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2866259 | Calculate [imath]\gcd(c^a+1,c^b+1)[/imath]
Given positive integers [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath], I'd like to ask whether we may adopt some variant of the Euclidean algorithm to calculate [imath]\gcd(c^a+1,c^b+1)[/imath] fast? | 2142932 | gcd of power plus one
I know that [imath]\gcd(a^b-1,a^c-1)=a^{\gcd(b,c)}-1[/imath] which can be seen by expanding both terms into geometric series. Is there any such simplification for [imath]\gcd(a^b+1,a^c+1)=?[/imath] EDIT: When working with polynomials instead of whole numbers, it is [imath]gcd(x^n+1,x^m+1)=\begin{cases}a^{\gcd(n,m)}+1 & \nu_2(n) =\nu_2(m)\\1 &\text{else}\end{cases}[/imath] But this gives only a lower bound for the whole-number case. |
1750317 | The smallest number that if multiplied by 2 forms a permutation of itself
I am looking for the smallest number larger than [imath]0[/imath] which when multiplied by [imath]2[/imath], forms a permutation of itself. I quickly remembered that the number [imath]142,857[/imath] does that, as well as with all numbers up to [imath]6[/imath] (including). This is the first cyclic number, formed by the fraction of [imath]\frac{1}7[/imath] But a computer program quickly showed that the smallest number that does what I asked for is [imath]125,874[/imath], because [imath]125,874*2 = 251,748[/imath]. Interestingly, this number is a permutation of [imath]142,857[/imath] too. But my question is, how can I find this smallest number without a computer program or pre-knowledge, but only with logic and algebra? Thanks! | 782334 | Interview Question Asked In yahoo
Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number. Source: http://gpuzzles.com/mind-teasers/very-hard-maths-riddle/ I understand the answer is [imath]125874 => 251748[/imath] [imath]251748[/imath] is twice the [imath]125874[/imath] and have same digits [imath]1,2,4,5,7[/imath] & [imath]8[/imath] but how to solve this non programmatic ? |
2863858 | Covariant derivative of inner product
I'm reading the Wikipedia page on Levi-Civita connections, and there's a step that I don't quite follow in one of the derivations. I was wondering if someone could help. Suppose [imath]X[/imath], [imath]Y[/imath], and [imath]Z[/imath] are vector fields on a manifold [imath]M[/imath], [imath]\nabla[/imath] is an affine connection on [imath]M[/imath] and [imath]g[/imath] is a Riemannian metric on [imath]M[/imath]. How can one show that [imath]X (g(Y, Z)) = g(\nabla_X Y, Z) + g(Y, \nabla_X Z)[/imath] using only the definition of [imath]\nabla[/imath]? Do we require compatibility with the metric? Edit: one of the comments has pointed out that this is the definition of compatibility with the metric. I don't see how to derive this relationship from the definition that I've seen, i.e. [imath]\nabla g = 0[/imath]. Thanks! Edit 2: Question is now a dupe of Equivalent definition of metric compatibility for a connection: what does [imath]\nabla g[/imath] mean? | 1749725 | Equivalent definition of metric compatibility for a connection: what does [imath]\nabla g[/imath] mean?
So the definition I know for metric compatibility is: [imath]Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ),[/imath] which make sense, as [imath]g(Y,Z)[/imath] is a smooth function from the manifold to reals and we think of [imath]X[/imath] as a derivation. So now I read this apparent equivalent definition that says [imath]\nabla g=0[/imath]. Can someone explain what this means? How can I do [imath]\nabla[/imath] of [imath]g[/imath] I thought [imath]g_p[/imath] is an element of [imath]T_p^*M\otimes T_p^*M[/imath] at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent? |
2866675 | Manipulation of Power Series
Show that ... [imath]\sum_{n=1}^\infty \frac{n^2} {2^n} = 6[/imath] How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ... f(x) = [imath]\sum {Cn}·{X^n}[/imath], g(x) = [imath]\sum {Dn}·{X^n}[/imath] f(x) ± g(x) = [imath]\sum_{n=b}^\infty {Cn}·{X^n}[/imath] ± [imath]\sum_{n=b}^\infty {Dn}·{X^n}[/imath] = [imath]\sum_{n=b}^\infty ({Cn ± Dn})·X^n[/imath] f [imath](K · {X^m})[/imath] = [imath]\sum_{n=b}^\infty {Cn}({K·{X^m}})^n[/imath] [imath]\sum_{n=b}^\infty {Cn} {X^{n+m}} [/imath] = [imath]{X^m} \sum_{n=b}^\infty {Cn}·{X^n}[/imath] if f(x) = [imath]\sum_{n=0}^\infty {Cn}·{X^n}[/imath] then f'(x) = [imath]\sum_{n=1}^\infty {Cn}·{nX^{n-1}}[/imath] if f(x) = [imath]\sum_{n=0}^\infty {Cn}·{X^n}[/imath] then [imath]\int f(x)[/imath] = [imath]\sum_{n=0}^\infty[/imath] [imath]\frac {{Cn}·X^{n+1}} {n+1}[/imath] [imath]\sum_{n=b}^\infty {Cn}·{X^n}[/imath] is continuous on Interval of Convergence. This is the completed work so far ... [imath]\sum_{n=1}^\infty \frac{n^2} {2^n} = \sum_{n=1}^\infty {(1/2)^n} ·{n^2}[/imath] let f(x) = [imath] \sum_{n=1}^\infty {X^n}{n^2}[/imath] then [imath] \int f(x) = \sum_{n=1}^\infty \frac {X^{n+1}} {n+1} · {n^2}[/imath] I assume the next step should be to cancel [imath]{n^2}[/imath] and n+1, but I'm not quite getting the connection. How would you proceed with this? | 62422 | Proof of the equality [imath]\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6[/imath]
Show that for [imath]k[/imath] running over positive integers [imath] \sum_{k=1}^\infty \frac{k^2}{2^k}=6 .[/imath] We can use finite calculus. |
2866859 | Determine whether [imath]\sum_{n=1}^{\infty}\frac{(-1)^n}{n \log^2(n+1)}[/imath] converges absolutely or conditionally.
Problem Let [imath]S = \sum_{n=1}^{\infty}\frac{(-1)^n}{n\log^2(n+1)}[/imath]. Determine the series converges absolutely or conditionally. Attempt [imath]S=\sum_{n=1}^{\infty}( -1)^n a_n[/imath] [imath]a_n[/imath] is monotonically decreasing and it approaches zero when [imath]n[/imath] approaches infinity. So series is convergent . Doubt How to check for absolute convergence? Ratio test fails here. Root test is of no use. I have attempted comparison tests using the fact that [imath]n>\log(n)[/imath], but no success there also. | 2866854 | Is [imath]\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}[/imath] absolutely convergent?
Consider the series [imath]\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}.[/imath] Determine whether it converges absolutely or conditionally. My attempt S=[imath]\sum_{n=1}^{\infty}( -1)^n[/imath] an an is monotonically decreasing and it approaches zero when n approaches infinity. So series is convergent . Doubt How to check for absolute convergence? Ratio test fails here. |
2866240 | Give a Bijection between [imath]\mathbb{R}\setminus\mathbb{Z}[/imath] and [imath]\mathbb{R}\setminus\mathbb{N}[/imath]
Give a Bijection between [imath]\mathbb{R}\setminus\mathbb{Z}[/imath] and [imath]\mathbb{R}\setminus\mathbb{N}[/imath] I got a bijection between [imath]\mathbb{N}[/imath] and [imath]\mathbb{Z}[/imath] Given by [imath]\phi(1)=0[/imath] [imath]\phi(2)=-1[/imath] [imath]\phi(3)=1[/imath] [imath]\phi(4)=-2[/imath] [imath]\phi(5)=2[/imath] [imath]\phi(6)=-3[/imath] But I cannot figure how to find a bijection between [imath]\mathbb{R}\setminus\mathbb{Z}[/imath] and [imath]\mathbb{R}\setminus\mathbb{N}[/imath] | 25241 | Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
Problem statement: Let A be an uncountable set, B a countable subset of A, and C the complement of B in A. Prove that there exists a one-to-one correspondence between A and C. My thoughts: There's a bijection between A and A (the identity function). There's a bijection between C and C (the identity function). There's a bijection between B and [imath]\mathbb{N}[/imath]. That's all I know. |
2866287 | Proving the non-differentiability of [imath]|x| + x[/imath].
For the function [imath]f(x) = |x| + x[/imath], how can one prove that it is not differentiable at the point [imath]x = 0[/imath]? Specifically, I wish to use the Definition of the Derivative [imath]\lim_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}[/imath] to attain two non-equal limiting values by taking [imath]\Delta x[/imath] to be first positive, and then negative. It was my original thought to rewrite [imath]f(x)[/imath] as [imath] f(x) = \begin{cases} 2x&\text{if}\, x\geq 0\\ 0&\text{if}\, x < 0 \end{cases} [/imath] and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative [imath]\Delta x[/imath], and so the cases method would fall out of line with that. Thank you. | 1794044 | Given [imath]f(x) = x + |x|[/imath] for what values of [imath]x[/imath] is [imath]f[/imath] differentiable
Problem : Given [imath]f(x) = x + |x|[/imath] for what values of [imath]x[/imath] is [imath]f[/imath] differentiable? For the sake of generality, let's assume that it is unknown to us that [imath]|x|[/imath] is not differentiable at [imath]x = 0[/imath] Attempted Solution : Using the definition of differentiability, a function is differentiable over an interval [imath]I[/imath] [imath]\text{iff}[/imath] [imath]f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \ \ \forall x\in I[/imath] Now implicitly this definition of differentiability requires [imath]\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}[/imath] Therefore [imath]f[/imath] will be differentiable only when [imath]\lim_{h\to 0^+} \frac{(x + h + |x+h|)-(x + |x|)}{h} = \lim_{h\to 0^-} \frac{(x - h + |x-h|)-(x + |x|)}{h}[/imath] But it is unclear what to do next as [imath]x[/imath] is just arbitrary [imath] \begin{equation} \begin{aligned} |x+h| &= \begin{cases}x+h & \text{if} & x \geq -h\\ -x-h & \text{if} & x < -h \end{cases} \\ &= \begin{cases}x+h & \text{if} & x \geq 0\\ -x-h & \text{if} & x < 0 \end{cases} \end{aligned} \end{equation}[/imath] A Wrong Solution : We could take the derivative of [imath]f[/imath], (and evaluate the domain of the derivative [imath]f'(x)[/imath] to values for which it is defined), using the rules of differentiation, but we would get a wrong answer. [imath]f'(x) = \frac{|x| + x}{|x|}[/imath] [imath]\implies f'(x) = \begin{cases} 2 & \text{if} & x \geq 0 \\ 0 & \text{if} & x < 0 \end{cases}[/imath] This implies the derivative [imath]f'(0)[/imath] exists, when it does not as per the definition of differentiability. Why is that so? Questions: How can a solution be found using the definition of differentiablity? Why does taking the derivative of [imath]f[/imath], [imath]f'(x)[/imath] and evaluating it's domain, not give the correct values of [imath]x[/imath] for which [imath]f[/imath] is differentiable? |
2867106 | Trigonometry : Find the value of [imath]\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7[/imath]
Find the value of [imath]\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7[/imath] My try : Converted it into Sin and then tried to apply series formula but failed | 470614 | Find the value of [imath]\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)[/imath]
What is the value of [imath]\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? [/imath] I tried to write [imath]\textrm{cosec}^2\left(\frac{4\pi}7\right)[/imath] as [imath]\textrm{cosec}^2\left(\frac{3\pi}7\right)[/imath]. Then converted in [imath]\sin[/imath]... But in vain.. Is there any other approach? |
2867307 | [imath]f(0)=0,\ f'(X)[/imath] is str. increasing: prove that [imath]f(X)/X[/imath] is also increasing function all [imath]X \in(0, \infty)[/imath]
[imath]f(0)=0, \ f'(X)[/imath] is a strictly increasing function: prove that [imath]f(X)/X[/imath] is also strictly increasing for all [imath]X \in(0, \infty)[/imath] | 572489 | Show that [imath]x \mapsto \frac{f(x)}{x}[/imath] is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
Let f : [0, 1] → R be a continuous function with f (0) = 0 and suppose that it is differentiable for all x ∈ (0, 1). Further assume that [imath]x \mapsto f'(x)[/imath] is a strictly increasing function on (0, 1). Show that [imath]x \mapsto \frac{f(x)}{x}[/imath] is also a strictly increasing function on (0, 1). |
2866937 | Calculation of sums using Fourier series
I will start my with an example, followed by questions based on this example, and then general questions that could apply to a general case. Calculate the sum : [imath]\sum_{n=1}^\infty \frac 1 {n^6}[/imath] The solution uses the function [imath]f(x)=\begin{cases} x(\pi-x),&x\in[0,\pi]\\ x(\pi+x) , &x\in[-\pi,0]\end{cases}[/imath] [imath]f[/imath] is an odd function, so all of the even fourier coefficients are [imath]0[/imath], and all of the coefficients of the odd terms will be imaginary. Calculating the coefficients gives : [imath]\hat f_0=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)dt=0[/imath] since [imath]f[/imath] is odd and the domain is symmetric around [imath]0[/imath], for [imath]n\neq0[/imath] we get : [imath]\hat f_n=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)(\cos(nt)+i\sin(nt))dt=\\ =\frac 1 {2\pi}\left[\int_{-\pi}^{\pi}f(t)(\cos(nt))dt+\int_{-\pi}^{\pi}f(t)(i\sin(nt))dt\right]=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)(i\sin(nt))dt[/imath] since [imath]f(t)\cos(nt)[/imath] is odd and the integral over a symmetric domain around [imath]0[/imath] is [imath]0[/imath]. Using integration by parts twice we get [imath]\hat f_n=\frac {2i((-1)^n-1)} {\pi n^3}[/imath] which means [imath]\hat f_n=\begin{cases}0 ,&\in\mathbb{N}_{odd} \\ -\frac {4i} {\pi n^3}&\text{otherwise}\end{cases}[/imath] and we get [imath]f(x) \sim \sum_{n=-\infty}^{\infty} e^{inx}\hat f_n[/imath] Using Parseval's identity we eventually get [imath]\|f\|^2=\sum_{n=-\infty}^{\infty}-\frac {4i} {\pi (2n+1)^3}[/imath] from here , we can evaluate the norm (it's a simple integral) and do several algebraic manipulations to the series and find the sum we wanted. This is a fairly straightforward and simple method if you have the right [imath]f(x)[/imath] for your sum. My questions are about the construction of [imath]f[/imath]: [imath]f[/imath] is an odd function. In my understanding, [imath]f[/imath] was constructed this way just to simplfy the calculations ? Could we choose [imath]f[/imath] to be an even function and the calculations would be as easy? Could we choose [imath]f[/imath] to be neither even nor odd and still be able to calculate it simply? [imath]f[/imath] is a polynomial of degree 2. This means we will integrate by parts twice,resulting in a fourier coefficient in a degree of [imath]\frac 1 {n^3}[/imath]. So in the case we want to calculate [imath]\frac 1 {n^6}[/imath] we will get the right degree from Parseval's identity. Does the fact that [imath]f[/imath] is even or odd affects this stage? For example, I've seen the even function [imath]f(x)=x^2[/imath] on the interval [imath][−π,π][/imath] gives coefficients which are of degree 2. What if we wanted to calculate [imath]\sum _{n=1}^\infty \frac 1 {n^{2k}}[/imath] Would we choose [imath]f(x)[/imath] to be of degree [imath]k[/imath] and use the same method? Would another way to calcultate [imath]\sum _{n=1}^\infty \frac 1 {n^{k}}[/imath] be to pick [imath]f[/imath] polynomial of degree [imath](k-1)[/imath] that converges pointwise at [imath]0[/imath] (or maybe other "nice" number like [imath]0.5\pi,\pi[/imath])? Is it even possible? The idea is to evaluate [imath]f(0)=\sum _{n=-\infty}^{\infty} \hat f_n e^{in0}[/imath] The next question is related to Paseval's Identity: Can Pasevals Identity be generalized to [imath]L_p[/imath] norms? [imath]||f||_{L_p}^p=\sum_{n=-\infty}^{\infty}|\hat f_n|^p[/imath] If so, if the fourier series converges uniformly[imath]^*[/imath] to [imath]f[/imath], can we calculate sums using this generalized identity? [imath](*)[/imath]-I haven`t gotten deep into Functional Analysis, but it does make sense to me that if the convergence is uniform then there is convergence to the function in any [imath]L_p[/imath] norm I hope my questions are clear and I`m sorry for the long question. Edit: Note that I`m not interested in help with the specific example I wrote down, I understand how to calculate the sum. I'm more interested in the general case and the construction of [imath]f[/imath], and the properties of [imath]f[/imath] that allows us to calculate the sum easily. The specific series I adressed to is only an example to show the method I was using. (And hopefully to be more clear with my question).Any answers on this specific case that can be generalized would be appriciated as well. | 1948206 | [imath]\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}[/imath] by Fourier series of [imath]x^2[/imath]
Prove that [imath]\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}[/imath] by the Fourier series of [imath]x^2[/imath]. By Parseval's identity, I can only show [imath]\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}[/imath]. Could you please give me some hints? |
2865832 | Must a degree [imath]6[/imath] field extension with Galois group [imath]S_3[/imath] be the splitting field of a cubic?
