just want to be a bit more algebraically savvy. |
115501 | [imath]\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}[/imath], or the limit of the sequence [imath]x_{n+1} = \sqrt{c+x_n}[/imath]
(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12) For [imath]c \gt 0[/imath], consider the quadratic equation [imath]x^2 - x - c = 0, x > 0[/imath]. Define the sequence [imath]\{x_n\}[/imath] recursively by fixing [imath]|x_1| \lt c[/imath] and then, if [imath]n[/imath] is an index for which [imath]x_n[/imath] has been defined, defining [imath]x_{n+1} = \sqrt{c+x_n}[/imath] Prove that the sequence [imath]\{x_n\}[/imath] converges monotonically to the solution of the above equation. Note: The answers below might assume [imath]x_1 \gt 0[/imath], but they still work, as we have [imath]x_3 \gt 0[/imath]. This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates. | 359145 | Which is the function that this sequence of functions converges
Prove that [imath] \left(\sqrt x, \sqrt{x + \sqrt x}, \sqrt{x + \sqrt {x + \sqrt x}}, \ldots\right)[/imath] in [imath][0,\infty)[/imath] is convergent and I should find the limit function as well. For give a idea, I was plotting the sequence and it's look like |
58677 | Limit of [imath](\sin\circ\sin\circ\cdots\circ\sin)(x)[/imath]
I'm trying to find this limit: [imath]\lim_{n \to \infty} \underbrace{\sin \sin \ldots \sin }_{\text{$n$ times}}x[/imath] Thank you | 1509850 | Limit of repeated [imath]\sin[/imath]
I was wondering if it was possible to formalize the following. If we let [imath]f_n(x)[/imath] denote the sequence of function so that [imath]f_1(x)=\sin(x)[/imath], [imath]f_2(x)=\sin(\sin(x))[/imath], [imath]f_3(x)=\sin(\sin(\sin(x)))[/imath] and so on. Does this sequence of functions have a limit? Is there a theory that could explain the behaviour of repeating trigonometric functions? |
41623 | Is that true that all the prime numbers are of the form [imath]6m \pm 1[/imath]?
Is that true that all the prime numbers are of the form [imath]6m \pm 1[/imath] ? If so, can you please provide an example? Thanks in advance. | 1971340 | Prime [imath]p \geq 5[/imath] is of the form [imath]6k + 1[/imath] or of the form [imath]6k + 5[/imath]
Let [imath]p[/imath] be an odd prime. Prove. [imath]p \geq 5[/imath] is of the form [imath]6k + 1[/imath] or of the form [imath]6k + 5[/imath] for some nonnegative integer [imath]k[/imath]. The solution says that there are 3 cases: [imath]a = 3k[/imath], [imath]a = 3k+ 1[/imath], and [imath]a = 3k + 2[/imath], but I can't figure out how they got those cases. Thanks for your help. |
26445 | Division by [imath]0[/imath]
I came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions. [imath]\dfrac {x}{0}[/imath] is Impossible ( If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both [imath]1.[/imath] and [imath]2.[/imath] are divided by zero, but only [imath]2. [/imath] has infinite solutions so as [imath]1.[/imath] has none solution, how and why ?) [imath]\dfrac {0}{0}[/imath] is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?) [imath]\dfrac {0}{x}[/imath] and [imath]x \ne 0[/imath], it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it. | 366574 | Dividing a number by zero
Why can't you divide a number by zero? It is possible to say [imath]\sqrt{-1}[/imath] is an imaginary number [imath]i[/imath], but why can't you say [imath]\frac{1}{0}[/imath] is also an imaginary number [imath]z[/imath] (for example)? |
76491 | Multiple-choice question about the probability of a random answer to itself being correct
I found this math "problem" on the internet, and I'm wondering if it has an answer: Question: If you choose an answer to this question at random, what is the probability that you will be correct? a. [imath]$25\%$[/imath] b. [imath]$50\%$[/imath] c. [imath]$0\%$[/imath] d. [imath]$25\%$[/imath] Does this question have a correct answer? | 1358035 | Is this a question on probability? Or not a question at all?
If you choose an answer to this question at random, what is the chance you will be correct? A) [imath]25\%[/imath] B) [imath]50\%[/imath] C) [imath]60\%[/imath] D) [imath]25\%[/imath] https://plus.google.com/+RaymondJohnson/posts/CSXeyftovTJ |
80889 | Proving uniform continuity on an interval
If I know that [imath]f:(0,\infty)\rightarrow \Bbb R[/imath] is uniformly continuous on the intervals [imath][a,\infty)[/imath] and [imath](0,a][/imath], where [imath]a[/imath] is in [imath](0,\infty)[/imath], how can I prove that it is uniformly continuous on [imath](0,\infty)[/imath]? I know the general definition of uniform continuity using epsilon-delta, but I am not sure how to apply it to the above. Thanks Edit: I meant Uniformly continuous on the first two intervals | 760658 | If [imath]f[/imath] is continuous on [imath][0, \infty)[/imath] and uniformly continuous on [imath][b, \infty)[/imath] for some [imath]b > 0[/imath] then [imath]f[/imath] is unif. continuous on $[0, \infty).
Prove that if [imath]f[/imath] is continuous on [imath][0, \infty)[/imath] and uniformly continuous on [imath][b, \infty)[/imath] for some [imath]b > 0[/imath] then [imath]f[/imath] is unif. continuous on [imath][0, \infty).[/imath] So far I have: Let A = [b, \infty)[imath] then [/imath]A^c = [0,b)[imath], my goal is to end up with a compact set from [/imath][0,b][imath] so I can use the theorem that "A function is continuous on a compact set [/imath]K[imath] is uniformly continuous on [/imath]K$" and then take the fact that of the preservation of connected sets. Am I going about this in the right way? |
297 | Simple numerical methods for calculating the digits of [imath]\pi[/imath]
Are there any simple methods for calculating the digits of [imath]\pi[/imath]? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits? | 445056 | Calculate value of [imath]\pi[/imath] to millionth digit?
** I wanted to calculate the value of [imath]\pi[/imath] till millionth digit precision ,but I really cannot think of any proper procedure to calculate it ** Can individual nth digit after decimal be calculated ? |
9505 | [imath]x^y = y^x[/imath] for integers [imath]x[/imath] and [imath]y[/imath]
We know that [imath]2^4 = 4^2[/imath] and [imath](-2)^{-4} = (-4)^{-2}[/imath]. Is there another pair of integers [imath]x, y[/imath] ([imath]x\neq y[/imath]) which satisfies the equality [imath]x^y = y^x[/imath]? | 355138 | Are there other powers which follow the rule [imath]a^b = b^a[/imath] than [imath]2^4[/imath]?
I was trying to find these powers, but to my disappointment I only found [imath]2^4 = 4^2[/imath]. Edit: [imath]a[/imath] must be different to [imath]b[/imath] of course. Is that the only possible setting, and why? If we assume the number and the power can have also decimal values. |
100978 | How many connected components does [imath]\mathrm{GL}_n(\mathbb R)[/imath] have?
I've noticed that [imath]\mathrm{GL}_n(\mathbb R)[/imath] is not a connected space, because if it were [imath]\det(\mathrm{GL}_n(\mathbb R))[/imath] (where [imath]\det[/imath] is the function ascribing to each [imath]n\times n[/imath] matrix its determinant) would be a connected space too, since [imath]\det[/imath] is a continuous function. But [imath]\det(\mathrm{GL}_n(\mathbb R))=\mathbb R\setminus\{0\},[/imath] so not connected. I started thinking if I could prove that [imath]\det^{-1}((-\infty,0))[/imath] and [imath]\det^{-1}((0,\infty))[/imath] are connected. But I don't know how to prove that. I'm reading my notes from the topology course I took last year and I see nothing about proving connectedness... | 1889662 | [imath]\mathcal{I} = \{X \in M_{n \times n} \mid X \text{ is invertible }\}[/imath] is not-connected.
Let [imath]\mathcal{I}[/imath] be the set of all invertible real matrix [imath]n \times n[/imath]. I have to prove that [imath]\mathcal{I}[/imath] is open and not-connected. My attempt: Take any [imath]A \in \mathcal{I}[/imath]. Puting [imath]\varepsilon = \frac{1}{\|A^{-1}\|}[/imath] we have that [imath]\|A-B\| < \varepsilon \Rightarrow B \in \mathcal{I}[/imath]. Therefore [imath]\mathcal{I}[/imath] is open. But I really don't know how I can show that [imath]\mathcal{I}[/imath] is not-connected. |
63102 | How to prove periodicity of [imath]\sin(x)[/imath] or [imath]\cos(x)[/imath] starting from the Taylor series expansion?
