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2167369 | Find the order of [imath]8/9, 14/5,48/28[/imath] in the additive group of [imath]\mathbb{Q}/\mathbb{Z}[/imath]
Find the order of [imath]8/9, 14/5,48/28[/imath] in the additive group of [imath]\mathbb{Q}/\mathbb{Z}[/imath] JUNK Guessing asking for [imath]8/9 * k = 0 [/imath] (being a whole integer) for smallest integer so for this one is 9 . another guess would be the smallest element of [imath]Q/Z[/imath] being something like [imath]9/8[/imath] [imath]14/5[/imath] would be 5 [imath]48/28=12/7[/imath] would be 7? | 2183983 | Order of elements in quotient Q/Z
So I have a question asking me to compute the order of: [imath]\frac 8 9 +\Bbb Z[/imath], [imath]\frac {14} 5 +\Bbb Z[/imath], and [imath]\frac {48} {28} +\Bbb Z[/imath] in the quotient [imath]\Bbb Q/\Bbb Z[/imath]. I think the answer is order 9, 5, and 7 respectively but would like some clarification as to why that is the case. Any help would be greatly appreciated as I am new to group theory. |
2169126 | how do you show this implication ?
[imath]\forall\; n \geq 3 \;\;\exists (x,y) \in \{2m+1;m\in \mathbb{N}\}^2[/imath] such that [imath]2^n=7x^2 + y^2 [/imath] I proved it by induction and I'm curious how you guys would you show it. | 1032340 | Find all solutions of the equation [imath]n^m=x^2+py^2[/imath] which satisfy the following properties
Prove or disprove that, There always exists a solution of the equation, [imath]n^m=x^2+py^2[/imath] with odd [imath]x[/imath] and [imath]y[/imath] and for all [imath]m\geq k[/imath] for some positive integral [imath]k[/imath]. Here [imath]p[/imath] is an odd prime and [imath]n\in 2\mathbb{N}[/imath]. Is [imath]k[/imath] dependent on [imath]n[/imath]? If so then find a way to calculate the value of [imath]k[/imath]. This is basically one of my conjecture which I am trying to prove for quite sometime. Unfortunately, I have progressed very little in this problem. So far I have been only able to prove that for all [imath]m\geq3[/imath] , [imath]n=2[/imath] and [imath]p=7[/imath] there is always a solution of the equation meeting the constraints. Any idea how to tackle the problem? Update After D. Burde's answer of the original problem (see below) I am now interested in finding all [imath](n,m,p)[/imath] triplets such that the equation holds. Any ideas regarding this problem? |
2169553 | Subgradient of nuclear norm
Show that the subgradient of the nuclear norm is given by [imath]$$\partial|X| = UV^T + W$$[/imath] where [imath]$X = U \Sigma V^T $[/imath] is the compact SVD of [imath]$X$[/imath], [imath]$W$[/imath] is a matrix such that [imath]U^T W = 0[/imath] and [imath]WV = 0[/imath], and [imath]$\|W\|_2 \le 1$[/imath]. This is exact that question. But I don't know how to get [imath]$W$[/imath]. Can some explain that paper explicitly? https://math.stackexchange.com/users/10117/lepidopterist | 354475 | Deriving the sub-differential of the nuclear norm
Let [imath]f(K)=||K||_*[/imath], the nuclear norm (sum of the singular values) of [imath]K=U\Sigma V^T[/imath]. How can one compute the subdifferential [imath]\partial F[/imath]. This may be a basic question, I'm trying to work my way through a paper in which minimizing [imath]f[/imath] over a convex set of matrices plays a central role. For what it's worth I have found papers that display the end result, but not the derivation. EDIT: This paper by Tao and Candes]1" derives an expression, but refers the proof to "Characterization of the subdifferential of some matrix norms" which does not prove it as far as I can tell. I also found a class homework assignment posted online that said this was easy to "grind out" with matrix derivatives, but that there was another way via projections. Any guidance would be greatly appreciated. |
2166890 | Elliptic Curve and Divisor Example help (Step 3)
I am reading this paper, specifically Example 2.3 on page 9, and am having a few problems understanding a part of it We construct an elliptic curve [imath]E[/imath] on [imath]\mathbb{F}_{11}[/imath] defined by [imath]y^2=x^3+4x[/imath] with a point at infinity [imath]\mathcal{O}[/imath] We then find a divisor of [imath]E[/imath] [imath]D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right][/imath] We can then say that: [imath]\begin{align}\text{div}(y-2x)&=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\end{align}[/imath] We can then replace various elements in the definition of [imath]D[/imath] with these divisors: [imath]\begin{align}D&=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]\\ &=\left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &=\left[(2,-4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(2,4)\right]+\text{div}\left(\frac{y+x+2}{x-2}\right)-2\left[\mathcal{O}\right]\\ &=\text{div}\left(x-2\right)+\text{div}\left(\frac{y-2x}{x-2}\right)+\text{div}\left(\frac{y+x+2}{x-2}\right)\end{align}[/imath] However, what I don't understand are the following steps: [imath]\begin{align} D&=\text{div}\left(x-2\right)+\text{div}\left(\frac{y-2x}{x-2}\right)+\text{div}\left(\frac{y+x+2}{x-2}\right)\\ &=\text{div}\left(\frac{(y-2x)(y+x+2)}{x-2}\right)\\ &=\text{div}\left(x^{2}-y\right)\end{align}[/imath] Can anyone help explain to me how they have come to this conclusion please This question comes in 3 parts: Step 1 - Elliptic Curve and Divisor Example help (Step 1) Step 2 - Elliptic Curve and Divisor Example help (Step 2) Step 3 - This one | 2168358 | Elliptic Curve and Divisor Example help (Step 1)
I have an elliptic curve [imath]E[/imath] over [imath]\mathbb{F}_{11}[/imath] defined by [imath]y^2=x^3+4x[/imath] with the point at infinity [imath]\mathcal{O}[/imath] I have a divisor of [imath]E[/imath], defined by [imath]D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right][/imath] I know that [imath]\text{deg}(D)=1+1+1+1-4=0[/imath] and it is clear that [imath]\text{sum}(D)=\infty[/imath] Therefore [imath]D[/imath] is the divisor of a function - we want to find this function I am given \begin{align}\text{div}(y-2x)&=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align} I am then told that we can express [imath]D[/imath] as follows: [imath]D = \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right][/imath] However, when I expand this back out, I get the following: \begin{align}D &= \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &= \left[(2,4)\right]+\text{div}(y-2x)-\text{div}(x-2)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &= \left[(2,4)\right]+\left(\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\right)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &= \left[(0,0)\right]+2\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right] \end{align} which clearly isn't equivalent to the original [imath]D[/imath] unless [imath]\left[(2,4)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right]=-4\left[\mathcal{O}\right][/imath] Can anyone either explain what I've done wrong or show me how I can obtain the above equation please This question comes in three parts: Step 1 - This one Step 2 - Elliptic Curve and Divisor Example help (Step 2) Step 3 - Elliptic Curve and Divisor Example help (Step 3) |
2169751 | Find [imath]\int_{-\infty}^{\infty} \frac{\sin^4{k}}{k^4}dk[/imath] using Parseval's Theorem
I'm having a little trouble with this one. So far I've found that [imath]g(k) = \sqrt{\frac{2}{\pi}}\frac{\sin{k}}{k}[/imath] when [imath]f(x) = 1, |x| < 1[/imath] and [imath]0[/imath] elsewhere. I've also found the convolution [imath](f*f)(x)[/imath] is [imath]\frac{1}{\sqrt{2\pi}}\begin{cases}0&x\le -2\\\\\int_{-1}^{x+1}(1)\,dx=(x+2)&-2<x<0\\\\\int_{x-1}^1(1)\,dx=2-x&0\le x<2\\\\0&x\ge 2\end{cases}[/imath] And the Fourier transform of this is [imath]\frac{2}{\pi}\frac{\sin^2{k}}{k^2}[/imath] I'm sure that from here the answer is quite simple, however I'm having a hard time figuring out the last step. Any help is appreciated. | 1226853 | Integral using Parseval's Theorem
How would I integrate [imath]\int_{-\infty}^{+\infty} \frac{\sin^{2}(x)}{x^{2}}\,dx[/imath] using Fourier Transform methods, i.e. using Parseval's Theorem ? How would I then use that to calculate: [imath]\int_{-\infty}^{+\infty} \frac{\sin^{4}(x)}{x^{4}}\,dx[/imath]? |
2167130 | The only proper subfield of [imath]\mathbb C[/imath] , which is path connected as a subset of [imath]\mathbb C[/imath] , must be [imath]\mathbb R[/imath]?
Is it true that the only proper subfield of [imath]\mathbb C[/imath] , which is path connected as a subset of [imath]\mathbb C[/imath] , must be [imath]\mathbb R[/imath] ? NOTE : The answer to this Subfields of [imath]\mathbb{C}[/imath] which are connected with induced topology indicates the claim stated is true . But the answer there only gives a reference and no proof ( or outline of proof) of the result , and the paper cited is highly technical . I would like to see if there is any elementary ( at least more elementary than the cited paper if possible ) proof . | 1831654 | Subfields of [imath]\mathbb{C}[/imath] which are connected with induced topology
The ring of continuous functions on [imath][0,1][/imath] to [imath]\mathbb{R}[/imath] has an interesting property: every maximal ideal of this ring is the subset of all functions vanishing at a common point. If we follow the argument in the proof of this property, it can be seen that [imath]\mathbb{R}[/imath] can be replaced by [imath]\mathbb{C}[/imath] and still similar property holds for new ring (we have to slightly modify argument). What if we place [imath]\mathbb{Q}[/imath]? Here [imath]\mathbb{Q}[/imath] is disconnected with induced topology from [imath]\mathbb{C}[/imath] and so the only continuous function from [imath][0,1][/imath] to [imath]\mathbb{Q}[/imath] are constant (am I correct?). So, the property no longer holds true with replacement of [imath]\mathbb{R}[/imath] by [imath]\mathbb{Q}[/imath]. Question: Are there subfields of [imath]\mathbb{C}[/imath] other than [imath]\mathbb{R}[/imath] which are connected as topological spaces with induced topology from [imath]\mathbb{C}[/imath]? |
1848202 | Show that there's no continuous function that takes each of its values [imath]f(x)[/imath] exactly twice.
I need to prove the following: There's no continuous function [imath]f:[a,b]\to \mathbb{R}[/imath] that takes each of its values [imath]f(x)[/imath], [imath]x\in [a,b][/imath] exactly twice. First of all, I didn't understand the question. For example [imath]x^2[/imath] takes [imath]1[/imath] twice, in the interval [imath][-1,1][/imath]. Is it saying that it does not occur for all [imath]x[/imath] in the interval? But what about [imath]f(x) = c[/imath]? Is it saying that it does not occur only exactly [imath]2[/imath] times, then? I have no idea about how to prove it. I know that for [imath]f(x)[/imath] such that [imath]f(a)<f(x)<f(b)[/imath], if [imath]f[/imath] is continuous then there is a [imath]c\in [a,b][/imath] such that [imath]f(c) = f(x)[/imath]. Now, there's the following proof in my book and I really wanted to understand it, instead of just getting a new proof Since the interval [imath][a,b][/imath] has only [imath]2[/imath] extreme points, then the maximum or minimum of [imath]f[/imath] must be in a point [imath]c\in int([a,b])[/imath] and and in another point [imath]d\in [a,b][/imath]. Then, there exists [imath]\delta>0[/imath] such that in the intervals [imath][c-\delta, c), (c,c+\delta)[/imath] (and if [imath]d[/imath] is not extreme of [imath][a,b][/imath], [imath][d-\delta, d][/imath]) the function takes values that are less than [imath]f(c) = f(d)[/imath]. Let [imath]A[/imath] be the greatest of the numbers [imath]f(c-\delta), f(c+\delta), f(d-\delta)[/imath]. By the intermediate value theorem, there are [imath]x\in [c-\delta, c), y\in (c, c+\delta][/imath] and [imath]z\in [d-\delta, d)[/imath] such that [imath]f(x)=f(y)=f(z)=A[/imath]. Contradiction. Well, why the last part? Why is it that I can apply the intermediate value theorem to these values? For example, [imath]<f(c-\delta)<p<f(c)[/imath], then by the theorem I know that there exists [imath]m\in [c-\delta, c)[/imath] such that [imath]f(m) = p[/imath]. Same for the other intervals. But what guarantees thhat the greatest of the values between [imath]x\in [c-\delta, c), y\in (c, c+\delta][/imath] will be inside the intervals [imath][c-\delta), c), (c,c+\delta), [d-\delta, d)[/imath]? | 2684245 | A better solution for an old problem
Problem: [imath]f[/imath] is a function from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath], here [imath]\mathbb{R}[/imath] is the set of real numbers. [imath]f[/imath] cannot be continuous if [imath]f[/imath] takes every value exactly twice. Now this question is really old and asked many times before on stackexchange. But every solution is by crude manipulation. I want to know if there exists an insightful solution of this. My problem: I have solved the problem before like this, first assume the function is continuous and then by applying intermediate value property lead to some contradiction. But if the problem is a more general one, like if the function takes every value thrice, the same argument works, but the solution becomes lengthier. That's why I am asking for some more nice solution, may be using some advanced tools. |
2168301 | Prove that for any integer [imath]a, a^{15}-a^3[/imath] is divisible by [imath]455[/imath].
The Problem Prove that for any integer [imath]a, a^{15}-a^3[/imath] is divisible by [imath]455[/imath]. Hint: [imath]455=5*7*13[/imath]. We proved that if [imath]n[/imath] is divisible by [imath]x[/imath], and [imath]n[/imath] is divisible by [imath]y[/imath], and [imath]\gcd(x,y) = 1[/imath], then [imath]n[/imath] is divisible by [imath]x*y[/imath]. The upshot of this is that if [imath]a^{15}-a^3[/imath] is divisible by [imath]5[/imath], and by [imath]7[/imath], then it is divisible by [imath]5*7=35[/imath]. And if [imath]a^{15}-a^3[/imath] is divisible by [imath]5[/imath], [imath]7[/imath], and [imath]13[/imath], then it is divisible by [imath]5*7*13[/imath]. So you need to prove that [imath]a^{15}-a^3[/imath] is divisible by [imath]5[/imath], by [imath]7[/imath], and by [imath]13[/imath]. Confusion My first guess was to try induction, although I wasn't sure if induction can even be used here since we are dealing with integers instead of natural numbers, so perhaps that is my first issue. Even if this were a statement about natural numbers, I was still unable to figure out how induction could work here. Secondly, the hint my instructor provided for this problem makes sense to me, but I haven't the slightest clue as to how to prove what's being asked. For instance, how do we prove that [imath]5|a^{15}-a^3[/imath]? Induction seems to get us nowhere because you end up with [imath](k+1)^{15}-(k+1)^3[/imath], and that seems like an insane amount of algebra. There has to be something I'm missing here. Any thoughts? | 2158766 | Show that [imath]7[/imath] divides [imath]a^{15}-a^3[/imath] for any integer [imath]a[/imath]
I showed that [imath]a^{15}-a^3=(a^5-a)[/imath] so by Fermat's, [imath]5|a^5-a[/imath] so [imath]5|a^{15}-a^3[/imath]. I also showed that [imath]a^{15}-a^3=a^3(a^{12}-1)[/imath] so by Euler, [imath]a^{12}-1 \equiv 0[/imath] mod 13. I am having a hard time showing that 7 divides [imath]a^{15}-a^3=(a^5-a)[/imath]. They talk about it some in this post but I am still not seeing it. Thank you. |
2073647 | Finding remainder when [imath]10^{10}+10^{10^2}+.........+10^{10^{10}}[/imath] is divided by [imath]7[/imath].
I have found a new problem which asks: Find the remainder when [imath]10^{10}+10^{10^2}+.........+10^{10^{10}}[/imath] is divided by [imath]7[/imath]. I am thinking to find the remainder using Fermet's theorem, but I think I am unable to do it. Please help. | 1334533 | Question of remainder on dividing by 7
Question : What is the remainder when [imath] 10^{10} + 10^{10^2} +10^{10^3} + \ldots + 10^{10^{100}} [/imath] is divided by [imath]7[/imath]? |
2170785 | Prove that [imath]x^2=-1[/imath] has no solution in [imath]\mathbb Z[/imath].
I have been asked to prove the following proposition: Proposition 2.10: The equation [imath]x^2=-1[/imath] has no solution in [imath]\mathbb Z[/imath]. Our professor told us that we will need to prove for two cases, when [imath]x=0[/imath] and when [imath]x \ne 0[/imath]. My question is that I don't understand why we need to prove for two different cases. If anybody could shed some light that would be great. | 1171287 | Prove that [imath]x^2 = - 1[/imath] has no solution in [imath]\mathbb{Z}[/imath]
I have this proposition to prove: The equation [imath]x^2 = -1[/imath] has no solution in [imath]\mathbb Z[/imath]. I was told that this is an opportunity for a proof by contradiction. I have already proven that for [imath]m \in\mathbb Z[/imath], if [imath]m \ne 0[/imath] then [imath]m^2 \in\mathbb N[/imath]. I have also proven that 1 is a natural number (hence, -1 isn't one). Here is my strategy: assume that there is a solution. Hence [imath]x^2[/imath] has 3 options: \begin{align*} x^2 \in\mathbb N\\ -(x^2) \in\mathbb N\\ x^2 = 0 \end{align*} Option 1 is not possible because -1 is not a natural number. Option 2 doesn't make sense because -[imath]\mathbb N \in \mathbb N[/imath] and option 3 doesn't make sense either because [imath]0 \ne -1[/imath]. What do you think? Thank you! Here is a simpler strategy based on the comments below. Proof: Assume that [imath]x^2 = -1[/imath] has a solution in [imath]\mathbb Z[/imath], then [imath]x^2[/imath] will either be [imath]0[/imath] or [imath]\in\mathbb N[/imath]. However, [imath]-1 \notin\mathbb N[/imath] and [imath]-1 \ne 0[/imath]. Hence, there is a contradiction. Here is how I have proven that [imath]-1 \notin\mathbb N[/imath]: Let [imath]m \in\mathbb N[/imath]. Hence: \begin{align*} -m \notin\mathbb N\\ (-1)m \notin\mathbb N \end{align*} If [imath]-1 \in\mathbb N[/imath], the product of [imath]-1[/imath] and [imath]m[/imath] should [imath]\in\mathbb N[/imath], which is not the case here. Hence, [imath]-1 \notin\mathbb N[/imath]. |
2171115 | When will series [imath]\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} [/imath] converge?
The question is to find the [imath]x[/imath] such that series [imath]\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} [/imath] converges. I can see if [imath]x = k \pi[/imath], then the series converges since it is alternating series. Also it converges when like [imath]x = k\pi/2,x = k\pi/3,x = k\pi/4,...[/imath] etc. I have a solution that actually [imath]x[/imath] can be any real number except for [imath]x=2k\pi,k=0,1,2,...[/imath]. Anyone can help explain why is this so? Thanks! | 1469162 | Is series [imath]\sum\limits^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}[/imath], for [imath]\alpha>0[/imath], convergent?