Prove (or disprove by example) the following: If [imath]F[/imath] is a field extension of [imath]K[/imath], not necessarily a Galois extension, with [imath][F:K]=6[/imath] and [imath]\mathrm{Aut}_K F \simeq S_3[/imath], then [imath]F[/imath] must be the splitting field of an irreducible cubic in [imath]K[x][/imath]. This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if [imath]F[/imath] doesn't have to be Galois over [imath]K[/imath], then there must be an example where [imath]F[/imath] isn't a splitting field at all, right? I've yet to come up with such an example though. | 2197561 | Does a Galois group being [imath]S_3[/imath] correspond to the extension being the splitting field of a cubic?
If [imath]f(x)[/imath] is an irreducible cubic, then [imath]\operatorname{Gal}(f(x))\cong S_3[/imath] or [imath]A_3[/imath]. But what about the converse? That is, if [imath]\operatorname{Gal}(K/F)\cong S_3[/imath], is it necessarily true that [imath]K[/imath] is the splitting field of some irreducible cubic in [imath]F[x][/imath]? |
2867534 | In higher-order derivatives, such as [imath]\frac{d^2y}{dx^2}[/imath], why is the exponent on [imath]d[/imath] in the numerator and [imath]x[/imath] in the denominator?
Why are the exponents in higher order derivatives the following: [imath] \frac{d^2y}{dx^2} [/imath] Why is the exponent on the [imath]d[/imath] in the numerator and [imath]x[/imath] in the denominator? | 2156821 | Multiple derivative notation
Official Leibniz notation for double derivative is: [imath]\frac{\mathrm d^2s}{\mathrm dt^2}[/imath] This term seems inconsistent. Two considerations: We have infinitesimal change in distance [imath]\mathrm ds[/imath] per infinitesimal change in time [imath]\mathrm dt[/imath]: [imath]\mathrm ds/\mathrm dt[/imath]. Both terms are a tiny value/interval. Because the [imath]\mathrm d[/imath] symbolizes difference, I would as a change of the change of the distance to time intuitively write: [imath]\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm ds^2/\mathrm dt^2)}{\mathrm dt}=\frac{\mathrm ds^2}{\mathrm dt^3}[/imath] where the extra [imath]\mathrm d[/imath] says that both terms are now "double" infinitesimal differences. Maybe more properly following mathematical logic and not my intuition, the [imath]\mathrm d[/imath] could be considered a "free" variable in itself that can be multiplied onto this [imath]\mathrm ds/\mathrm dt[/imath] fraction numerator: [imath]\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm d^2s/\mathrm dt)}{\mathrm dt}=\frac{\mathrm d^2 s}{\mathrm dt^2}[/imath] That agrees with the actual notation but doesn't really make physical sense now. [imath]\mathrm d[/imath] means (infinitesimal) difference, so that [imath]\mathrm ds=s_{final}-s_{start}[/imath], and therefore it makes no physical sense to consider the [imath]\mathrm d[/imath] and the [imath]s[/imath] separate. The [imath]\mathrm ds[/imath] is physically just a "name"/"symbol" for one term, which could just as well have been called [imath]x[/imath] or [imath]a[/imath] or anything else. Now, while searching for an explanation, the answers always tend to consider [imath]\frac{\mathrm d}{\mathrm dt}[/imath] as one symbol in itself, so that a double derivative is [imath]\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d}{\mathrm dt}s=\frac{\mathrm d^2}{\mathrm dt^2}s=\frac{\mathrm d^2s}{\mathrm dt^2}[/imath] - which makes even less physical sense, since the [imath]\mathrm dt[/imath] term has to be a separable term before we can treat [imath]\frac{\mathrm ds}{\mathrm dt}[/imath] as a normal fraction (as done in integration e.g.). [imath]\frac{\mathrm d}{\mathrm dt}[/imath] can't possibly be just "a symbol". Why is [imath]\frac{\mathrm d^2s}{\mathrm dt^2}[/imath] the correct one in a physical context, where [imath]\mathrm ds[/imath] actually means the infinitesimal difference in [imath]s[/imath]? Are my considerations in point 2 correct, and I just can't figure out that splitting [imath]\mathrm d[/imath] and [imath]s[/imath] is allowed? Update The answers already given at this time both indicate the use of [imath]$\mathrm d/\mathrm dt$[/imath] as merely a symbol. So, neither of my two suggestions mentioned above are the case. Sure, I can accept that. But the question still remains of why as well of how come we still treat them as variables then, e.g. in integration? Let me clarify those two points: Firstly, if it indeed is the case that [imath]$\mathrm d/\mathrm dt$[/imath] is merely a symbol and should be thought of as just a symbol, then I do not understand the motivation for this symbol. Why did Leibniz choose [imath]$\mathrm d/\mathrm dt$[/imath] as a symbol, which causes the confusion and inconsistency described in the question above? Why not, say, [imath]$\mathrm d/\mathrm d$[/imath], in which case we would get a writing-style that at least looks a bit more "consistent": [imath]$$\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm ds}{\mathrm dt}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^2s}{\mathrm d^2t}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^3s}{\mathrm d^3t}\qquad \cdots$$[/imath] Or even better yet, if the two [imath]$\mathrm d$[/imath] involved in this [imath]$\mathrm d/\mathrm dt$[/imath] symbol have no meaning as neither a variable nor an indicator of a change in the parameter, then why use this letter at all? Why not stick to, say, the prime-notation throughout and never jump into the Leibniz notation: [imath]$$s_t'\qquad s_t''\qquad s_t'''\qquad \cdots$$[/imath] And secondly, if the [imath]$\mathrm d/\mathrm dt$[/imath] really just is a symbol, and that's it, then how come we suddenly can treat it as a fraction again containing a set of variables [imath]$\mathrm ds$[/imath] and [imath]$\mathrm dt$[/imath] that we can split apart during for instance integration? Such as here: [imath]$$\frac{\mathrm ds}{\mathrm dt}=v\quad\Leftrightarrow\quad \mathrm ds=v\,\mathrm dt\quad\Leftrightarrow\quad \int 1 \,\mathrm ds=\int v\,\mathrm dt \quad\Leftrightarrow\quad s=\int v\,\mathrm dt$$[/imath] I hope to get this notion cleared out and appreciate all comments and answers that can help. |
2866701 | Polya’s urn model
Polya’s urn model supposes that an urn initially contains [imath]r[/imath] red and [imath]b[/imath] blue balls. At each stage a ball is randomly selected from the urn and is then returned along with m other balls of the same color. Let [imath]X_k[/imath] be the number of red balls drawn in the first [imath]k[/imath] selections. (a) Find [imath]\mathbb{E}[X_1][/imath]. (b) Find [imath]\mathbb{E}[X_2][/imath]. (c) Find [imath]\mathbb{E}[X_3][/imath]. (d) Conjecture the value of [imath]\mathbb{E}[X_k][/imath], and then verify your conjecture by a conditioning argument. (e) Give an intuitive proof for your conjecture. | 378810 | a problem on Polya's urn scheme
In an urn with [imath]b[/imath] blue and [imath]r[/imath] red balls, each time (call it a trial) a ball is chosen at random and then put again in the urn along with [imath]c[/imath] extra balls of the same color. Now probability of getting a blue ball in the 1st trial = [imath]\frac{b}{b+r}[/imath]. Surprisingly, I also see that probability of getting a blue ball in the [imath]n[/imath]-th trial is also [imath]\frac{b}{b+r}[/imath]. I don't understand the intuition behind this. Another question is the intuitive argument to prove that the number of red balls in the first [imath]n[/imath] trials follow an uniform distribution between [imath]0[/imath] and [imath]n[/imath] when [imath]b=r=c[/imath]. |
2867884 | Let G be [imath]\mathbb Z_p\times\dots\times \mathbb Z_p[/imath]. Find [imath]A(G)[/imath] (the Automorphism group of [imath]G[/imath]).
Let G be [imath]\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}[/imath]. Find [imath]A(G)[/imath] (the Automorphism group of [imath]G[/imath]). ([imath]\mathbb Z_p[/imath] is the integres modulo p, for example [imath]\mathbb Z_2=\{0,1\},\ \ \mathbb Z_5=\{0,1,2,3,4\}[/imath]) I thoght about: Think about [imath](\mathbb Z_p)^n[/imath] as vector space over [imath]\mathbb Z_p[/imath] of dimension n. Then to think about the automorphisms as linear transformations (do I need to explain it more?). So [imath]A(\mathbb Z_p)^n[/imath] is equal to to the space of all invertible linear transformations of [imath](\mathbb Z_p)^n[/imath] to itself. Then we get that [imath]A(G)\cong GL_n(\mathbb Z_p)[/imath]. is that enough or I need to add something? | 2866008 | Let G be [imath]\mathbb Z_p\times\dots\times \mathbb Z_p[/imath] . Find A(G).
Let G be [imath]\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}[/imath]. Find [imath]A(G)[/imath]. I know that [imath]A(G)\cong GL_n(\mathbb Z_p)[/imath]. I prove it by taking [imath]\varphi[/imath] from [imath]A(G)[/imath] and show that it's like matrix multiplication, than I took [imath]\theta[/imath] from [imath]A(G)[/imath] and I show that the composition of those two is a again matrix multiplication. Than I show that because we take matrix from [imath]A(G)[/imath] then there kernl is trivial so the matrix need to be not reversible. I don't know if this is good way or not, I know it's good for [imath]\mathbb Z_p\times\mathbb Z_p[/imath] but I don't know if it's enough for [imath]\mathbb Z_p\times \mathbb Z_p\times\dots\times\mathbb Z_p[/imath] (n times). If you have other wat please help me, there is a hint to think about vector space [imath](\mathbb Z_p)^n[/imath] but I don't know it could help. |
2864418 | Difficulty in understanding the "worth" of the Archimedean property
I'm taking an introductory, undergraduate course in calculus. I'm having some difficulty in understanding the "point"(significance) of the proof of the Archimedean property (For every real number [imath]x[/imath], there exists a natural number [imath]n[/imath], such that [imath]x<n[/imath]) which is done using the completeness axiom. Why isn't the following proof acceptable? [imath]x=\lfloor x\rfloor + \{x\}[/imath] Let [imath]y=1-\{x\}[/imath] Now[imath]1\geq y>0[/imath] So [imath]x+y>x[/imath] Now [imath]x+y[/imath] is clearly a natural number greater than [imath]x[/imath]. | 423107 | Archimedean Proof?