With the Taylor series representation of [imath]\sin[/imath] or [imath]\cos[/imath] as a starting point (and assuming no other knowledge about those functions), how can one: a. prove they are periodic? b. find the value of the period? | 658261 | Proving periodicity of sine and cosine
If we define the sine and cosine functions by their Maclaurin expansions, how do we prove they are periodic with period [imath]2\pi[/imath]? |
57398 | How to show [imath](a^b)^c=a^{bc}[/imath] for arbitrary cardinal numbers?
One of the basic (and frequently used) properties of cardinal exponentiation is that [imath](a^b)^c=a^{bc}[/imath]. What is the proof of this fact? As Arturo pointed out in his comment, in computer science this is called currying. | 773591 | Prove that [imath]|A^{B×C}| = |(A^B)^C|[/imath]
I'm having trouble proving that [imath]|A^{B×C}| = |(A^B)^C|[/imath] , where [imath]M^N[/imath] is the set of all the functions [imath]f:N \to M[/imath]. My thoughts: to prove this, I need to find a bijection between [imath]|A^{B×C}|[/imath] and [imath]|(A^B)^C|[/imath], so I need a bijection between the set of all functions [imath]g:B×C \to A[/imath] and the set of all functions [imath]h: C \to A^B[/imath]. |
132722 | Probability problem
I have [imath]3[/imath] coins, [imath]1[/imath] coin has [imath]2[/imath] heads (HH), 1 coin has [imath]2[/imath] tails (TT), [imath]1[/imath] coin has [imath]1[/imath] head and [imath]1[/imath] tail (HT). I toss the coin, it fells on my hand, and the side i see is a tail. What's the chance that the other side is also a tail? I got this as a teaser from a friend, possible from here, as you can see he is insisting on 1/2 as not being the correct answer, I got 1/3 as my answer, am I right? | 1527009 | Probability of card color
Task: In box are three cards. First card is red on both sides. Second card is black on both sides. Third card is red on one side and black on other side. I random choose one card and I see that color on one side is red. What is the probability that second side is red? My solution is [imath]P=\frac{1}{2}[/imath] because on other side should be black or red color. Is this correct? |
260111 | Proving Cauchy condensation test
I have to prove the condensation test of Cauchy by tomorrow and I am really unconfident about what I did: [imath]\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}[/imath] I did the following: Let [imath](b_n)[/imath] be a sequence as follow: [imath]b_{2^k+m}:=a_{2^k}[/imath] with [imath]k\in\mathbb N_0[/imath] and [imath]0\leq m<2^k[/imath]. It's [imath]a_{n+1}\leq a_n[/imath] and so [imath]0\leq a_{n+p}\leq a_n[/imath] for all [imath]n,p\in\mathbb N[/imath]. So [imath]\sum\limits_{n=1}^\infty b_n[/imath] converges by the majorizing series [imath]\sum\limits_{n=1}^\infty a_n[/imath]. And it's [imath]\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}[/imath] so [imath]\Rightarrow[/imath] is done. For [imath]\Leftarrow[/imath] consider [imath]c_{2^k+m}:=a_{2^k}[/imath] with [imath]k\in\mathbb N_0[/imath] and [imath]0\leq m<2^k[/imath]. It's [imath]|a_n|\leq c_n[/imath] and [imath]\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}[/imath] and so [imath]\sum\limits_{n=1}^\infty a_n[/imath] converges by the majorizing series [imath]\sum\limits_{n=0}^\infty c_n[/imath]. Is this in form and content correct? | 1544342 | [imath]\sum_{n=1}^\infty a(n)[/imath] converges if and only if [imath]\sum_{n=1}^\infty 2^na(2^n)[/imath] converges
Let [imath]\sum_{n=1}^\infty a(n)[/imath] be a series such that [imath](a(n))[/imath] is a decreasing sequence of positive numbers. Prove that [imath]\sum_{n=1}^\infty a(n)[/imath] converges if and only if [imath]\sum_{n=1}^\infty 2^na(2^n)[/imath] converges. |
13054 | How to show [imath]e^{e^{e^{79}}}[/imath] is not an integer
In this question, I needed to assume in my answer that [imath]e^{e^{e^{79}}}[/imath] is not an integer. Is there some standard result in number theory that applies to situations like this? After several years, it appears this is an open problem. As a non-number theorist, I had assumed there would be known results that would answer the question. I was aware of the difficulty in proving various constants to be transcendental -- such as [imath]e + \pi[/imath], which is not known to be transcendental at present. However, I was looking at a question that seems simpler, naively: whether a number is an integer, rather than whether it is transcendental. It seems that what appeared to be possibly simpler is actually not, with current techniques. The main motivation for asking about this particular number is that it is very large. It is certainly possible to find a pair of very large numbers, at least one of which is transcendental. But the current lack of knowledge about this particular number is even an integer shows just how much progress remains to be made, in my opinion. Any answers that describe techniques that would suffice to solve the problem (perhaps with other, unproven assumptions) would be very welcome. | 2613384 | Why is this number : [imath]e^{e^{e^{79}}}[/imath] conjectured to be an integer number which is a skew number?
skew number is defined as : [imath]e^{e^{e^{79}}}[/imath] , I seek for the mathematical reasons which let [imath]e^{e^{e^{79}}}[/imath] conjectured to be an integer ? |
16175 | On sort-of-linear functions
Background A function [imath] f: \mathbb{R}^n \rightarrow \mathbb{R} \ [/imath] is linear if it satisfies [imath] (1)\;\; f(x+y) = f(x) + f(y) \ , \ and [/imath] [imath] (2)\;\; f(\alpha x) = \alpha f(x) [/imath] for all [imath] x,y \in \mathbb{R}^n [/imath] and all [imath] \alpha \in \mathbb{R} [/imath]. A function satisfying only (2) is not necessarily linear. For example* [imath] f: \mathbb{R}^2 \rightarrow \mathbb{R} \ [/imath] defined by [imath] f(x) = |x| \ [/imath] (where [imath] |x| \ [/imath] is the [imath] L^2 [/imath] norm) satisfies (2) but is not linear. However, a function satisfying (1) does satisfy a weaker version of (2), namely [imath] (2b)\;\; f(ax)=af(x) [/imath] for all [imath] a \in \mathbb{Q} [/imath]. *Edit: As pointed out in the comments this example doesn't quite work since |ax|=|a||x|. When [imath] f [/imath] is continuous it's relatively straight-forward to show that under the extra hypothesis that [imath] f [/imath] is continuous, (2b) implies (2). I want to say that continuity is a necessary condition for (1) to imply (2), or at least (worst) there is some extra hypothesis required (possibly weaker than continuity), but I'm not sure how to show it. My question is therefore two-fold: -Is continuity a necessary condition for (1) to imply (2) and how could I go about proving it. -What are some examples (if there are any) of a function satisfying (1) but not (2) This can be stated in a slightly more general context as follows: Suppose [imath] V\ [/imath] is a vector space over [imath] \mathbb{R}\ [/imath] and [imath] f: V \rightarrow \mathbb{R}\ [/imath] satisfies [imath] (1') \;\; f(x+y) = f(x)+f(y) [/imath] for all [imath] x,y \in V [/imath]. Under what conditions is [imath] f\ [/imath] a vector space homomorphism? The reason I believe continuity is necessary is because of the similarity to the fact that [imath] x^{\alpha} x^{\beta} = x^{\alpha + \beta} [/imath] for all [imath] \alpha,\beta \in \mathbb{R} [/imath]. Irrational powers can be defined either via continuity (i.e. if [imath] \alpha \ [/imath] is irrational, then [imath] x^{\alpha}:= \lim_{q\rightarrow \alpha} x^q \ [/imath] where q takes on only rational values) or by using the exponential and natual log functions, and either way proving the desired identity boils down to continuity. I have come up with one example that satisfies (something similar to) (1) and not (2), but it doesn't quite fit the bill: [imath] \ [/imath] Define [imath] \phi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q} \ [/imath] defined by [imath] \phi(a+b\sqrt{2}) = a+b [/imath]. Then [imath] \phi(x+y) = \phi(x)+\phi(y) \ [/imath] but if [imath] \alpha=c+d\sqrt{2} \ [/imath] then [imath] \phi(\alpha(a+b\sqrt{2})) = ac+2bd + ad+bc \neq \alpha \ \phi(a+b\sqrt{2}) [/imath]. [imath] \ [/imath] The problem is that even though [imath] \mathbb{Q}(\sqrt{2}) \ [/imath] is a vector space over [imath] \mathbb{Q} [/imath], the [imath] \alpha \ [/imath] is coming from [imath] \mathbb{Q}(\sqrt{2}) \ [/imath] instead of the base field [imath] \mathbb{Q} [/imath]. | 415832 | Examples of group morphisms which are not linear applications? (In [imath]\mathbb R[/imath].)