I've done the following exercise: Is series [imath]\displaystyle\sum^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}[/imath], for [imath]\alpha>0[/imath], convergent? My approach: We're going to use the Dirichlet's criterion for convergence of series. Let [imath]\displaystyle\ \{a_n\}=\frac{1}{n^{\alpha}}[/imath] and [imath]\{b_n\}=\cos(nx)[/imath]. We see that [imath]\{a_n\}[/imath] is decreasing and has limit [imath]0[/imath]. We have to see now that [imath]\sup_{N}{\left| \sum_{n=1}^{N}{\cos(nx)} \right|}<\infty.[/imath] [imath]\displaystyle \sum_{n=0}^{N}e^{inx}=\sum_{n=0}^{N}(e^{ix})^{n}= \frac{1-e^{ix(n+1)}}{1-e^{ix}}, \text{if x} \neq 2k\pi. [/imath] So [imath]{\left| \sum_{n=0}^{N}{\cos(nx)} \right|}={\left| \Re\left(\sum_{n=0}^{N}{e^{inx}}\right) \right|}={\left| \Re\left(\frac{1-e^{ix(N+1)}}{1-e^{ix}}\right) \right|} \leq {\left| \frac{1-e^{ix(N+1)}}{1-e^{ix}} \right|}\leq {\frac{\left|1\right|+\left| e^{i(N+1)x} \right|}{\left|1-e^{{ix}} \right|}}=\frac{2}{\left|1-e^{{ix}} \right|}.[/imath] So, by Dirichlet, if [imath]x\neq 2k\pi[/imath], the series is convergent. What happens if [imath]x= 2k\pi[/imath]? [imath]\displaystyle\sum^{\infty}_{n=1}\frac{\cos(n(2k\pi))}{n^\alpha}=\displaystyle\sum^{\infty}_{n=1}\frac{1}{n^\alpha}[/imath] and we know that this series converges when [imath]\alpha>1[/imath] and diverges if [imath]\alpha\leq 1[/imath]. Have I done any mistake/s? Is my approach correct? Thank you. |
689853 | Show that [imath]\limsup(a_n)\geq1[/imath] if [imath]a_n[/imath] is bounded, positive, and [imath]\lim_{n\to\infty} a_na_{n+1}=1[/imath]
Let [imath]\{a_n\}[/imath] be a bounded positive sequence ([imath]a_n>0[/imath]). [imath]\lim_{n\to\infty} a_na_{n+1}=1.[/imath] Show that [imath]\limsup(a_n)\geq1[/imath]. Thanks. | 2169986 | Prove [imath]\limsup_{n \rightarrow \infty} a_n\geq1[/imath] if [imath]\lim_{n \rightarrow \infty} a_na_{n+1}=1[/imath]
Let [imath]\{ a_n \}_{n=1}^{\infty}[/imath] be a bounded sequence s.t. [imath]a_n>0,\ \forall n \in \mathbb{N}[/imath]. Suppose [imath]\lim_{n \rightarrow \infty} a_na_{n+1}=1[/imath], prove [imath]\limsup_{n \rightarrow \infty} a_n\geq1[/imath]. I tried to prove by contradicition using subsequence, but couldn't work it out. Any help appreciated. |
2171329 | What is the length of the intercept made by the parabola [imath]{ y }^{ 2 }=4x[/imath] on the line [imath]x+y=3[/imath]?
I did not understand what is meant by length of intercept? Isn't intercept the point where the line cuts the axis? How can a point have a length? The answer given for length of intercept is [imath]8\sqrt { 2 } [/imath] | 2171298 | If the line [imath]y=x-8[/imath] intersects the parabola [imath]{ y }^{ 2 }=4x[/imath] at A and B, what's the chord's length?
How to find the coordinates of the points A and B? If one gets that, then distance formula can be used to find the length. |
2149190 | Countable ordinals property
I was wondering how to prove that an ordinal [imath]\alpha[/imath] is countable if and only if the next holds: there is an order-preserving injection from [imath](\alpha,\leq)[/imath] into [imath](\mathbb Q,\leq)[/imath]. | 665692 | Order-preserving injections of ordinals into [imath][0,1][/imath]
The usual "matchstick" representation of an ordinal number can be thought of as an order-preserving injection of that ordinal into the interval [0,1]. For example, here's a representation of [imath]\omega^2[/imath]: One can use this technique to visualize even larger and larger ordinals; for instance, [imath]\omega^3[/imath] would look the same as the above, but each individual matchstick would be replaced with yet another infinite triangle leading to the next matchstick, so that there are three layers of infinite triangles. So long as there's an order-preserving injection from your chosen ordinal into [imath][0,1][/imath], this technique will work to help you visualize it. So my question is: what is the supremum of all ordinals for which an order-preserving injection exists in this way? Is it the first uncountable ordinal? The initial ordinal of [imath]2^{\aleph_0}[/imath]? Something smaller? Something larger? |
2171801 | Formal Proof for [imath]\textbf{Vect}_k\cong\textbf{Mat}_k[/imath]?
Where this question came from: I was researching about the graphical languages for monoidal categories involving the category [imath]\textbf{PROP}[/imath] and wanted to see if I could create a graphical language for [imath]\textbf{Vect}_k[/imath] because of [imath]\textbf{Mat}_k[/imath]. (probably not the exact reason however this is what I remember) Let [imath]\textbf{Vect}_k[/imath] be the category of vector spaces for some field [imath]k[/imath] and [imath]\textbf{Mat}_k[/imath] be the category of matrices under the same field [imath]k[/imath]. I know that we can define an equivalence between categories between [imath]\textbf{Vect}_k[/imath] and [imath]\textbf{Mat}_k[/imath] because given vector spaces [imath]V=k^n[/imath] consisting of all [imath]n[/imath]-tuples over [imath]k[/imath], and [imath]U=k^m[/imath] consisting of all [imath]m[/imath]-tuples over [imath]k[/imath] where [imath]U,V\in\text{Obj}(\textbf{Vect}_k)[/imath], then [imath]\text{Hom}_k(V,U)[/imath] creates the vector space [imath]k^{m\times n}[/imath] representing the set of all [imath]m\times n[/imath] matrices over [imath]k[/imath]. Thus showing that [imath]k^{m\times n}\in\text{Obj}(\textbf{Mat}_k)[/imath]. If [imath]X=k^{m\times n}[/imath] and [imath]Y=k^{n\times q}[/imath] (set of all [imath]n\times q[/imath] matrices over [imath]k[/imath]), then [imath]\text{Hom}(X,Y)=X\otimes Y[/imath] where the tensor product, [imath]\otimes[/imath], is the cross product between the set of matrices [imath]X[/imath] and [imath]Y[/imath]. The cross product rules still apply where [imath]\text{Hom}(Y,X)=\emptyset[/imath]. Now lets define the functor [imath]F:\textbf{Vect}_k\to\textbf{Mat}_k[/imath]. We have a morphism [imath]f:V\to U[/imath] and the image of [imath]f[/imath] in [imath]\textbf{Mat}_k[/imath], [imath]Ff[/imath], will connect the image of [imath]V[/imath] to the image of [imath]U[/imath]. Now I am pretty sure I might be thinking about this wrongly and I hope you could help me with this but I think that [imath]Ff:FV\to FU[/imath] is a morphism from the set of all [imath]m\times1[/imath] matrices to the set if all [imath]1\times n[/imath] matrices. Thus makes [imath]\text{Hom}_{Ff}(FV\to FU)=FV\otimes FU[/imath] which means the cross product between matrices [imath]FV=k^{m\times1}[/imath] and [imath]FU=k^{1\times n}[/imath]. I know this connection however I do not know how to form a formal equivalence proof for [imath]\textbf{Vect}_k\cong\textbf{Mat}_k[/imath]. If I could get some help I would greatly appreciate it. Sources: https://unapologetic.wordpress.com/2008/06/24/the-category-of-matrices-iv/ http://www.math.uiuc.edu/Software/magma/text387.html http://www.math.uiuc.edu/Software/magma/text388.html http://users.ecs.soton.ac.uk/ps/papers/ih.pdf https://www.cs.ox.ac.uk/qpl2015/slides/sobocinski-tutorial.pdf | 1088411 | What's the explicit categorical relation between a linear transformation and its matrix representation?
There several questions about linear transformations and its respective matrices in some basis, but I'm particularly interested in the explicit definition of this relation in the category [imath]Vect[/imath] (of vector spaces and linear transformations.) Of course, a inevitable question is if there is always an matrix (or tensor, perhaps) representation (representable Functor?) for every morphism in an arbitrary category. Sorry if it is a silly question, but I'm learning CT by myself and it is difficult for me to understand this relation in categorical terms. |
2171363 | Integral inequality problem (cauchy-schwarz)
Prove that any continuously differentiable function [imath]f : [a,b] \rightarrow R[/imath] for which [imath]f(a) = 0[/imath] satisfies the following inequality [imath]\int^b_a f(x)^2dx \leqslant (b-a)^2 \int_a^b f^{'}(x)^2dx[/imath] By looking at it, I think we need to apply the Cauchy-Schwarz inequality but I am unable to go any further. Any help is appreciated. Thanks in advance. | 121818 | Integral inequality (Cauchy-Schwarz)
Let [imath]u\in \mathcal{C}^1[a,b][/imath] be such that [imath]u(a)=u(b)=0[/imath]. Show that [imath]\int_a^b u^2(x)dx\leq (b-a)^2\int_a^b (u')^2(x)dx[/imath] using the Schwarz's inequality. |
2172102 | If [imath]p[/imath] is a prime then [imath]p^2+26[/imath] is not a prime
I have tested for all primes less than [imath]40,000[/imath] and it seems that, if [imath]p[/imath] is a prime then [imath]p^2+26[/imath] is not. I would like to see a proof or a counter-example. | 390231 | On the primality of integers of the form [imath]p^2+k[/imath]
I am not able to find an answer to the following question: For which positive even integers [imath]k[/imath] is the integer [imath]p^2+k[/imath] prime, where [imath]p[/imath] is a prime number [imath]\gt5[/imath]? |
2171428 | Prove that the following is vector space
Prove that for any [imath]n\ge 1[/imath], [imath](K^n/K,+,\cdot)[/imath] is a vector space, where [imath](K^n,+)[/imath] is an abelian group and [imath](K,+,\cdot)[/imath] is a commutative field. I've been having trouble proving this lately. Any help is welcome! | 982744 | What does "[imath]\mathbb{F^n}[/imath] is a vector space over [imath]\mathbb{F}[/imath]" mean?
I am self learning Linear Algebra using "Linear Algebra Done Right" a book by Sheldon Axler. And for me its important to understand the smallest thing mentioned in the book. Can someone explain to me what the author means by OVER R/C? And there should be a reason he mentioned this (an importance), can someone explain me that? This is what the book says : [imath]\mathbb{R^n}[/imath] is a vector space over [imath]\mathbb{R}[/imath], and [imath]\mathbb{C^n}[/imath] is a vector space over [imath]\mathbb{C}[/imath] |
2171390 | The chain rule for derivatives - actual meaning
Consider the derivative of y with respect to t: dy/dt, according to the chain rule: [imath]\frac{dy}{dt} = (\frac{dy}{dx})(\frac{dx}{dt}) [/imath] This means that the derivative of [imath]y[/imath] with respect to a variable [imath]t[/imath] is the coordinate of [imath]y[/imath] with respect to [imath]x[/imath](i.e. the slope at [imath]x[/imath]) times the coordinate of [imath]x[/imath] with respect to [imath]t[/imath]. How would it make sense more intuitively? Intuitively in the sense, that it would make more sense physically or well, intuitively rather than just analytically, i.e. just using algebraic laws, it would be rather meaningful, if it made sense in a sort of a logical argument. And how do we take this further to the second order deriative of parametric equations: d2y/dx2 = (dy'/dt)/(dx/dt) | 2059282 | Intuitive Proof of the Chain Rule in 1 Variable
Is there a simple and intuitive way to prove the chain rule, that is, if [imath]y[/imath] is a function of [imath]u[/imath] and [imath]u[/imath] is a function of [imath]x[/imath], then why is [imath]\frac{dy}{dx}[/imath] = [imath]\frac{dy}{du}[/imath] [imath]\cdot[/imath] [imath]\frac{du}{dx}[/imath] ? This could just be an intuitive argument. PS: The only proofs I found were based off of confusing definitions. |
2171426 | [imath](a,n)=1\Rightarrow \exists [b]\in J_n [/imath] such that [imath][a][b]=1[/imath]
I'm stuck on the following exercise from Herstein's "Topics in Algebra": "Show that if [imath](a,n)=1[/imath] then you can find [imath][b]\in J_n[/imath] such that [imath][a][b]=1[/imath]" (note: [imath][a][/imath] is the congruence class (mod [imath]n[/imath]) of [imath]a[/imath], and [imath]J_n[/imath] is the set of congruence classes mod [imath]n[/imath]) I've been thinking about using the property [imath](a,n)=1\Rightarrow \exists h,k\in\mathbb{Z}[/imath] such that [imath]ha+kn=1[/imath] and also the Euclidean algorithm by writing [imath]a=qn+b[/imath] [imath]0\leq b<a[/imath] (since [imath]J_n=\{[0],...[n-1]\})[/imath] but I haven't been able to come up with a proof. So, I would appreciate any comment/hint about how to prove this fact. Best regards, lorenzo | 1204444 | $ a\in \Bbb Z_n$ is invertible $\,\Rightarrow\gcd (a,n) =1$
I need help with this: If a is an element of [imath]$\mathbb Z_n$[/imath] and [imath]$\gcd(a,n) > 1$[/imath] , then a is not invertible. First you show that if [imath]$a$[/imath] is an element of [imath]$\mathbb Z_n$[/imath] and [imath]$\gcd(a,n)>1$[/imath], then there is an element [imath]$b$[/imath] of [imath]$\mathbb Z_n$[/imath] and ([imath]$b$[/imath] is not equal to the zero element) from which [imath]$ab = 0$[/imath]. The second part I cant get. i.e. show if [imath]$b$[/imath] is not the zero element and [imath]$ab=0$[/imath] then [imath]$a$[/imath] is not invertible. |
2172119 | Find the number of triangles formed by n lines.
You are given [imath]n[/imath] lines with equations [imath]ax+by+c=0[/imath]. Moreover, no [imath]3[/imath] lines pass through the same point. You have to find the number of triangles that can be formed using these lines. Input First line of input contains [imath]N[/imath] - number of lines. [imath]N[/imath] lines follow, each line contains three integers [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] corresponding to the formula of [imath]i[/imath]th line. Moreover, [imath]a[/imath] and [imath]b[/imath] will not be simultaneously [imath]0[/imath] for a line. Output Output a single integer corresponding to the number of triangles formed by these [imath]N[/imath] lines. Constraints 1<=N<=300000 |a|,|b|,|c|<=10^9 Sample Input - 6 0 1 0 -5 3 0 -5 -2 25 0 1 -3 0 1 -2 -4 -5 29 Sample Output - 10 | 260037 | Number of triangles formed by [imath]m[/imath] lines
Problem: How many triangles do [imath]m[/imath] lines form if a) Every two lines intersect and no three lines intersect at one point. b) There are [imath]n[/imath] lines among [imath]m[/imath] lines that are parallel to each other. No other line is parallel to these [imath]n[/imath] lines, and no other two lines are parallel to each other. Again no three lines intersect at one point. Thank you. |
2171666 | How does a value not part of the domain satisfy the function?
Consider [imath]y= \frac{(x^2-9)}{x-3}[/imath] For y to be meaningful x must not equal 3 , So, 3 is not included within the domain of the function y (Call it f(x) if you will) It follows that, [imath]yx-3y = x^2 -9 [/imath] Plug in 3 to x out of curiosity, I find, [imath] 0=0 [/imath] A perfectly valid relation! How is this possible? This problem is even more predominant in partial fraction decomposition. Have a look at this example I found on the web, Partial fraction decomposition x cannot equal -1 or 2. How is it giving us the right answers then? What is the math behind this? When I do these kind of problems it feels like something's is happening mathematically, hidden from my view All help is appreciated | 1313454 | Is this a valid partial fraction decomposition?
Write [imath]\dfrac{4x+1}{x^2 - x - 2}[/imath] using partial fractions. [imath] \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}[/imath] [imath]4x+1 = A(x-2)+B(x+1)[/imath] [imath]x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3[/imath] [imath]x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1[/imath] Thus, [imath]\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}[/imath] The substitution of [imath]x[/imath] ([imath]x = 2, -1[/imath]) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying [imath](x+1)(x-2)[/imath], which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at [imath]x=-1,2[/imath]. How can we substitute these value for [imath]x[/imath]? So is this method valid and rigorous? How to modify it so that it is rigorous? |
2106663 | How does one prove that proof by contradiction has completely and utterly failed?
to clarify the difference between this question and the supposed duplicate, absolutely nowhere do I mention the proof being attempted in the other question. It merely serves as an example related to this question. Now, the way I view proof by contradiction is like this: We have two logical statements [imath]a[/imath] and [imath]b[/imath] and we claim that "if [imath]a[/imath] then [imath]b[/imath]". We assume [imath]b[/imath] is false (given [imath]a[/imath]) and we attempt to find some kind of contradiction with something we assume is true at the beginning (generally it tends to be proving [imath]a[/imath] is false). Now, proof by contradiction can "fail" by resulting in no contradiction (i.e. that [imath]a[/imath] ends up being true when [imath]b[/imath] is false). However, this seems like a proof that the original statement is false. What I'm asking is more like how one detects or proves that when [imath]b[/imath] is false the truthfulness of [imath]a[/imath] is unprovable. Basically, that [imath]a[/imath] and [imath]b[/imath] are incapable of proving each other. A good example of this is the parallel postulate of geometry. From what I heard (paraphrasing) "a failed proof by contradiction showed that there existed consistent systems where it was false and that it was unprovable from the other postulates". I'm not quite sure how that worked. Could someone please explain (in general) how one proves a proof by contradiction fails this brutally? I'm not asking specifically how it was done for the parallel postulate, but just how one goes about doing it in general? | 1888042 | Is Gödel's modified liar an illogical statement?
In a previous question I relied on the notion of an "illogical statement" which led to some debate and I ended up making its definition an addendum to the question. I'd like to ask whether the notion of an "illogical statement" as defined here is a worthwhile concept, is it a concept that exists already, is it logical or not, and is it fair to apply the notion to Gödel's modified liar paradox? It would seem to me to have merit to define an illogical statement as follows: An illogical statement is a statement whose truth value is the inverse of its own truth value. 1: (Sentence [imath]s_1[/imath] is illogical if [imath]s_1\implies\neg s_1[/imath] or [imath]\neg s_1\implies s_1[/imath]) 2: Let [imath]s_1[/imath] be the Liar Paradox "This sentence is false.". 3: Rewriting 2: [imath]s_1\implies \neg s_1[/imath] 4: From 3,1: [imath]\implies s_1[/imath] is illogical. And for the modified Liar: Modified Liar: This sentence is unprovable: 5: Let [imath]s_2[/imath] be "This sentence is unprovable." 6: [imath](s_2\implies s_2)\implies\neg s_2[/imath] 7: [imath]s_2\implies\neg s_2[/imath] 8: From 7,1: [imath]\implies s_2[/imath] is illogical. It would seem to me that we should only permit true and not false statements to be accepted as true. This rule would appear to be the implicit basis of proof by contradiction. In a proof by contradiction, we suppose a proposition to be true, show that this supposition leads us to the conclusion that a statement is "both true and false" and therefore it is not accepted to be true on the grounds that it isn't "true and not false". And it would seem that as a basic rule, illogical statements should not be permitted by the metatheory since they introduce a contradiction to the logical system. The modified liar, the moment we say it, is both true and false and therefore is not "true and not false" so it would seem to me that by the same rule as any proof by contradiction, the moment we suppose the modified liar, we can immediately reject it by contradiction. As surely it's rejected by contradiction regardless of the system into which it is introduced. So the conclusion that the decidability of the modified liar is independent of some set of axioms does not imply that it is undecidable in the combination of those axioms and the metatheory, since the metatheory in isolation already provides for decidability that the modified liar must be rejected. The problem seems to arise for most logicians, when they extend this conclusion by saying that since the modified liar is rejected, it is not true, so it must be false. But this step is only valid for sentences whose truth and falsehood are mutually exclusive. The modified liar does not possess that property. Were it true, it would not be "true and not false" and were it false, it would not be "false and not true". Therefore with statements like this it is incorrect to make the leap that if it is not false it must be true. I'm no maths professor (far from it) and I accept I'm naively proposing a challenge to a deeply-founded concept here so in anticipation of the inevitable cascade of downvotes please accept my sincere apologies for any misconception on my part. UPDATE (Post accepting the answer): My current thinking is that it would be best to define an illogical statement as: An illogical statement is a statement which is incapable of being true without being contradictory in any system in which it is provable and in which [imath]s\implies s[/imath] |
2126840 | Does anyone know how to reduce this sum of sums into something simpler in order to find a special value?
to clarify the difference between this and the supposed duplicate, these two questions talk about completely different functions with completely different purposes I was given this from a friend. They asked me to deduce what the equation is of. I played around with trying to compute alpha for some time. Plugging it into f(x), the function appeared to equal [imath]0[/imath] almost everywhere. I could never find the actual value of alpha. It appears to be an infinite irrational decimal number starting with [imath]1.1973...[/imath]. Let [imath]\alpha = \sum_{m=1}^{\infty} m * (2^{-(\sum_{j=2}^{m} (\lfloor \frac {(j-1)! + 1}{j} \rfloor - \lfloor \frac {(j-1)!}{j} \rfloor))^2}) \\* \left(\left\lfloor \frac {\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1}{\sqrt {(\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1)^2 + 1}} \right\rfloor + 1\right) * \left(\left\lfloor \frac {-\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)+1}{\sqrt {(\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1)^2 + 1}} \right\rfloor + 1\right)[/imath] Then let [imath]f(x) = \lfloor 2^{x^2}* \alpha \rfloor - 2^{2x-1}*\lfloor 2^{(x-1)^2}*\alpha \rfloor[/imath]. What is the function f(x)? Is there any way to reduce alpha to something simpler? I think this equation isn't something trivial. It doesn't appear to be [imath]0[/imath]. It seems bizarre that alpha consists of so many sums. How do I reduce them into something simpler? | 1850401 | Reducing an indicator function summation into a simpler form.