I've been struggling with a concept concerning the Archimedean property proof. That is showing my contradiction that For all [imath]x[/imath] in the reals, there exists [imath]n[/imath] in the naturals such that [imath]n>x[/imath]. Okay so we assume that the naturals is bounded above and show a contradiction. If the naturals is bounded above, then it has a least upper bound (supremum) say [imath]u[/imath] Now consider [imath]u-1[/imath]. Since [imath]u=\sup(\mathbb N)[/imath] , [imath]u-1[/imath] is an element of [imath]\mathbb N[/imath]. (here is my first hiccup, not entirely sure why we can say [imath]u-1[/imath] is in [imath]\mathbb N[/imath]) This implies (again not confident with this implication) that there exists a [imath]m[/imath] in [imath]\mathbb N[/imath] such that [imath]m>u-1[/imath]. A little bit of algebra leads to [imath]m+1>u[/imath]. [imath]m+1[/imath] is in [imath]\mathbb N[/imath] and [imath]m+1>u=\sup(\mathbb N)[/imath] thus we have a contradiction. Can anyone help clear up these implications that I'm not really comfortable with? Thanks! |
2868536 | find the the limit points of this sequence..
Given [imath]a_1 \in \mathbb{R}[/imath] , consider the sequence {[imath]a_n[/imath]} defined by [imath] a_{n+1} = \begin{cases} \frac{a_n}{2} &\text{ for even n} ,\\ \frac{ 1+ a_n}{2} & \text {for odd n} \end{cases}[/imath]. find the the limit points of this sequence.. My attempt : For n odd i get [imath] a_{n+1}= \frac{ 1+ a_n}{2}[/imath] nad putting [imath]a_{n+1} = {a_n} = l[/imath] Now i get [imath]l= \frac{ 1+ l}{2}[/imath] as i get [imath]l = 1[/imath] for odd.. as I don't know how to find the [imath]a_{n+1}[/imath] when n will even Pliz help me,,,,, | 2722971 | Find the sub-sequential limits of a recursive sequence
Given [imath]a_1 \in \mathbb R[/imath] . Consider the sequence [imath]\{a_n\}[/imath], which is recursively defined by[imath] a_{n+1} =\left\{\begin{array}{ccc} \dfrac {a_n}{2} & \text{if $n$ is even,}\\ \dfrac {1 + a_n }{2} & \text{if $n$ is odd}.\end{array} \right. [/imath] Find the sub-sequential limit of this sequence. I think for odd [imath]n[/imath] the limit point will be [imath]l= \frac{1+l}{2}[/imath] now [imath]2l = 1+l[/imath] , here [imath] l = 1[/imath] I don't know the limit point for0 even [imath]n[/imath]. Please help me. |
2869033 | If a sequence is such that [imath]a_n = \frac{a_{n-1}+a_{n-2}}{2}[/imath], then find the limit of sequence
If a sequence is such that [imath]a_k[/imath] is arithmetic mean of its two immediately preceding terms. Show that the sequence converge. Find the limit of sequence I got to see that the odd sub sequence is increasing and bounded above by [imath]a_2[/imath] and even sub sequence decreasing and bounded below by [imath]a_1[/imath]. Now the sequence is convergent. Bow how to find the limit of sequence. | 2276402 | Limit of sequence in which each term is defined by the average of preceding two terms
We have a sequence of numbers [imath]x_n[/imath] determined by the equality [imath]x_n = \frac{x_{n-1} + x_{n-2}}{2}[/imath] The first and zeroth term are [imath]x_1[/imath] and [imath]x_0[/imath].The following limit must be expressed in terms of [imath]x_0[/imath] and [imath]x_1[/imath] [imath]\lim_{n\rightarrow\infty} x_n [/imath] The options are: A)[imath]\frac{x_0 + 2x_1}{3}[/imath] B)[imath]\frac{2x_0 + 2x_1}{3}[/imath] C)[imath]\frac{2x_0 + 3x_1}{3}[/imath] D)[imath]\frac{2x_0 - 3x_1}{3}[/imath] Since it was a multiple choice exam I plugged [imath]x_0=1[/imath] and [imath]x_1=1[/imath]. Which means that all terms of this sequence is [imath]1[/imath],i.e, [imath]x_n=1, n\in \mathbb{N} [/imath] From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit? |
2868769 | How to prove that [imath]x^2+1=5^y[/imath] has no positive integer solutions for [imath]y\geq 2[/imath]?
I am sure I saw a similar question like this one before but I can't find it now. I tried using "order" but failed. It is quite obvious when [imath]y[/imath] is an even number. The real problem is when [imath]y[/imath] is odd. Is there any easy way to solve this? Thanks. | 1697058 | On equations [imath]m^2+1=5^n[/imath]
I am looking for integer solutions of Diophantine equation [imath]m^2+1=5^n[/imath]. I found that [imath]m=0,n=0[/imath] and [imath]m=2,n=1[/imath]. I could not find any other solutions. I try to prove this but I could not. Could anyone help me to solve this equation? |
2857391 | How can I prove that [imath]\log^k(n) = O(n^\epsilon)[/imath]?
How can I prove that [imath]\log^k(n) = O(n^\epsilon)[/imath]? (for any [imath]\epsilon > 0 [/imath] and for any integer and positive [imath]k[/imath]). I tried to do it by - definition , and to prove that exists [imath]c,n_0 > 0[/imath] so that for any [imath]n > n_0 [/imath] exists: [imath]c \cdot n^\epsilon > \log^k(n) [/imath], but I don't have idea how to continue for this point. I will be happy to listen ideas :). | 2927297 | Prove that for sufficiently large [imath]x[/imath], [imath](\log x)^p where p is a positive integer.[/imath]
Ok so I was playing with GeoGebra by plotting various functions, then I come up with the following kind of graph when I plot [imath]\log(x)[/imath] to various positive powers and [imath]f(x)=x[/imath]. The green one is the function [imath]f(x)=x[/imath] and the blue one is [imath]g(x)=(\log(x))^4[/imath]. So it seems for sufficiently large [imath]x[/imath], any positive power of [imath]\log(x)[/imath] is less than [imath]x[/imath]. But I am unable to prove this fact. Can anyone give me a hint as how to approach to this problem? |
1880198 | The numbers [imath]1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{100}[/imath] are written on a white board
The numbers [imath]1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n}[/imath] are written on a white board. In every step we pick two numbers [imath]a[/imath] and [imath]b[/imath] and put [imath]a+b+ab[/imath] besides. Prove that the last number that remains is [imath]n[/imath]. By induction it is easy to prove: It is true for [imath]1[/imath]. Think it is true for [imath]n[/imath] Prove it is true for [imath]n+1[/imath] [imath]n+\frac{1}{n+1}+\frac{n}{n+1}=n+1[/imath] But I want a non-induction prove.Thanks. | 1703768 | Function of a sequence of numbers
You're given a sequence of number: [imath]1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{64}[/imath]. A function takes two numbers in the sequence, say [imath]a,b[/imath] and replaces them both with [imath]a+b+ab[/imath]. For example, we could take the initial sequence and apply the operation to the numbers [imath]1[/imath] and [imath]1/64[/imath] to obtain the new list [imath]\frac{1}{2}, \frac{1}{3},\cdots, \frac{1}{63}, \frac{33}{32}, \frac{33}{32}[/imath] Is there a solution so that after some number of steps, all terms in the sequence are equal? How many unique solutions are there? Note that this question is different from the question asked here, because the number of terms in that question changes, while in this question, the number of terms remains constant ([imath]64[/imath]). By the reasoning in the linked question, we can observe that the operation [imath]a*b = a + b + ab = (a+1)(b+1)-1[/imath] on [imath]\mathbb{Q}[/imath] is identical to the usual multiplication, under the isomorphism [imath](\mathbb{Q}, *) \to (\mathbb{Q}, \times)[/imath] given by [imath]x \mapsto x+1[/imath]. |
2868703 | An exercise of commutative algebra.
Let [imath]\phi:A\rightarrow B[/imath] be a ring homomorphism. Let [imath]X＝ \operatorname{Spec}(A)[/imath], [imath]Y= \operatorname{Spec}(B)[/imath]. If [imath]q\in Y[/imath], then [imath]\phi^{-1}(q)[/imath] is a prime ideal of [imath]A[/imath]. Hence [imath]\phi[/imath] induces a mapping [imath]\phi^* :Y\rightarrow X[/imath]. Show that: If [imath]b[/imath] is an ideal of [imath]B[/imath], then [imath]\overline{\phi^*(V(b))}＝V(b^c).[/imath] Here [imath]b^c[/imath] is the ideal [imath]\phi^{-1}(b)[/imath]. I can prove that the left side is contained in the right side. But I can't prove the right side is contained in the left side. | 1286845 | A proof for Atiyah-Macdonald Exercise I.21.iii
The following is exercise I.21.(iii) of Atiyah-Macdonald: Let [imath]\phi \colon A \to B[/imath] be a ring homomorphisms. Let [imath]X = \operatorname{Spec} A[/imath] and [imath]Y = \operatorname{Spec} B[/imath] [and let [imath]\phi^\ast \colon Y \to X[/imath] be the induced mapping, [imath]\phi^\ast (\mathfrak q) = \phi^{-1} (\mathfrak q)[/imath]]. Show that [...] iii) If [imath]\mathfrak b[/imath] is an ideal of [imath]B[/imath], then [imath]\overline{\phi^\ast (V (\mathfrak b))} = V (\mathfrak b^c)[/imath]. [...] After trying myself for a while, I looked at solutions posted on the internet, one of which presented the following solution. It is clear that [imath]\mathfrak p \in \overline{\phi^\ast (V (\mathfrak b))} \color{blue} \Leftrightarrow r (V (\mathfrak b)^c) \subset \mathfrak p[/imath] and [imath]\mathfrak p \in V (\mathfrak b^c) \Leftrightarrow \mathfrak b^c \subset \mathfrak p[/imath]. Then it suffices to show that [imath]\mathfrak b^c \subset \mathfrak p[/imath] if and only if [imath]r(V (\mathfrak b)^c) \subset \mathfrak p[/imath], which is true because [imath]r(V (\mathfrak b)^c) = r(V (\mathfrak b))^c \color{blue} = r(\mathfrak b)^c = r(\mathfrak b^c)[/imath]. where I have marked the things I don't understand in blue. In the meanwhile I have found a different proof for the exercise, but I still don't understand this proof. In particular I don't understand the connection between taking the closure and taking the radical. I don't know if [imath]r(V (\mathfrak b)^c)[/imath] is a shorthand for the set of elements contained in the contractions of prime ideals contained [imath]\mathfrak b[/imath] or for the ideal generated by these elements or something else. I don't see why the marked equality holds. |
2869017 | field are domains.
I want to prove that field are domain. Let [imath]K[/imath] a field. Suppose it's not a domain. Then there are [imath]a,b\in K\setminus\{0\}[/imath] s.t. [imath]ab=0[/imath]. Let [imath]a'[/imath] and [imath]b'[/imath] s.t. [imath]a'a=aa'=1[/imath] and [imath]b'b=bb'=1[/imath]. Then [imath]1=(a'a)(bb')=a'(ab)b'=0,[/imath] which is impossible. Does it work ? | 1121319 | Unital subrings of fields are all domains
Prove that any subring of a field which contains the identity is an integral domain. Do i need to show that [imath]ab = 0[/imath] where [imath]a = 0[/imath] or [imath]b = 0[/imath]. Or in general What do you need to show in order something is an integral domain |
2868877 | Proving [imath]\lim_{z\to \infty}f'(z)=0[/imath]
Let assume [imath]f(z)[/imath] is defined and analytic at [imath]|z|>1[/imath] and such that [imath]\lim_{z\to \infty}f(z)=L[/imath] when [imath]L<\infty[/imath] Prove: [imath]\lim_{z\to \infty}f'(z)=0[/imath] [imath]f(z)[/imath] is defined and analytic at [imath]|z|>1[/imath] so it can be expanded to a power series [imath]f(z)=\sum_{n=0}^{\infty}\frac{a_n}{z^n}[/imath] at [imath]|z|>1[/imath] now we know that [imath]lim_{z\to \infty}f(z)=\lim_{z\to \infty}\sum_{n=0}^{\infty}\frac{a_n}{z^n}=L[/imath] Taking the derivative we get: [imath]\lim_{z\to \infty}f'(z)=\lim_{z\to \infty}\sum_{n=1}^{\infty}-n*\frac{a_n}{z^{n-1}}=0[/imath] Or am I missing something? | 2094408 | The analytic function at infinity
Prove that if [imath] f\mathrm{(}z\mathrm{)} [/imath] is analytic at the infinity ,then : [imath] {\lim}_{{z}\mathrm{\rightarrow}\mathrm{\infty}}{f}\prime{\mathrm{(}}{z}{\mathrm{)}}\mathrm{{=}}{0} [/imath] |
2869221 | Group structure on arbitrary topological spaces
Definition. Let [imath](G,\ast)[/imath] be any group. Then [imath]G[/imath] will be said to be a topological group if there exists a topology on [imath]G[/imath] such that the map [imath]f:G\times G\to G[/imath] defined by [imath]f(x,y)=xy^{-1}[/imath] for all [imath](x,y)\in G\times G[/imath] is continuous. Now observe that in this definition it is important to note that we first have a group and then the topology which makes [imath]f[/imath] cotinuous. In other words, we impose topological stricture on a group. My question Is it possible to impose group structure on a topological space? I don't think that it is possible always. So my question is, if [imath](X,\tau)[/imath] be a topological space then under which condition(s) does there exists a binary operation [imath]\ast:X\times X\to X[/imath] such that [imath](X,\ast)[/imath] is a group? If there is any related research in mathematics literature then can you please let me know about some of those? It has been pointed out (see below) that this question is a duplicate of this question. However, I don't think that they are for the reason as explained in this comment. It is also not clear to me that imposing a group structure on a topological space (if it's possible of course) is expected to result in a topological group. | 2707066 | On which topological spaces, can we give a group structure to make it a topological group?