I was wondering whether there are group morphisms in [imath]\mathbb R[/imath] which are not linear applications. I would have guessed that it exists but I cannot think of an example. Would someone have some examples? |
11150 | Zero to the zero power – is [imath]0^0=1[/imath]?
Could someone provide me with a good explanation of why [imath]0^0=1[/imath]? My train of thought: [imath]x>0[/imath] [imath]0^x=0^{x-0}=0^x/0^0[/imath], so [imath]0^0=0^x/0^x=\,?[/imath] Possible answers: [imath]0^0\cdot0^x=1\cdot0^0[/imath], so [imath]0^0=1[/imath] [imath]0^0=0^x/0^x=0/0[/imath], which is undefined PS. I've read the explanation on mathforum.org, but it isn't clear to me. | 340566 | What is [imath]0^0[/imath]? Should we define [imath]0^0[/imath] on its correctness or convenience?
What is [imath]0^0[/imath] ? I have read many debates about this question. The debate has been going on at least since the early 19th century. At that time, most mathematicians agreed that [imath]0^0[/imath] = 1, until in 1821 Cauchy listed [imath]0^0[/imath] along with expressions like [imath]\frac{0}{0}[/imath] in a table of undefined forms. On Wikipedia, there is a thread for the same. On Quora, i read a discussion about the same. The question is not to explain for [imath]0^0[/imath]=1 or [imath]0^0[/imath]=0 or it is undefined. What sholuld we conclude for defining [imath]0^0[/imath] ? |
49229 | Why can ALL quadratic equations be solved by the quadratic formula?
In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't. Why can I use [imath]x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}[/imath] to solve all quadratic equations? | 832881 | Where does the quadratic formula come from?
Everywhere I look, the [imath]ax^2+bx+c[/imath] portion of the quadratic formula is listed as given. Does anyone know where this comes from? Edit How can we prove that (x+y)^2 = ax^2+bx+c? |
26772 | Birthday-coverage problem
I heard an interesting question recently: What is the minimum number of people required to make it more likely than not that all 365 possible birthdays are covered? Monte Carlo simulation suggests 2287 ([imath]\pm 1[/imath], I think). More generally, with [imath]p[/imath] people, what is the probability that for each of the 365 days of the year, there is at least one person in the group with that birthday? (Yes, ignoring the leap-day.) | 1254160 | Approximation for Coupon Collector Probability
I was calculating the probability to draw all items in a list of N items, by picking one randomly, replacing it in the list, etc. (Edit: It's the Coupon collector Problem) I found this formula after n picking. [imath] \sum_{i=0}^N (-1)^{N-i}{\binom{N}{i}}\left(\frac{i}{N}\right)^n [/imath] But, I have to calculate this formula with N > 1000000000 in a java program so to avoid a huge loop (the sum). I'm looking for a simplification, any approximation formula will help me. |
8379 | [imath]|G|>2[/imath] implies [imath]G[/imath] has non trivial automorphism
Well, this is an exercise problem from Herstein which sounds difficult: How does one prove that if [imath]|G|>2[/imath], then [imath]G[/imath] has non-trivial automorphism? The only thing I know which connects a group with its automorphism is the theorem, [imath]G/Z(G) \cong \mathcal{I}(G)[/imath] where [imath]\mathcal{I}(G)[/imath] denotes the Inner- Automorphism group of [imath]G[/imath]. So for a group with [imath]Z(G)=(e)[/imath], we can conclude that it has a non-trivial automorphism, but what about groups with center? | 677939 | Every finite group of order more than two has a nontrivial automorphism
I want to prove that every finite group [imath]G[/imath] of order more than 2 has a nontrivial automorphism. I've seen this question answered on this site for infinite groups, but the proofs given use the fact that if [imath]g^2=1[/imath] for every [imath]g[/imath] in [imath]G[/imath], then [imath]G[/imath] is a vector space over [imath]\mathbb{Z}_2[/imath]. This is an exercise in Herstein's text that appears before the section on (the fundamental theorem of) finite abelian groups. I think I can prove this result using that theorem, but was wondering if there are more elementary proofs. Here is my proof: If [imath]G[/imath] is nonabelian, then [imath]\exists x \in G[/imath] such that the map [imath](T_x: g \mapsto x^{-1}gx)[/imath] is a nontrivial automorphism of [imath]G[/imath]. So suppose [imath]G[/imath] is abelian. Then the map [imath]g \mapsto g^{-1}[/imath] is an automorphism of [imath]G[/imath]; this automorphism is nontrivial if some element in [imath]G[/imath] has order at least 3. If every element in [imath]G[/imath] has order 2, then by the fundamental theorem on finite abelian groups, [imath]G \cong C_2 \times \cdots \times C_2[/imath] is the direct product of [imath]k[/imath] copies of [imath]C_2[/imath] for some [imath]k \ge 2[/imath]. A map that interchanges the generators of the first two copies and fixes the remaining [imath]k-2[/imath] copies yields a nontrivial automorphism of [imath]G[/imath]. QED. |
237009 | An operator has closed range if and only if the image of some closed subspace of finite codimension is closed.
Let [imath]B[/imath] be a Banach space, [imath]H,K[/imath] be closed subspaces and let [imath]K[/imath] be finite dimensional. Suppose [imath]B = H\oplus K[/imath] and [imath]T:B\to B[/imath] is a bounded linear operator. How do I show that [imath]T(B)[/imath] is closed [imath]\iff[/imath] [imath]T(H)[/imath] is closed ? | 1979712 | The image of bounded linear operator is closed iff the image of its subspace is closed
Banach space [imath]E[/imath] is the direct sum of its closed subspaces [imath]L[/imath] and [imath]M[/imath]. [imath]M[/imath] is finite dimensional. [imath]T:E \rightarrow E[/imath] is a bounded linear operator. I'm asked to prove that [imath]T(E)[/imath] is closed iff [imath]T(L)[/imath] is closed. I have only learned the open mapping theorem and closed graph theorem, and have no idea how to prove that the image is closed. Any hint will help. Thank you. |
255 | Why does the series [imath]\sum_{n=1}^\infty\frac1n[/imath] not converge?
Can someone give a simple explanation as to why the harmonic series [imath]\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots [/imath] doesn't converge, on the other hand it grows very slowly? I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks. | 369774 | Why does Σ1/x diverge?
Why does the following series diverge? [imath]\sum_{n = 1}^\infty\frac{1}{n}[/imath] I've tried to make sense of it, but can't seem to wrap my head around it. Thanks. |
107336 | Why doesn't [imath]d(x_n,x_{n+1})\rightarrow 0[/imath] as [imath]n\rightarrow\infty[/imath] imply [imath]{x_n}[/imath] is Cauchy?
What is an example of a sequence in [imath]\mathbb R[/imath] with this property that is not Cauchy? I know that Cauchy condition means that for each [imath]\varepsilon>0[/imath] there exists [imath]N[/imath] such that [imath]d(x_p,x_q)<\varepsilon[/imath] whenever [imath]p,q>N[/imath]. | 1985521 | [imath]\{x_n\}[/imath] be a bounded sequence of distinct real numbers , [imath]|x_{n+1}-x_n|<|x_n-x_{n-1}|,\forall n\in \mathbb N[/imath] , then is [imath]\{x_n\}[/imath] convergent?
Let [imath]\{x_n\}[/imath] be a bounded sequence of distinct real numbers such that [imath]|x_{n+1}-x_n|<|x_n-x_{n-1}|,\forall n\in \mathbb N[/imath] , then is it true that [imath]\{x_n\}[/imath] converges ? The motivation for this comes from the fixed point theorem that if [imath]X[/imath] is compact metric space and [imath]f:X\to X[/imath] is a function such that [imath]d(f(x),f(y))<d(x,y),\forall x,y \in X , x\ne y[/imath] then [imath]f[/imath] has a fixed point. |
145065 | use contradiction to prove that the square root of [imath]p[/imath] is irrational
On a practice exam, our teacher provides us with this question and this answer. Let [imath]p[/imath] be a prime number. Use contradiction to prove that [imath]\sqrt{p}[/imath] is irrational. ANSWER: By way of contradiction, assume [imath]\sqrt{p}[/imath] is rational. Then there exist [imath]a, b \in \mathbb{Z}[/imath] with [imath]b\neq 0[/imath] such that [imath]\sqrt{p} = \frac{a}{b}[/imath]. Without loss of generality, we may assume [imath]\text{gcd}(a,b) = 1[/imath]. Then [imath]p = \frac{a^2}{b^2}[/imath]. Thus [imath]p | a^2[/imath] which implies [imath]p | a[/imath], i.e., [imath]\exists k \in \mathbb{Z}[/imath] such that [imath]p k = a[/imath]. We now have [imath]p b^2 = (\pi k)^2 = p(p k^2)[/imath], so [imath]p b^2 = a^2[/imath]. Since [imath]p \neq 0[/imath], [imath]b^2 = p k^2[/imath], which means [imath]p|b[/imath]. Thus [imath]p[/imath] is a common factor of [imath]a[/imath] and [imath]b[/imath]. This is a contradiction as [imath]a[/imath] and [imath]b[/imath] are relatively prime. My question is, how do you know to assume that the [imath]\text{gcd} (a,b)=1[/imath]? It seems really random and I don't know why the proof jumps there. | 2682280 | prove that if [imath]p[/imath] is a prime number, then [imath]\sqrt{p}[/imath] is an irrational number.