I want to simplify the following: [imath]\sum_{n=0}^x 1 - (1 - I[a \ne 0] * I[a \ge b]) * (1 - I[b \ne 0] * I[b \ge c]) * (1 - I[c \ne 0] * I[c \ge d]) * (1 - I[d = 0])[/imath] |
2107100 | Can variables refer to equations and what are the operations governing said variables.
Now, I know that equations are generally thought of as relationships between variables. They can refer to functions (in the case of differential equations), numbers (in the case of algebraic equations), or boolean values (in the case of logical equations). However, can one say something like: "Let [imath]a[/imath] be an arbitrary differential equation and let [imath]y[/imath] be the set of solutions that satisfy it." Of course, the above is just an example. Furthermore, there are many different types of equations. Is there any standard sense of algebra for equations? I know that equations can obviously be manipulated in numerical algebra, but is there a more direct sense of algebraic manipulation of actual equations rather than numbers in equations (i.e. some kind of operations upon equations and the manipulation of them). I know this is kind of two questions in one, but I think they kind of go hand in hand. After all, if equations can be referred to by variables then one should naturally conclude how those variables can be manipulated rigorously in an operational setting. | 1568098 | bound and free variables
I have a question that has been bothering me for quite some time. In second order logic sometimes there is an indication that a variable can be both bound and free. The simplest example I can give is using the variable [imath]v_0[/imath] and the constants [imath]c_0, c_1[/imath] who are not equal to one another. How does one interpret the following formula. [imath](v_0=c_0)\wedge \left[ \exists v_0 (v_0 = c_1)\right][/imath] I see a few possible interpretations: The statement is not actually valid because variables cannot be both bound and free. The formula will always be false because [imath](v_0=c_0)[/imath] restricts [imath]v_0[/imath] such that there cannot exist a [imath]v_0=c_1[/imath] The formula is true when [imath]v_0=c_0[/imath] because one essentially treats the free version of a variable differently from the bound version such that even though [imath]v_0=c_0[/imath] in this particular instance, it still theoretically could be that a variable [imath]v_0[/imath] equals the constant [imath]c_1[/imath] and thus the existence clause is satisfied as well. I also recognize that it could be something completely different from any of these that I have not thought of yet, I was just curious if there was an accepted way to handle variables which are both free and bound because I have not found any satisfying answers as of yet. |
2172255 | [imath]P(n) = n^{ \phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)} [/imath]
Let [imath]P(n)[/imath] be the product of the positive integers which are [imath]\leq n[/imath] and relatively prime to [imath]n[/imath]. Prove that [imath]P(n) = n^{\phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)}. [/imath] | 7531 | Proving [imath]P(n) =n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}[/imath]
Actually, I posted this long ago in MO but did not get a reply as it was unfit. Now this is an exercise in some textbook (I think Apostol), and I would be happy to receive some answers. Let [imath]P(n)[/imath] be the product of positive integers which are [imath]\leq n[/imath] and relatively prime to [imath]n[/imath]. Prove that [imath] \displaystyle P(n) = n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}.[/imath] |
2169904 | An example for two finitely generated and residually finite whose quotients are finite and Isomorphic, but themselves are not Isomorphic
I'm Trying to find an example for two non isomorphic, finitely generated, residually finite groups [imath]G_1[/imath] and [imath]G_2[/imath] such that there exists [imath]N_1\unlhd G_1[/imath],[imath]N_2\unlhd G_2[/imath] while [imath]G_1/N_2 \cong G_2/N_2[/imath] and [imath]|[G_1:N_1]|=|[G_2:N_2]|<\infty[/imath]. It's my understanding that residually finite is not the same as finitely presented. Thank's in advance to all responders. | 2171305 | Example of [imath]2[/imath] non-isomorphic groups that have the same quotients
I'm looking for an example of [imath]2[/imath] non isomorphic groups [imath]G_1,G_2[/imath] that are finitely generated and presented that have the same finite quotients (up to isomorphism) . Thanks |
2159355 | Finding equivalent of an iterative sequence
Define [imath](u_n)_{n\in\mathbb{N}^*}[/imath]: [imath]u_1 =1 , u_{n+1}=1+\dfrac{n}{u_n}[/imath] Find an asymptotic equivalent of [imath]u_n[/imath] when [imath]n\to+\infty[/imath]. I guess that the answer should be [imath]\sqrt{n}[/imath], but I couldn't prove it... | 2166928 | Find equivalent of recurrent sequence [imath]u_1=1[/imath], [imath]u_{n+1}=1+\frac n {u_n}[/imath]
The sequence is defined: [imath]u_1=1, u_{n+1}=1+\dfrac n {u_n}[/imath] The question asks to find an asymptotic development of 2 terms for [imath]n\to+\infty[/imath]. I have got [imath]u_n\sim_\infty\sqrt{n} [/imath], but how to derive the 2nd term? |
2172874 | [imath]n[/imath] is prime [imath]\Leftrightarrow [a][b]=0\Rightarrow [a]=[b]=0 [/imath]
I'm stuck on the following exercise from Herstein's "Topics in Algebra": "show that ([imath]n[/imath] is prime) [imath]\Leftrightarrow ([a][b]=[0]\Rightarrow [a]=[b]=[0]) [/imath] in [imath]J_n[/imath]". for the rightward implication I have: [imath][a][b]=[ab]=[0]\Rightarrow n|ab\Rightarrow n|a[/imath] or [imath]n|b[/imath] (or both) [imath]\Rightarrow [a]=[0][/imath] or [imath][b]=[0][/imath] and I'm stuck on the leftward implication. Why do I get [imath][a]=[0][/imath] OR [imath][b]=[0][/imath] and not [imath][a]=[0][/imath] AND [imath][b]=[0][/imath] in the leftward implication? Where is it that I go wrong? I'd also appreciate any comment/hint about how to prove the remaining left implication. | 885211 | Avoiding negative integers in proof of Fundamental Theorem of Arithmetic
There is an exercise on page 44 of Amann's book Analysis, Vol I which stuck me so much. I quoted it here: Ex7: Let [imath]p\in\mathbb{N}[/imath] with [imath]p>1.[/imath] Prove that [imath]p[/imath] is a prime number if and only if, for all [imath]m,n\in\mathbb{N},[/imath] [imath]p\mid mn\implies p\mid m \quad \text{or} \quad p\mid n.[/imath] Since this exercise occurs only after the Euclid's Division Algorithm (Page 34, Amann, Analysis Vol 1) and the Fundamental Theorem of Arithmetic (Page 36), I can not use the result of greatest common divisor of [imath]m[/imath] and [imath]n,[/imath] because that concerns the definition of the negative numbers, which is not coming into being in that section. Therefore, I ask the following question: Can Euclid's Division Algorithm and/or Fundamental Theorem of Arithmetic implies Ex7? Or in another word, Can we prove Ex7 by only using Euclid's Division Algorithm and/or Fundamental Theorem of Arithmetic? |
2172927 | Partial Fraction Decomposition When Denominator Has an Irreducible Factor of Degree 2
When doing partial fraction decomposition, there are three cases: Case 1: Denominator has distinct linear factors. [imath]f(x) = \dfrac{P(x)}{(x - a_1)...(x - a_k)} = \dfrac{A}{x - a_2} + \dfrac{B}{x - a_k}[/imath] where [imath]a_1, ..., a_k[/imath] are pairwise distinct. Case 2: Denominator has repeated linear factors. [imath]f(x) = \dfrac{P(x)}{(x - a)^c} = \dfrac{B_1}{x - a} + ... + \dfrac{B_c}{(c - a)^c}[/imath] Case 3: Denominator has an irreducible factor of degree 2: [imath]f(x) = \dfrac{P(x)}{(x - a)(x^2 + bx + c)} = \dfrac{A_1}{x - a} + \dfrac{C_1x + C_2}{x^2 + bx + c}[/imath] In cases 1 and 2, the numerators are all given constant placeholders ([imath]A, B[/imath]), but in case 3, the numerator is given a polynomial of degree one as the placeholder ([imath]C_1x + C_2[/imath]). I'm curious as to why this is the case. What is the difference between case 1 and 2 and 3 that necessitates this difference in the numerator? What is the reasoning behind this? I would greatly appreciate it if people could please take the time to elaborate on this. | 672430 | Partial Fraction Decomposition?
For repeated, linear factors, (the term [imath](x-r)^m[/imath] appears in the denominators for some integer [imath]r[/imath] and [imath]m[/imath], where [imath]m > 1[/imath]) there is a partial fraction for each power upto and including [imath]m[/imath]. Why? Why can't you just have one partial fraction for the term with the [imath]m^{th}[/imath] power? Could someone give me an explanation about why you need a partial fraction for each power? |
2172521 | [imath]S[/imath] is a subring of an integral domain [imath]R[/imath] which is finitely generated algebra over [imath]S[/imath], Jac(S)=0 implies Jac(R)=0
This is a problem in Atiyah Macdonald Commutative Algebra.(Problem 5.22) [imath]S[/imath] is a subring of an inregral domain [imath]R[/imath]. [imath]R[/imath] is finitely generated [imath]S[/imath] algebra. If Jacobson radical of [imath]S[/imath] is 0, then Jacobson radical of [imath]R[/imath] is 0. I followed the hint provided the book as following. The purpose is to show for each element of [imath]R[/imath], there is a maximal ideal of [imath]R[/imath] avoiding it. Pick a [imath]0\neq v\in R[/imath]. Since we have injective maps [imath]S\to R[/imath] and [imath]R\to R_v[/imath], we embed [imath]S[/imath] in [imath]R_v[/imath] as a subring and this [imath]R_v[/imath] is finitely generated as well over [imath]S[/imath]. It is clear that there is [imath]0\neq s\in A[/imath] such that we can extend ring homomorphism [imath]\phi:S\to\Omega[/imath] to [imath]\tilde{\phi}:R_v\to\Omega[/imath] where [imath]\Omega[/imath] is some algebraically closed field. Here [imath]\Omega[/imath] is chosen to be the algebraically closed field of [imath]S/m[/imath] where [imath]s\not\in m[/imath] as jacobson radical of [imath]S[/imath] is 0. It is clear that [imath]v[/imath] under the map [imath]R_v\to\Omega[/imath] is not zero or we will have [imath]1[/imath] being sent to [imath]0[/imath]. However, I am stucked at showing [imath]Ker(R_v\to\Omega)\subset R_v[/imath] is maximal ideal since this will lead to [imath]Ker(R_v\to\Omega)\cap B[/imath] is maximal. It is not even clear that this will be maximal. However, I can show if [imath]R[/imath]'s jacobson radical is trivial, [imath]R[x][/imath]'s is trivial. I also noticed that quotient and localization do not commute with infinite intersection. What should I do at this point? | 2172775 | Atiyah Macdonald Exercise 5.22
This is a problem in Atiyah Macdonald, Commutative Algebra. Problem 5.22: [imath]S[/imath] is a subring of an integral domain [imath]R[/imath]. [imath]R[/imath] is a finitely generated [imath]S[/imath] algebra. If the Jacobson radical of [imath]S[/imath] is 0, then the Jacobson radical of [imath]R[/imath] is 0. I followed the hint provided in the book as follows. The purpose is to show for each element of [imath]R[/imath], there is a maximal ideal of [imath]R[/imath] avoiding it. Pick a [imath]0\neq v\in R[/imath]. Since we have injective maps [imath]S\to R[/imath] and [imath]R\to R_v[/imath], we embed [imath]S[/imath] in [imath]R_v[/imath] as a subring and this [imath]R_v[/imath] is finitely generated as well over [imath]S[/imath]. It is clear that there is [imath]0\neq s\in A[/imath] such that we can extend the ring homomorphism [imath]\phi:S\to\Omega[/imath] to [imath]\tilde{\phi}:R_v\to\Omega[/imath] where [imath]\Omega[/imath] is some algebraically closed field. Here [imath]\Omega[/imath] is chosen to be the algebraically closed field of [imath]S/m[/imath] where [imath]s\not\in m[/imath] as jacobson radical of [imath]S[/imath] is 0. It is clear that [imath]v[/imath] under the map [imath]R_v\to\Omega[/imath] is not zero or we will have [imath]1[/imath] being sent to [imath]0[/imath]. However, I am stuck at showing [imath]Ker(R_v\to\Omega)\subset R_v[/imath] is maximal ideal since this will lead to [imath]Ker(R_v\to\Omega)\cap B[/imath] is maximal. It is not even clear that this will be maximal. However, I can show if [imath]S[/imath]'s Jacobson radical is trivial, then [imath]R=S[x][/imath]'s is trivial. It is possible that [imath]R=S[x]/Q[/imath] where [imath]Q[/imath] is some prime ideal and Jacobson radical of [imath]R[/imath] is the intersection of maximal ideals containing [imath]Q[/imath]. I also noticed that quotient and localization do not necessarily commute with infinite intersection. What other options do I have here? |
2173271 | Evaluating [imath]\sum_{i=1}^k \frac1d_i[/imath]
Let [imath]d_1, d_2,..., d_k[/imath] be all factors oc a positive integer [imath]n[/imath] including 1 and n. Suppose [imath]d_1+d_2+... +d_k=72[/imath]. The question is to evaluate [imath]\sum_{i=1}^k \frac1d_i[/imath] I found this question in Test of Maths at 10+2level. I tried applying inequality which made it ascertain that the summation is greater than [imath]72/k[/imath] but i couldn't find it's value. Any ideas? Thanks. . | 1330298 | Find the sum of reciprocals of divisors given the sum of divisors
Let [imath]d_1, d_2, \cdots d_k[/imath] be all the factors of a positive integer '[imath]n[/imath]' including [imath]1[/imath] and [imath]n[/imath]. Suppose [imath]d_1 + d_2 + d_3+\cdots+d_k = 72[/imath]. Then find the value of [imath]\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}[/imath]. Attempt: Consider, [imath]d_1, d_2, \cdots d_k[/imath] are the factors of [imath]n[/imath] ascending order. Then [imath]d_1d_k=n[/imath], [imath]d_2d_{k-1}=n[/imath], [imath]\cdots[/imath], [imath]d_kd_{1}=n[/imath] ............... (1) Assume, [imath]\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}=P[/imath] Then [imath]\frac{n}{d_1}+\frac{n}{d_2}+\cdots + \frac{n}{d_k}=nP[/imath] i.e [imath]d_k+d_{k-1}+\cdots +d_2+d_1=nP[/imath] (using 1) i.e [imath]72=nP[/imath], or [imath]P=72/n[/imath]. Problem: I have considered that number of divisors of [imath]n[/imath] are even as every divisor [imath]d[/imath] of [imath]n[/imath] has a twin divisor [imath]n/d[/imath]. But it is not true if [imath]n[/imath] is a perfect square i.e [imath]n=d^2[/imath] as in this case [imath]n[/imath] has odd number of divisors (for example, [imath]4=2^2[/imath], [imath]4[/imath] has three divisors [imath]1,2 ~\&~ 4[/imath]). |
2173163 | What are the stepps to this shortcut?
I looked at an answer for a question and I don't know how they came to this conclusion. [imath](k+1)!-1 + (k+1)!(k+1) = (k+2)! - 1[/imath] What are the stepps behind it? | 349474 | What are the rules for factorial manipulation?
I know that [imath](k+1)! - 1 + (k+1)(k+1)! = (k+2)! - 1[/imath] thanks to wolframalpha, but I don't understand the steps for simplification, and I can't seem to find any rules about factorial manipulations on google. Can someone explain this please? |
2173959 | How to change [imath]\dfrac{k(k+1)(2k+1)}6+(k+1)^2[/imath] into [imath]\dfrac{(k+1)(k+2)(2k+3)}6[/imath]
Prove [imath]1^2+2^2+3^2+4^2...+n^2=\dfrac{n(n+1)(2n+1)}6[/imath] by induction. I am trying to show true for [imath]n=k+1[/imath]. I now have the equation [imath]1^2+2^2+3^2+4^2...+k^2+(k+1)^2=\dfrac{k(k+1)(2k+1)}6+(k+1)^2[/imath]. I know that the result should be [imath]1^2+2^2+3^2+4^2...+k^2+(k+1)^2=\dfrac{(k+1)(k+2)(2k+3)}6[/imath] How can I achieve this? | 2137292 | How can you prove [imath]\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}[/imath] without much effort?
I will keep it short and take only an extract (most important part) of the old task. [imath]\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}[/imath] What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : / If anyone wants see, here is my long solution which I'm not happy with: [imath]\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow[/imath] [imath]\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow[/imath] [imath]\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow[/imath] [imath]\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}[/imath] |
2174237 | Prove that the countable union of sets of cardinality [imath]c[/imath] again has cardinality [imath]c[/imath].
Where [imath]c = 2^{\aleph_0}[/imath]. I have no idea how to tackle this problem. Any help is greatly appreciated! | 1181761 | Countable union of sets of cardinality [imath]c[/imath] has cardinality [imath]c[/imath]
The book Theory of Functions of a Real Variable by I. P. Natanson, proves that a denumerable or finite union of pairwise disjoint sets of cardinality [imath]c[/imath] has cardinality [imath]c[/imath]. The proofs given in the book are fairly easy, using the axiom of choice. But then it is left as an exercise a generalization of the above theorems for not necessarily pairwise disjoint sets. I have searched all over the web, but all the proofs found so far are for pairwise disjoint sets. The proof given in the book relies on the fact that each set [imath]A_k[/imath] can be matched with a set like [imath][a_{k-1},a_k)[/imath], with [imath]a_k[/imath] point of a partition of [imath][0,1][/imath], such that [imath]\bigcup_{k=1}^{\infty\lor n}A_k=[0,1)[/imath]. So it is easy to guess the bijective function, but I can't generalize this proof. Any help is highly appreciated. Thanks and regards. |
2174380 | Prove [imath]f(x)\geq e^x[/imath]
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be a differentiable function, such that [imath]f^{\ '}\geq f[/imath] and [imath]f(0)=1.[/imath] Prove [imath]f(x)\geq e^x, \forall x\geq0[/imath]. So I started by defining a function [imath]g(x)=f(x)-e^x[/imath], and my intuition tells me that [imath]f^{\ '}\geq f[/imath] means [imath]f[/imath] is increasing, but I'm not sure how to continue. Any help appreciated. | 2129435 | If [imath]f(x)[/imath] is continuous and differentiable function , prove that [imath]f(x) \geq e^{3x}, \forall x \geq 0[/imath]
[imath]f(x)[/imath] is continuous and differentiable function defined in [imath]x \geq 0[/imath]. If [imath]f(0)=1[/imath] and [imath]f'(x) >3 f(x)[/imath], then prove that [imath]f(x) \geq e^{3x}, \forall x \geq 0[/imath] Could someone give me little hint as how to proceed in this question? I am not able to initiate. |
2174881 | Argue by contradiction to show that there is no surjection...
from [imath]A[/imath] to [imath]\{0,1\}^A[/imath]. When [imath]A[/imath] is any (possibly infinite) set How does one go about this? | 1414007 | Showing a function [imath]f[/imath] cannot be surjective
Good day all! So I have a question about the problem: Let [imath]E[/imath] be a set, and [imath]f[/imath] be a mapping from [imath]E[/imath] to [imath]P(E)[/imath]. Consider a set [imath]A[/imath] such that [imath]x[/imath] is in [imath]E[/imath] but [imath]x[/imath] in NOT in [imath]f(x)[/imath] Show [imath]f[/imath] cannot be surjective. My question is: How is this? I see that if [imath]x[/imath] is in [imath]A[/imath] then [imath]x[/imath] is not in [imath]f(x)[/imath] which means [imath]x[/imath] is not in [imath]P(E)[/imath]. How is [imath]x[/imath] in [imath]E[/imath] but [imath]x[/imath] not in [imath]P(E)[/imath]? Is my understanding flawed? |
1088469 | Is the complement of a prime ideal closed under both addition and multiplication?
Let [imath]P[/imath] be a prime ideal in a commutative ring [imath]R[/imath] and let [imath]S=R-P[/imath] ,i.e. [imath]S[/imath] is the complement of [imath]P[/imath] in [imath]R[/imath]. Then, justify with reason which of the following(s) are correct: [imath]S[/imath] is closed under addition. [imath]S[/imath] is closed under multiplication. [imath]S[/imath] is closed under both addition and multiplication. The following argument provides a partial answer: Let [imath]P=3 \mathbb Z[/imath] and [imath]R= \mathbb Z[/imath] the [imath]2,4[/imath] in [imath]S[/imath] but [imath]2+4[/imath] in [imath]P[/imath], so option 1. and 3. are incorrect. But I don't know about 2. | 2176188 | Complement of a prime ideal is a subset closed under multiplication, containing 1
Let [imath]R[/imath] be a commutative ring and let [imath]I[/imath] be a prime ideal of [imath]R[/imath]. Prove that [imath]S = R \setminus I[/imath] is a multiplicative closed subset containing [imath]1[/imath]. Please can anyone help me out here? |
2176264 | Suppose n is a product of distinct odd primes. Show [imath]a^{\phi(n)+1}\equiv a [/imath] (mod [imath]n[/imath]).