Let [imath]X[/imath] be a non-empty set. It is known that we can give a group structure on [imath]X[/imath]. Now let [imath]X[/imath] be a non-empty topological space. Then can we give a group structure on [imath]X[/imath] so that it becomes a topological group w.r.t. its original topology ? |
2869533 | If [imath](x_n)[/imath] is bounded and diverges, then there exist two subsequences of [imath](x_n)[/imath] that converge to different limits.
Is the following argument correct? If [imath](x_n)[/imath] is bounded and diverges, then there exist two subsequences of [imath](x_n)[/imath] that converge to different limits. Proof. Assume [imath](x_n)[/imath] is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that [imath](x_n)[/imath] contains a subsequence [imath](x_{n_k})[/imath] such that [imath](x_{n_k})\to\beta[/imath] for some [imath]\beta\in\mathbf{R}[/imath]. Now assume that every subsequence converges to the same limit, but then [imath](x_n)\to\beta[/imath], a contradiction, thus we must have another subsequence [imath](x_{n_r})[/imath] such that [imath](x_{n_r})\to\alpha[/imath] and [imath]\alpha\neq\beta[/imath]. [imath]\blacksquare[/imath] | 298817 | Let [imath](x_n)[/imath] be a bounded but not convergent sequence. Prove that [imath](x_n)[/imath] has two subsequences converging to different limits.
Let [imath](x_n)[/imath] be a bounded but not convergent sequence. Prove that [imath](x_n)[/imath] has two subsequences converging to different limits. My attempt is: Since the sequence is bounded , there exists [imath]M>0[/imath] such that [imath]x_n \in [-M,M][/imath] for all [imath]n \in \mathbb{N}[/imath]. Since the sequence does not converge to [imath]x[/imath], there exists [imath]\epsilon_0>0[/imath] such that [imath] \forall N \in \mathbb{N}[/imath], there exists [imath]n \geq N[/imath] such that [imath]|x_n-x| \geq \epsilon_0[/imath]. Then we have [imath]x_n \in [-M,x-\epsilon_0] \cup x_n \in [x+\epsilon_0,M][/imath]. By Bolzano-weierstrass theorem, there exists a convergent subsequence in the two intervals. Is my proof valid? [imath]{}{}[/imath] |
2869189 | Det of matrix [imath]4\times 4[/imath]
Is the method of calculating determinant of [imath]3\times 3[/imath] matrix by diagonals, apply also on [imath]4\times 4[/imath] matrix? for example: [imath]\begin{matrix}2&2&1&3|\\1&4&4&5|\\5&1&1&6|\\7&1&4&5|\end{matrix}\begin{matrix}2&2&1\\1&4&4\\5&1&1\\7&1&4\end{matrix}[/imath] [imath]\det = 2\cdot4\cdot1\cdot5+\dotsb+3\cdot1\cdot1\cdot4 - 7\cdot1\cdot4\cdot3-\dotsb-5\cdot5\cdot4\cdot1 = 171[/imath] Is this valid? | 1224390 | Can we use Sarrus' method for finding the determinant of matrix greater than [imath]3\times3[/imath]?
Can Sarrus' method of finding the determinant be used for finding the determinant of matrices greater than [imath]3\times3[/imath], as I am unable to find any example of a matrix greater than [imath]3\times3[/imath] whose determinant is found by Sarrus' method? I have tried many questions and Sarrus' method only works for [imath]2\times2[/imath] and [imath]3\times3[/imath] matrices. I have not read about Sarrus' method in any book but only from the internet, so I'm not sure about it. |
2869652 | Where's my error in this partial derivatives problem?
Let [imath]u(x, y)=x+y[/imath]. What is [imath]\displaystyle\frac{\partial u}{\partial x}[/imath] and [imath]\displaystyle\frac{\partial u}{\partial y}[/imath]? My answers are [imath]1[/imath] and [imath]1[/imath]. Suppose I now told you that [imath]y=x[/imath], so that [imath]u=2x[/imath]. Now it appears that [imath]\displaystyle\frac{\partial u}{\partial x}=2[/imath] and [imath]\displaystyle\frac{\partial u}{\partial y}=0[/imath]. Where have I gone wrong? Note: My question is completely different from "Partial derivatives paradox". I have renamed my question. | 521058 | Partial derivative paradox
Okay, perhaps not a paradox, but somewhat of a lack of understanding on my part. Let [imath]z[/imath] equal some function of [imath]x[/imath] and [imath]y[/imath], i.e. [imath]z = f(x, y)[/imath] and take partial derivatives [imath]\frac{\partial z}{\partial x} = f_x[/imath] and [imath]\frac{\partial z}{\partial y} = f_y[/imath] all and good. But now say I do partial differentiation with respect to z. [imath]\frac{\partial z}{\partial z} = f_z[/imath] which equals [imath]0[/imath] because [imath]f(x, y)[/imath] is a function of [imath]x[/imath] and [imath]y[/imath], but not [imath]z[/imath], so all [imath]x[/imath] and [imath]y[/imath] are held constant and the derivative of a constant is zero. But that's not the right answer is it? [imath]\frac{\partial z}{\partial z}[/imath] should be equal to 1. Where did my logic go wrong? |
280961 | Extending a continuous function on the unit sphere to the unit ball
I'm trying to solve a problem in Lee's topology book. In this book, a closed n-cell [imath]D[/imath] is any space homeomorphic to the closed unit ball. Then [imath]\mathrm{Int} D[/imath] , resp. [imath]\partial D[/imath] are the images of the open unit ball, resp. the boundary of the closed unit ball under such a homeomorphism. Given a continuous function [imath]f: \partial D \rightarrow \left [ 0,1 \right ][/imath], he wants one to show that [imath]f[/imath] extends to a continuous function [imath]F: D \rightarrow \left [ 0,1 \right ][/imath] that is strictly positive in [imath]\mathrm{Int} D[/imath]. Obviously, it suffices to show this for a closed unit ball and I was thinking something like [imath]\left \|x \right \|f\left(\frac{x}{\left \| x \right \|}\right) + \dfrac{1 -\left \| x \right \|}{2},[/imath] but that does not quite do it. Tietze's extension theorem is not in that book (and in fact is a weaker result). Any ideas? Thanks in advance. | 1120533 | Extending a continuous map between the boundary of two cells.
I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes. The problem is: Let [imath]D[/imath] and [imath]D'[/imath] be two closed cells not necessarily of the same dimension. Show that any continuous map [imath]f:\partial D \to \partial D'[/imath] can be extended to a continuous map [imath]F:D \to D'[/imath] such that [imath]F(Int \; D) \subseteq Int \; D'[/imath]. Given points [imath]p \in Int \; D[/imath] and [imath]p' \in Int \; D'[/imath], show that [imath]F[/imath] can be chosen so [imath]F(p) = p'[/imath]. Show that if [imath]f[/imath] is a homeomorphism then [imath]F[/imath] can be chosen to be a homeomorphism. I proved the first part roughly as follows: First suppose [imath]D[/imath] and [imath]D'[/imath] are convex and each contain [imath]0[/imath] in their interior (in their respective ambient spaces). Then every element other than [imath]0[/imath] in [imath]D[/imath] can be expressed uniquely in the form [imath]\lambda q[/imath] where [imath]q \in \partial D[/imath] and [imath]\lambda\in (0,1][/imath] (an equivalence relation on the chords connecting [imath]0[/imath] to boundary points partitions [imath]D- \{0\}[/imath]). I then define the map [imath]F(\lambda q) = \lambda f(q)[/imath] which is continuous since if [imath]\lambda_n q_n \to \lambda q[/imath] in [imath]D[/imath] then [imath]F(\lambda_n q_n) \to F(\lambda q)[/imath]. Lastly [imath]\lambda q \in Int \; D[/imath] implies [imath]\lambda <1[/imath] and so [imath]\lambda f(q)[/imath] is an interior point of [imath]D'[/imath] since [imath]f(q)[/imath] is a boundary point and [imath]D'[/imath] is convex. Now if we suppose [imath]D[/imath] and [imath]D'[/imath] are arbitrary closed cells there are homeomorphisms [imath]g_1: \overline{\mathbb{B}^n} \to D[/imath] and [imath]g_2: \overline{\mathbb{B}^m} \to D'[/imath] (where possibly [imath]m=n[/imath]) then we have that [imath]g_2^{-1}\circ f \circ g_1[/imath] is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map [imath]F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}[/imath]. The mapping [imath]g_2 \circ F \circ g_1^{-1}[/imath] is a continuous map that satisfies the desired claim. My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed. Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem. |
2869753 | Is [imath]∅[/imath] proper subset of [imath]\{\{∅\}\}[/imath]?
I get that [imath]∅[/imath] is subset of every set thus [imath]∅ ⊆ \{\{∅\}\}[/imath]. However, I'm not sure if [imath]∅ ⊂ \{\{∅\}\}[/imath]. From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one. What I'm confused is, does [imath]\{\{∅\}\}[/imath] have an element that [imath]∅[/imath] doesn't have? | 2590405 | If null set is an element of a set then will it belongs to set or subset?
if [imath]A=\left\{\varnothing \right\}[/imath] we know that [imath]A[/imath] is a Singleton Set whose single element is a Null set. That is [imath]A[/imath] is a set containing another set which is Null set. Now are these two statements True? [imath]1.[/imath] [imath]\varnothing \in A[/imath] [imath]2.[/imath] [imath]\varnothing \subset A[/imath] |
2869941 | [imath]f:[0,1] \to [0,1][/imath] be a continuous function. Prove there exists a point p an element of [imath][0,1][/imath] such that [imath]f(p)=p[/imath]
[imath]f:[0,1] \to [0,1][/imath] be a continuous function. Prove there exists a point p an element of [imath][0,1][/imath] such that [imath]f(p)=p[/imath] I have no idea how to do this but is it asking me for a proof if the intermediate value theorem? i'm not entirely sure how the IVT would solve this because it doesn't specify [imath]f(p)=p[/imath]. | 425033 | Show that any continuous [imath]f:[0,1] \rightarrow [0,1][/imath] has a fixed point [imath]\zeta[/imath]
Be a continuous function [imath]f:[0,1] \rightarrow [0,1][/imath]. Show that there is a [imath]\zeta \in [0,1][/imath] with [imath]f(\zeta)=\zeta[/imath] ([imath]\zeta[/imath] is called fixed point). Consider the function [imath]g:[0,1] \rightarrow [-1,1][/imath], [imath]g(x):= f(x)-x[/imath]. [imath]g[/imath] is continuous. Because of [imath]f(0),f(1) \in [0,1][/imath] is [imath]g(0)\geq0[/imath] and [imath]g(1)\leq 0[/imath]. Because [imath]f(0)[/imath] has a value between [imath]0[/imath] and [imath]1[/imath], [imath]f(0)\geq 0[/imath]. [imath]g(0) = f(0)-0= f(0) \geq 0 - 0 = 0[/imath]. Because [imath]f(1)[/imath] has a value between [imath]0[/imath] and [imath]1[/imath], [imath] f(1) \leq 1 [/imath]. [imath]g(1)=f(1)-1 \leq 1 - 1 = 0[/imath] [imath]\Leftrightarrow g(0)\geq 0[/imath] and [imath]g(1)\leq 0 [/imath] After the IVT: [imath]\exists \zeta \in [0,1]:g(\zeta)=0 \Leftrightarrow f(\zeta) = \zeta [/imath] [imath]\zeta[/imath] is a fixed point of [imath]f[/imath]. [imath]\Box[/imath] My questions are: Is this proof done in the correct way or have I missed something? Is there something I can improve? |
2870010 | Mean value theorem for multivariable functions
I`m trying to generalize Lagrange's mean value theorem for [imath]f:A \to \mathbb{R}^n[/imath] , differentiable in [imath]A \subset \mathbb{R}^m[/imath]. What I have so far is that if [imath]n=1[/imath], we can write [imath]g_i(t)=f(x_1,..tx_i,..,x_n[/imath]), where [imath]x_j[/imath] are any real numbers and [imath]x_i[/imath] is 1. We have [imath]g'(t)=D_f(t)\cdot(0,..,1,...0)=\frac {\partial f} {\partial x_i}(t)[/imath], And since [imath]g_i(t):\mathbb{R} \to \mathbb{R}[/imath] , we can apply the single variable mean value theorem to say that there exists a point [imath]c_{x_i}\in [a,b] [/imath] s.t [imath]\frac {\partial f} {\partial x_i}(c_{x_i})=g_i'(c_{x_i})=\frac {g_i(a)-g_i(b)} {a-b}[/imath] , for any partial derivative. However, For any partial derivative we get a different [imath]c_{x_i}[/imath], And I couldn`t find a way to relate [imath]g[/imath] back to [imath]f[/imath]. Also, if [imath]n>1[/imath], And I`d know how to handle the [imath]n=1[/imath] case, I know I can break down [imath]f[/imath] to [imath]n[/imath] different functions, apply the [imath]n=1[/imath] to each of them seperately, but then again, we get [imath]n[/imath] different points [imath]c[/imath] which aren't necessarily the same,So I got stuck here too. I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking? | 1462746 | Generalization of Mean Value Theorem to functions, [imath]f: \mathbb{R}^n \rightarrow \mathbb{R}^n[/imath]
I was thinking of a generalization of the mean value theorem and I think I found a simple way to generalize the MVT to differentiable functions defined such that, [imath]f: I \rightarrow \mathbb{R}^n[/imath] where [imath]I[/imath] is a compact subset of [imath]\mathbb{R}^n[/imath]. Here's my idea: 1) Every function defined on [imath]\mathbb{R}^n[/imath] may be decomposed into n functions using the orthogonal basis [imath]\{x_i\}_{i=1}^n[/imath] where [imath]x_i[/imath] are unit vectors. 2) So for any function [imath]f[/imath] on [imath] \mathbb{R}^n[/imath], we have [imath]f=\sum_{i=1}^{n} f_i(x|x_i)[/imath] where [imath]x|x_i[/imath] is the projection of the vector [imath]x[/imath] onto the unit vector [imath]x_i[/imath]. It follows that since each since [imath]\forall x \in I, f_i(x|x_i)[/imath] is a function mapping [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath], the usual mean value theorem applies for each [imath]f_i[/imath]. This means that [imath]\exists p \in I[/imath] such that [imath]f'(p)= \sum_{i=1}^{n} \frac{f_i(sup(I|x_i))-f_i(inf(I|x_i))}{sup(I|x_i)-inf(I|x_i)}[/imath] where [imath]I|x_i[/imath] is the projection of the set I onto the unit vector [imath]x_i[/imath]. Update: on second thought, the variables of the vector valued function must be linearly independent and the space they occupy must be of dimension n. |
2870382 | Uniform Convergence for [imath]x^n(1-x)^n[/imath] over [0,1]
I am fairly new to the concept of supremum/infimum and slightly confused over idea of uniform convergence using these notations. According to the wiki for uniform convergence we say that any [imath]f_n(x)[/imath] uniformly converges to [imath]f(x)[/imath] if [imath]\lim\limits_{n\to\infty} sup|(f_n(x)-f(x))| = 0[/imath]. Now my question is that for [imath]f_n(x) = x^n(1-x)^n[/imath] over [0,1] , If i take [imath]f(x)[/imath] = 1/4 how can I show that this equation holds. What I can see though that [imath]\lim\limits_{n\to\infty} sup(|f_n(x)|) - f(x) = 0[/imath] but won't value of [imath]\lim\limits_{n\to\infty} sup|(f_n(x)-f(x))|[/imath] will be equal to 1/4? | 2868006 | Convergence of [imath]f_n(x)=x^n(1-x)^n[/imath]
Does [imath]f_n(x)=x^n(1-x)^n[/imath] converges uniformely on [imath][0,1][/imath] and why? If [imath]x=0[/imath] or [imath]x=1[/imath] we have [imath]f_n(x)=0[/imath]. If [imath]x\in (0,1)[/imath] we have [imath]x^n\rightarrow 0[/imath] and [imath](1-x)^n\rightarrow 0[/imath]. So [imath]f_n\rightarrow 0[/imath]. To check uniform convergence, we have to check that [imath]\forall\varepsilon>0\ \exists N\in\mathbb{N}[/imath] such that [imath]\forall n\geq N, x\in [0,1]: |f_n(x)|<\varepsilon[/imath]. How do I show that? |
2870530 | Show that [imath]S=\mathbb{R}\setminus \left \{-1 \right \}[/imath] is abelian.