[imath]\sqrt{p}[/imath] is rational. [imath]\sqrt{p}=\frac{a}{b}[/imath], where [imath]a,b[/imath] are integers with [imath]\gcd(a,b)=1[/imath]. [imath]a^2=b^2p[/imath]. Since [imath]p[/imath] divides [imath]a^2[/imath], [imath]p[/imath] divides [imath]a[/imath]. [imath]a=kp[/imath]. [imath]a^2=k^2p^2=b^2p[/imath] [imath]p=\frac{b^2}{k^2}\Rightarrow\sqrt{p}=\frac{b}{k}\Rightarrow\frac{a}{b}=\frac{b}{k}[/imath] [imath]b^2=ak\Rightarrow b=\sqrt{ak}[/imath]. [imath]\sqrt{p}=\frac{a}{b}=\frac{a}{\sqrt{ak}}=\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{k}}[/imath] which has [imath]\gcd(a,b)=\sqrt{a}\neq 1[/imath] , a contradiction. ([imath]a\neq 1[/imath] because the only number that divides 1 is 1, but 1 is not a prime from [imath]a=kp[/imath]) Hence, [imath]p[/imath] is irrational. Is this legit? |
12139 | Number of relations that are both symmetric and reflexive
Consider a non-empty set A containing n objects. How many relations on A are both symmetric and reflexive? The answer to this is [imath]2^p[/imath] where [imath]p=[/imath] [imath]n \choose 2[/imath]. However, I dont understand why this is so. Can anyone explain this? | 359093 | Determine the number of binary relations on [imath]A \times A[/imath] that satisfy the following:
Let [imath]|A| = 8[/imath]. Determine the number of binary relations on [imath]A \times A[/imath] that satisfy the following: A) Symmetric B) Neither reflexive or irreflexive C) Reflexive and symmetric D) Irreflexive and anti-symmetric ... I know that a relation [imath]R[/imath] on a set [imath]A[/imath] is reflexive if [imath](a,a) \in R[/imath] for every element [imath]a \in A[/imath] and that it's symmetric if [imath](b,a) \in R[/imath] whenever [imath](a,b) \in R[/imath] for all [imath]a,b \in R[/imath]. This is what I have so far: Reflexive: [imath]2^{[n(n+1)]/2} = 2^{(8*9)/2} = 2^{36}[/imath] Symmetric: [imath]2^{(n^2 - n)} = 2^{56}[/imath] A) = [imath]2^{56}[/imath] B)[imath]2^{64}[/imath] - [imath]2^{57}[/imath] C) [imath]2^{28}[/imath] D)...? Any help would be greatly appreciated. |
97054 | Suppose [imath]H[/imath] is the only subgroup of order [imath]o(H)[/imath] in the finite group [imath]G[/imath]. Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath].
Suppose [imath]H[/imath] is the only subgroup of order [imath]o(H)[/imath] in the finite group [imath]G[/imath]. Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. I've been trying this problem for quite a while but to no avail. What I can't understand is, how do you relate the subgroup being normal/abnormal to its order? This question is from I.N.Herstein's book Topics in Algebra Page 53, Problem no. 9. This is NOT a homework problem!! I'm studying this book on my own. | 511930 | If [imath]G[/imath] has only one subgroup of order [imath]n[/imath], then that Subgroup is Normal
How can I show that if some group [imath]G[/imath] has only one subgroup [imath]K[/imath] of order [imath]n[/imath], then [imath]K[/imath] is a normal subgroup? Would that mean that it only has one subgroup total? If so then I guess that makes sense. |
160847 | Polynomials irreducible over [imath]\mathbb{Q}[/imath] but reducible over [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath]
Let [imath]f(x) \in \mathbb{Z}[x][/imath]. If we reduce the coefficents of [imath]f(x)[/imath] modulo [imath]p[/imath], where [imath]p[/imath] is prime, we get a polynomial [imath]f^*(x) \in \mathbb{F}_p[x][/imath]. Then if [imath]f^*(x)[/imath] is irreducible and has the same degree as [imath]f(x)[/imath], the polynomial [imath]f(x)[/imath] is irreducible. This is one way to show that a polynomial in [imath]\mathbb{Z}[x][/imath] is irreducible, but it does not always work. There are polynomials which are irreducible in [imath]\mathbb{Z}[x][/imath] yet factor in [imath]\mathbb{F}_p[x][/imath] for every prime [imath]p[/imath]. The only examples I know are [imath]x^4 + 1[/imath] and [imath]x^4 - 10x^2 + 1[/imath]. I'd like to see more examples, in particular an infinite family of polynomials like this would be interesting. How does one go about finding them? Has anyone ever attempted classifying all polynomials in [imath]\mathbb{Z}[x][/imath] with this property? | 2665525 | converse of reduction criterion?
By the reduction criterion, I mean the following test for the irreducibility of polynomial with Dedekind domain coefficients. Let [imath]\mathfrak{m}[/imath] be maximal in Dedekind domain A and [imath]f(X)\in A[X] [/imath]. If [imath]f[/imath] reduced modulo [imath]\mathfrak{m}[/imath] is irreducible in [imath]A/\mathfrak{m}[/imath], then [imath]f[/imath] is irreducible in [imath]A[/imath]. I know that the converse doesn't generally hold. There are irreducible polynomials with integer coefficients that is reducible when reduced modulo [imath]p[/imath] for some prime [imath]p[/imath]. So I conjecture the following for polynomials with integer coefficients. For monic polynomial [imath]f(X)\in \mathbb{Z}[X][/imath], if [imath]f(X)[/imath] modulo [imath]p[/imath] is reducible for all prime [imath]p[/imath], then [imath]f(X)[/imath] is reducible in [imath]\mathbb{Z}[X][/imath]. I have thought of some galois theoretic approach but couldn't quite reach the conclusion. How would I prove or disprove this statement? |
21581 | How does one actually show from associativity that one can drop parentheses?
I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this? [imath](a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?[/imath] Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks. | 1555868 | Induction with an associative operator
I've been starting to play around with some properties of groups, and I wanted to prove this claim to make things simpler for me in the future. If [imath]G[/imath] is closed under and operation [imath]\cdot[/imath] that is also associative, I want to show that [imath]\forall a_1, a_2, ..., a_n \in G[/imath], no matter how I choose to bracket [imath]a_1 \cdot a_2 \cdot ... \cdot a_n[/imath] while retaining the order, I will end up with the same element in [imath]G.[/imath] My friend told me that induction on the number of terms would probably be my best bet, and I see how the [imath]n = 1 \wedge n = 2[/imath] cases are trivial with the [imath]n = 3[/imath] case implied by associativity. However I am having some problems Figuring out how to work out the induction step. It seems to me that my induction hypothesis is not helping me when proving for brackets that include the [imath]n+1[/imath]th element together with the [imath]n[/imath]th element. |
305991 | Why is Grassmanian a projective variety?