Suppose n is a product of distinct odd primes. Show [imath]a^{\phi(n)+1}\equiv a [/imath] (mod [imath]n[/imath]). Let [imath]\phi(n)[/imath] be the Euler Totient function. I let [imath]n = p_1p_2\cdot\cdot\cdot p_k[/imath] for [imath]p_i > 2[/imath] and each of the [imath]p_i[/imath] are distinct. I let [imath]a = q_1^{r_1}q_2^{r_2}\cdot\cdot\cdot q_m^{r_m}[/imath] for primes [imath]q_i[/imath]. There are two cases: either gcd(a, n) = 1 or gcd(a, n) is a product of some of the primes. I can easily figure Case 1 by using Euler's Theorem and the Chinese Remainder Theorem. I'm having problems figuring out the second case. Any suggestions of where I might start? | 786526 | Proof of an alternative form of Fermat-Euler's theorem.
I want to know a proof of an alternative form of Fermat-Euler's theorem [imath]a^{\phi (n) +1} \equiv a \pmod n[/imath] when [imath]a[/imath] and [imath]n[/imath] are not relatively prime. I searched some number theory books and a cryptography book and internet, but there were only proofs of the original theorem [imath]a^{\phi (n)} \equiv 1 \pmod n[/imath] or in case n=pq which is used for RSA. So I would be very thankful if someone show me a proof or book I should read. |
2176083 | Variation of Dirichlet's function
I am to prove that the variation of Dirichlet's function is infinite. I would really appreciate any help, because I do not know how to start. [imath]f(x)= \begin{cases} 1 & \text{if $x\in\mathbb{Q}$}\\ 0 & \text{if $ x \notin\mathbb{Q}$} \\ \end{cases}[/imath] | 330147 | Some questions about functions of bounded variation: Jordan's theorem
I was trying to do some of these questions to check my understanding about the topic, but I'm not sure if they're correct. Here are my answers. 1) Suppose [imath]f[/imath] is continuous on [imath][0,1][/imath]. Must there be a nondegenerate closed subinterval [imath][a,b][/imath] of [imath][0,1][/imath] for which the restriction of [imath]f[/imath] to [imath][a,b][/imath] is of bounded variation? No, because [imath]f(x)[/imath] might be a constant function. In this case, [imath]f(x)[/imath] will not be the difference of two increasing functions...and so will not be of bounded variation. 2) Let [imath]f[/imath] be the Dirichlet function, the characteristic function of the rationals in [imath][0,1][/imath]. Is [imath]f[/imath] of bounded variation on [imath][0,1][/imath]? No, because we cannot express it as the difference of two increasing functions. 3) Let [imath]f[/imath] be a step function on [imath][a,b][/imath]. Find a formula for its total variation. If we partition [imath][a,b][/imath] in such a way that for each [imath][x_{i-1}, x_i][/imath] (which is part of the partition), [imath]x_{i-1}[/imath] and [imath]x_i[/imath] each correspond to the midpoint of a step. I know this is not a formal answer, but I'm just trying to check my understanding. |
2175888 | How to show that a theorem holds for isomorphic vector spaces?
Theorem: Any two norms on a finite-dimensional vector space [imath]V[/imath] are equivalent. My book begins the proof by noting that we can identify [imath]V[/imath] with [imath]\mathbb R^n[/imath] for some [imath]n\in\mathbb N[/imath], and therefore we only need to prove the theorem for [imath]\mathbb R^n[/imath]. However, I don't see why this is true. I know that there is a bijection between [imath]V[/imath] and [imath]\mathbb R^n[/imath] for some [imath]n\in\mathbb N[/imath], or equivalently, there is a bijection between a basis in [imath]\mathbb R^n[/imath] and a basis in [imath]V[/imath]. But how to proceed from there on? I've read somewhere that an isomorphism is basically just the relabelling of the elements... but how do I know that things like inequalities and the like stay preserved between two identified elements? Things like "the vector space structure is preserved" sounds slightly vague to me, so is there something more rigorous to say? | 243612 | n-dimensional vector space isomorphic to the n-th product of a field as abelian groups
Let [imath]V[/imath] be a finite dimensional vector space (of dimension n) over a field [imath]F[/imath]. I need to show that [imath]V[/imath] is isomorphic to [imath]F^n[/imath] as abelian groups. However, I don't really understand what does "isomorphic as abelian groups" mean, my (poor) attempt to solve this was: Let [imath]f:V\to F^n[/imath], [imath]f((v_1,v_2,\ldots,v_n)):=v_1+\ldots+v_n[/imath], clearly [imath]f[/imath] is a group homomorphism, but it is not biyective since, for example, in the case where [imath]dim(V)=2[/imath] and [imath]F=\mathbb{R}[/imath] we have [imath]f((2,-2))=0=f((3,-3))[/imath], I've also tried it by defining [imath]f[/imath] to be the product of a vector's entries but obviously it didn't work so I'm stuck! I think this should be an easy problem but since I don't quite understand it, it's giving me problems. Thank you for your help. |
1857939 | Sections of tensor bundle are tensor product of sections
Given [imath]E,F[/imath] vector bundles over a manifold [imath]M[/imath], I would like to know a proof of [imath]\Gamma(E\otimes F) = \Gamma(E) \otimes_{C^\infty(M)} \Gamma(F)[/imath]. Where [imath]\Gamma[/imath] denotes the smooth sections over [imath]M[/imath]. Thanks | 492166 | Global sections of a tensor product of vector bundles on a smooth manifold
This question is similar to Conditions such that taking global sections of line bundles commutes with tensor product? and Tensor product of invertible sheaves except that I am concerned with smooth manifolds. If [imath]M[/imath] is a smooth manifold and [imath]V[/imath] and [imath]W[/imath] are (smooth, real) vector bundles on [imath]M[/imath], are there reasonable conditions that ensure that the map [imath]\Gamma(M,V)\otimes_{\mathscr{O}_M(M)}\Gamma(M,W)\rightarrow\Gamma(M,V\otimes W)[/imath] is an isomorphism? Equivalently, are there any reasonable conditions under which taking global sections of finite locally free [imath]\mathscr{O}_M[/imath]-modules commutes with tensor product? I don't really know much smooth manifold theory, and my intuition is mostly derived from algebraic geometry, where the analogous question, as is indicated in the questions I linked to, is not a simple one, even for invertible sheaves. I'd always assumed the situation for smooth manifolds would be the same, but I was just looking at the Wikipedia page on vector-valued differential forms, and it seems (unless I'm misunderstanding which is certainly possible) to claim there that for [imath]V[/imath] an arbitrary vector bundle and [imath]W[/imath] the [imath]p[/imath]-th exterior power of the cotangent bundle, [imath]p\geq 1[/imath], the answer to my question is "yes," with the justification being that "[imath]\Gamma(M,-)[/imath] is a monoidal functor." Looking at the definition of a monoidal functor, it seems to me that all this implies is that there is a map from the tensor product of the global sections to the global sections of the tensor product (which is clear from the definitions) and not that this map is an isomorphism. |
2169545 | Integral of [imath]\sec x[/imath] using certain identity
Use the identity [imath]\sec x=\frac{\cos x}{1-\sin^2x}[/imath] to prove that the integral of [imath]\sec x[/imath] equals [imath]\frac12\ln\frac{1+\sin x}{1-\sin x}[/imath]. I tried to manipulate it but nothing seems to work. Help. | 1492942 | Verify [imath]\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C[/imath]
Question says it all, how can I verify the following? [imath]\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C[/imath] |
2178532 | How do you graph f(x) = (-1)^x
What does the graph of the function [imath] f(x) = (-1)^x [/imath] look like? It seems as if on all even [imath]x[/imath], [imath]y = 1[/imath], and on all odd [imath]x[/imath], [imath]y = -1[/imath], but between these values (fractional indices) y's value is ambiguous. | 1073024 | How to visualize [imath]f(x) = (-2)^x[/imath]
Background I teach Algebra and second year Algebra to middle school students. We are currently studying Exponential, Power, and Logarithmic functions. We study exponential functions (of the form [imath]f(x) = ab^x[/imath] where [imath]a \neq 0, b > 0[/imath]). I've taught the basics of exponential functions before and I've wondered about the function [imath]f(x) = (-2)^x[/imath]. At first I was confused as to why the calculator would not graph any exponential function with a negative base. Then I realized that depending on the roots, say for [imath](-2)^{1.25}[/imath] and [imath](-2)^{1/3}[/imath] some very different calculations would need to be performed. I'm still lost as to how you would explain [imath](-2)^\pi[/imath] but I suppose that's just being annoying. I want to know how to explain the output of [imath]f(x) = (-2)^x[/imath]. Comments I know for example that there are a host of values for which the output would be imaginary. If we limit the domain of [imath]x[/imath] to rational numbers with odd denominators Here are two examples [imath](-2)^{2/5} = \left(\sqrt[5]{-2}\right)^2 \approx 1.3195[/imath] [imath](-2)^{3/5} = \left(\sqrt[5]{-2}\right)^3 \approx -1.5157[/imath]) we should be able to compute [imath]f(x)[/imath] for all such numbers. I would then think the graph would look something like f(x) = 2^x and [imath]f(x) = -(2^x)">[/imath] Curious how to visualize the plot including the imaginary outputs when x \in \mathbb{R}[imath][/imath] Wolfram Computes the result using the principal roots Question: How could you visualize and interpret f(x) = (-2)^x[imath] while incorporating the real-root based output I presented above? Or, alternatively, what's wrong with this?[/imath] Wolfram|Alpha says that it has no properties as a real function which makes sense given the restricted domain necessary to compute real outputs. Is there any mistake in thinking that with the domain restrictions I specified above (x[imath] all rational numbers such that [/imath]x = a/b[imath] where [/imath]b[imath] is odd) the output would resemble my plot above?[/imath] Is there a way to visualize -- or even better -- explain the behavior of this function for for all real numbers x$? If we added in an imaginary axis, could we use it to show the results when we need to compute an imaginary output? |
2177126 | Proving the predicate [imath]\forall n \in \mathbb{N} , n \geq 2[/imath] using the Pigeonhole principle
The version of the pigeonhole principle we’ll look at is the following: “For all natural numbers [imath]n[/imath] greater than [imath]1[/imath], and all subsets [imath]S, T[/imath] of [imath]\mathbb{N}[/imath] where [imath]|S| = n[/imath] and [imath]|T| = n − 1[/imath], there does not exist a one-to-one function from [imath]S[/imath] to [imath]T[/imath].” We define the predicate [imath]PHP(n)[/imath]: [imath]∀S, T ⊆ \mathbb{N}, |S| = n ∧ |T| = n − 1 ⇒ (∀f : S → T, ∃s_{1}, s_{2} ∈ S, s_{1} \neq s_{2} ∧ f(s_{1}) = f(s_{2}))[/imath] How do we prove that [imath]∀n ∈ N, n ≥ 2 ⇒ PHP(n)[/imath]. I'm having a very hard time with this proof and don't even know where to begin with. | 2174207 | Proving the predicate [imath]\forall n \in \mathbb{N} , n \geq 2[/imath] using the Pigeonhole principle
The version of the pigeonhole principle we’ll look at is the following: “For all natural numbers [imath]n[/imath] greater than [imath]1[/imath], and all subsets [imath]S, T[/imath] of [imath]\mathbb{N}[/imath] where [imath]|S| = n[/imath] and [imath]|T| = n − 1[/imath], there does not exist a one-to-one function from [imath]S[/imath] to [imath]T[/imath].” We define the predicate [imath]PHP(n)[/imath]: [imath]∀S, T ⊆ \mathbb{N}, |S| = n ∧ |T| = n − 1 ⇒ (∀f : S → T, ∃s_{1}, s_{2} ∈ S, s_{1} \neq s_{2} ∧ f(s_{1}) = f(s_{2}))[/imath] How do we prove that [imath]∀n ∈ N, n ≥ 2 ⇒ PHP(n)[/imath]. I'm having a very hard time with this proof and don't even know where to begin with. |
1799491 | compute the L^2-distance of a given function to the set of Gaussian functions
I am faced with the following question: given a probability density function [imath]f[/imath] over [imath]\mathbb{R}[/imath] with [imath]\int_{\mathbb{R}}f(x)x^2dx=\sigma^2[/imath] given, I am trying to find the "nearest" Gaussian to [imath]\sqrt{f}[/imath] in [imath]L^2[/imath], i.e. compute [imath]\inf_{a\in \mathbb{R}, b > 0}\|\sqrt{f}-G_{a,b}\|_2^2[/imath] where [imath]G_{a,b}(x)=a\exp(-bx^2)[/imath] and the [imath]\|\cdot\|_2[/imath] is the classical [imath]L^2[/imath]-norm over [imath]\mathbb{R}.[/imath] If one replaces the [imath]L^2[/imath]-distance by the relative entropy [imath]H(f|G_{a,b})[/imath], it is easy to check that the infimum is a minimum given by the centered gaussian with variance [imath]\sigma^2.[/imath] For the [imath]L^2[/imath]-distance, it seems more involved...... | 1799248 | How to compute the L^2-distance of a given function to the set of Gaussian functions
I am faced with the following question: given a probability density function [imath]f[/imath] over [imath]\mathbb{R}[/imath] with [imath]\int_{\mathbb{R}}f(x)x^2dx=\sigma^2[/imath] given, I am trying to find the "nearest" Gaussian to [imath]\sqrt{f}[/imath] in [imath]L^2[/imath], i.e. compute [imath]\inf_{a\in \mathbb{R}, b > 0}\|\sqrt{f}-G_{a,b}\|_2^2[/imath] where [imath]G_{a,b}(x)=a\exp(-bx^2)[/imath] and the [imath]\|\cdot\|_2[/imath] is the classical [imath]L^2[/imath]-norm over [imath]\mathbb{R}.[/imath] If one replaces the [imath]L^2[/imath]-distance by the relative entropy [imath]H(f|G_{a,b})[/imath], it is easy to check that the infimum is a minimum given by the centered gaussian with variance [imath]\sigma^2.[/imath] For the [imath]L^2[/imath]-distance, it seems more involved...... |
2178650 | The idea of the proof of a theorem.
Could anyone explain the contradiction that he will use to prove the theorem? Especially when he said that:" the idea now is to construct a family of trigonometric functions [imath]{p_{k}}[/imath] that peaks at 0, and so that [imath]\int p_{k}( \theta) f(\theta) d\theta \rightarrow \infty as\ k\ \rightarrow \infty.[/imath] This will be our contradiction, since those integrals tends to zero by assumption." I do not see what is the assumption that makes them tends to zero and why the integral above tends to [imath]\infty.[/imath] Thanks. they are attached in the following photos. | 1670464 | Difficulty in understanding a part in a proof from Stein and Shakarchi Fourier Analysis book.
Theorem 2.1 : Suppose that [imath]f[/imath] is an integrable function on the circle with [imath]\hat f(n)=0[/imath] for all [imath]n \in \Bbb Z[/imath]. Then [imath]f(\theta_0)=0[/imath] whenever [imath]f[/imath] is continuous at the point [imath]\theta_0[/imath]. Proof : We suppose first that [imath]f[/imath] is real-valued, and argue by contradiction. Assume, without loss of generality, that [imath]f[/imath] is defined on [imath][-\pi,\pi][/imath], that [imath]\theta_0=0[/imath], and [imath]f(0) \gt 0[/imath]. Since [imath]f[/imath] is continuous at [imath]0[/imath], we can choose [imath] 0\lt \delta \le \frac \pi2[/imath], so that [imath]f(\theta) \gt \frac {f(0)}2[/imath] whenever [imath]|\theta| \lt \delta[/imath]. Let [imath]p(\theta)=\epsilon + \cos\theta,[/imath] Where [imath]\epsilon \gt 0[/imath] is chosen so small that [imath]|p(\theta)| \lt 1 - \frac \epsilon2[/imath], whenever [imath]\delta \le |\theta| \le \pi[/imath]. Then, choose a positive [imath]\eta[/imath] with [imath]\eta \lt \delta[/imath], so that [imath]p(\theta) \ge 1 + \frac \epsilon2[/imath], for [imath]|\theta| \lt \eta[/imath]. Finally, let [imath]p_k(\theta)=|p(\theta)|^k[/imath], and select [imath]B[/imath] so that [imath]|f(\theta)| \le B[/imath] for all [imath]\theta[/imath]. This is possible since [imath]f[/imath] is integrable, hence bounded. By construction, each [imath]p_k[/imath] is a trigonometric polynomial, and since [imath]\hat f(n)=[/imath] for all [imath]n[/imath], we must have [imath]\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\,d\theta=0[/imath] for all [imath]k[/imath]. I understood the first paragraph clearly. But the rest is not making it's way into my head. In the beginning of second paragraph, how does the given range works for choosing [imath]\delta[/imath]? If the continuity is used to get the range, then how? How can we choose [imath]\epsilon[/imath] so small such that, [imath]|p(\theta)| \lt 1 - \frac \epsilon2[/imath], whenever [imath]\delta \le |\theta| \le \pi[/imath]? How can we choose positive [imath]\eta[/imath] with [imath]\eta \lt \delta[/imath], so that [imath]p(\theta) \ge 1+ \frac \epsilon2[/imath], for [imath]|\theta| \lt \eta[/imath]. Why do we must have [imath]\int_{-\pi}^{\pi}f(\theta)p_k(\theta)\,d\theta=0[/imath] for all [imath]k[/imath]? |
2178627 | Linear Algebra vectors proof?
I have looked at this question and don't understand almost everything about it. Looking for a detailed answer. For any two vectors a and b, Show that [imath]|a+b|^2 + |a-b|^2 = 2|a|^2 +2|b|^2[/imath] | 1919758 | The Parallelogram Identity for an inner product space
For a given inner product space [imath]V[/imath], I would like to prove the inequality [imath]\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).[/imath] We see that [imath]\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,[/imath] which by linearity [imath]=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.[/imath] I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement? |
2180129 | Does it imply that [imath]\int _0^\infty f_n\to \int _0^\infty f[/imath]?
If [imath]f_n[/imath] is a sequence of continuous functions such that [imath]f_n[/imath] converges to [imath]f[/imath] uniformly . Does it imply that [imath]\int _0^\infty f_n\to \int _0^\infty f[/imath]? I think the answer is no though I could not find a counter-example. Because if [imath]f_n\to f[/imath] uniformly then [imath]\int _0^\infty |f_n- f|<\epsilon \times \infty [/imath] which is undefined. | 843776 | Does $\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x)\,dx\,$?