Can someone please give me a hint on how to solve this? Let [imath]S=\mathbb{R}\setminus \left \{ -1 \right \}[/imath] and define a binary operation on [imath]S[/imath] by [imath]a\circ b=a+b+ab[/imath]. Prove that [imath](S,\circ)[/imath] is an abelian group. I cannot just say [imath]a\circ b=a+b+ab=b+a+ba=b\circ a[/imath], so what should I do? | 2860195 | Properties of the composition law [imath]a @ b= ab + b + a[/imath] on the real numbers
The operation [imath]@[/imath] is defined on the real numbers as [imath]a @ b= ab + b + a[/imath] a) Show that [imath]0[/imath] is an identity for the operation. b) Show that some real numbers have inverses under the operation. c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers. I have so far came to understand as far as: [imath]a, b,c \in \mathbb{R}[/imath] Identity: Ia= aI= a @ b I= @ b + b + I I @ b = I b + b + I |
1133794 | Arbitrary circles equal to polar coordinates
[imath]\alpha[/imath] is an arbitrary (random) circle that passes through the origin. Show that there are real numbers [imath]s[/imath] and [imath]t[/imath] such that [imath]\alpha[/imath] is the graph of [imath]r = 2s \cos (\theta + t)[/imath]. I believe that [imath]r = 2s \cos (\theta +t)[/imath] can be simplified down to [imath]r = x^2+y^2[/imath] using Cartesian coordinates. What next? | 1317750 | Polar form to cartesian
Let [imath]\Gamma[/imath] be a circle that passes through the origin. Show that we can find real numbers [imath]s[/imath] and [imath]t[/imath] such that [imath]\Gamma[/imath] is the graph of [imath]r = 2s \cos (\theta + t).[/imath] I know this has to be converted to a cartesian equation, but how do I do this, and what do I do after? Thanks |
2871198 | Does every iterative definition (or at least the one describe in the details) require the use of the Axiom of Choice?
One of my books gives the following proof that if the image of a sequence of real numbers [imath]x_n[/imath] has an accumulation point [imath]x_0[/imath], then there exists a subsequence [imath]x_{n_k}[/imath] convergent to [imath]x_0[/imath]. The sequence having finitely many values is a trivial case, so let's consider the case where there are infinitely many points in it. We consider the intersections between the sequence and the balls of radius [imath]\frac{1}{k}[/imath], [imath]k \in \mathbb{N}[/imath], centered in [imath]x_0:[/imath] [imath]U_k =B_{\frac{1}{k}}(x_0) \ \cap \ \{x_n\}_{n \in \mathbb{N}}[/imath] Since [imath]x_0[/imath] is an accumulation point, [imath]U_k[/imath] contains infinitely many points of the sequence, for every [imath]k[/imath]. We define a subsequence as follows: we choose any point in [imath]U_1[/imath] as our first point, call its index [imath]n_1[/imath]; since there are infinitely many points in [imath]U_2[/imath], infinitely many of them have an index greater than [imath]n_1[/imath]. Choose one and call its index [imath]n_2[/imath]; ... since there are infinitely many points in [imath]U_{k+1}[/imath], infinitely many of them have an index greater than [imath]n_k[/imath]. Choose one and call its index [imath]n_{k+1}[/imath]; and so on. The subsequence [imath]x_{n_k}[/imath] converges to [imath]x_0[/imath]. Is the Axiom of Choice needed to justify this definition procedure? I'm used to books always pointing out when they use AC, but in this case my book doesn't. The book also doesn't mention it, but by the well-ordering of the naturals, the iterative process can be made more "canonical" and less "choosy" by always defining [imath]n_{k+1}[/imath] as the minimal element of the set [imath]\{n \ | \ n>n_k \land x_n \in U_{k+1} \}[/imath]. Would this make AC unneeded? Was it unneeded in the first place, even with the book's procedure? | 1277087 | Does the proof of Bolzano-Weierstrass theorem require axiom of choice?
When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of explanation. I want to prove this statement: Let [imath]a_1, a_2, \ldots \in \mathbf{R}[/imath], and [imath](a_n)_{n\in\mathbf{N}}[/imath] is bounded, then [imath](a_n)[/imath] has some convergent subsequence. The proof is as follows. Since [imath](a_n)[/imath] is bounded, for all [imath]n\in\mathbf{N}[/imath], [imath]a_n \in I = [b, c][/imath]. Now, let [imath]I_0 = I[/imath] and if [imath]I_n = [b_n, c_n][/imath], we define [imath]d_n = (b_n+c_n)/2[/imath] and if infinite terms of [imath](a_n)[/imath] is included in [imath][b_n, d_n][/imath](resp. [imath][d_n, c_n][/imath]), we will define [imath]I_{n+1} = [b_n, d_n][/imath](resp. [imath][d_n, c_n][/imath]).If both intervals contain infinite terms, let [imath]I_{n+1}[/imath] be [imath][d_n, c_n][/imath]. For all [imath]n\in \mathbf{N}[/imath], infinite numbers of [imath]m \in \mathbf{N}[/imath] exist such that [imath]a_m \in I_n[/imath] suffices. We take the sequence of natural numbers [imath](n(k))_{k\in\mathbf{N}}[/imath]which suffices [imath]n(0) < n(1) < \cdots < n(k) < \cdots[/imath] following this procedure: Now we have already selected [imath]a_{n(1)}, \ldots, a_{n(k)}[/imath], there are infinite numbers of [imath]m\in \mathbf{N}[/imath] which suffices [imath]n(k)<m, a_m \in I_{k+1}[/imath], so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?). I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times. |
2868285 | If [imath]A\subset B[/imath] are integral domains with the same field of fractions and [imath]B[/imath] is faithfully flat over [imath]A[/imath], then [imath]A=B[/imath].
I found this exercise in Matsumura, Commutative Ring Theory, (Exercise 7.2). If [imath]A\subset B[/imath] are integral domains with the same field of fractions and [imath]B[/imath] is faithfully flat over [imath]A[/imath], then [imath]A=B[/imath]. Can anyone give me a proof for this statement or tell me where to find it? | 482908 | Birational and faithfully flat [imath]\implies[/imath] isomorphism
Let [imath]A \subseteq B[/imath] be integral domains with the same field of fractions. Assume that [imath]A \to B[/imath] is faithfully flat. Why do we have [imath]A=B[/imath]? This is an exercise in Matsumura's book. Here is my idea: If [imath]b \in B[/imath], consider [imath]I = \{a \in A : ab \in A\}[/imath]. This is an ideal of [imath]A[/imath]. By asumption [imath]I \neq 0[/imath], and our goal is to show that [imath]I=A[/imath]. It suffices to prove [imath]IB=B[/imath]. But how can we achieve this? |
2871445 | Find all roots for the equation
I am trying to solve [imath]2^x = 4x[/imath]. Have taken logs on both sides, represented as an exponent and haven't got it close to the form from which I could find a solution. | 1399050 | Solve for [imath]x[/imath]: [imath]2^x=4x[/imath]
Given that [imath]x[/imath] is a positive integer. By using methods of trial and error as well as plotting two lines: [imath]y=2^x[/imath], [imath]y=4x[/imath] on a graph and find their intersection point, we can easily solve for [imath]x[/imath] which is equal to 4. However, I do not know how to solve this using only equation. Can anyone help me? |
1105248 | Sequence of rational number converging to a irrational number [imath]\beta[/imath].
Here is the sequence of irrational number that converges to [imath]\alpha\in \mathbb{Q}[/imath]. Take [imath]x_n=\alpha -\frac{\sqrt{2}}{n}[/imath]. Clearly [imath]\{x_n\}\rightarrow\alpha .[/imath] But I'm trying to find a sequence of rationals that converges to arbitrary irrational number [imath]\beta[/imath]. Can you give me such a sequence. | 1751047 | Sequence of rational numbers converging to any irrational number
I attempting to solve the following question: "Show there is a sequence of rational numbers converging to any irrational number". I don't understand how to tackle this problem because it says any irrational number. I know for example that the sequence [imath]s_n[/imath] = [imath](1 + 1/n)^n[/imath] converges to [imath]e[/imath], but I can't think of a sequence that will converge to any rational number. As a caveat I have looked at other similar questions on stack exchange but none seem to cover this any aspect. Thank you. |
2871392 | Continuity of an exponential series
Let [imath](a_n)_{n \in \mathbb{N}}[/imath] be an infinite positive sequence of integers. Assume that [imath]f(\lambda) = \sum\limits_{n=0}^{\infty} a_n \lambda^n[/imath] is finite for any [imath]\lambda \in [0, \lambda_0)[/imath]. Does this necessarily imply that [imath]f(\lambda)[/imath] is also continuous in [imath](0, \lambda_0)[/imath], or might something strange happen? | 153841 | continuity of power series
I want to prove that every power series is continuous but I am stuck at one point. Let [imath]\sum\limits_{n=0}^\infty a_n(x-x_0)^n[/imath] a power series with a radius of convergence [imath]r>0[/imath] and let [imath]D:=\{x\in\mathbb R:|x-x_0|<r\}[/imath]. Then [imath]S(x)=\sum\limits_{n=0}^\infty a_n(x-x_0)^n[/imath] is continuous on [imath]D[/imath]. Proof: Let [imath]x\in D[/imath] and [imath]r_0\in\mathbb R^+[/imath] such that [imath]|x-y_0|<r_0<r[/imath] and for any positive integer [imath]N[/imath] let [imath]S_N(y)=\sum\limits_{n=0}^{N-1} a_n(y-x_0)^n[/imath] and [imath]\phi_N(y)=\sum\limits_{n=N}^\infty a_n(y-x_0)^n[/imath]. Since the power series converges uniformly on the closed disk [imath]|y-x_0|\leq r_0[/imath], we may choose a positive integer [imath]N_\varepsilon[/imath] such that [imath]|\phi_{N_\varepsilon}|<\frac\varepsilon3[/imath] for all [imath]|y-x_0|\leq r_0[/imath]. Since [imath]S_{N_\varepsilon}[/imath] is polynomial we can choose a [imath]\bar\delta>0[/imath] such that [imath]|S_{N_\varepsilon}(y)-S_{N_\varepsilon}(x)|<\frac\varepsilon3[/imath]. So we get [imath]|S(y)-S(x)|<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon[/imath] for [imath]|y-x|<\delta[/imath] Why is in the violet term [imath]|\phi_{N_\varepsilon}|<\frac\varepsilon3[/imath] correct? Don't I have two terms in the absolute value? Thanks for helping! |
2871409 | Any finite group is a subgroup of an orthogonal group
Prove that any finite group of order [imath]n[/imath] is isomorphic to a subgroup of [imath]\mathbb{O}(n)[/imath], the group of [imath]n\times n[/imath] orthogonal real matrices. Attempt: Let [imath]G[/imath] be a group of order [imath]n[/imath]. Then [imath]G[/imath] is isomorphic to a subgroup of the symmetric group [imath]S_n[/imath]. But how to go further? | 1396031 | Can we embed every finite group in some special orthogonal group or special linear group of some order , over [imath]\mathbb R[/imath]?