The grassmanian [imath] \mathbf G(r,n)[/imath] is the set of all [imath]k[/imath]-dimensinal subspaces of a [imath]n[/imath]-dimensional vector space. I understand how [imath] \mathbf G(r,n)[/imath] can be embebbed in the projective space [imath]\mathbb P^{{n \choose r }-1}[/imath].Let [imath]\theta[/imath] be a full rank [imath]d \times n[/imath] matrix.We denote the submatrix with column indices [imath]\sigma \subset \{1,..,n\}[/imath] and [imath] |\sigma|=n[/imath] by [imath]\theta_\sigma[/imath] , so the corresponding minor is [imath]det(\theta_\sigma ).[/imath] Then list [imath](det(\theta_\sigma ) | σ ⊆ [n])[/imath] minors up to scale identifies the row span of Θ uniquely.So, each r-dimensional subspace can be thought of as a point in [imath]\mathbb P^{{n \choose r }-1}[/imath]. But, what is the set of homogenous polynomials in [imath]K[x_1,x_2,...,x_{n \choose r}][/imath] that [imath] \mathbf G(r,n)[/imath] satisfies? | 49190 | Plücker Relations
Let [imath]K[/imath] be a field, [imath]1 \leq d \leq n[/imath] integers and [imath]V[/imath] an [imath]n[/imath]-dimensional vector space. The Plücker relations are quadratic forms on [imath]\wedge^d V[/imath] whose zero set is exactly the set of decomposable vectors in [imath]\wedge^d V[/imath] (i.e. which are of the form [imath]v_1 \wedge ... \wedge v_d[/imath]), thus describing the ideal corresponding to the Plücker embedding [imath]\text{Gr}_d(V) \to \mathbb{P}(\wedge^d V)[/imath]. But in every book I've read so far, these Plücker relations are constructed by means of many identifications between duals, exterior powers, etc. so that I am not able to write them down explicitely. Although I've tried it, many signs and sums confuse me. Question. Is it possible to write down these Plücker relations explicitely as a set of polynomials in the ring [imath]K[\{x_H\}][/imath], where [imath]H[/imath] runs through the subsets of [imath]\{1,...,n\}[/imath] with [imath]d[/imath] elements? (Of course it is possible, but I wonder how do this in general) Edit: Following the answer below, here is the Answer: Instead of using these subsets [imath]H[/imath], use indices [imath]1 \leq i_1 < ... < i_d \leq n[/imath], and extend the definition of [imath]x_{i_1,...,i_d}[/imath] to all [imath]d[/imath]-tuples in such a way that [imath]x_{i_1,...,i_d}=0[/imath] if these [imath]i_j[/imath] are not pairwise distinct, and otherwise [imath]x_{i_1,....,i_d} = sign(\sigma) \cdot x_{i_{\sigma(1)},...,i_{\sigma(d)}}[/imath], where [imath]\sigma[/imath] is the unique permutation of [imath]1,...,d[/imath] which makes [imath]i_{\sigma(1)} < ... < i_{\sigma(d)}[/imath]. Then the Plücker relations are [imath]\sum\limits_{j=0}^{d} (-1)^j x_{i_1,...,i_{d-1},k_j} * x_{k_0,...,\hat{k_j},...,k_d} = 0[/imath] for integers [imath]i_1,...,i_{d-1},k_0,...,k_d[/imath] between [imath]1,...,n[/imath]. |
297770 | Prove that [imath]\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1[/imath]
Prove that [imath]\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1[/imath] Hints- [imath]1[/imath]. Use Euclids Lemma [imath]2[/imath]. [imath]2^a=2^{a\%c}\mod (2^c)-1[/imath] [imath]3[/imath]. If [imath]a=q\cdot b+c[/imath] then [imath]2^a=(2^c)^q\cdot 2^r[/imath] | 225289 | Proving that [imath]\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1[/imath]
Somewhere on Stack Exchange I saw the equation [imath]\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.[/imath] I had never seen this before, so I started trying to prove it. Without success... Can anyone explain me (so actually prove) why this equation is true? And can we say the same when replacing the '[imath]2[/imath]' by any integer number '[imath]a[/imath]'? |
40996 | Prove that if [imath](ab)^i = a^ib^i \forall a,b\in G[/imath] for three consecutive integers [imath]i[/imath] then G is abelian
I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4): If [imath]G[/imath] is a group such that [imath](ab)^i = a^ib^i[/imath] for three consecutive integers [imath]i[/imath] for all [imath]a, b\in G[/imath], show that [imath]G[/imath] is abelian. I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this. My approach: Let [imath]j=i+1, k=i+2[/imath] for some [imath]i\in \mathbb{Z}[/imath]. Then we have that [imath](ab)^i = a^ib^i[/imath], [imath](ab)^j = a^jb^j[/imath] and [imath](ab)^k = a^kb^k[/imath]. If [imath](ab)^k = a^kb^k[/imath], then [imath]a^jb^jab =a^jab^jb[/imath]. We cancel on the left and right and we have [imath]b^ja = ab^j[/imath], that is [imath]b^iba = ab^j[/imath]. Multiply both sides by [imath]a^i[/imath] on the left and we get [imath]a^ib^iba = a^jb^j[/imath], so [imath](ab)^iba = (ab)^j[/imath]. But that is [imath](ab)^iba = (ab)^iab[/imath]. Cancelling on the left yields [imath]ab=ba[/imath], which holds for all [imath]a,b \in G[/imath], and therefore, [imath]G[/imath] is abelian. Thanks! | 781051 | showing the group is abelian
I need someone to guide me to solve the following problem Let [imath]a, b \in G[/imath]. a. f [imath](a *b)^i = a^i * b^i[/imath] for three consecutive natural numbers, then show that [imath]G[/imath] is Abelian. b. If the above property holds only for two numbers, then it is not necessarily Abelian. |
24413 | Is there a function with infinite integral on every interval?
Could give some examples of nonnegative measurable function [imath]f:\mathbb{R}\to[0,\infty)[/imath], such that its integral over any bounded interval is infinite? | 1572138 | Nowhere integrable function
Does there exist a function [imath]f : I \to \mathbb{R}[/imath] defined on an interval [imath]I \subseteq \mathbb{R}[/imath] that is measurable but not integrable on any compact subinterval [imath][a,b] \subseteq I[/imath]? One can try to call for Lusin's continuity theorem: If [imath]m[/imath] denotes the Lebesgue measure then for each [imath]\varepsilon > 0[/imath] there exists a compact set [imath]K_\varepsilon \subseteq I[/imath] such that [imath]m(I \setminus K_\varepsilon) < \varepsilon[/imath] and [imath]f|_{K_\varepsilon}[/imath] is continuous. Now the question is whether for some [imath]\varepsilon > 0[/imath] the compact set [imath]K_\varepsilon[/imath] contains an interval or equivalently has non-empty interior. In general, a compact subset of [imath][0,1][/imath] (or [imath]\mathbb{R}[/imath]) can have both empty interior and positive Lebesgue measure and even more: for each [imath]\varepsilon \in (0,1)[/imath] the fat Cantor set is compact, has Lebesgue measure [imath]\varepsilon[/imath] and empty interior. |
77155 | Irreducible polynomial which is reducible modulo every prime
How to show that [imath]x^4+1[/imath] is irreducible in [imath]\mathbb Z[x][/imath] but it is reducible modulo every prime [imath]p[/imath]? For example I know that [imath]x^4+1=(x+1)^4\bmod 2[/imath]. Also [imath]\bmod 3[/imath] we have that [imath]0,1,2[/imath] are not solutions of [imath]x^4+1=0[/imath] then if it is reducible the factors are of degree [imath]2[/imath]. This gives that [imath]x^4+1=(x^2+ax+b)(x^2+cx+d)[/imath] and solving this system of equations [imath]\bmod 3[/imath] gives that [imath]x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3[/imath]. But is there a simpler method to factor [imath]x^4+1[/imath] modulo a prime [imath]p[/imath]? | 427439 | Why is [imath]X^4+1[/imath] reducible over [imath]\mathbb F_p[/imath] with [imath]p \geq 3,[/imath] prime
I have proven that in [imath]\mathbb F_{p^2}^*[/imath] exists an element [imath]\alpha[/imath] with [imath]\alpha^8 = 1[/imath]. Let [imath]f(X) := X^4+1 \in \mathbb F_p[X][/imath]. How can I prove that [imath]f[/imath] is reducible over [imath]\mathbb F_p[/imath]? Has [imath]f[/imath] a zero in [imath]\mathbb F_p[/imath] ? |
133578 | When can we find holomorphic bijections between annuli?