If [imath]\displaystyle\{f_n(x)\}[/imath] is sequence of continuous function on [imath]\mathbb{R}[/imath] converging uniformly to [imath]f(x)[/imath], then does [imath]\displaystyle\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x) dx\,[/imath] hold? I know it holds if we integrate over an interval of the form [imath][a,b][/imath] but not sure about improper integral! |
2161518 | limit : [imath]\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n=?[/imath]
fine the limit : [imath]\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n=?[/imath] [imath]a,b\in \mathbb{R}[/imath] My Try : [imath]\lim_{n\to \infty} (1)+\lim_{n\to \infty}(\frac{\sqrt[n]{a}-1}{b})^n=1+0=1[/imath] Because: [imath]\lim_{n\to \infty} \sqrt[n]{a}=1 [/imath] is it right ? | 2160401 | Limit [imath]\lim (1+(a^{1/n}-1) / b)^n[/imath]
How can I find the [imath]\lim (1+(a^{1/n}-1) / b)^n[/imath] when [imath]n \to \infty[/imath]? I would greatly appreciate it if you kindly give me some hints. |
670535 | Formally proving that [imath]\lim\left[ n^2/2^n \right] = 0[/imath]
Not sure how to formally prove this (specifically regarding the choice of [imath]\epsilon[/imath] in the formal limit definition)... Any suggestions? | 2179947 | How can I prove [imath]\lim_{n \to \infty} \frac{n^2}{2^n}=0[/imath]
How can I prove [imath]\lim_{n \to \infty} \frac{n^2}{2^n}=0[/imath] I tried to use [imath]\mid \frac{n^2}{2^n} - 0 \mid <\epsilon [/imath] However, because of [imath]n^2[/imath] I cannot use it. Also, I tried to use ratio test and I got [imath]\lim_{n->\infty}\frac{(n+1)^2}{2}\frac{1}{n^2}[/imath], but after that one, I dont know how to get limit < 0 I have no idea to solve this problem. I can also use a squeez lemma, but dont know how to apply to this question. |
2180785 | A question in real anaylsis
If we let [imath]f_n[/imath] be a sequence of functions on [imath][a,b][/imath] such that, -there exists a point [imath]x_0\in[a,b][/imath] where [imath]lim_{n->\infty}f_n(x_0)[/imath] exists -each [imath]f_n[/imath] is differentiable -the sequence [imath](f'_n)[/imath] converges uniformly on [imath][a,b][/imath] I want to prove that [imath]f_n[/imath] converges uniformly on [imath][a,b][/imath] to some function [imath]f[/imath] My logic: if I use the triangle inequality to write, [imath]|f_n(x)-f_m(x)|<=|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|+|f_n(x_0)-f_m(x_0)|[/imath] Then I can try and apply the mean value theorem to prove that [imath]f_n[/imath] converges uniformly on [imath][a,b][/imath] to some function [imath]f[/imath]. Applying the MVT to see the right side above can be written [imath]|(f_n-f_m)'(c)(x-x_0)|+|f_n(x_0)-f_m(x_0)| \le \sup_{[a,b]}|f_n'-f_m'|(b-a)+|f_n(x_0)-f_m(x_0)|.[/imath] from here how can I use "uniformly Cauchy" criterion for uniform convergence to prove my problem? | 2179683 | [imath]f_n[/imath] converges uniformly on [imath][a,b][/imath] to some function [imath]f[/imath]
If we let [imath]f_n[/imath] be a sequence of functions on [imath][a,b][/imath] such that, -there exists a point [imath]x_0\in[a,b][/imath] where [imath]lim_{n->\infty}f_n(x_0)[/imath] exists -each [imath]f_n[/imath] is differentiable -the sequence [imath](f'_n)[/imath] converges uniformly on [imath][a,b][/imath] I want to prove that [imath]f_n[/imath] converges uniformly on [imath][a,b][/imath] to some function [imath]f[/imath] My logic: if I use the triangle inequality to write, [imath]|f_n(x)-f_m(x)|<=|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|+|f_n(x_0)-f_m(x_0)|[/imath] Then I can try and apply the mean value theorem to the first term on the left hand side to prove that [imath]f_n[/imath] converges uniformly on [imath][a,b][/imath] to some function [imath]f[/imath]. The mean value theorem states, Suppose f(x) is a function that satisfies both of the following conditions, -[imath]f(x)[/imath] is continuous on the closed interval [imath][a,b][/imath] -[imath]f(x)[/imath] is differentiable on the open interbal [imath](a,b)[/imath] Then there is a number [imath]c[/imath] such that [imath]a<c<b[/imath] and, [imath]f'(c)=\frac{f(b)-f(a)}{b-a}[/imath] |
2180815 | [imath]A[/imath] is a countable subset of [imath]X[/imath]
how can we prove countable function like: If [imath]f : X \to Y[/imath] is a function and [imath]A[/imath] is a countable subset of [imath]X[/imath], then [imath]f(A)[/imath] is countable. | 2163912 | Prove that functions map countable sets to countable sets
One of my homework problems asks to prove that if [imath]f:X\to Y[/imath] is a function and [imath]A[/imath] is a countable subset of [imath]X[/imath], then [imath]f(A)[/imath] is countable. I believe I have a valid proof by partitioning the set A into equivalency classes based in image under [imath]f[/imath]. Then using that partition as domain for a choice function [imath]C_f[/imath] back to [imath]A[/imath]. So now the function [imath]f[/imath] restricted to the domain (range [imath]C_f[/imath]) [imath]f|_{range(C_f)}[/imath] maps to [imath]f(A)[/imath] 1-1 and onto. Therefore [imath]|f(A)|=|range(C_f)|\leq |A|=\aleph_0[/imath] [imath]\therefore f(A)[/imath] is a countable set. [imath]\quad \blacksquare[/imath] My problem is that I used the axiom of countable choice [imath]AC_\omega[/imath] in my proof but that isn't a part of the course I am taking. Does anyone have a simpler proof or one that does not involve Choice? |
2180674 | If a polynomial [imath]f(X,Y)[/imath] vanishes on all [imath](x,y)[/imath], can we conclude [imath]f(X,Y)=0[/imath]?
Let [imath]k[/imath] be an infinite field and [imath]f(X,Y) \in k[X,Y][/imath] be a polynomial with two variables. If [imath]f(X,Y)[/imath] vanishes on all points [imath](x,y) \in k^2[/imath], i.e., [imath]f(x,y) = 0, \forall (x,y) \in k^2[/imath] can we conclude that [imath]f(X,Y) = 0[/imath]? Put another way, let [imath]k[/imath] be an infinite field, and let [imath]k[X,Y]^* := \{f : k^2 \longrightarrow k, (x,y) \mapsto f(x,y) \mid f(X,Y) \in k[X,Y]\}[/imath] denote the ring of polynomial functions. Then do we have [imath]k[X,Y] \cong k[X,Y]^*[/imath]? | 102182 | Polynomials vanishing on an infinite set
I'd like some help making this argument complete and rigorous (if it's correct - if not, help with that would be nice). Here [imath]k[/imath] is a field. Let [imath]A_1,\ldots,A_n \subseteq k[/imath] be infinite subsets. Then any polynomial in [imath]k[x_1,\ldots,x_n][/imath] that vanishes on [imath]A_1\times\cdots\times A_n\subseteq k^n[/imath] must be [imath]0[/imath] (as a polynomial). This is what I have ... For the case [imath]n=1[/imath], a non-constant polynomial can only have as many roots as its degree, and in particular, it must have a finite number of roots. The only polynomial in one variable that has an infinite number of roots is [imath]0[/imath], so if a polynomial in [imath]k[x_1,\ldots,x_n][/imath] vanishes on an infinite subset then it must be [imath]0[/imath]. For the inductive step, suppose the proposition is true for less than [imath]n[/imath] subsets and variables. Let [imath]p\in k[x_1,\ldots,x_n][/imath] vanish on [imath]A_1\times\cdots\times A_n[/imath]. Fix [imath]x_n[/imath] as some [imath]a\in A_n[/imath], and we have a polynomial in [imath]n-1[/imath] variables that vanishes on the set [imath]A_1\times\cdots\times A_{n-1}[/imath], so by the inductive hypothesis it must be identically [imath]0[/imath]. (Now it gets sketchy). Since this is true for any of the infinite values in [imath]A_n[/imath], and , [imath]p[/imath] must be [imath]0[/imath]. |
2180983 | Integration By Parts of Gamma Function
A textbook I'm self-studying - Introduction to Mathematical Statistics by Hogg - has the following text: T(a) = [imath]\int_{0}^{\infty} y^{\alpha-1}e^{-y} dy[/imath] [gamma function] If [imath]\alpha > 1[/imath], an integration by parts shows that [imath]T(\alpha) = (\alpha - 1)\int_{0}^{\infty} y^{\alpha-2}e^{-y} dy[/imath]. I'm having trouble deriving this for myself. How does an Integration by Parts lead to this result? Thank you. | 314527 | Help evaluating a gamma function
I need to do a calculus review; I never felt fully confident with it and it keeps cropping up as I delve into statistics. Currently, I'm working through a some proof theory and basic analysis as a sort of precursor to the calc review, and I just hit a problem that requires integration. Derivatives I'm ok with, but I really don't remember how to take integrals correctly. Here's the problem: [imath]\Gamma (x) = \int_0^\infty t^{x-1} \mathrm{e}^{-t}\,\mathrm{d}t[/imath] If [imath]x \in \mathbb{N}[/imath], then [imath] \Gamma(x)=(x-1)![/imath] Check that this is true for x=1,2,3,4 I did a bit of reading on integration, but it left my head spinning. If anyone wants to lend a hand, I'd appreciate it. I should probably just push this one to the side for now, but part of me wants to plow through it. Update: Thanks for the responses. I suspect this will all make more sense once I've reviewed integration. I'll have to revisit it then. |
2181269 | Summation of [imath]n/3^n[/imath]
[imath]\sum_{n=1}^\infty \frac{n}{3^n}[/imath] How do you find the sum? I don't know how to start this problem and no other website I found talks about a problem like this. | 1774607 | Identifying the series [imath]\sum\limits_{k=-\infty}^{\infty} 2^k x^{2^k}[/imath]
I came across following bi-infinite sum: [imath]\sum_{k=-\infty}^{\infty} 2^k x^{2^k}[/imath] Is this a known series? After some plotting I have the feeling that it could be equal or very similar to [imath]-\frac{1}{\ln(2) \ln(x)}[/imath] for all [imath]x\in(0,1)[/imath]. Is my assumption correct? |
2180349 | Nonexistence of [imath]f:\mathbb{Z}\to \{1,2,3\}[/imath] with certain properties
Show that there does not exist a function [imath]f:\mathbb{Z}\to \{1,2,3\}[/imath] satisfying [imath]f(x)\neq f(y)[/imath] for all [imath]x,y\in \mathbb{Z}[/imath] such that [imath]|x-y|\in \{2,3,5\}[/imath]. Proof: Suppose by contradiction that such function exists and denote it by [imath]f(x)[/imath]. Let [imath]f(0)=a, f(2)=b, f(5)=c[/imath] and then [imath]a\neq b, a\neq c, b\neq c[/imath] and also [imath]a,b,c\in \{1,2,3\}[/imath] and WLOG we put that [imath]a=1, b=2, c=3[/imath]. Consider [imath]f(3)[/imath] and since [imath]f(3)\neq 1, f(3)\neq 3[/imath] then [imath]f(3)=2[/imath]. Similarly, considering [imath]f(7)[/imath] and since [imath]f(7)\neq 2, f(7)\neq 3[/imath] then [imath]f(7)=1[/imath]. Also [imath]f(8)\neq f(3)=2, f(8)\neq f(5)=3[/imath] then [imath]f(8)=1[/imath]. Also [imath]f(6)\neq f(3)=2, f(6)\neq f(8)=1[/imath] and hence [imath]f(6)=3[/imath]. Finally, examining [imath]f(4)[/imath] we see that [imath]f(4)\neq f(2)=2[/imath] and [imath]f(4)\neq f(6)=3[/imath] and so [imath]f(4)=1[/imath]. But [imath]f(4)\neq f(7)=1[/imath] and here is a contradiction. Is it suitable proof? | 2051936 | No function [imath]f:\mathbb{Z}\rightarrow \{1, 2, 3\}[/imath] satisfying [imath]f(x)\ne f(y)[/imath] for all integers [imath]x,y[/imath] and [imath]|x-y|\in\{2, 3, 5\}[/imath]
Show there does not exist a function [imath]f:\mathbb{Z}\rightarrow \{1, 2, 3\}[/imath] satisfying [imath]f(x)\ne f(y)[/imath] for all integers [imath]x,y[/imath] and [imath]|x-y|\in\{2, 3, 5\}[/imath]. This is taken from "Putnam and Beyond" and I cannot seem to understand why this cannot be true. I let [imath]x[/imath] be fixed and assigned [imath]f(x)=1[/imath], then assumed [imath]f(y)=2, |x-y|=2.[/imath] Then I proceeded to substitute [imath]y-1[/imath] and [imath]y-3[/imath] but did not get anywhere in finding a contradiction. I am looking for comments about my approach and an explanation of the solution. |
1810286 | Any integral or series to prove [imath]\frac{1}{\sqrt{3}}>\gamma[/imath]?
I recently noticed that these two numbers are remarkably close: [imath]\frac{1}{\sqrt{3}}-\gamma=0.000135\dots[/imath] Are there any integrals or series which can prove that [imath]\frac{1}{\sqrt{3}}>\gamma[/imath]? Meaning that (as usual in such cases) the function under the integral has to be non-negative and the value should be proportional to the difference of these numbers. The same goes for the series (strictly non-negative terms). A good overview for the inequalities with [imath]\pi[/imath], like [imath]\frac{22}{7}>\pi[/imath], can be found in this question. A related question. But I don't consider my question a duplicate, because the linked question is more general. | 1069685 | Showing [imath]\gamma < \sqrt{1/3}[/imath] without a computer
In 1735 Euler gave the value of [imath]\gamma[/imath] as [imath]0.577218.[/imath] The constant is generally defined as the limit of the difference between the harmonic series and [imath]\log n:~\gamma= \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k}-\log n.[/imath] Euler apparently relished this sort of calculation and must have taken quite a few terms to get such a good approximation. My question is whether without a computer one can now prove that [imath]\gamma < \sqrt{1/3}[/imath] with at least some savings in terms of the type of work Euler apparently expended? I don't think there's any point in raising the bar to [imath]\gamma < \ln 2\sqrt{\frac{23}{29}} [/imath] because it seems to be the same sort of question. My own thought was to compare [imath]\frac{1}{2}\int_0^{1/3}\frac{dx}{\sqrt{x}}[/imath] to something like [imath](-1)\cdot\int_0^\infty e^{-u}\ln u~du[/imath] but I expect there's a better way. |
2180835 | If [imath]f[/imath] is continuous on [imath]\left[ a,b\right][/imath] then [imath]f[/imath] is uniformly continuous on [imath]\left[ a,b\right][/imath].
Let [imath]\left[ a,b\right]\rightarrow \mathbb{R}[/imath]. If [imath]f[/imath] is continuous on [imath]\left[ a,b\right][/imath] then [imath]f[/imath] is uniformly continuous on [imath]\left[ a,b\right][/imath]. Proof-trying. Assume [imath]f[/imath] on [imath]\left[ a,b\right][/imath]. We need to show that [imath]f[/imath] is uniformly continuos on [imath]\left[ a,b\right][/imath], i.e., for any [imath]\varepsilon >0[/imath] there exist [imath]\delta >0[/imath] such that [imath]\left| f\left( x\right) -f\left( c\right) \right|<\varepsilon[/imath] whenever [imath]x,c\in\left[ a,b\right][/imath]. How can I show this, can you help? | 2086124 | Continuous with compact support implies uniform continuity
This might be a duplicate but I tried googling the MSE site and could not find a satisfactory answer. Let [imath](X, d)[/imath] be a metric space and [imath]f[/imath] be a real valued continuous function on [imath]X[/imath]. Suppose [imath]f[/imath] has a compact support. Does this imply the uniform continuity of [imath]f[/imath]? I tried proving this statement, but only for locally connected spaces have I succeeded in doing so. Is thus true for general metric spaces? I just couldn't provide a proof (or a counterexample) by myself. Please enlighten me. |
2181848 | Prove that unit disk/~ is homeomorphic to [imath]S^2[/imath]
I find this example in Topology 2nd ed. by James Munkres P.139 Example 4 Let [imath]X = \{x \times y \mid x^2+y^2\leq1\}[/imath], [imath]X^*[/imath] be the partition of X consisting of all the one-point sets [imath]\{x \times y \}[/imath] for which [imath] x^2+y^2< 1 [/imath] along with the set [imath]S^1=\{\{x \times y\}\mid x^2+y^2=1\}[/imath] Show that [imath]X^*[/imath] is homeomorphic to [imath]S^2[/imath] where [imath]S^2=\{(x,y,z) \mid x^2+y^2+z^2=1\}[/imath] [imath]X^*[/imath] just likes gluing the boundary together to form a sphere.So I identify [imath]\{0 \times 0\}[/imath] to [imath](0,0,-1)[/imath], which means I match the centre to the south pole. Then I divide the disc into two parts: For [imath]x^2+y^2=r^2[/imath] where [imath]r<0.5[/imath], I map [imath](x,y)[/imath] maps into [imath](\alpha x,\alpha y, r)[/imath] where [imath]\alpha ^2 = \frac{1-r^2}{r^2}[/imath] . It means I map the inner ball into southern hemisphere. I do the similar thing for the remaining annulus. Finally, I map the [imath]S^1[/imath] into the north pole[imath]=(0,0,1)[/imath] Am I doing it right? Thank you! | 660880 | Quotient space of closed unit ball and the unit 2-sphere [imath]S^2[/imath]
This is an example from Munkres's Topology (Example 4 in Section 22 titled "The Quotient Topology", 2nd edition). Example 4: Let [imath]X[/imath] be the closed unit ball [imath]\{ x \times y \mid x^2 + y^2 \le 1\}[/imath] in [imath]\mathbb{R}^2[/imath], and let [imath]X^{\ast}[/imath] be the partition of [imath]X[/imath] consisting of all the one-point sets [imath]\{ x \times y \}[/imath] for which [imath]x^2 + y^2 < 1[/imath], along with the set [imath]S^1= \{ x \times y \mid x^2 + y^2 = 1 \}[/imath]. Typical saturated open sets in [imath]X[/imath] are pictured by the shaded regions in the figure below. One can show that [imath]X^{\ast}[/imath] is homeomorphic with the subspace of [imath]\mathbb{R}^3[/imath] called the unit 2-sphere, defined by [imath]S^2 = \{ x \times y \times z \mid x^2 + y^2 + z^2 =1 \}.[/imath] I am confused about two points in the example. Problem 1: About the saturated open sets in [imath]X[/imath]. In the example, the typical ones are pictured by the shaded regions ([imath]U, V[/imath]) in the figure. However, I am not sure whether the boundaries (visually, by picture) of [imath]U[/imath] and [imath]V[/imath] are included. Particularly, there are two boundaries for [imath]U[/imath]. Are they both contained in [imath]U[/imath]? And why? Problem 2: How to show that [imath]X^{\ast}[/imath] is homeomorphic with [imath]S^2[/imath]? What is the mapping between them? The following is my understanding: Solution to problem 1: For [imath]V[/imath], the boundary is not included. For [imath]U[/imath], the outer boundary is included, while the inner one not. Is this solution right? |
2181627 | True or false? The linear system of equations [imath]Ax=b[/imath] is uniquely solvable for every [imath]b \in \mathbb{R}^{n}[/imath] if and only if det(A) [imath]\neq[/imath] 0
Let [imath]n \in \mathbb{N}[/imath], let [imath]A \in \mathbb{R}^{n \times n}[/imath] True or false? The linear system of equations for [imath]Ax=b[/imath] is uniquely solvable for every [imath]b \in \mathbb{R}^{n}[/imath] if and only if [imath]\text{det(A)} \neq[/imath] 0 I think the statement is true. Making [imath]\text{det(A)} \neq 0[/imath] should avoid us getting non-unique results such as [imath]0=0[/imath]. Remebering this and knowing [imath]\text{det(A)} \neq 0[/imath] gives linearly independent vectors, we can only get unique solution. Edit: We don't know anything about [imath]x[/imath]. If [imath]x \neq 0[/imath], then the statement will be true, right? Please correct me if I'm wrong / unclear, these tasks will be asked in my exam and I want pass it :) | 1718441 | Are all of the following statements equivalent? (vectors and matrices)
I know there's a theorem in linear algebra that has a list of statements that are equivalent, as follows: [imath]\det(A) \ne 0[/imath] [imath]Ax = 0[/imath] has only the trivial solution [imath]Ax = b[/imath] is consistent for every matrix [imath]b[/imath] [imath]Ax = b[/imath] has exactly one solution for every [imath]n \times 1[/imath] matrix [imath]b[/imath] (and several other which I do not need for the question that I'd like to ask). So we've been learning about linear independence, span and bases, and I was wondering the following: Is statement 2 the same as saying a set of vectors are linearly independent? Is statement 3 the same as saying the set of vectors span a vector space? (Meaning that every single vector [imath]b[/imath] in the vector space can be made from some combination of [imath]Ax[/imath], where [imath]A[/imath] would represent that vectors and [imath]x[/imath] would be all of the scalar multiples of the vectors? Is statement 4 the same as saying the set of vectors is a basis? (since that one solution must be the trivial solution, it must be linearly independent, and since it is consistent for every [imath]b[/imath], it spans the vector space?) It seems pretty obvious that it's all related to each other, but I'm just wondering if they're all exactly the same. |
2175531 | The Fourier Coefficient Calculation.