For every finite group [imath]G[/imath] , does there exist [imath]n \in \mathbb Z^+[/imath] such that [imath]G[/imath] can be embedded in [imath]SO_n(\mathbb R)[/imath] ? Can every finite group be embedded in [imath]SL_n(\mathbb R)[/imath] for some [imath]n[/imath] ? |
2871402 | Rows of a Matrix is divisible by 19, show that its Determinant is also divisible by 19
I came across the following problem while self studying: Let \begin{equation} A = \begin{bmatrix} 2 & 1 & 3 & 7 & 5\\ 3 & 8 & 7 & 9 & 8\\ 3 & 4 & 1 & 6 & 2\\ 4 & 0 & 2 & 2 & 3\\ 7 & 9 & 1 & 5 & 4\\ \end{bmatrix} \end{equation} Use the fact that 21375, 38798, 34162, 40223, and 79154 are divisible by 19 to show, without evaluating, that [imath]\det[A][/imath] is divisible by 19. I noticed that each of these numbers are the entries in the rows of A, but I don't see how that helps me. | 2558312 | On determinants and common divisors
Let [imath]n\in\mathbb N[/imath] and let [imath]a_1,\ldots,a_n[/imath] be natural numbers smaller than [imath]10^n[/imath]. Write each [imath]a_k[/imath] in base [imath]10[/imath] and add [imath]0[/imath]'s to the left of each decimal expansion, if needed, so that each [imath]a_k[/imath] is written with [imath]n[/imath] digits. Consider the matrix such that the entries of the [imath]k[/imath]th row are the digits of [imath]a_k[/imath] (written in the same order). Let [imath]d\in\mathbb N[/imath] be such that [imath]d[/imath] divides each [imath]a_k[/imath]. Prove that [imath]d\mid\det A[/imath]. For instance, suppose that [imath]n=4[/imath] and that your numbers are [imath]3876[/imath], [imath]2784[/imath], [imath]684[/imath], and [imath]8388[/imath], each of which is a multiple of [imath]12[/imath]. Then[imath]A=\begin{bmatrix}3 & 8 & 7 & 6 \\ 2 & 7 & 8 & 4 \\ 0 & 6 & 8 & 4 \\ 8 & 3 & 8 & 8\end{bmatrix}[/imath]and [imath]\det A=-360[/imath], which is, in fact, a multiple of [imath]12[/imath]. I learned about this problem yesterday. I found it quite cute and I decided to share it with all of you. Note: I know how to prove it. |
2868246 | Show that [imath]S^{-1}A = B[/imath] for integral domains and [imath]S = \{x\in A\setminus\{0\}: x^{-1}\in B\}[/imath]
Let [imath]A \subset B[/imath] be commutative integral domains with [imath]\operatorname{Quot}(A) = \operatorname{Quot}(B).[/imath] Now consider the multiplicatively closed subset [imath]S = \{x\in A\setminus\{0\}: x^{-1}\in B\}[/imath]. I want to show that [imath]S^{-1}A = B[/imath]. I would appreciate any help. | 287245 | Subrings of fraction fields
Let [imath]R[/imath] be an integral domain and let [imath]S[/imath] be a ring with [imath]R \le S \le \text{Frac}(R)[/imath] (fraction field). Question: Is there a multiplicatively closed subset [imath]U \subseteq R\setminus \{0\}[/imath] such that [imath]S=R[U^{-1}][/imath] ? |
2872103 | How to prove "A matrix commutating with all Idempotent matrices is necessarly scalar"
I'm stuck to prove this problem, Probably I have done for [imath]2[/imath] order matrices. MY TRY: Let [imath]P[/imath] be any idempotent and we are to find [imath]A[/imath] such that [imath]AP=PA[/imath]. We have three precise cases. CASE [imath]1[/imath]: If all the eigenvalues of [imath]P[/imath] are [imath]0[/imath].As [imath]P[/imath] is diagonalizable it has eigenspace of dimension [imath]2[/imath] and it follows that [imath]P=O[/imath], zero matrix. So any [imath]A[/imath] will commute with [imath]P[/imath]. CASE [imath]2[/imath]: If all the eigenvalues of [imath]P[/imath] are [imath]1[/imath], as [imath]P[/imath] is diagonalizable it has two independent vectors , say [imath]v_1,v_2[/imath] such that [imath]Pv_1 =v_1, Pv_2 =v_2[/imath]. Now any vector can be expressed a , say [imath]v[/imath] can be expressed as [imath]v =c_1v_1 +c_2v_2[/imath] [imath]\implies[/imath] [imath] Pv = c_1Av_1 [/imath] [imath]+c_2Av_2[/imath] [imath]\implies[/imath] [imath]Pv =c_1v_1 + c_2v_2 [/imath] [imath]\implies[/imath] [imath]Pv =v[/imath], for any [imath]v\in R^2[/imath]. So [imath]P[/imath] becomes an identity and so any matrix [imath]A[/imath] will commute with [imath]P[/imath]. CASE [imath]3[/imath]: Let [imath]0,1[/imath] are the two eigenvalues of [imath]P[/imath]. as [imath]P[/imath] is diagonalizable we can write any vector [imath]v \in R^2[/imath] as direct-sum of two vectors [imath]v_1,v_2[/imath] i.e. [imath]v =v_1 +v_2[/imath] with [imath]Pv_1 =v_1, Pv_2 =0[/imath]. Now [imath]AP(v_1) =A(v_1) =c_1v_1 +c_2v_2[/imath] ( I assume [imath]Av_1 =c_1v_1 +c_2v_2, c_1,c_2[/imath] are scalers) [imath]PA(v_1) = P(c_1v_1 +c_2v_2) =c_1v_1[/imath], Now as [imath]AP =PA[/imath] ,we get [imath]c_2 = 0[/imath], Similarly if I take [imath]Av_2 =d_1v_1 +d_2v_2[/imath] then I find that [imath]d_1 =0[/imath]. Hence [imath]Av_1 =c_1v_1 ;Av_2 =d_2v_2[/imath] and [imath]A[/imath] becomes diagonalizable . Suppose [imath]c_1 \ne d_2[/imath], I'm choosing [imath] \begin{pmatrix} c_1 & 0 \\ 0 & d_2 \end{pmatrix} [/imath] as [imath]A[/imath].If I choose [imath]P[/imath] as [imath] \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} [/imath] . [imath]AP \ne PA[/imath]. So I have to take those eigenvalues of [imath]A[/imath] equal such that [imath]A[/imath] commutes with all the idempotent [imath]P[/imath]'s Then [imath]v=c_1v_1 + c_2v_2[/imath] [imath]\implies[/imath] [imath]Av = c_1v[/imath] ,(as [imath]c_1 = d_2[/imath]). Clearly ,[imath]A[/imath] becomes scalar matrix and I know that scalar matrix commutes with any matrix , it will commute with any Idempotent matrix. So finally considering all those three cases i'm claiming [imath]A[/imath] is scalar matrix in [imath]M_2(R)[/imath] . But How can I prove it in [imath]n[/imath] ordered matrices.please preovide me any hint. Thanks for reading! | 2054084 | Show that the collection of matrices which commute with every idempotent matrix are the scalar matrices
Show that the collection of [imath]n\times n[/imath] real matrices which commute with every idempotent matrix are the scalar matrices . Let [imath]\mathcal P[/imath] denote set of all idempotent matrices . Let [imath]A=\{B:BP=PB\forall P\in \mathcal P\}[/imath]. So I need to show that [imath]A=\{cI:c\in \Bbb R\}[/imath]. I am feeling totally confused on this. Will you kindly give some hints here? |
2872115 | Equi-consistency of inaccessible cardinals
I have trouble understanding this wiki page, concerning inaccessible cardinals. First "These statements are strong enough to imply the consistency of ZFC". I guess what they mean by that is if [imath]\kappa[/imath] is a strongly inaccessible cardinal, then [imath]V_\kappa[/imath] is a model of ZFC, where [imath]V[/imath] is the Von Neumann hierarchy. Then they argue that a proof in ZFC, assuming ZFC is consistent, of the consistency of ZFC+[imath]\kappa[/imath] is impossible. Because it would prove the consistency of ZFC, violating Gödel's incompleteness theorem. I don't understand this, I imagine a relative consistency proof would rather be a model transformation : given a model of ZFC as input, produce a model of ZFC+[imath]\kappa[/imath] as output. Then we could continue producing [imath]V_\kappa[/imath] to get another model of ZFC. Where is the contradiction in that ? To me it seems we first assumed a model of ZFC, then produced another one. ZFC didn't produce a model of itself from scratch. | 1523282 | Proving that the existence of strongly inaccessible cardinals is independent from ZFC?
ZFC can't prove that strongly inaccessible cardinals exist, or else it would prove that a model of itself exists and hence [imath]Con(ZFC)[/imath]. So this leaves us with two options: ZFC proves there are no strongly inaccessible cardinals The existence of strongly inaccessible cardinals is independent from ZFC I've heard, however, that you can't actually prove that it's #2. That is, you can't prove the independence of large cardinal axioms from ZFC. Why is this? At first, I thought it was because proving independence would mean that ZFC proves the consistency of a larger theory that contains ZFC as a model. However, I don't think my reasoning was right. That wouldn't prove that a model exists - just that a model could exist. And isn't it proven that Con(ZFC) is independent from ZFC, and hence that ZFC + Con(ZFC) is consistent iff ZFC is? So how do you show that you can't show the independence of large cardinal axioms? |
2872246 | Show that [imath]x-1[/imath] is a factor of [imath]P(x)[/imath] where [imath]P(x^5)+xQ(x^5)+x^2R(x^5)=\big(x^4+x^3+x^2+x+1\big)S(x)[/imath]
For the problem below, my first step was to rewrite the right side as [imath]\big(x^4+x^3+x^2+x+1\big)S(x)=\frac{x^5-1}{x-1}S(x)[/imath] From there I can isolate [imath]P(x^5)[/imath] which gives me [imath]P(x^5)=\dfrac{S(x)(x^5-1)-x(x-1)\big(Q(x^5)-xR(x^5)\big)}{x-1}[/imath] I'm not sure how to move forward. To get [imath]P(x)[/imath] should I rewrite all the [imath]x^5[/imath] as [imath]x[/imath] and take the [imath]5^{th}[/imath] root of each [imath]x[/imath]? The questions prior used the idea of roots of unity so perhaps I can use that somehow. Thanks | 2586440 | If [imath]P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)[/imath] , then prove that [imath]P(x)[/imath] is divisible by [imath]x-1[/imath]
[imath]P,Q,R,S[/imath] are polynomials such that: [imath]P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)[/imath] , then prove that [imath]P(x)[/imath] is divisible by [imath]x-1[/imath] I thought a lot on this but no result!! By the way,one idea is to insert some values for [imath]x[/imath] and try to produce a system of equations for the given polynomials,but I'm not sure it works. |
2872664 | Prove that there exists a decreasing sequence [imath]b_n \rightarrow 0[/imath] such that [imath]\sum_{n=1}^{\infty} a_n b_n[/imath] diverges.
It is given that [imath]a_n>0[/imath] and [imath]\sum_{n=1}^{\infty} a_n[/imath] is divergent. Prove that there exists a decreasing sequence [imath]b_n \rightarrow 0[/imath] such that [imath]\sum_{n=1}^{\infty} a_n b_n[/imath] diverges. I thought of an example: Let [imath]b_1=1[/imath] and [imath]b_{n+1}=\min (1/(2^n a_{n+1}), b_n, 1/n)[/imath] for [imath]n \geq 1[/imath]. Will this work? | 1304460 | If a [imath]\{a_n\}[/imath] diverges, so [imath]a_n \rightarrow + \infty[/imath], how to find sequence [imath]\{b_n\}[/imath] such that [imath]\sum |b_n|<\infty[/imath] but [imath]\sum |a_n||b_n|[/imath] diverges?
If we are given any sequence of real numbers [imath]\{a_n\}[/imath] diverges, so [imath]a_n \rightarrow + \infty[/imath], how can we find a sequence [imath]\{b_n\}[/imath] such that [imath]\sum |b_n|[/imath] converges but [imath]\sum |a_n||b_n|[/imath] diverges? I want to use this fact in another problem but don't immediately see how to prove it. |
2872702 | If [imath](2x_{n+1}-x_n)[/imath] converges to [imath]x[/imath], then show that [imath](x_n)[/imath] converges to [imath]x[/imath].
Let [imath](x_n)[/imath] be a sequence in [imath]\mathbb{R}[/imath]. Show that if [imath](2x_{n+1}-x_n)[/imath] converges to [imath]x[/imath], then [imath](x_n)[/imath] converges to [imath]x[/imath]. We don't know if the sequence [imath](x_n)[/imath] is a convergent sequence or a cauchy sequence. If we are able to prove any one, then the problem is simple. Please provide any clue on how to go about. | 606172 | If [imath] \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x [/imath], then is it true that [imath] \lim_{n \to \infty} x_{n} = x [/imath]?