I'm self-studying some complex analysis, and apparently holomorphic bijections between two annuli exist precisely when the ratios of the radii are the same. More exactly, if [imath]A_{\sigma,\rho}=\{z\in\mathbb{C}:\sigma<|z|<\rho\}[/imath], then there is a holomorphic bijection between [imath]A_{\sigma,\rho}[/imath] and [imath]A_{\sigma',\rho'}[/imath] iff [imath]\rho/\sigma=\rho'/\sigma'[/imath]. Is there a reference where this fact is proven? Or can a proof be included here if it's not overly involved? Thanks. | 1870987 | Proof of a necessary and sufficient condition between annuli centered at the origin
What is a simple way to prove that two annuli [imath]A_1 = {z: r_1 < |z| < R_1}[/imath] and [imath]A_2 = {z: r_2 < |z| < R_2}[/imath] are conformally equivalent if and only if [imath]R_1/r_1 = R_2/r_2[/imath], using standard results of complex analysis? I'm not looking for a concise proof so much as one that uses elementary concepts. |
202072 | Finiteness of the Algebraic Closure
Let [imath]\mathbb R[/imath] be the field of real numbers. Its algebraic closure, the field [imath]\mathbb C[/imath], is a finite extension of [imath]\mathbb R[/imath], which has degree 2. Are there other examples of fields (not algebraic closed) such that its algebraic closure is a finite extension? | 1743361 | Infinite extensions of "finite degree under [imath]\mathbb{Q}[/imath]"
Consider an algebraic extension [imath]K[/imath] of [imath]\mathbb{Q}[/imath]. The degree [imath][K:\mathbb{Q}][/imath] of [imath]K[/imath] is defined as the dimension of the extension considered as a vector space. Now, let [imath]\overline{\mathbb{Q}}[/imath] be algebraic closure of [imath]\mathbb{Q}[/imath]. My question is, Can we built an arbitrary algebraic extension [imath]F[/imath] of [imath]\mathbb{Q}[/imath] such that [imath][\overline{\mathbb{Q}}:F]=n[/imath], for any [imath]n\in \mathbb{N}[/imath]? How? |
112067 | How discontinuous can a derivative be?
There is a well-known result in elementary analysis due to Darboux which says if [imath]f[/imath] is a differentiable function then [imath]f'[/imath] satisfies the intermediate value property. To my knowledge, not many "highly" discontinuous Darboux functions are known--the only one I am aware of being the Conway base 13 function--and few (none?) of these are derivatives of differentiable functions. In fact they generally cannot be since an application of Baire's theorem gives that the set of continuity points of the derivative is dense [imath]G_\delta[/imath]. Is it known how sharp that last result is? Are there known Darboux functions which are derivatives and are discontinuous on "large" sets in some appropriate sense? | 485350 | The continuity of the derivative of a differentiable function
Let [imath]f:[a, b] \rightarrow \mathbb{R}[/imath] be a function that is differentiable everywhere. What can we say about the continuity of [imath]f^{(1)}[/imath]? The only results that is related to this I that I can find is that if [imath]f^{(1)}(a) < \lambda < f^{(1)}(b)[/imath], then there is [imath]x\in (a,b)[/imath] s.t. [imath]f(x) = \lambda[/imath] and so [imath]f^{(1)}[/imath] may not have any discontinuities of the first kind. An example of a function that has a derivative that is discontinuous is [imath]f(x) = x^{2}\sin(\frac{1}{x})[/imath] [imath] \text{for } x\neq{0}[/imath] and [imath]f(x)=0[/imath] if [imath]x=0[/imath] which has a derivative that has a discontinuity of the second kind at [imath]0[/imath]. I'm not sure how to use this example to create other examples, say [imath]f^{(1)}[/imath] is discontinuous on a: dense subset of [imath][a,b][/imath], on a subset of measure [imath]\frac{b-a}{2}[/imath], etc. Also what happens if we replace " differentiable everywhere" by "differentiable a.e."? |
287 | How come [imath]32.5 = 31.5[/imath]? (The "Missing Square" puzzle.)
Below is a visual proof (!) that [imath]32.5 = 31.5[/imath]. How could that be? (As noted in a comment and answer, this is known as the "Missing Square" puzzle.) | 1097219 | How do the dimensions of an "infinite" bar of chocolate change?
Here is a GIF image illustrating a supposedly "infinite" supply of white chocolate. After watching this repeatedly, I can't definitively say why it doesn't add up. It clearly can't be infinite and the sizes of the pieces don't seem to be changed/edited. My guess is that the volume of the spaces between pieces somehow adds up to the final piece's volume. However, the real question I have is: how have the dimensions of the array of chocolate changed? That is, if you start with a [imath]6\times 4[/imath] grid of chocolate pips, what are the final dimensions of the almost complete grid? I figure that height need not be considered because the cuts are made normal to the table surface. |
3510 | How to prove Euler's formula: [imath]e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)[/imath]?
Could you provide a proof of Euler's formula: [imath]e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)[/imath]? | 754370 | Why is [imath]sinx[/imath] the imaginary part of [imath]e^{ix}[/imath]?
Most of us who are studying mathematics are familiar with the famous [imath]e^{ix}=cos(x)+isin(x)[/imath]. Why is it that we have [imath]e^{ix}=cos(x)+isin(x)[/imath] and not [imath]e^{ix}=sin(x)+icos(x)[/imath]? I haven't studied Complex Analysis to know the answer to this question. It pops up in Linear Algebra, Differential Equations, Multivariable Equations and many other fields. But I feel like textbooks and teachers just expect us students to take it as given without explaining it to a certain extent. I also couldn't find any good article that explains this. |
92147 | Show [imath]\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = \|f\|_{\infty}[/imath] for [imath]f \in L^{\infty}[/imath]
I have a question that I need help with getting started (possibly I would be back for more help). I have a measure space [imath](X,A,\mu)[/imath] that is finite, and [imath]f \in L^{\infty}(\mu)[/imath]. Also, defined is [imath]a_n = \int_X\,|f|^n\,d\mu[/imath]. I need to show that the limit is: [imath]\lim_{n\to\infty}\,\frac{a_{n+1}}{a_n} = \|f\|_{\infty} .[/imath] I am stuck on getting started, anybody have any suggestions? thanks much | 469212 | Limit of consecutive Lp norms
I've been wrestling with the following proof off and on for a number of days, and I'm in need of a nudge in the right direction. Let [imath](E,\mathcal{M},\mu)[/imath] be a measure space with [imath]0 < \mu(E) < \infty[/imath]. Consider [imath]f \in L^\infty(E)[/imath] with [imath]\|f\|_\infty > 0[/imath]; show that [imath] \lim_{n\to\infty} \|f||_n = \lim_{n\to\infty} \frac{\|f\|_{n+1}^{n+1}}{\|f\|_n^n} = \|f\|_\infty [/imath] Now I'm familiar with the result and proof that under these circumstances [imath]\lim_{n\to\infty} \|f||_n = \|f\|_\infty[/imath], so I've been focusing on somehow showing the two limits to be the same. |
978 | How to prove and interpret [imath]\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))[/imath]?