I want to calculate the Fourier Cofficient for the function defined by: [imath]f(\theta) =\frac {(\pi - \theta)^{2}}{4}[/imath] for [imath]0 \leq\theta \leq 2\pi.[/imath] but the definition for the Fourier Coefficient is from [imath]-\pi[/imath] to [imath]\pi[/imath] for the functions that is integrable on this interval.So, in my function shall I change the boundaries in the definition of the Fourier Coeffient to be from 0 to [imath]2\pi.[/imath] But I do not think so because if I do this then I found that [imath]\hat{f}(0) = \frac{\pi^{2}}{4}[/imath] But it is answered in the book as:[imath]\hat{f}(0) = \frac{\pi^{2}}{12}[/imath] Could anyone give me an advice? thanks. | 2181222 | Computing the Fourier series of [imath]f(\theta)=(\pi-\theta)^2/4[/imath] for [imath]0\leq\theta\leq{2\pi}[/imath]
According to p. 36 of Stein and Shakarchi's Fourier Analysis, the Fourier series is [imath]\frac{\pi^2}{12} + \sum_{n=1}^{\infty} \frac {\cos{n\theta}}{n^2}[/imath]. But to calculate the Fourier coefficients, I computed [imath]\int_{-\pi}^{\pi}\frac{(\pi-\theta)^2}{4}e^{-in\theta}d\theta = \frac {1} {4} \Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{2}{in}\int_{-\pi}^{\pi}e^{-in\theta}(\pi-\theta)d\theta\Bigg]=\frac {1} {4}\Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{4\pi{e^{in\pi}}+2e^{-in\pi}-2e^{in\pi}}{n^2}\Bigg]=\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}[/imath] This means the Fourier series is [imath]\sum_{n=-\infty}^{\infty}\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}e^{in\theta}=\sum_{n=-\infty}^{\infty}\frac{\pi^2e^{in(\pi+\theta)}}{in}+\frac{e^{in(\theta-\pi)}}{2n^2}[/imath], but I can't seem to get this to equal that simple thing from the book. Can anyone help? |
2181631 | In what sense is the kernel of a group homomorphism a universal?
Question: In what sense is "the kernel of a group homomorphism a universal"? Do we here mean that a group homomorphism is a universal arrow in some comma category? If so, what comma category? Attempt: If we let [imath]K[/imath] be a normal subgroup of [imath]G[/imath] and [imath]i: K \rightarrow G[/imath] being the inclusion homomorphism, then for any other normal subgroup [imath]T[/imath] of [imath]K[/imath] with [imath]k : T \rightarrow K[/imath] being another inclusion homomorphism and likewise with [imath]j : T \rightarrow G[/imath], then we have that [imath]j = i \circ k[/imath]. Now let [imath]C[/imath] be the category of the subgroups of [imath]G[/imath] with inclusion homomorphisms as morphisms. Let [imath]S: C \rightarrow C[/imath] be the identity functor on [imath]C[/imath]. Then do we not here have that [imath](K, i)[/imath] is a universal arrow in [imath](C \downarrow G)[/imath]? And if so, is this the sense in which "the kernel of a group homomorphism is a universal"? EDIT: Answers in another thread on this site have confused me in the sense that they don't discuss explicitly about universal arrows and comma categories. | 1226916 | Identify the universal property of kernels
I'm reading Mac Lane's "Categories for the Working Mathematician". I found the following sentence in page 59 of it: Similarly, the kernel of a homomorphism (in [imath]\mathbf{Ab}[/imath], [imath]\mathbf{Grp}[/imath], [imath]\mathbf{Rng}[/imath], [imath]R[/imath]-[imath]\mathbf{Mod}[/imath], . . .) is a universal, more exactly, a universal for a suitable contravariant functor. I know how to express a quotient by a universal property in a category. But I really have trouble understanding how to express the kernel of a homomorphism as having a universal property. |
2182048 | Class number of [imath]\mathbb{Q}(\sqrt{6})[/imath]
I want to compute the class number of [imath]K=\mathbb{Q}(\sqrt{6})[/imath], the result should be [imath]1[/imath]. The discriminant [imath]d_K[/imath] is [imath]24[/imath], so the Minkowski bound is [imath]M_K = \sqrt{6} < 3[/imath]. So every ideal class contains an ideal in [imath]\mathcal{O}_K[/imath] with ideal norm less than or equal to [imath]M_K[/imath]. I know that [imath]2[/imath] is a norm of the principal ideal [imath](2 + \sqrt{6})[/imath], in fact [imath]2 = -N(2 + \sqrt{6})[/imath]. This is a principal ideal, so the class number is [imath]1[/imath]. Now I wonder if one can, after computing [imath]M_K[/imath], see that the class number is [imath]1[/imath] by knowing that [imath]2[/imath] is ramified (because [imath]6 \equiv 2 \bmod 4[/imath])? That means [imath]2 \mathcal{O}_K = P^2[/imath] for a prime ideal [imath]P[/imath]. | 1255698 | Computing class group of [imath]\mathbb Q(\sqrt{6})[/imath]
I am calculating the class group of [imath]\mathbb Q(\sqrt 6)[/imath]. My working is as follows: The Minkowski bound is [imath]\lambda(6)=\sqrt 6<3[/imath] so we only need to look at prime ideals of norm [imath]2[/imath]. [imath]2[/imath] divides the discriminant, [imath]24[/imath], of [imath]\mathbb Q(\sqrt 6)[/imath], so [imath]2[/imath] ramifies, and so for any prime ideal [imath]\mathfrak p[/imath] of norm [imath]2[/imath], [imath]\mathfrak p^2[/imath] is principal. But [imath](2, \sqrt 6)^2=(2)[/imath], so [imath]\mathfrak p=(2, \sqrt 6)[/imath] has norm [imath]2[/imath]. If I can show [imath]\mathfrak p[/imath] is not principal, then I can conclude that [imath]Cl(\mathbb Q(\sqrt 6))\cong C_2.[/imath] But if there was an element of [imath]\mathcal O_{\mathbb Q(\sqrt 6)}[/imath] of norm 2, this would correspond to a solution of the diophantine equation [imath]x^2-6y^2=2[/imath]. Clearly [imath]x[/imath] is even, so writing [imath]x=2n[/imath] we get [imath]2n^2-3y^2=1[/imath]. Now reducing modulo [imath]3[/imath], we get [imath]2n^2\equiv 1 \pmod 3[/imath] which has no soultions. Hence [imath]\mathfrak p[/imath] is not principal and [imath]Cl (\mathbb Q(\sqrt 6)) \cong C_2.[/imath] However, this link says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is [imath](2, \sqrt 6)[/imath] principal? |
2183141 | Prove that there exist integers [imath]a[/imath] and [imath]b[/imath], so [imath]am + bn = 1[/imath].
When [imath]m,n\in \mathbb{Z} [/imath] and [imath]a,b \in \mathbb{Z}[/imath], show that there exist integers [imath]a[/imath] and [imath]b[/imath], so [imath]am + bn = 1[/imath] which also means [imath]gcd(m,n) = 1[/imath]. I tried proof through contradiction but I didn't succeed. Any suggestion would be helpful. | 1377440 | If $\gcd(a,b)=1$, then there exists integers $x$ and $y$ such that [imath]xa + yb = 1[/imath]
Did not find this from this website... If [imath]$$ \gcd(a,b)=1,$$[/imath] then there exists integers [imath]$x$[/imath] and [imath]$y$[/imath] such that [imath]xa+yb=1.[/imath] Now, the tip is to use particular corollary, that states: The class [imath][m]_{n}[/imath] generates [imath]$\mathbb{Z}/n\mathbb{Z}\Leftrightarrow \gcd(m,n)=1.$[/imath] I am totally lost with the corollary. Let's assume that [imath]$\gcd(m,n)=1$[/imath]. Then, [imath][m]_{n}[/imath] generates [imath]\mathbb{Z}/n\mathbb{Z}[/imath]. OK! Then what? There is also follow up, where I have to prove the converse. I am familiar with Bezout's lemma. |
2183066 | Exercise in Atiyah's book: Prove that an ideal is prime.
Atiyah's Introduction to Commutative Algebra, chapter 1 exercise 14: In a commutative ring [imath]A[/imath] with identity, let ∑ be the set of all ideals in which every element is a zero-divisor. Show that the set ∑ has maximal elements and that every maximal element of ∑ is a prime ideal. Hence the set of zero-divisors in A is a union of prime ideals. The existence of maximal elements is guaranteed by Zorn's lemma. To prove that every maximal element of ∑ is prime: Given [imath]\mathfrak m[/imath] a maximal element of ∑ and [imath]xy \in \mathfrak m[/imath]. There exists a nonzero [imath]z[/imath] such that [imath]xyz=0[/imath]. If [imath]xz=0[/imath], then [imath](x)\in[/imath] ∑. If [imath]xz\neq0[/imath], then [imath](y)\in[/imath] ∑. So we suppose [imath](x)\in[/imath] ∑. HERE IS THE PROBLEM：Intuitively I want to prove [imath](x,\mathfrak m)\in[/imath] ∑ and by maximality of [imath]\mathfrak m[/imath], [imath]x\in\mathfrak m[/imath]. But how to prove that every element in [imath](x,\mathfrak m)[/imath] is a zero-divisor? For example, if given [imath]ax+bz[/imath] where [imath]a,b\in A, z\in \mathfrak m[/imath] and [imath]xs=zt=st=xz=0, xt\neq0,zs\neq0 [/imath], how to prove that [imath]ax+bz[/imath] is a zero-divisor? | 627655 | Is an ideal which is maximal with respect to the property that it consists of zero divisors necessarily prime?
This is in follow-up to this question. Let [imath]R[/imath] be a commutative ring with identity and consider the set [imath]Z \subset R[/imath] of zero divisors. If the ideal [imath]I\subset Z[/imath] is maximal with respect to the constraint, need it be prime? Since Zorn's Lemma applies equally well to proving that either minimal prime ideals exist or that maximal ideals contained in [imath]Z[/imath] exist, this would provide an alternative proof of that question if true. A naive approach to proving this is to assume [imath]ab \in I[/imath] with [imath]a, b \notin I[/imath]. Since [imath]ab \in I[/imath], it follows that [imath]ab[/imath] is a zero divisor, so [imath]abx = 0[/imath] for some [imath]x \neq 0[/imath], and therefore either [imath]a[/imath] is a zero divisor annihilated by [imath]bx \neq 0[/imath] or [imath]b[/imath] is a zero divisor annihilated by [imath]x[/imath]. In either case, the obvious thing to consider would be [imath]aR+I[/imath] or [imath]bR+I[/imath], but there's no obvious reason why either of these ideals should consist of zero divisors. According to rschwieb, the answer is yes for reduced Noetherian rings. |
2183466 | Must an irreducible polynomial over a field extension divide the irreducible polynomial over the subfield?
Let [imath]k \subset E[/imath] be a field extension, and let [imath]\alpha[/imath] be algebraic over [imath]k[/imath]. Let [imath]f[/imath] be the irreducible polynomial of [imath]\alpha[/imath] over [imath]k[/imath], and let [imath]g[/imath] be the irreducible polynomial of [imath]\alpha[/imath] over [imath]E[/imath]. Is it true that [imath]g[/imath] always divides [imath]f[/imath]? | 443424 | Minimal polynomial over an extension field divides the minimal polynomial over the base field
I need help proving this theorem: Given the field extension: [imath]\mathbf{K} \subseteq \mathbf{L}[/imath], for [imath]\alpha \in \mathbf{L}[/imath] and [imath]g(x) \in \mathbf{K}[x][/imath], [imath]\alpha[/imath]'s minimal polynomial over [imath]K[/imath], and [imath]f(x) \in \mathbf{L}[x][/imath], [imath]\alpha[/imath]'s minimal polynomial over [imath]L[/imath], then the degree of [imath]g[/imath] is bigger than the degree of [imath]f[/imath] and [imath]f(x)[/imath] divides [imath]g(x)[/imath]. |
2177682 | What is the value of the limit?
I need to solve the limit [imath]\lim_{x\to0}\frac{e^{-\frac{1}{x^2}}}{x}[/imath] If I replace [imath]x[/imath] with [imath]0[/imath] I get [imath]\frac00[/imath] so that is not what I should do. Do you have any idea how to do it? | 2168549 | Compute [imath]\lim_{x\to 0} \frac{e^{-1/x^2}}{x}[/imath] without using L'Hopital?
Purely out of curiosity, is it possible to compute the following limit without using L'Hopital? [imath]\lim_{x\to 0} \frac{e^{-1/x^2}}{x}[/imath] Using L'Hopital and realizing that we cannot get what we want if we simply take derivatives of [imath]e^{-1/x^2}[/imath] and [imath]x[/imath] directly, we have that [imath]\lim_{x\to 0}\frac{e^{-1/x^2}}{x}=\lim_{x\to 0}\frac{1}{x}\frac{1}{e^{1/x^2}} = \lim_{x\to 0} \frac{1/x}{e^{1/x^2}}=0[/imath] I am curious about whether there is a different way of computing this. |
1106116 | Estimating population variance
Why is population variance estimated to be [imath]\frac{1}{N-1}\Sigma_{1 \leq i\leq N}(x_i-m)^2[/imath] as opposed to sample variance which is [imath]\frac{1}{N}\Sigma_{1 \leq i\leq N}(x_i-m)^2[/imath], where m is the mean? I know that this is the unbiased estimate of population mean. But i am not able to grasp the intuition behind it. What does the term degree of freedom mean in this context? | 823111 | Why is there a difference between a population variance and a sample variance
Sorry if this answer is simple but I was wondering why is there a difference between a population variance and a sample variance? I understand The variance is calculated as: [imath]\text{Var} = \frac{1}{N}(x_i-\mu)^2[/imath] and the sample variance is computed as [imath]\text{Var}_s = \frac{1}{N-1}(x_i-\mu)^2[/imath] In real world data sets would you use the sample variance most of the time? What if the population is also a sample? How would you know? Also does the mean change when you are looking at a population or a sample? |
2184037 | Proving the product of any # of commutators can be written as an "extended commutator" (commutator of 3+ elements)
Suppose G is a group and as per usual, the commutator is [imath][g,h] := ghg^{-1}h^{-1}[/imath] for [imath]g,h \in G[/imath]. I need to show that the product of any number of commutators can be written in the following defined form: [imath]g_1 \cdot g_2 \cdot \dots \cdot g_n \cdot g_1^{-1} \cdot g_2^{-1} \cdot \dots \cdot g_n^{-1}[/imath] where all [imath]g_i \in G[/imath] for [imath]1 \geq i \geq n[/imath] Every identity I know with commutators uses 2-3 elements (and their inverses), so I'm having trouble proving this since the starting point I'm at involves 4 elements that are not necessarily related. I figured the best thing to do would be to use induction but for the base case I have, for example, [imath][g_1,g_2][g_3,g_4]=(g_1g_2g_1^{-1}g_2^{-1})(g_3g_4g_3^{-1}g_4^{-1})[/imath] and I can't seem to manipulate that in anyway. | 725144 | About product of [imath]k[/imath] commutators
I need help in folloving lemma. Lemma: Any product of [imath]k[/imath] commutators is expressible in the form [imath]a_1^{-1}a_2^{-1}...a_{2k}^{-1}a_1a_2...a_{2k}[/imath] I guess author means that any product [imath][x_1,y_1][x_2,y_2]...[x_k,y_k][/imath] can be expressible in that form [imath]a_1^{-1}a_2^{-1}...a_{2k}^{-1}a_1a_2...a_{2k}[/imath] where [imath][x_i,y_i]=x_i^{-1}y_i^{-1}x_iy_i[/imath]. I tried to prove it by induction on [imath]k[/imath] but I failed. If anyone can prove it,I would be thankful. |
2182414 | If [imath]\sum_{n=1}^\infty{a_n}[/imath] and [imath]\sum_{n=1}^\infty{b_n}[/imath] converge then If [imath]\sum_{n=1}^\infty{a_n*b_n}[/imath] converge
My attempt; Since [imath]a_n[/imath] converges then its sequence of partial sums [imath](x_n)[/imath] converges to a limit x. Also, since [imath]b_n[/imath] converges then its sequence of partial sums [imath](y_n)[/imath] converges to a limit y. Observe that by using the multiplication limit theorem: If [imath]x_n \to x[/imath] and [imath]y_n \to y[/imath] then [imath]x_n*y_n \to x*y[/imath]. We know that the product of its partials sums converge, thus the product of the series converges.$ Is this correct? Any guidance is appreciated ! | 163491 | Example of two convergent series whose product is not convergent.
Could someone give me an example of two convergent series [imath]\sum_{n=0}^\infty a_n[/imath] and [imath]\sum_{n=0}^\infty b_n[/imath] such that [imath]\sum_{n=0}^\infty a_nb_n[/imath] diverges? |
2183777 | Calculating limit using non-obvious way
The limit in question is [imath]\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}[/imath] I set [imath]z=0[/imath] and converted this to polar coordinates: [imath]\lim_{r\to0}\frac{r^2\cos \theta \sin \theta}{r^2\cos^2 \theta +r^2\sin^2 \theta}[/imath] This can be reduced to [imath]\lim_{r\to0}\cos \theta \sin \theta[/imath] if I did everything properly. Does this mean that the limit on the origin doesn't exist? Plugging different values of [imath]\theta[/imath] in the limit returns different values, but I'm not entirely sure if this is correct. I know there are easier ways to do this, and several of them are already on MSE, but I have a feeling a trick like this (if it is correct) might be useful for some other more complicated limit. | 418063 | Evaluate the limit: [imath]\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}[/imath]
Could someone give me a hint? I would like to continue to attempt it. The limit to evaluate that I would like a hint on is: [imath]\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}[/imath] |
2184244 | Example of pairwise but not mutually independent
Give an example of a probability space (Ω,Pr) and pairwise independent events A, B, and C which are not mutually independent. This is my understanding of what pairwise independent events are: Events [imath]A_1, A_2,..., A_k[/imath] are pairwise independent if for all i,j, [imath]A_i [/imath] and [imath]A_j[/imath] are independent: [imath]Pr(A_i\cap A_j) = Pr(A_i)Pr(A_j)[/imath] Events [imath]A_1, A_2,..., A_k[/imath] are mutually independent if for all [imath]I\subset 1,2,...,k, Pr(\bigcap ._{i\subset I} A_i) = \prod\limits_{i\subset I} Pr(A_i) [/imath] so [imath] Pr(A_1)\cap Pr(A_2) ... Pr(A_k) = Pr(A_1)Pr(A_2)...Pr(A_k)[/imath] | 1143256 | What is an example of pairwise independent random variables which are not independent?
I've just read in a stochastics textbook: Let [imath](\Omega, P)[/imath] be a discrete probability space. (a) The events [imath]A_i \subseteq \Omega, i=1,2, \dots[/imath] are called independent, if [imath]P(A_{i_1} \cap A_{i_2} \cap \dots \cap A_{i_k}) = P(A_{i_1}) \cdot \dots \cdot A_{i_k}[/imath] for all finite sets [imath]\{i_1, \dots, i_k\} \subsetneq \{1,2, \dots\}[/imath] and all [imath]k \geq 2[/imath]. (b) The events [imath]A_i \subseteq \Omega, i=1,2,\dots[/imath] are called pairwise independent, if [imath]P(A_{i_1} \cap A_{i_2}) = P(A_{i_1}) \cdot P(A_{i_2})[/imath] for all pairs [imath]\{i_1, i_2\} \subsetneq \{1,2,\dots\}[/imath]. After these two definitions it states that pairwise independent events are not always independent. Do you have an example? |
1333664 | If [imath]f'[/imath] is increasing and [imath]f(0)=0[/imath], then [imath]f(x)/x[/imath] is increasing
Let [imath]a>0[/imath] and [imath]f:[0,a] \to \mathbb{R}[/imath] continuous function that is twice differentiable on [imath](0,a).[/imath] Also [imath]f(0)=0[/imath] and [imath]f'[/imath] is strictly increasing function on [imath](0,a).[/imath] I have to show that the function [imath]g[/imath] defined as [imath]g(x) = \frac{f(x)}{x}[/imath] is strictly increasing on [imath](0,a].[/imath] Progress: I computed [imath]g'[/imath] and got [imath]g'(x) = \frac{f'(x)x-f(x)}{x^2}.[/imath] I have to show that [imath]g'(x)>0 \ \ \forall x \in (0,a].[/imath] Since [imath]x^2[/imath] is always positive for [imath]x \in (0,a][/imath] I have to show that [imath]f'(x)x-f(x) >0 \ \ \forall x \in (0,a].[/imath] I know both terms are positive, but I don't know how to show that [imath]f'(x)x>f(x) \ \ \forall x \in (0,a].[/imath] | 1715047 | Show that if [imath]f'[/imath] is strictly increasing, then [imath]\frac{f(x)}{x}[/imath] is increasing over [imath](0,\infty)[/imath]
Suppose [imath]f[/imath] is a differentiable function over [imath]\mathbb{R}[/imath] satisfying [imath]f(0) = 0[/imath]. Show that if [imath]f'[/imath] is strictly increasing, then [imath]\frac{f(x)}{x}[/imath] is increasing over [imath](0,\infty)[/imath]. Attempt: Since [imath]f'[/imath] is increasing we know that [imath]f'' > 0[/imath] for all [imath]x[/imath]. Now let [imath]g(x) = \frac{f(x)}{x}[/imath] and thus [imath]f(x) = xg(x)[/imath]. Then we wish to show that [imath]g'(x) > 0[/imath] on [imath](0,\infty)[/imath] and so since [imath]f'(x) = xg'(x)+g(x)[/imath] and [imath]f''(x) = xg''(x)+2g'(x) > 0[/imath], we have [imath]g''(x) > \frac{-2g'(x)}{x}[/imath]. I am not sure how to use this though to prove the statement. |
2183668 | Complex exponent with real part = 1/2 should convert x into -x
Complex [imath]z[/imath] is given as [imath]\frac{1}{2} + yi[/imath] What is [imath]y[/imath] if [imath] 2^{\frac{1}{2} + yi} = -2 [/imath] What is [imath]y[/imath] if [imath] 3^{\frac{1}{2} + yi} = -3 [/imath] How to calculate it for other values? | 2183565 | Which complex exponent can turn x into a -x?