If [imath] (x_{n})_{n \in \mathbb{N}} [/imath] is a sequence in [imath] \mathbb{R} [/imath] and [imath] \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x [/imath], then is it necessarily true that [imath] \lim_{n \to \infty} x_{n} = x? [/imath] Could anyone kindly offer suggestions on how to approach this problem? Thanks! |
2872731 | Harmonic function which is zero on the boundary is zero in the domain
Let [imath]u[/imath] be harmonic ([imath]\Delta u=0[/imath]) on the unit circle, if [imath]u\equiv 0[/imath] on [imath]\partial G[/imath] then [imath]u\equiv0[/imath] on [imath]G[/imath] We set [imath]P=-u\frac{\partial u}{\partial y}, Q=u\frac{\partial u}{\partial x}[/imath] Using Green we get [imath]Q_x-P_y=(\frac{\partial u}{\partial x})^2+u \frac{\partial^2 u}{\partial x^2}+u\frac{\partial^2 u}{\partial y^2}+(\frac{\partial u}{\partial y})^2=(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2[/imath] By green [imath]\iint_{G}(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2=\oint_{\partial G}u(-\frac{\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy)=0[/imath] how can I an conclude that [imath]u\equiv 0[/imath]? must it be on the unit circle? | 1536965 | Using Green's identity to show that a harmonic function with zero boundary values is identically zero
I am confused how to do this question. I need to use Green's first identity and if [imath]\nabla(f)=0[/imath] then [imath]f[/imath] is constant on [imath]\Omega[/imath] since [imath]\Omega[/imath] is path connected. I have subbed in the information into green's identity but I don't get anything useful. [imath]\iiint_\Omega \nabla f\cdot \nabla g \,dV =\iint_{\partial\Omega} f\nabla g\cdot n \,dA - \iiint_\Omega f\cdot \Delta g \,dV[/imath] We get: [imath]\iiint_\Omega \nabla f\cdot \nabla g \,dV =-\iiint_\Omega f\cdot \Delta g \,dV[/imath] [imath]\iiint_\Omega \nabla. (f \nabla g) \,dV =0[/imath] [imath]\iint_{\partial\Omega} f\nabla g\cdot n \,dA=0[/imath] |
1661906 | Show that [imath]|z^2| + Re(a \cdot z) +b = 0[/imath] has only solutions for [imath]|a^2|\geq 4b[/imath]
The Question: show that [imath]|z^2| + Re(a \cdot z) +b = 0[/imath] only has a solution for [imath]|a^2|\geq 4b[/imath] [imath]a \in \Bbb{C}[/imath] [imath]b \in \Bbb{R}[/imath] Question 1.10 If I substitute [imath]z=r_z e^{i\phi_z}[/imath] and [imath]a=r_a e^{i \phi_a}[/imath] I get [imath]r_z^2 + Re(r_z r_a e^{i(\phi_a+\phi_z)}) + b= 0[/imath] [imath]r_z^2 + r_z r_a cos(\phi_a+\phi_z) + b= 0[/imath] Since [imath]r_z[/imath] has to be real, the root of [imath](r_a cos(\phi_a+\phi_z))^2 - 4b[/imath] has to be real. Therefore [imath](r_a cos(\phi_a+\phi_z))^2 - 4b \geq 0[/imath] But I have no idea how to get from this to [imath]|a^2|\geq 4b[/imath]. Did I go down the wrong path, did I make a mistake or am I just missing a last step? Thanks | 992055 | Show complex solutions exist
Let A be a complex number and B a real number. Show that the equation [imath]\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,[/imath] has a solution iff [imath]\,\lvert A^2\rvert \geq 4B[/imath]. If this is so, show that the solution set is a circle or a single point. Well i am trying to do the first part first. So assuming the equation has a solution that would mean [imath]z = x+iy[/imath] satisfies the equation. I was going to let [imath]A = s+it[/imath] for a complex number, but it is not working out for me. Wrong step? |
2873395 | Prove the existence of an [imath](x_n)_{n\in\mathbb N}[/imath] in A such that [imath]\lim_{n->\infty}[/imath] [imath]x_n = supA[/imath]
Let [imath]\emptyset = A \subset \mathbb R [/imath] where A is bounded above, and A has a supremum that is not contained in A. Prove the existence of an [imath](x_n)_{n\in\mathbb N}[/imath] in A such that [imath]\lim_{n->\infty}[/imath] [imath]x_n = supA[/imath]. I'm guessing we can construct an example with the information given. I have been playing around with the idea the that we can order A such that [imath]a_i \leq a_j[/imath] for all [imath]i \leq j[/imath] and define [imath](x_n)_{n\in\mathbb N}[/imath] such that [imath]x_l = a_l[/imath] for all [imath]l \in \mathbb N[/imath]. Hence [imath]x_i\leq x_j[/imath]. I also believe, the actual definition of the limit will have to come in, but here I do not know how to continue. Any help is appreciated. | 581128 | Prove that subsequence converges to limsup
Given a sequence of real numbers, [imath]\{ x_n \}_{n=1}^{\infty}[/imath], let [imath]\alpha =[/imath] limsup[imath]x_n[/imath] and [imath]\beta = [/imath] liminf[imath]x_n[/imath]. Prove that there exists a subsequence [imath]\{ x_{n_k}\}[/imath] that converges to [imath]\alpha[/imath] as [imath]k \rightarrow \infty[/imath]. Not sure how to start this without since I'm not given that the subsequence is bounded.. |
2861045 | Hyperbola asymptotes from conic general equation
If I have the coefficients of the following equation: [imath]AX^2 + BXY + CY^2 + DX + EY + F = 0[/imath] And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F? i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas. | 898005 | Finding the asymptotes of a general hyperbola
I'm looking to find the asymptotes of a general hyperbola in [imath]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/imath] form, assuming I know the center of the hyperbola [imath](h, k)[/imath]. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I was wondering if I could run it by someone and see if it works. It's mostly algebraic, and I'm prone to making tiny errors in algebra that throw off the entire problem. So to start, since we know the center [imath](h, k)[/imath], we can first translate the hyperbola by [imath](-h, -k)[/imath] using the transform [imath]x_0 = x - \Delta{x}, y_0 = y - \Delta{y}[/imath] with [imath]\Delta{x} = -h[/imath] and [imath]\Delta{y} = -k[/imath]. Assuming [imath]F'[/imath] is the translated [imath]F[/imath], we can divide the entire equation by [imath]-F'[/imath] to put it in the following form: [imath] ax^2 + bxy + cy^2 + dx + ey = 1 [/imath] With [imath]a = -A'/F'[/imath] and [imath]A'[/imath] the translated [imath]A[/imath], [imath]b = -B'/F'[/imath] and [imath]B'[/imath] the translated [imath]B[/imath], and so on. Next we convert to polar coordinates to get the following: [imath] r^2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) + r(d\cos{\theta} + e\sin{\theta}) - 1 = 0 [/imath] Solving for [imath]r[/imath] will give us [imath] r = \frac{-d\cos{\theta} - e\sin{\theta} \pm \sqrt{(d\cos{\theta} + e\sin{\theta})^2 + 4(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})}}{2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})} [/imath] Now assume [imath]\theta_0 = 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})[/imath], this means that as [imath]\theta_0 \rightarrow 0^{\pm}[/imath], [imath]r \rightarrow \pm\infty[/imath]. The angles at which [imath]r \rightarrow \pm\infty[/imath] are the asymptotes of the hyperbola, so now it's just a matter of solving for where [imath]\theta_0 = 0[/imath]. This is where the majority of the algebra takes place and this is where I'm worried I made some miniscule mistake. [imath] \begin{align} & 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) = 0\\ & \Longleftrightarrow a(1 - \sin^2{\theta}) + b\cos{\theta}\sin{\theta} + c\sin^2{\theta} = 0 \\ & \Longleftrightarrow \sin^2{\theta}(c - a) + b\cos{\theta}\sin{\theta} = -a \\ & \Longleftrightarrow \frac{1 - \cos{2\theta}}{2}(c - a) + \frac{b}{2}\sin{2\theta} = -a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + b\sin{2\theta} = -2a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + 2a = -b\sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow \frac{a - c}{b}(1 - \cos{2\theta}) - \frac{2a}{b} = \sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow (\frac{a - c}{b}(1 - \cos{2\theta}))^2 - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow (\frac{a - c}{b})^2(1 - 2\cos{2\theta} + \cos^2{2\theta}) - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + [2(\frac{a - c}{b})(\frac{a}{b} - \frac{a - c}{b})]\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})^2 - 4a\frac{a - b}{b^2}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})(\frac{-3a - c}{b}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [\frac{a^2 + 2ac + c^2}{b^2} - 1] = 0 \\ \end{align} [/imath] Now let [imath] U = [(\frac{a - c}{b})^2 + 1] \\ V = 2(\frac{a - c}{b})(\frac{c}{b}) \\ W = \frac{a^2 + 2ac + c^2}{b^2} - 1 [/imath] So that the above equation becomes [imath] U\cos^2{2\theta} + V\cos{2\theta} + W = 0. [/imath] Solving for [imath]\cos{2\theta}[/imath] gives us [imath] \cos{2\theta} = \frac{-V \pm \sqrt{V^2 - 4UW}}{2U}. [/imath] Finally we may solve for theta like so, [imath] \theta = \frac{1}{2}\arccos{\frac{-V \pm \sqrt{V^2 - 4UW}}{2U}}. [/imath] This gives us two numbers, [imath]\theta_1[/imath] and [imath]\theta_2[/imath], each corresponding to the slopes [imath]m_1[/imath] and [imath]m_2[/imath] of the asymptotes. The relationship between the slope of a line [imath]m[/imath] and the angle [imath]\theta[/imath] between the line and the positive x-axis is [imath]m = \tan{\theta}[/imath]. You can use the identity [imath]\tan{(\frac{1}{2}\arccos{x})} = \sqrt{\frac{1 - x}{x + 1}}[/imath] to solve for [imath]m_1[/imath] and [imath]m_2[/imath] in terms of non-trig functions, but I think this answer is sufficient enough. Now that we have the slopes of the asymptotes, we can find the y-intercepts [imath]b_1[/imath] and [imath]b_2[/imath] of each line by simply plugging in the original center [imath](h, k)[/imath] into each equation for the line and solving. [imath] b_1 = k - m_1h \\ b_2 = k - m_2h [/imath] Thus, the asymptotes of the hyperbola with general equation [imath]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/imath] have equations [imath]y = m_1x + b_1[/imath] and [imath]y = m_2x + b2[/imath]. Does this look correct? Also, did I overcomplicate things? Was there an easy solution all along that I was missing? |
2872489 | Homology of quotient space of [imath]S^2[/imath] under the identifications [imath]x \sim -x[/imath] for [imath]X[/imath] in the equator [imath]S^1[/imath].
Apparently this question has been asked many times on here, but my question is more specific. Let [imath]X[/imath] be the quotient space of [imath]S^2[/imath] under the identifications [imath]x \sim -x[/imath] for [imath]x[/imath] in the equator [imath]S^1[/imath]. We are asked to compute the homology groups [imath]H_i(X)[/imath]. The method I chose is to use cellular homology. Give [imath]X[/imath] the following cell structure: one 0-cell [imath]e^0[/imath], one 1-cell [imath]e^1[/imath], and two 2-cells [imath]e_1^2[/imath] and [imath]e_2^2[/imath]. The 2-cells are attached via the quotient projection of the boundary [imath]x \sim -x[/imath]. In particular, the 0 and 1 skeletons are [imath]\mathbb{R}P^0[/imath] and [imath]\mathbb{R}P^1[/imath] respectively and the 2-cells are attached in the same way as in Hatcher's computation of the homology of [imath]\mathbb{R}P^n[/imath] using cellular homology. The nontrivial part of cellular chain complex is [imath]0 \rightarrow \mathbb{Z}^2 \xrightarrow[]{d_2} \mathbb{Z} \xrightarrow[]{0} \mathbb{Z} \rightarrow 0[/imath]. I don't understand why [imath]d_2[/imath] is not multiplication by [imath]2[/imath] here since they are attached in the same way as in Hatcher's Example 2.42. According to another answer [imath]d_2(e_1^2)=2e^1[/imath] and [imath]d_2(e_2^2)=-2e^1[/imath] "by the cellular boundary formula". I do not see why this is the case since [imath]e_1^2[/imath] and [imath]e_2^2[/imath] have the same attaching map. | 1341182 | Homology of [imath]S^2/x\sim -x[/imath] for [imath]x[/imath] on the equator
Let [imath]X[/imath] be the quotient space of [imath]S^2[/imath] under the identifications [imath]x\sim -x[/imath] for [imath]x[/imath] in the equator [imath]S^1[/imath]. Compute the homology groups [imath]H_i(X)[/imath]. I wrote my solution/attempt below and I would like to know if it it correct and, more importantly, if there is a simpler solution. I tried using cellular homology but I'm not very experienced with it and kept getting nonsense. Here is my attempt: [imath]X[/imath] can be written as the following pushout: [imath]\begin{align*} \require{AMScd} \begin{CD} P^1 @>{i}>> P^2\\ @V{i}VV @VVV \\ P^2 @>>> X \end{CD} \end{align*}[/imath] where [imath]i:P^1\hookrightarrow P^2[/imath] is the inclusion. So by a Mayer Vietoris argument, we have [imath]H_i(X,P^2) \approx H_i(P^2, P^1)[/imath]. This can be simplified to [imath]\tilde{H}_i(P^2/P^1) \approx \tilde{H}_i(S^2)[/imath]. Now we consider the LES of the pair [imath](X, P^2)[/imath]. [imath]0\to H_2(P^2)\to H_2(X)\to H_2(X, P^2)\to H_1(P^2)\to H_1(X) \to 0[/imath] We know all homology groups involved except those of [imath]X[/imath], so the LES reads: [imath]0\to \Bbb{Z}\to H_2(X) \to \Bbb{Z}\to \Bbb{Z}_2\to H_1(X)\to 0[/imath] The problem lies in identifying the map [imath]\Bbb{Z}\to \Bbb{Z}_2[/imath], call it [imath]f[/imath]. The LES tells us that [imath]H_1(X)\approx \Bbb{Z}_2/\text{im }f[/imath]. On the other hand, by the Van Kampen theorem for attaching 2-cells, [imath]X[/imath] can be viewed as [imath]P^2[/imath] with a 2-disk attached along the path [imath]aa[/imath], hence [imath]\pi_1(X)\approx \Bbb{Z}_2[/imath]. This is abelian, so it coincides with [imath]H_1(X)[/imath]. Then [imath]f[/imath] must be the [imath]0[/imath] map. Using that, from the LES we can also compute [imath]H_2(X) \approx \ker f \oplus \Bbb{Z}\approx \Bbb{Z}^2[/imath]. |
2862756 | no of non negative solutions of equation a+b+c+...n variables=N where a<=b<=c<=d<=....