Let [imath]A[/imath] and [imath]B[/imath] be two matrices which can be multiplied. Then [imath]\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).[/imath] I proved [imath]\operatorname{rank}(AB) \leq \operatorname{rank}(B)[/imath] interpreting [imath]AB[/imath] as a composition of linear maps, observing that [imath]\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)[/imath] and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information. How to manage [imath]\operatorname{rank}(AB) \leq \operatorname{rank}(A)[/imath]? Is there a nice interpretation like the previous one? | 419922 | [imath]\operatorname{rank}AB\leq \operatorname{rank}A, \operatorname{rank}B[/imath]
Prove that if [imath]A,B[/imath] are any such matrices such that [imath]AB[/imath] exists, then [imath]\operatorname{rank}AB \leq \operatorname{rank}A,\operatorname{rank}B[/imath]. I came across this exercise while doing problems in my textbook, but am not sure where to start for the proof of this. I think columnspace might be involved in the proof, although I am not sure. |
77662 | Proof for convergence of a given progression [imath]a_n := n^n / n![/imath]
"Examine whether the given progressions converge (eventually improperly) and determine their limits where applicable. (a) [imath](a_n)_{n\in\mathbb{N}}:=\frac{n^n}{n!}[/imath] [...]" I am having problems getting this homework done as I have no clue about convergency at all. Through Mathematica I know that [imath]\lim\limits_{n\to\infty}a_n=+\infty[/imath] however I don't know how to proove it! During my research I found out that for sufficient big [imath]n[/imath] the inequality [imath]x^n\leq n!\leq n^n[/imath] for [imath]x\in\mathbb{R}[/imath] and [imath]n\in\mathbb{N}[/imath] is true. However this doesn't help me at all. Having a look at our lecture notes I developed the following statement: every [imath]x^n[/imath] with [imath]|x| \geq 1[/imath] diverges as [imath]x^n[/imath] is unbounded and therefore has the improper limit [imath]\lim\limits_{n\to\infty}x^n=+\infty[/imath]. Furthermore it seems obvious to me that I need to express the terms in a simpler way, like fractions converging to 0 etc, but I don't know how. Neither some work with the [imath]\varepsilon[/imath]-definition of limits in combination with the triangle inequality helped me out. Any suggestions or hints? Information: This is "Analysis for computer scientist" and therefore we haven't "officially" learned any fancy tricks like L'Hôpital's rule etc. just basic properties of [imath]\mathbb{R}[/imath] and some inequalities as this is an introductory lecture. | 676658 | Limit of [imath]\frac {n^n}{n!}[/imath]
I have to prove that [imath]\lim_{n\to \infty} \frac {n^n} {n!}=\infty[/imath] I've tried to look for a lower bound that also converges to [imath]\infty[/imath] (I don't know if I'm explainig myself correctly), but I haven't found one yet. Applying L'Hôpital is way too complicated in [imath]n![/imath], and the epsilon proof does not work as I have no way whatsoever of finding N. Any ideas? |
180169 | Summation equation for [imath]2^{x-1}[/imath]
Since everyone freaked out, I made the variables are the same. [imath] \sum_{x=1}^{n} 2^{x-1} [/imath] I've been trying to find this for a while. I tried the usually geometric equation (Here) but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need: 1, 3, 7, 15, 31, 63 If my math is correct. | 1921892 | Why is it intuitively "obvious" that: [imath]2^0+2^1+2^2+...+2^n=2^{n+1}-1[/imath]
Why is it "obvious" that: [imath]2^0+2^1+2^2+...+2^n=2^{n+1}-1[/imath] And how about: [imath]2^{-1}+2^{-2}+2^{-3}+...+2^{-n}=1-2^{-n}[/imath] I am looking for some intuition about these facts. Is there a similar pattern when the base is [imath]3[/imath]? |
154900 | Indefinite integral of secant cubed
I need to calculate the following indefinite integral: [imath]I=\int \frac{1}{\cos^3(x)}dx[/imath] I know what the result is (from Mathematica): [imath]I=\tanh^{-1}(\tan(x/2))+(1/2)\sec(x)\tan(x)[/imath] but I don't know how to integrate it myself. I have been trying some substitutions to no avail. Equivalently, I need to know how to compute: [imath]I=\int \sqrt{1+z^2}dz[/imath] which follows after making the change of variables [imath]z=\tan x[/imath]. | 299886 | Integration by parts question,, possibly a circular example
I am having trouble figuring this out. [imath]\int_0^{1/3} \sec^3(\pi x) \, dx[/imath] We are currently doing integration by parts,, so I set [imath]g(x)=\sec^3(\pi x)[/imath] and [imath]f'(x)=1[/imath]. I arrived at: [imath]x\sec^3(\pi x) - 3\pi \int x\tan(\pi x)\sec^3(\pi x) dx[/imath] I am lost here, I tried u substitution for [imath]\tan(\pi x)[/imath] so that I can get rid of [imath]\sec^2(\pi x)[/imath] but that doesnt help. |
68120 | Why is [imath]f(x) = x\phi(x)[/imath] one-to-one?
I noticed that [imath]f(x) = x\phi(x)[/imath] seems to be one-to-one, where [imath]\phi(x)[/imath] is Euler's Phi function. In particular, I'm writing some numerical python code and the line I have looks something like sorted([n*phi(n) for n in range(1,1000)]) and there are no duplicates in the list. First, is it one-to-one? Second, if it is, is there a simple proof sketch? | 707658 | Totient function: show that numbers are equal
I am a bit lost. It seems to be true, but I am not sure how to prove this to myself. If [imath]m*ϕ(m)=n*ϕ(n)[/imath] then [imath]m=n[/imath] It is clear that this property would hold if m and n were prime, but I am not sure how to show this for the general situation. Help is greatly appreciated. Thank you. |
87756 | When is [imath]\sin(x)[/imath] rational?
Obviously, there are some points (e.g. [imath]\pi[/imath], [imath]30^\circ[/imath]) but I am unsure if there are more. How can it be proved that there are no more points, or, if there are, what those points will be? EDIT: I largely meant to ask what is supposed in the first comment. Put another way, are there numbers such that [imath]\sin x[/imath] and [imath]\arcsin x[/imath] are both rational? | 1713188 | Is it possible to calculate [imath] \sin(\alpha) [/imath] (and other trigonometric functions) as a rational number?
I am creating a computer library for arbitrary-precision calculations, by expressing numbers as rationals (with an arbitrary-precision numerator and denominator). Now, I am exploring the possibility to add trigonometric functions to this library. I know from college that certain values like [imath]\sin(\frac{1}{2}) = \frac{\pi}{2} [/imath] and [imath]\sin( [/imath] are defined and easy to remember. Is there a way to find rational solutions of [imath]\sin(\alpha)[/imath] for all possible angles [imath]\alpha[/imath] ? |
7473 | Prove that [imath]\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1[/imath]
For all [imath]a, m, n \in \mathbb{Z}^+[/imath], [imath]\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1[/imath] | 428004 | A Greatest Common Divisor Property
Show that: If [imath]c|a^m-1[/imath] and [imath]c|a^n-1[/imath] then [imath]c|a^{gcd(m,n)}-1[/imath] |
30867 | Show that [imath]\mathbb{R}^2[/imath] is not homeomorphic to [imath]\mathbb{R}^2 \setminus\{(0,0)\}[/imath]
Show that [imath]\mathbb{R}^2[/imath] is not homeomorphic to [imath]\mathbb{R}^2 \setminus \{(0,0)\} [/imath]. | 1037804 | [imath]\Bbb{R}^2[/imath] not homeomorphic to [imath]\Bbb{R}^2\setminus \{0\}[/imath]
I would like to show that [imath]\Bbb{R}^2[/imath] and [imath]\Bbb{R}^2\setminus \{0\}[/imath] are not homeomorphic without using Algebraic Topology. Is there an elementary way to do this? |
87735 | Is the integral [imath]\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}[/imath] equal for all [imath]a \neq 0[/imath]?
Let [imath]a[/imath] be a non-zero real number. Is it true that the value of [imath]\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}[/imath] is independent on [imath]a[/imath]? | 696269 | Integral [imath]\int_0^\infty \frac{1}{(1+x^m)(1+x^2)}\,dx[/imath]
I saw somewhere that the above integral is equal to [imath]\pi/4[/imath] for all real number [imath]m[/imath]. This seems to be surprising. Does anyone have a nice proof? |
34351 | Simpler way to compute a definite integral without resorting to partial fractions?
I found the method of partial fractions very laborious to solve this definite integral : [imath]\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx[/imath] Is there a simpler way to do this ? | 379953 | How to compute the integral [imath]\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx[/imath]
I want to compute this integral [imath]\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx[/imath] What I did was the following. I substituted [imath]x=t^{6}[/imath], so that my [imath]dx= 6t^{5} \ dt[/imath] and so the integral changes to [imath]\int_{0}^{\infty} \frac{t^2}{1+t^{12}} \cdot 6 t^{5} \ dt =6 \cdot \int_{0}^{\infty} \frac{t^7}{1+t^{12}} \: dt[/imath] Now If I substitute [imath]t^{4}=v[/imath] then what I will be having is the following integral [imath]\frac{6}{4}\cdot \int_{0}^{\infty} \frac{v}{1+v^{3}} \ dv[/imath] Now I can write [imath]1+v^{3} = (1+v) \cdot (1-v+v^{2})[/imath] and so I have \begin{align*} \int_{0}^{\infty} \frac{v}{1+v^{3}}\: dv &= \int_{0}^{\infty}\biggl[\frac{1}{3}\cdot \frac{v+1}{1-v+v^{2}} - \frac{1}{3} \cdot \frac{1}{1+v}\biggr]\: dv \end{align*} Now the point is that the integral of [imath]1/(1+v) \to \infty[/imath], so I am not sure if this is the right way to do. Can anyone suggest anything? |
10261 | Inverse of [imath]y=xe^x[/imath]
I feel like finding the inverse of [imath]y=xe^x[/imath] should have an easy answer but can't find it. | 1681715 | How can I solve this equation: [imath]xe^{ax}=b[/imath]?
How can I solve this type of equation? [imath]xe^{ax}=b[/imath]? |
60340 | Fibonacci divisibilty properties $ F_n\mid F_{kn},\,$ [imath]\, \gcd(F_n,F_m) = F_{\gcd(n,m)}[/imath]
Can any one give a generalization of the following properties in a single proof? I have checked the results, which I have given below by trial and error method. I am looking for a general proof, which will cover the all my results below: Every third Fibonacci number is even. 3 divides every 4th Fibonacci number. 5 divides every 5th Fibonacci number. 4 divides every 6th Fibonacci number. 13 divides every 7th Fibonacci number. 7 divides every 8th Fibonacci number. 17 divides every 9th Fibonacci number. 11 divides every 10th Fibonacci number. 6, 9, 12 and 16 divides every 12th Fibonacci number. 29 divides every 14th Fibonacci number. 10 and 61 divides every 15th Fibonacci number. 15 divides every 20th Fibonacci number. | 972693 | Prove that for each Fibonacci number [imath]f_{4n}[/imath] is a multiple of [imath]3[/imath].