Actually that is, I would like to use a complex exponent to turn a number x into an -x just by applying some complex exponent to x. I think I would have to use some ln, π and so one, but I am actually stack and cannot find the correct path to solving it. For example, let's say for 2 we choose the real part to be equal 1, what would be the solution for y in this special case. [imath] 2^{1 + iy}=-2 [/imath] What would be the solution for the imaginary part of z if the real part must be 0.5? [imath] x^{0.5 + iy} = -x [/imath] [imath] x^{0.5} * x^{iy} = \sqrt{x} * -\sqrt{x} = -x [/imath] What is y in this case? Is there maybe a general solution for [imath] x^{z}=-x [/imath] |
2184019 | how to prove that every cauchy sequence is a convergent sequence in [imath]\mathbb{R}[/imath]
I just want to know how can I prove that: Let [imath]\{x_n\}\subset\mathbb{R}[/imath] a real numbers sequence, if [imath]\{x_n\}[/imath] is a cauchy sequence, then [imath]x_n\longrightarrow x\in\mathbb{R}[/imath] | 437904 | Confused with proof that all Cauchy sequences of real numbers converge.
First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, [imath]\{a_{n_{k}}\}[/imath] that converges to a limit, say [imath]L[/imath]. Now we use this to prove that all Cauchy sequences are convergent. So an [imath]N_1[/imath] exists such that [imath]\left|a_{n_{k}}-L\right|<\frac{\epsilon}{2}[/imath] for all [imath] k > N_1[/imath], and an [imath]N_2[/imath] exists such that [imath]\left|a_m-a_n\right|<\frac{\epsilon}{2}[/imath] for all [imath]n,m > N_2 [/imath]. Pick any [imath]k > N_1[/imath] such that [imath]n_k > N_2[/imath]. Then for every [imath]n > N_2[/imath], [imath]\left| a_n - L \right | \leq \left| a_n - a_{n_{k}} \right| + \left| a_{n_{k}} - L \right| < \epsilon/2 + \epsilon/2 = \epsilon[/imath]. So [imath]\lim_{n \rightarrow \infty}a_n = L[/imath] I'm fine with this proof until the last part - I'm confused as to why we can pick an abitrary [imath]k[/imath] like we do? Should the limit not depend only on [imath]n[/imath]? Now it appears like it also depends on [imath]k[/imath] and if we pick a [imath]k < N_1[/imath], the inequality isn't true. Can someone clarify this for me please? |
2184747 | Showing Aut of G is a group
I held back for a while before deciding to pose this question. I've done this before but couldn't quite 're-reasoned'. Question: If G is a group, prove that Aut(G) is a group. Let [imath]\alpha, \beta \in Aut\left ( G \right )[/imath]. By definition: the group operation is preserved. [imath]\alpha \left ( x_{1}x_{2} \right )=\alpha \left ( x_{1} \right )\alpha \left ( x_{2} \right )[/imath] [imath]\beta \left ( x_{1}x_{2} \right )=\beta \left ( x_{1} \right )\beta \left ( x_{2} \right )[/imath] [imath]\alpha, \beta[/imath] is a bijection from G to G. By the two-step subgroup test, it suffices to show that [imath]\alpha \beta \in Aut\left ( G \right )[/imath] whenever, [imath]\alpha, \beta \in Aut\left ( G \right )[/imath] and [imath]\alpha^{-1} \in Aut\left ( G \right )[/imath] whenever [imath]\alpha[/imath] is. A bit of help to get me going? | 267666 | Prove that the set [imath]\mathrm{Aut}(G)[/imath] of all automorphisms of the group [imath]G[/imath] with the operation of taking the composition is a group
Let [imath]G[/imath] be a group. Say what it means for a map [imath]\varphi: G \rightarrow G[/imath] to be an automorphism. Show that the set-theoretic composition [imath]\varphi \psi = \varphi \circ \psi[/imath] of any two automorphisms [imath]\varphi, \psi[/imath] is an automorphism. Prove that the set [imath]\mathrm{Aut}(G)[/imath] of all automorphisms of the group [imath]G[/imath] with the operation of taking the composition is a group. I have said: A map is an automorphism of a group [imath]G[/imath] if it is an isomorphism to itself. a) For the next bit, I want to show if [imath]\varphi, \psi[/imath] is bijective, then [imath]\varphi \circ \psi[/imath] is bijective: For two elements [imath]a, b \in G[/imath] we have [imath]\varphi \circ \psi (ab) = \varphi(\psi(ab)) = \varphi(\psi(a)\psi(b))[/imath] as [imath]\psi[/imath] is an isomorphism. Also, as [imath]\varphi[/imath] is an isomorphism, we have [imath]\varphi(\psi(a) \psi(b)) = \varphi \circ \psi(a) \varphi \circ \psi(b)[/imath] Showing [imath]\varphi \circ \psi[/imath] is an isomorphism iff [imath]\varphi, \psi[/imath] are isomorphisms. For the group bit, we want to prove the 3 group axioms. 1) Associativity: [imath]\varphi \circ (\psi \circ \zeta) = (\varphi \circ \psi) \circ \zeta[/imath]. So for some [imath]x \in G[/imath], we get: [imath]\varphi \circ (\psi \circ \zeta)(x) = (\varphi \circ \psi) \zeta(x) = \varphi(\psi(\zeta(x))) [/imath] 2) Identity: If we let the identity automorphism, [imath]e: G \rightarrow G[/imath], be the map [imath]e(x) = x[/imath], then clearly we get that [imath]e \circ \psi = \psi \circ e = \psi[/imath]. 3) Inverse: As the automorphisms are bijective (already proved) then we know that by definition of a bijection, there is well defined inverse such that [imath]\psi^{-1}: G \rightarrow G[/imath] exists. (Second edit to correct proof for inverse): For any two elements [imath]a,b \in G[/imath], we want to see if [imath]\psi^{-1}(ab) = \psi^{-1}(a)\psi^{-1}(b)[/imath]. Apply [imath]\psi[/imath] to both sides gives us [imath]\psi \circ \psi^{-1}(ab) = \psi(\psi^{-1}(ab)) = ab[/imath] Doing the same on RHS gives us [imath]ab[/imath] and so we have proved the inverse exists and is unique. Is this right and enough to prove this? EDIT: Actually, can I just say that by definition of two bijective maps, the composition is also bijective and this is enough? |
2185263 | [imath](AB)^t = B^tA^t[/imath] for all [imath]A ∈ Mm×k(K)[/imath], [imath]B ∈ M_{k×n}(K)[/imath]
Recall that the transpose of a matrix [imath]A[/imath] is denoted by [imath]A^t[/imath]. Let [imath]K[/imath] be a field and denote the [imath]m[/imath]x[imath]n[/imath]-matrices over [imath]K[/imath] by [imath]M_{m×n}(K)[/imath]. Prove that [imath](AB)^t = B^tA^t[/imath] for all [imath]A ∈ Mm×k(K)[/imath], [imath]B ∈ M_{k×n}(K)[/imath]. You should use the formula for the [imath](i,j)[/imath]-entry of the product of two matrices, and you should use that the [imath](k, l)[/imath]-entry of the transpose of a matrix [imath]C[/imath] is the [imath](l, k)[/imath]-entry of [imath]C[/imath]. I know I would begin by stating [imath]AB[/imath] is an [imath]m×n[/imath]- matrix then [imath](AB)^t[/imath] must be too, but don't know how to continue. What is the next step? | 1440305 | How to prove [imath](AB)^T=B^T A^T[/imath]
Given an [imath]$m\times n$[/imath]-matrix [imath]A[/imath] and an [imath]$n\times p$[/imath]-matrix [imath]B[/imath]. Prove that [imath]$(AB)^T = B^TA^T$[/imath]. Here is my attempt: Write the matrices [imath]$A$[/imath] and [imath]$B$[/imath] as [imath]$A = [a_{ij}]$[/imath] and [imath]$B = [b_{ij}]$[/imath], meaning that their [imath]$\left(i,j\right)$[/imath]-th entries are [imath]$a_{ij}$[/imath] and [imath]$b_{ij}$[/imath], respectively. Let [imath]C=AB=[c_{ij}][/imath], where [imath]c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}[/imath], the standard multiplication definition. We want [imath](AB)^T = C^T = [c_{ji}][/imath]. That is the element in position [imath]j,i[/imath] is [imath]\sum_{k=1}^n a_{ik}b_{kj}[/imath]. For instance, if [imath]i=2, j=3[/imath], then the element in [imath]2,3[/imath] of [imath]C[/imath] is that sum, but the element in position [imath]3,2[/imath] of the transpose is that sum. I need to get the same value for the element in position [imath]3,2[/imath] of the right side. The transpose matrices are [imath]B^T=[b_{ji}], A^T=[a_{ji}][/imath]. They are size [imath]p \times n[/imath] and [imath]n \times m[/imath]. That is, they switch rows and columns. Let [imath]D = B^T A^T = [d_{ji}][/imath]. I write the indices backwards because if I want the element in position [imath]3,2[/imath], that is, [imath]i=2, j=3[/imath] just like on the other side. So I need the summation for [imath]d_{ji}[/imath]. But I get as [imath]d_{ji} = \sum_{k=1}^n b_{jk}a_{ki}[/imath], which does not match. |
532960 | showing that [imath]n[/imath]th cyclotomic polynomial [imath]\Phi_n(x)[/imath] is irreducible over [imath]\mathbb{Q}[/imath]
I studied the cyclotomic extension using Fraleigh's text. To prove that Galois group of the [imath]n[/imath]th cyclotomic extension has order [imath]\phi(n)[/imath]( [imath]\phi[/imath] is the Euler's phi function.), the writer assumed, without proof, that [imath]n[/imath]th cyclotomic polynomial [imath]\Phi_n(x)[/imath] is irreducible over [imath]\mathbb{Q}[/imath]. I know that for n=p, p is prime, [imath]\Phi_n(x)[/imath] is irreducible over [imath]\mathbb{Q}[/imath] by Eisenstein's criterion. But I don't know how [imath]\Phi_n(x)[/imath] is irreducible over [imath]\mathbb{Q}[/imath] when n is not prime. | 2884673 | If [imath]\zeta_n[/imath] is a primitive [imath]n[/imath]th root of unity, why is [imath]\text{dim}_{\Bbb Q}\Bbb Q[\zeta_n]=\phi(n)[/imath]?
I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated. |
2184802 | Maximum Value of a Poisson Distribution
Let [imath]X[/imath] be a Poisson random variable with parameter [imath]\lambda[/imath]. What is the most likely outcome of [imath]X[/imath]? Please help me solve this exercise. Thanks in advance! :) | 246496 | The mode of the Poisson Distribution
Lately, I am doing an investigation on Stirling's formula and its applications. So I thought I could use it to prove that the mode of the Poisson model is approximately equal to the mean. Of course, you do that by considering the curve that is formed by connecting the points of the probabilities of occurrence and the different values of the discrete random variable. Then you differentiate the p.d.f. where for [imath]x![/imath] you use Stirling's formula [imath]x!\approx \sqrt{2\pi x}~x^xe^{-x}[/imath]. The result is [imath]\lnλ-1/(2x)-\ln x[/imath] whose roots cannot be found analytically, but by iterative methods we find that as λ is larger and larger, the mode~mean. Problem is, I found the following paper online, which seems to be the solution from a Harvard's undergraduate problem set. http://www.physics.harvard.edu/academics/undergrad/probweek/sol84.pdf It reads "You can also show this by taking the derivative of eq. (2), with Stirling’s expression in place of the [imath]x![/imath]. Furthermore, you can show that [imath]x = a[=λ ~\text{in my case}~]-1/2[/imath] leads to a maximum [imath]P(x)[/imath] value of [imath]P_\max\approx1/\sqrt{2\pi a}[/imath]." Does this puzzle you as much as it puzzles me? My main concern is over the "=" sign: how does this hold? The derivative=0 equation cannot have such an exact solution. Furthermore at such x, how does [imath]P(X=a-1/2)[/imath] give [imath]1/\sqrt{2\pi a}[/imath]? Am I (and my professor) missing something rather obvious or is the solution wrong? Discuss! PS: This sort of question might have been asked before, but still, I am really curious that somebody reads the paper in the link above, so that I can figure out what's going on. |
2184873 | True or false? [imath](\mathbb{Z} \setminus \left\{0\right\}, \cdot)[/imath] is a group
True or false? [imath](\mathbb{Z} \setminus \left\{0\right\}, \cdot)[/imath] is a group I need to know this because it's from an old exam and it might be asked in my exam, or a task like that. I'm not entirely sure but I would say that the statement is false. The associative property will work, there will be an identity element too. But we will fail when we are looking for inverse elements as these would require rational numbers. (So even if we had something like [imath](\mathbb{Z}, \cdot)[/imath], it wouldn't be a group still.) I hope I did it alright? | 1374808 | Prove or disprove: [imath](\mathbb{Z}^*, \cdot)[/imath] and/or [imath](\mathbb{Z}^*, \div)[/imath] is a group.
I am teaching myself information about groups, but don't really understand how to work through this problem. Here is what I have been thinking so far (please note that I do not need to work through showing any closure): [imath]\mathbb{Z}^*[/imath] denotes the nonzero integers. Prove or disprove: [imath](\mathbb{Z}^*, \cdot)[/imath] and/or [imath](\mathbb{Z}^*, \div)[/imath] is a group. For [imath](\mathbb{Z}^*, \cdot)[/imath]: Group. Associative Law: Let [imath]a,b,c \in \mathbb{Z}^*[/imath]. Then [imath](a\cdot b)\cdot c=a\cdot (b \cdot c)[/imath], and the associative law holds. Existence of Identity: Let [imath]a \in \mathbb{Z}^*[/imath], then [imath]a \cdot 1 = 1 \cdot a = a[/imath], and the identity holds. Existence of Inverses: Let [imath]a \in \mathbb{Z}^*[/imath]. Then the inverse of [imath]a^{-1}=-a[/imath], and the inverse holds. Note that this only works because we are ommitting zero. Hopefully this looks like I am on the right track. I don't know to what extent I can explain logic vs showing it with mathematical operations, so I would appreciate guidance in this respect. For [imath](\mathbb{Z}^*, \div)[/imath]: Not a group. Disprove by counterexample. Yeah, I dunno about this one. I don't think it's true, because we will be introduction rational expressions that won't allow this to hold, but I'm not sure where I can use this information to disprove the statement. |
2176339 | If [imath]E(XY)=E(X) \cdot E(Y)[/imath], why is not [imath]E(X^2)=[E(X)^2][/imath]
My book states that the expected value of the product of two independent random variables equal the product of their expected values. Why cannot one use this to give the following? [imath]E(X \cdot X) = E(X) \cdot E(X) = [E(X)]^2[/imath] Of course, this is incorrect, as if this were true the variance would always be 0. | 149723 | Why is the expected value [imath]E(X^2) \neq E(X)^2[/imath]?
I wish to use the Computational formula of the variance to calculate the variance of a normal-distributed function. For this, I need the expected value of [imath]X[/imath] as well as the one of [imath]X^2[/imath]. Intuitively, I would have assumed that [imath]E(X^2)[/imath] is always equal to [imath]E(X)^2[/imath]. In fact, I cannot imagine how they could be different. Could you explain how this is possible, e.g. with an example? |
2185701 | Use the definition to prove that [imath]\lim_{n \rightarrow \infty} b^{\frac1n} = 1[/imath], where [imath]b>0[/imath].
Let [imath]b > 0[/imath]. Use the definition to prove [imath]\lim_{n \rightarrow \infty} b^{\frac1n} = 1[/imath]. please help me with that. | 1867269 | Use [imath]\delta-\epsilon[/imath] to show that [imath]\lim_{n\to\infty} a^{\frac{1}{n}} = 1[/imath]?
Hope this is a meaningful question, but I'm curious if is possible to show that [imath]\lim_{n\to\infty} a^{\frac{1}{n}}=1, \text{where }a>0[/imath] using [imath]\delta-\epsilon[/imath] directly or other methods. One method that I am aware of is to use the following: If [imath]\{s_n\}[/imath] is a nonzero sequence, then [imath]\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|[/imath] |
2185911 | Proof that a derived set is closed in a [imath]T_1[/imath]-space
Let [imath]X[/imath] be a [imath]T_1[/imath]-space and [imath]A\subset X[/imath]. Claim: [imath]A'[/imath] is closed, where [imath]A'=\{x\in X|(U\setminus\{x\})\cap A\ne\emptyset[/imath] for all open neighbourhoods of [imath]x\}[/imath]. Proof: Let [imath]x\in (A')^c[/imath] be arbitrary. Then [imath]x\notin A'[/imath] so there is an open set [imath]U[/imath] containing [imath]x[/imath] such that [imath]U\setminus\{x\}\cap A=\emptyset.[/imath] Since [imath]X[/imath] is [imath]T_1,[/imath] it follows that [imath]\{x\}[/imath] is closed and so [imath]U\setminus\{x\}[/imath] is open. Thus, using the facts that [imath]U\setminus\{x\}\cap A=\emptyset[/imath] and [imath]U\setminus\{x\}[/imath] is open, we have [imath]U\setminus\{x\}\subset (A')^c.[/imath] And since [imath]x\in (A')^c[/imath] it follows that [imath]U\subset (A')^c.[/imath] So we have shown that for any [imath]x\in (A')^c,[/imath] there is an open set [imath]U[/imath] containing [imath]x[/imath] such that [imath]U\subset (A')^c,[/imath] and so [imath](A')^c[/imath] is open. Does this prove the claim? EDIT: I know the claim is true. I just wanted to verify whether my reasoning was correct. | 1483574 | The set [imath]A'[/imath] is closed in a [imath]T_1[/imath] space
Let [imath]E[/imath] be a [imath]T_1[/imath] space, and [imath]A\subset E[/imath] , I need to prove that [imath]A'[/imath] is closed, so my idea is to prove that [imath]C_{E}A'[/imath] is open i.e a neighborhood of all it's point. But how to do this ? |
2186518 | Integration by parts sine
Let [imath]f(t)[/imath] be a continuous function. so-[imath]\frac{\pi}{2} \int_{0}^{\pi}f(\sin x)dx=\int_{0}^{\pi}xf(\sin x)dx[/imath] I tried many times Integration in parts but didn't succeed. | 971888 | Show that [imath]\int_{0}^{\pi} xf(\sin(x))\text{d}x = \frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))\text{d}x[/imath]
Show that [imath]\int_{0}^{\pi} xf(\sin(x))\text{d}x = \frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))\text{d}x[/imath] this is so confusing. i have no idea how to even start. im thinking integration by parts but that doesnt seem right. hope you guys can help!! |
2186658 | Basic fact used in proofs in Eisenbud's commutative algebra book.
Page 31. Corollary 1.5. Let [imath]k[/imath] be a field, and let [imath]S = k[x_1, \dots, x_r][/imath] be a polynomial graded by degree. Let [imath]R[/imath] be a [imath]k[/imath]-subalgebra of [imath]S[/imath]. If [imath]R[/imath] is a summand of [imath]S[/imath], in the sense that there is a map of [imath]R[/imath]-modules [imath]\varphi: S \to R[/imath] that preserves degrees and takes each element of [imath]R[/imath] to itself, then [imath]R[/imath] is a finitely generated [imath]k[/imath]-algebra. Proof. Let [imath]\hat{m} \subset R[/imath] be the ideal generated by the homogeneous elements of [imath]R[/imath] of strictly positive degree. Since [imath]S[/imath] is Noetherian, the ideal [imath]\hat{m}S[/imath] has a finite set of generators, which may be chosen to be homogeneous elements [imath]f_1, \dots, f_s[/imath] of [imath]\hat{m}[/imath]. We shall show that these elements generate [imath]R[/imath] as a [imath]k[/imath]-algebra. Why is the bolded part true? | 1616002 | Can we always choose the generators of an ideal of a Noetherian ring to be homogeneous?