The Fibonacci numbers are defined as follows: [imath]f_0 = 0[/imath], [imath]f_1 = 1[/imath], and [imath]f_n = f_{n-1} + f_{n-2}[/imath] for [imath]n \ge 2[/imath]. Prove that for each [imath]n \ge 0[/imath], [imath]f_{4n}[/imath] is a multiple of [imath]3[/imath]. I've tried to prove to by induction. So, my basis is [imath]f(0)[/imath], which is true. Then, i assume that it's true for some integer [imath]k[/imath], and I am stuck at this point, can't really get what to do |
90177 | Subgroup of [imath]\mathbb{R}[/imath] either dense or has a least positive element?
Let's say [imath]G[/imath] is some additive subgroup of [imath]\mathbb{R}[/imath] that has at least two elements. From what I understand, [imath]G[/imath] is then either dense in [imath]\mathbb{R}[/imath], or has some least positive element. What is the reason for this? | 1211750 | Show that there are only two types of subgroups in R , either Discrete or Dense?
Show that there are only two types of subgroups in [imath]\mathbb{R}[/imath] , either Discrete or Dense? |
299333 | Let G be finite and let [imath]p[/imath] be the smallest prime dividing [imath]|G|[/imath]. Let [imath]H \le G[/imath] be of index [imath]p[/imath]. Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath].
This is a problem from Herstein that I have been stuck upon for ages. I am becoming increasingly disappointed and disillusioned about my abilities due to this problem. Let G be finite and let [imath]p[/imath] be the smallest prime dividing [imath]|G|[/imath]. Let [imath]H \le G[/imath] be of index [imath]p[/imath]. Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. I am nowhere near getting a solution. Assume the statement is not true, so [imath]N(H) \ne G[/imath] then [imath]H \le N(H) \le G[/imath] implies that [imath]N(H)=H[/imath]. How can I get a contradiction from this? I am trying to find a homomorphism which has H has its kernel, but getting nowhere. I would like to see, atleast a couple of different solutions for this. | 164244 | Normal subgroup of prime index
Generalizing the case [imath]p=2[/imath] we would like to know if the statement below is true. Let [imath]p[/imath] the smallest prime dividing the order of [imath]G[/imath]. If [imath]H[/imath] is a subgroup of [imath]G[/imath] with index [imath]p[/imath] then [imath]H[/imath] is normal. |
9692 | When [imath]L_p = L_q[/imath]?
As we know that [imath]L_p \subseteq L_q[/imath] when [imath]0 < p < q[/imath] for probability measure, I was wondering when [imath]L_p = L_q[/imath] is true and why. Is it to impose some restriction on the domain space? Thanks! | 318533 | When is it the case that [imath]L^p(X,\mu)\subset L^r(X,\mu)[/imath]?
Given a measure space [imath](X,\Sigma,\mu)[/imath], when is it the case that [imath]L^p(X,\mu)\subset L^r(X,\mu)[/imath] for [imath]p>r[/imath], or for [imath]p<r[/imath] . Thanks. |
15556 | Is zero odd or even?
Some books say even numbers start from two but if you consider the number line concept, I think zero ([imath]0[/imath]) should be even because it is in between [imath]-1[/imath] and [imath]+1[/imath] (i.e in between two odd numbers). What is the real answer? | 2378898 | Is [imath]0[/imath] an even number?
I have noticed that it is useful to treat [imath]0[/imath] as an even number, and do so. Especially for patterns, puzzles, etc, if I develop a formula that works for something, and uses the parity of the number, then for my formula to work for something, I usually need to treat [imath]0[/imath] as even. Is [imath]0[/imath] treated as an even number? |
139035 | How do I show that the sum [imath](a+\frac12)^n+(b+\frac12)^n[/imath] is an integer for only finitely many [imath]n[/imath]?
Show that if [imath]a[/imath] and [imath]b[/imath] are positive integers, then [imath]\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n[/imath]is an integer for only finitely many positive integers [imath]n[/imath]. I tried hard but nothing seems to work. :( | 682343 | Show that if a and b are positive integers, then [imath](a + \frac12)^n + (b + \frac12)^n[/imath] is an integer for only finitely many positive integers n.
I stumbled upon this problem when reading the resource provided by AoPS, on Number Theory. Here is the problem: Show that if [imath]a[/imath] and [imath]b[/imath] are positive integers, then [imath]\left(a + \frac12\right)^n + \left(b + \frac12\right)^n[/imath] is an integer for only finitely many positive integers [imath]n[/imath]. I'm unsure on how to begin - any help would be much appreciated! Thanks |
129301 | What's the meaning of a set to the power of another set?
[imath]{ \mathbb{N} }^{ \left\{ 0,1 \right\} }[/imath] and [imath]{ \left\{ 0,1 \right\} }^{ \mathbb{N} }[/imath] to be more specific, and is there a countable subset in each one of them? How do I find them? | 130287 | Set of subsets notation.
Why is it that we denote the set of all subsets of [imath]A[/imath] by [imath]2^A[/imath]? Is there any historical or logical cause that motivated this notation? |
78546 | Conditional expectation for a sum of iid random variables: [imath]E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}[/imath]
I don't really know how to start proving this question. Let [imath]\xi[/imath] and [imath]\eta[/imath] be independent, identically distributed random variables with [imath]E(|\xi|)[/imath] finite. Show that [imath]E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}[/imath] Does anyone here have any idea for starting this question? | 407346 | Average of two i.i.d. random variables and their conditional expectations.
I would like to see some indications on how to approach this: Let [imath]\xi[/imath] and [imath]\eta[/imath] be independent, identically distributed random variables, with [imath]E\xi[/imath] defined (that is, [imath]\xi[/imath] is semi-integrable). Show that [imath]E(\xi | \xi + \eta)=E(\eta | \xi + \eta ) = (\xi + \eta) / 2[/imath] almost surely. |
10760 | A vector space over [imath]R[/imath] is not a countable union of proper subspaces
I was looking for alternate proofs of the theorem that "a vector space [imath]V[/imath] of dimension greater than [imath]1[/imath] over an infinite field [imath]\mathbf{F}[/imath] is not a union of fewer than [imath]|\mathbf{F}|[/imath] proper subspaces" and possible generalizations. A simple measure-theoretic proof over [imath]\mathbb{R}[/imath] is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction. I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings. | 359956 | The union of kernels is a proper subspace
I was asked by someone about the confusion when reading the following proof: Let [imath]V[/imath] be a vector space over a field [imath]F[/imath] of characteristic [imath]0[/imath], and let [imath]f_1,f_2,\ldots, f_n[/imath] be linear mappings from [imath]V[/imath] to [imath]F[/imath] none of them being the zero mapping. Then there exists [imath]a \in V[/imath] so that [imath]f_i(a) \neq 0[/imath] for [imath]i=1,\ldots, n[/imath]. And the proof goes something like this: Take the union of the kernels of the [imath]f_i[/imath]. It claims that (which I found confusing) this union is a proper subset of [imath]V[/imath], so there must exist such an [imath]a[/imath] as stated in the question. Can someone give me some explanation of what is going on here? Any help is appreciated. |
29980 | Evaluating [imath]\int P(\sin x, \cos x) \text{d}x[/imath]
Suppose [imath]\displaystyle P(x,y)[/imath] a polynomial in the variables [imath]x,y[/imath]. For example, [imath]\displaystyle x^4[/imath] or [imath]\displaystyle x^3y^2 + 3xy + 1[/imath]. Is there a general method which allows us to evaluate the indefinite integral [imath] \int P(\sin x, \cos x) \text{d} x[/imath] What about the case when [imath]\displaystyle P(x,y)[/imath] is a rational function (i.e. a ratio of two polynomials)? Example of a rational function: [imath]\displaystyle \frac{x^2y + y^3}{x+y}[/imath]. This is being asked in an effort to cut down on duplicates, see here: Coping with *abstract* duplicate questions. and here: List of Generalizations of Common Questions. | 832911 | Evaluate [imath]\int_{0}^{\pi}\sin^5{\theta}\cos^2{\theta}\ d\theta[/imath]
I'm trying to find the mass of a spherical object with a given density function, and to do so I must solve this integral [imath]\int_{0}^{\pi}\sin^5{\theta}\cos^2{\theta}\ d\theta,[/imath] but no matter which method I choose (integration by parts, substitution, etc) I can't for the life of my figure out the anti-derivative. |