Let [imath]R[/imath] be a [imath]k[/imath]-subalgebra of [imath]S=k[x_1,x_2,\dots,x_n][/imath]. Let [imath]m\subseteq R[/imath] be the ideal generated by the homogeneous elements of [imath]R[/imath] of positive degree. As [imath]S[/imath] is Noetherian, the ideal [imath]mS[/imath] has a finite set of generators, which may be chosen to be homogeneous elements of [imath]m[/imath]. Why can we choose homogeneous elements of [imath]m[/imath] to be the generators? |
2187109 | Finding taylor series in Real Analysis
In the following question I am trying to find the Taylor series for, [imath]f(x)=x^3\sin(x^2)[/imath] centered at [imath]x=0[/imath]. My first thought was to find the Taylor series for [imath]\sin x[/imath] then derive it to get my series I wanted but I am not quite sure how to do that. I worked out the Taylor series for [imath]f(x)=\sin(x)[/imath] centered at [imath]x=0[/imath] and got, [imath]f(x)=\sin x, f(0)=0[/imath] [imath]f'(x)=\cos x, f'(0)=1[/imath] [imath]f"(x)=-\sin x, f''(0)=0[/imath] [imath]f'''(x)=-\cos x, f'''(0)=-1[/imath] [imath]f''''(x)=\sin x, f''''(0)=0[/imath] So the Taylor series for [imath]\sin x[/imath] centered at [imath]x=0[/imath] is, [imath]\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...[/imath] But how do I use this to find the Taylor series for [imath]f(x)=x^3\sin(x^2)[/imath] | 2173553 | Maclaurin series question
What is the Maclaurin series of [imath]z^3\sin(z^2)[/imath] I'm able to differentiate it but can't figure out how to write it in general series form, can somebody help me please. |
2186026 | Provide infinite sequence of coprime numbers
Provide an infinite sequence of natural numbers [imath]x_1,x_2,x_3,\ldots[/imath] such that I'm not sure if I'm on the right track with that. any help would be great | 169439 | Approximating next prime number
Suppose that there is a prime number. Now I want to approximate the next prime number. (It does not have to be exact.) What would be the time-efficient way to do this? Edit: what happens if we limit the case to the prime number of the form [imath]4k+1[/imath] where k is a natural number? Edit: it's fine to replace approximate prime number with finding any prime number that is bigger than the given prime number in a time-efficient way. |
2187137 | Multivariable Optimization Rectangular Box Volume
What are the dimensions for the cheapest possible rectangular box with a volume of [imath]490 cm^3[/imath] if the material for the bottom costs [imath]\[/imath]6/cm^2[imath], material for the sides costs [/imath]\[imath]7/cm^2[/imath], and material for the top costs [imath]\[/imath]14/cm^2[imath] ?[/imath] I can't seem to get anywhere close to the answer in x,y,z$ form. | 858378 | Rectangular Box Optimization Problem
A rectangular box is to have a square base and a volume of 40 ft3. If the material for the base costs \[imath]0.31 per square foot, the material for the sides costs [/imath]0.05 per square foot, and the material for the top costs \$0.19 per square foot, determine the dimensions of the box that can be constructed at minimum cost. Ive been trying to solve this for a while but I can't figure out what the cost function should be. or even how i should construct the volume equation. |
2179975 | Given [imath]f(\frac{x_1 + x_2}{2})[/imath] [imath]\le\frac{(f(x_1) + f(x_2))}{2}[/imath]
Let [imath]f[/imath] be bounded in [imath][a,b][/imath]. Assume [imath]f[/imath] satisfies [imath]f\left(\frac{x_1 + x_2}{2}\right) \le\frac{f(x_1) + f(x_2)}{2}[/imath] for all [imath]a\le x_1 \le x_2 \le b [/imath] . Prove that [imath]f[/imath] is continous on [imath]x[/imath] when [imath]a< x < b[/imath]. I'm more concerned with how I can derive the prove of this question | 137909 | Proving continuity of [imath]f[/imath]
Here is a problem I encountered some time back: Suppose [imath]f[/imath] is bounded for [imath]a\leq x\leq b[/imath] and for every pair of values [imath]x_1[/imath] and [imath]x_2[/imath] with [imath]a\leq x_1\leq x_2 \leq b[/imath], [imath]f(\frac{1}{2}(x_1+x_2))\leq \frac{1}{2}(f(x_1)+f(x_2))[/imath]. Prove that [imath]f[/imath] is continuous for [imath]a<x<b[/imath]. I tried to solve it but I could not really come with anything... Here's an attempt.The idea is not due to me but to a friend; By the condition given, [imath]f(\frac{2x+2\delta}{2})\leq \frac{1}{2}(f(x+2\delta)+f(x))[/imath],i.e. [imath]f(x+\delta)-f(x)\leq \frac{1}{2}f(x+2\delta)-f(x)[/imath] and in this manner, [imath]f(x+\delta)-f(x)\leq \frac{1}{2}f(x+2\delta)-f(x)\leq \frac{1}{2^2}(f(x+4\delta)-f(x))\leq[/imath] [imath] \dots \dots \dots \dots [/imath] [imath]\frac{1}{2^n}(f(x+2^n\delta)-f(x))[/imath] where [imath]a<x+2^n\delta<b[/imath] As [imath]\delta\to 0[/imath],[imath]f(x+2^k\delta)\to f(x)[/imath] for [imath]k=1,2,\dots n[/imath] i.e. [imath]f[/imath] is continuous in the interval [imath](a,b)[/imath]. I tried posting this flawed attempt on Aops , but no one has suggested how to finish off the proof using what I used.I will be happy if someone could suggest something.Thanks! |
2187466 | Sharpening an established inequality?
Nesbitt's inequality states that [imath]\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \;\ge\;\frac{3}{2}\tag{Nes}[/imath] holds for all [imath]\,x,y,z\in\mathbb{R}^{>0}[/imath]. Applying Cauchy-Schwarz to it yields [imath]\left(x^2+y^2+z^2\right) \left[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right] \;\ge\;\frac{9}{4}\tag{1}[/imath] and my question: Does this remain true upon replacing the first factor by [imath]xy+yz+zx\,[/imath]? Please prove or reject [imath](xy+yz+zx)\left[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right] \;\ge\;\frac{9}{4}\tag{2} [/imath] with [imath]\,x,y,z>0[/imath] as before. There's the estimate [imath]\,xy+yz+zx\le\left(x^2+y^2+z^2\right)\,[/imath] obtained from [imath]\operatorname{(AM\ge GM)}[/imath]. It can get very coarse, e.g. choose [imath]x=1[/imath] and both [imath]y,z[/imath] close to zero, but it appears as if the other factor in brackets can compensate such that [imath](2)[/imath] still holds true. This is a pro, but not a proof. When trying to prove [imath](2)[/imath] along the same lines as above I am stuck: Applying Cauchy-Schwarz to [imath]\sum_{\text{cyc}}\frac{\sqrt{xy}}{x+y}[/imath] gives the needed LHS of [imath](2)[/imath], but by [imath]\operatorname{(AM\ge GM)}[/imath] the cyclic sum is bounded above by [imath]\frac{3}{2}[/imath]. So this is a con, but not a counterexample. | 959688 | Hard inequality [imath] (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} [/imath]
I need to prove or disprove the following inequality: [imath] (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} [/imath] For [imath]x,y,z \in \mathbb R^+[/imath]. I found no counter examples, so I think it should be true. I tried Cauchy-Schwarz, but I didn't get anything useful. Is it possible to prove this inequality without using brute force methods like Bunching and Schur? This inequality was in the Iran MO in 1996. |
1901981 | A field extension that is a finitely generated algebra is an algebraic extension
Given a field [imath]F[/imath] and a field extension [imath]K[/imath] such that [imath]K[/imath] is a finitely generated [imath]F[/imath]-algebra, show that [imath]K[/imath] is algebraic over [imath]F[/imath]. I think the result should follow by induction on the cardinality of a minimal generating set for [imath]K[/imath] as an [imath]F[/imath]-algebra, but I haven't been able to proceed with that line of argument. I am guessing one needs to use some results about integral extensions, but I am not sure. | 2187507 | Is this field extension finite?
Let [imath]\mathbb{F}, \mathbb{F'}[/imath] be fields, [imath]\mathbb{F} \subset \mathbb{F'}[/imath], and [imath]\alpha \in \mathbb{F'}[/imath]. Assume [imath]\mathbb{F(\alpha)}[/imath] is finitely generated algebra over [imath]\mathbb{F}[/imath]. Is this true: [imath](\mathbb{F}(\alpha):\mathbb{F}) < \infty[/imath] ? First i thought it's a consequence of this theorem: Finitely Generated Algebraic Extension is Finite But we don't know if [imath]\alpha[/imath] is algebraic or not. |
2187870 | How to solve [imath]\tan(x+100)=\tan(x+50)\tan(x)\tan(x-50)?[/imath]
The question is The minimum value of x for which [imath]\tan(x+100)=\tan(x+50)\tan(x)\tan(x-50)[/imath] Here is what I've tried so far: In the R.H.S- [imath]\frac {\sin(x+50)\sin(x-50)\sin(x)}{\cos(x+50)\cos(x-50)\cos(x)}\implies \frac {(\sin^2x-\sin^250)\sin x}{(\cos^2x-\sin^250)\cos x}[/imath] But, I think I am proceeding wrong because my process is not yielding any result. Kindly help me out. | 621213 | Determine the smallest positive value of x(in degrees) for which: [imath]\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})[/imath]
Determine the smallest positive value of x(in degrees) for which: [imath]\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})[/imath] I tried to apply the formula of [imath]\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}[/imath] but that led me nowhere resulting in a huge equation. Please help. |
2188431 | Properties of log function
Suppose a function [imath]L:(0,\infty )\rightarrow \mathbb{R} [/imath] satisfies the following properties [imath]L(1)=0, L'(x)=1/x.[/imath] How can we show that [imath]L(xy)=L(x)+L(y)?[/imath] | 2176741 | Prove that [imath]F(xy) =F(x) + F(y)[/imath] when [imath]F(x)[/imath] is not [imath]\ln(x)[/imath]
So we have this function for [imath]x > 0[/imath]. [imath]\int_{1}^{x} \frac{1}{t}\text{d}t[/imath] Show that [imath]F(xy) = F(x) + F(y)[/imath] without assuming [imath]F(x) = \ln(x)[/imath]. I came so far as this point, but I can't crack the last step. [imath]\int_{1}^{x} \frac{1}{t}\text{d}t + \int_{x}^{xy} \frac{1}{t}\text{d}t[/imath] which is: [imath]F(x) + \int_{x}^{xy} \frac{1}{t}\text{d}t[/imath] However, I am having difficulties proving that: [imath]\int_{x}^{xy} \frac{1}{t}\text{d}t = F(y)[/imath] Help would be very very appreciated! |
2183176 | The image of a polynomial curve [imath]p:F\to F^2[/imath] lies in the zero set of another polynomial [imath]q:F^2\to F[/imath]
Suppose [imath]\mathbb{F}[/imath] is an infinite field. Define [imath]p:\mathbb{F}\to\mathbb{F}^2[/imath] as [imath]p(t)=(p_1(t),p_2(t))[/imath] where [imath]p_1,p_2:\mathbb{F}\to\mathbb{F}[/imath] are polynomials. Show that the image of [imath]p[/imath] lies in the zero set of some polynomial [imath]q:\mathbb{F}^2\to\mathbb{F}[/imath]. (Bonus) Give a lower bound to the degree of [imath]q[/imath]. Any polynomial [imath]q:\mathbb{F}^2\to\mathbb{F}[/imath] of degree [imath]D[/imath] looks like [imath]q(x,y)=\sum_{0\leq i+j\leq D}c_{ij}x^iy^j\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)[/imath] where [imath]c_{ij}\in\mathbb{F}[/imath]. Now finding [imath]q[/imath] is solving the equation [imath]q(p(t))=q(p_1(t),p_2(t))=0.(2)[/imath] Plugging (1) in (2) we will have the equation [imath]\sum_{0\leq i+j\leq D}c_{ij}p_1(t)^ip_2(t)^j=0,\;\;\text{ for all }t\in\mathbb{F}.[/imath] Since this is true for all [imath]t\in\mathbb{F}[/imath] and the field is infinite, we can choose [imath]t_1,t_2,\ldots,t_{N=\frac{(D+1)(D+2)}{2}}[/imath] values for [imath]t[/imath] from [imath]\mathbb{F}[/imath] so that we get a system \begin{cases} \sum_{0\leq i+j\leq D}c_{ij}p_1(t_1)^ip_2(t_1)^j=0\\ \vdots\\ \sum_{0\leq i+j\leq D}c_{ij}p_1(t_N)^ip_2(t_N)^j=0 \end{cases} So now we want to show that there exists those [imath]t_1,t_2,\ldots,t_N[/imath] that make this system not invertible. How to do that? | 1287140 | All polynomial parametric curves in [imath]k^2[/imath] are contained in affine algebraic varieties
I have started working through the textbook Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea and I am stuck on one part of an introductory question. The question begins by getting one to show (1) If [imath]f(t)[/imath] and [imath]g(t)[/imath] are polynomials of degree [imath]\le n[/imath] in [imath]t[/imath], then for [imath]m[/imath] large enough, the monomials [imath][f(t)]^e[g(t)]^f[/imath] with [imath]e+f\le m[/imath] form a linearly dependent set in [imath]k[t][/imath]. The next part of the question asks one to deduce (2) If [imath]C\,:\,x=f(t),\,y=g(t)[/imath] is any polynomial parametic curve in [imath]k^2[/imath], then [imath]C[/imath] is contained in [imath]\textbf{V}(F)[/imath] for some [imath]F\in k[x,y][/imath]. I have demonstrated (1), but I am at a loss as to how to use that fact to prove (2). It doesn't seem like it is supposed to be a hard problem, but never the less I am having trouble wrapping my head around it. Additionally I assume that for (2) we are not interested in the trivial solution [imath]F=0[/imath]. [Added for completeness] To do part (1), we recognize that if [imath]\deg(f),\deg(g)\le n[/imath], then [imath]\deg\left([f(t)]^e[g(t)]^f\right)\le ne+nf\le nm[/imath] Hence if we have more than [imath]nm+1[/imath] polynomials of the form [imath][f(t)]^e[g(t)]^f[/imath] then we have a dependent set (since [imath]d+1[/imath] is the dimension of the vector space of polynomials of degree less than or equal to [imath]d[/imath]). Now we simply count the number of polynomials in this form (this was an exercise before the problem) - it is [imath]\binom{m+2}{2}=\frac{1}{2}(m+1)(m+2)[/imath] from a standard application of stars and bars. Ensuring [imath]m[/imath] large enough (specifically [imath]m>2n-3[/imath]) gives us the conclusion. |
2188653 | Contour integration of [imath]1/z[/imath] around 1.
What happens if we try to integrate a function like [imath]\frac{1}{z}[/imath] over the circle radius one centered at one, since the function is holomorphic in the interior, but at the point on the contour [imath]z=0[/imath], the function does not exist. What about the general case of a function being holomorphic on a circular domain and we take the contour integral on the boundary. How do we do it? | 1123070 | Contour Integration where Contour contains singularity
There are many theorems in complex analysis which tell us about integration [imath]\int_{\gamma} f[/imath] where [imath]f[/imath] is continuous (or even differentiable) in the interior of [imath]\gamma[/imath] except finitely many points. I would like to see how should we solve [imath]\int_{\gamma}f(z)dz[/imath] where [imath]\gamma[/imath] contains a singular point of [imath]f[/imath]. For example, in solving [imath]\int_{|z|=1} \frac{1}{z-1}dz[/imath], I slightly pulled the curve [imath]|z|=1[/imath] near [imath]1[/imath] towards left, so that [imath]\gamma[/imath] will not contain [imath]1[/imath]; then the singularity of [imath]\frac{1}{z-1}[/imath] will not be inside this deformed curve, so integration over deformed curve is zero, so taking limit, I concluded that [imath]\int_{|z|=1} \frac{1}{z-1}dz=0[/imath]. However, if I pull the curve near [imath]1[/imath] on the right side, then the singularity of [imath]\frac{1}{z-1}[/imath] will be inside this deformed curve, and integration is then non-zero. I confused between these two processes; can you help me what is the correct way to proceed for finding integration along curves, in which singularity of the function is on the curve? |
2189211 | Evaluate the remainder of [imath]2^{100}[/imath] upon division by [imath]7[/imath].
Not sure how to find the remainder. Any help would be much appreciated. Thanks Checked out the solution for [imath]\frac{2^{2014}}{7}[/imath] but, I wasn't 100% sure of the actual working for finding the remainder of such equations. Therefore requested help for a different value so that i could perhaps know how the whole process worked for a different value. | 2184733 | Find the remainder of [imath]\frac{2^{2014}}{7}[/imath]
Find the remainder of [imath]\dfrac{2^{2014}}{7}[/imath] I am new to the modular arithmetic, Any suggestions to solve this question? |
2189447 | Evaluation of the limit [imath]\lim_{n\to\infty}\sum_{k=1}^n\frac k{k^2+n^2}[/imath]
[imath]\lim_{n\to\infty}\sum_{k=1}^n\frac k{k^2+n^2}[/imath] I need to evaluate this limit, but I don't know how to start. Should i take [imath]1/n^2[/imath] out? Help required. Thank you. | 879611 | How do you calculate this limit [imath]\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{n^2+k^2}[/imath]?
How to find the value of [imath]\lim_{n\to\infty}S(n)[/imath], where [imath]S(n)[/imath] is given by [imath]S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}[/imath] Wolfram alpha is unable to calculate it. This is a question from a questions booklet, and the options for the answer are-- [imath]\begin{align} &A) \dfrac{\pi}{2} \\ &B) \log 2 \\ &C) \dfrac{\pi}{4} \\ &D) \dfrac{1}{2} \log 2 \end{align}[/imath] |
2189390 | If [imath]a+b=1[/imath], show that [imath]a^{4b^2} + b^{4a^2} \leq 1[/imath]
If [imath]a+b=1[/imath], show that [imath]a^{4b^2} + b^{4a^2} \leq 1[/imath]. It's given that [imath]a,b \in \mathbb{R^+}[/imath]. I couldn't get the equality case actually. I couldn't get a rigorous way to get the equality case. EDIT: New method For the inequality, WLOG assume [imath]a>b[/imath]. Then [imath]a^{4b^2} + b^{4a^2} \leq a^{4a^2} + b^{4a^2} \leq (a+b)^{4a^2} \leq 1[/imath] Is it correct? Thanks a lot StackExchangers!! | 2176571 | If [imath]a+b=1[/imath] so [imath]a^{4b^2}+b^{4a^2}\leq1[/imath]
Let [imath]a[/imath] and [imath]b[/imath] be positive numbers such that [imath]a+b=1[/imath]. Prove that: [imath]a^{4b^2}+b^{4a^2}\leq1[/imath] I think this inequality is very interesting because the equality "occurs" for [imath]a=b=\frac{1}{2}[/imath] and also for [imath]a\rightarrow0[/imath] and [imath]b\rightarrow1[/imath]. I tried to work with function of one variable, but the derivative is not easy. I also don't get something solvable by Taylor series. |
2186270 | Is it known whether a triangular number can be any power with exponent greater than [imath]2[/imath]?
Here : Who can prove that a triangular number cannot be a cube, fourth power or fifth power? I asked how it can be proven that a triangular number (a number of the form [imath]\frac{n(n+1)}{2}[/imath]) cannot be a cube, fourth power or fifth power and lulu posted very useful results which probably solve the problem completely. But even more seems to be true : CONJECTURE : A triangle number cannot be a power [imath]k^s[/imath] with [imath]k\ge 2[/imath] , [imath]s\ge 3[/imath] The conjecture is true for [imath]n\le 10^9[/imath] and I currently verify the range [imath][10^9,10^{10}][/imath] without having found a counter-example yet. Is it known whether this conjecture is true ? | 2185585 | Who can prove that a triangular number cannot be a cube, fourth power or fifth power?
Triangular numbers (See https://en.wikipedia.org/wiki/Triangular_number ) are numbers of the form [imath]\frac{n(n+1)}{2}[/imath] In ProofWiki I found three claims about triangular numbers. The three claims are that a triangular number cannot be a cube, not a fourth power and not a fifth power. Unfortunately, neither was a proof given nor did I manage to do it myself. Therefore my qeustions : Does someone know a proof that a triangular number cannot be a cube, a fourth power or a fifth power ? |
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