I need to find the number of non negative solutions of equation [imath]\underbrace{a+b+c+ \dotsb }_{n \text{ variables}}=N[/imath] where [imath]a\le b \le c \le d \le \dotsb [/imath]. What I have tried is: Suppose [imath]S(n)[/imath] gives us the required solution. For [imath]n=1[/imath], only one solution [imath](a=N)[/imath], [imath]S(1)=1[/imath] for [imath]n=2[/imath], [imath]a+b=N[/imath]. [imath]a[/imath] can take either [imath]0[/imath] or non zero. if a takes 0 then total solutions = [imath]S(1)=1[/imath]; otherwise, a can take values from 1 to N/2 so total floor(N/2) possible solutions. Hence [imath]S(2)=S(1)+N/2 = 1 + N/2[/imath] for [imath]n=3, a+b+c=N[/imath], a can take either 0 or non zero if a takes 0 then total solutions = [imath]S(2)[/imath] otherwise , a can take values from 1 to N/3 so total floor(N/3) possible solutions. Hence [imath]S(3)= S(2) + N/3 = 1 + N/2 + N/3[/imath] Similarly for [imath]n=n[/imath], [imath]S(n)=1+ N/2 + N/3 + N/4 + ....... + N/n[/imath] Can anyone tell whether the above formula is right or wrong? | 2237261 | Partitions with at most k parts, each of at most l
I'm taking a course on Analytics Combinatorics (based on Flajolet's and Sedwick's book), and i'm trying to solve note I.16 of the book, that is, verify: [imath] P^{(\leq l, \{1,\ldots,k\})}(z) = \frac{(1-z)\ldots(1-z^{k+l})}{((1-z)\ldots(1-z^{k}))\cdot((1-z)\ldots(1-z^{l}))} [/imath] Where the left hand means the generating function of the class of partitions with at most k parts, each [imath] \leq l [/imath] Truth is I already tried by multiple ways, but with no success: First, looking at the formula I guessed that: [imath] P^{(\leq l, \{1,\ldots,k\})} \times P^{(\{1,\ldots,k +l\})} \tilde = P^{(<l)} \times P^{(\{1,\ldots,k\})} [/imath] but i couldn't understand why. Secondly, using the same reasoning as to find that p(k,l,n) = p(l,k-1,n) + p(l-1,k,n -k), I thought that: [imath] P^{(\leq l, \{1,\ldots,k\})} \tilde = P^{(\leq l, \{1,\ldots,k-1\})} +\mathcal Z^k \times P^{(\leq l - 1, \{1,\ldots,k\})} [/imath] but i could't solve the recurrence. and so on... As you can see, I'm quite confused... can someone give a tip that put me in my way? thank you in advanced! |
2873691 | Does [imath]K \cong K_0[/imath] and [imath]F \cong F_0[/imath] imply [imath][K : F] = [K_0 : F_0][/imath]?
Let [imath]K[/imath] and [imath]K_0[/imath] be two isomorphic fields and let [imath]F[/imath] and [imath]F_0[/imath] be two isomorphic fields such that [imath]K / F[/imath] and [imath]K_0 / F_0[/imath] are two extensions. I have furthermore that [imath]K_0 / F_0[/imath] is monogenic and his degree is [imath]n \in \mathbb{N}[/imath]. Can I state that [imath][K : F] = n[/imath]? Thank you very much in advance. | 964689 | Fields extensions over isomorphic fields of different degrees
What are the simplest examples of situations where in a field [imath]F[/imath] there are two subfields [imath]L_1[/imath] and [imath]L_2[/imath] such that extensions [imath]F/L_1[/imath] and [imath]F/L_2[/imath] are finite, degrees are different [imath] [F:L_1] \neq [F:L_2], [/imath] but fields [imath]L_1[/imath] and [imath]L_2[/imath] are isomorphic as abstract fields. |
1892355 | Clarification of solution for Spivak Calculus on Manifolds Problem 2-35
The question in the title is the following: If [imath]f : \mathbb{R}^n \to \mathbb{R}[/imath] is differentiable and [imath]f(0) = 0[/imath], prove that there exist [imath]g_i : \mathbb{R}^n \to \mathbb{R}[/imath] such that [imath]f(x) = \sum_{i = 1}^{n} x_i g_{i}(x).[/imath] And the hint provided is the following: If [imath]h_x(t) = f(tx)[/imath], then [imath]f(x) = \int_{0}^{1} h_x^{'}(t)dt.[/imath] I followed the hint and got the result below, [imath]f(x) = \int_{0}^{1} \sum_{i = 1}^{n}x_i D_if(tx)dt.[/imath] Now I'd like to bring the sum outside due to linearity of the integral, but for that I need to show that the functions [imath]h_i : \mathbb{R} \to \mathbb{R}[/imath] defined by [imath]h_i(t) = D_i f(tx)[/imath] are integrable. How do I show this? I know that there are differentiable real valued functions whose derivatives are not integrable, so I don't know if this result is true at all (by this I mean I think it's possible to slightly tweak such a function with non integrable derivative to get a counterexample). Am I right in thinking so? All the solutions I can find to this problem (for example, here: http://vision.caltech.edu/~kchalupk/spivak.html) just expand the integral without considering integrability of the [imath]h_i[/imath]s so I might be missing something basic here. Any help is appreciated. | 1107081 | [imath]f:\mathbb{R}^n \to \mathbb{R}[/imath] has expansion [imath]\sum_i g_i(x)x^i[/imath]
Problem 2-35 on page 34 of Spivak's Calculus on Manifolds states If [imath]f: \mathbb{R}^n: \to \mathbb{R}[/imath] is differentiable and [imath]f(0) =0[/imath], prove that there exist [imath]g_i: \mathbb{R}^n \to \mathbb{R}[/imath] such that [imath]f(x) = \sum_{i=1}^n x^ig_i(x)[/imath]. Hint: if [imath]h_x(t) = f(tx)[/imath], then [imath]f(x) = \int_0^1 h_x'(t)[/imath]. I understand the answer he wants, but hasn't he left out a hypothesis that would ensure [imath]h_x'(t)[/imath] is integrable? (For instance the continuity of [imath]df[/imath].) Secondly, I'm wondering if this theorem is used anywhere. Perhaps in differential geometry it might be useful to write locally [imath]f \in C^\infty (M)[/imath] as [imath]\sum_i g_i(x) x_i[/imath] for some coordinates [imath]x_i[/imath]. |
2874169 | Sum of series [imath]1 - 1/3 + 1/6 - 1/10 + \ldots[/imath]?
I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed. I know the series 1 - 1/2 + 1/3 - 1/4 + ... is equal to ln 2. But the series 1 + 1/2 - 1/3 - 1/4 + 1/5 + ... is equal to pi/4 + (1/2 ln 2) - what a strange result! I also tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872. Can someone please help me figure this out? (I don't have any experience with calculus) | 2874148 | Sum of an infinite series: [imath]1 + \frac12 - \frac13- \frac14 + \frac15 + \frac16 - \cdots[/imath]?
I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed. Now I want to look at a series that doesn't involve powers of two. For example, I already know that the series 1 + 1/2 + 1/3 + 1/4 + ... does not converge, but the series 1 - 1/2 + 1/3 - 1/4 + ... converges to ln(2), and that's not too hard to prove. But what about the series 1 + 1/2 - 1/3 - 1/4 + ... ? That has two alternating plus and minus signs. When I ran 1,000,000 iterations of the sequence in Python, I obtained a result approximately equal to 1.131971. I've checked a few things, and it's not ln(3), ln(4), ln(pi), or anything else (it's about ln(3.1017), but as far as I am concerned, the number 3.1017 does not have any significance here). Can someone please help me figure this out? (I don't have any experience with calculus) Also, I tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872. Thanks everyone! |
2874288 | Alternating sum of inverse prime numbers
It is well known that the sum of all inverse primes is divergent. But the alternating sum is convergent by the Leiniz criterion. To which known constant "a" does the sum converge? [imath]a = \frac{1}{2} - \frac{1}{3} +\frac{1}{5}-\frac{1}{7}+\frac{1}{11} -+ ...[/imath] | 2327896 | Does the alternating sum of prime reciprocals converge?
[imath]\sum_{n = 1}^\infty \frac{(-1)^n}{p_n}[/imath] where [imath]p_n[/imath] is the [imath]n[/imath]th prime. I have computed this to 10000 rather than infinity. My results suggest that convergence does happen but it's very slow. But I can't even be sure about the first few digits: [imath]-0.26959[/imath]? I have looked at the "questions that may already have your answer" and some of the "similar questions," but they involve somewhat different formulas. |
2875022 | Why is [imath]\binom{n}{2} = \sum_{i=0}^{n} i = \frac{n(n+1)}{2}[/imath]?
I can see this easily by definition. But will it have combinatorial meaning of [imath]\binom{n}{2} = \sum_{k=1} ^{n-1} (k)?[/imath] | 413987 | Is [imath]\sum_{i=1}^{n-1}i=\binom{n}{2}[/imath]?
How can I show that [imath] \sum_{i=1}^{n-1}i=\binom{n}{2}? [/imath] This is what I have tried, but I do not know if it is correct: Proof. Let [imath]n=2[/imath]. Then, [imath] \begin{align} \sum_{i=1}^{1}i&=1\text{, and}\\ \binom{2}{2}&=1. \end{align} [/imath] Hence, it holds. Assume that this is true for [imath]k[/imath]. Then, we show that it is also true for [imath]k+1[/imath]: [imath] \begin{align} \sum_{i=1}^{k}i=k+\sum_{i=1}^{k-1}i&=k+\binom{k}{2}\\ &=k+\frac{k!}{2(k-2)!}\\ &=k+\frac{k(k-1)}{2}\\ &=\frac{k(k+1)}{2}\\ &=\frac{(k+1)!}{2!(k-1)!}=\binom{k+1}{2}.\square \end{align} [/imath] Also, proofs by induction are confusing to write. Is there a standard "skeleton" for them I can use? |
2875094 | Alternating sum with ever increasing range of equal signs
Consider this special form of an alternating sum [imath]s=1-(\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})-(\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11})-(\frac{1}{12}+...+\frac{1}{16})+...[/imath] Notice that the ranges of equal signs increase linearly. I wonder if the sum is convergent or not, and if it is convergent what is the result. Edit by Rushba Mehta: The explicit form of the series is [imath]1 - \sum\limits_{n=2}^\infty (-1)^{\lfloor\frac{\sqrt{8n-15}-1}2\rfloor}\cdot\frac1n[/imath] My Edit I find that the sum can be written as [imath]s=1+\sum_{n=0}^\infty (-1)^{n+1}(H_{n(n+3)/2+2}-H_{(n^2+n+2)/2})[/imath] Where [imath]H_{k}=1+1/2+1/3+...+1/k[/imath] is the harmonic number. Convergence follows from the asymptoptic expansion of [imath]H_{k}[/imath] which leads to a series similar to that of log(2). | 2430060 | How to test for convergence for the harmonic series with irregular (binomial) sign changes?
How would you test for convergence with a series such as this using the alternating series test? [imath]1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}...-\frac{1}{10}+\frac{1}{11}...+\frac{1}{15}-....[/imath] I have attempted to find a summation equation for this by grouping in different ways in order to use the alternating series test but I can't find a way to group these in a logical way. Is there a grouping that would allow for analysis of this series using the alternating series test? |
2865814 | Rate of convergence for sum of random variables
I'm working on the following problem and I'm stuck. My plan was to use SLLN, but this gives me an expression for [imath]\sum_{i=1}^n \xi_{i}^p/n[/imath] and it's not clear how to convert it to having denominator [imath]n^{1/p}[/imath]. I'm wondering if maybe something involving borel cantelli is useful? source: Feb 1, 2012 | 1144729 | How to show that [imath]p[/imath]th moment being finite is equivalent to a limit existing
Let [imath]p \in (0,2)[/imath] and let [imath]\xi_n, n \geq 1[/imath], be iid random variables. Show that the following two conditions are equivalent: With probability one, the limit [imath] \lim_{n \rightarrow \infty} \frac{1}{n^{1/p}} \sum_{k=1}^n \xi_k[/imath] exists and is finite. [imath]\mathbb{E}\lvert \xi_i \rvert^p < \infty[/imath] AND either [imath]\mathbb{E} \xi = 0[/imath] or [imath]p \leq 1[/imath]. I have tried using Holder's inequality but didn't really get anywhere. I don't really have any idea how to approach the problem for either direction... |
2875483 | Evaluation of [imath]\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n^2}{n^3+1}[/imath]
I have accrossed this sum :[imath]\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n^2}{n^3+1}[/imath] in my textbook , However i have tried to evaluate it using Frobenius method and also the standard sum [imath]\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=\log 2[/imath] , But i didn't suceeded to get the same closed form which it assumed by Wolfram alpha here using hyperbolic function, then any way for evaluation ? | 576212 | FoxTrot Bill Amend Problems
So I found this on the Wolfram website today: So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, [imath]\LaTeX[/imath]ed: [imath] \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} [/imath] [imath] \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} [/imath] [imath] \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] [/imath] [imath] \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta[/imath] Ideas The degree of the numerator is [imath]\frac{3}{2}[/imath]. The degree of the denominator is [imath]\frac{3}{2}[/imath]. The expression is of an indeterminate form, namely [imath]\frac{\infty}{\infty}[/imath], so we use l'Hoptial's Rule. No ideas, really. Derivatives are always pretty easy, although this one is a bit bashy. Basically bash Product and Chain Rules, etc. Integration by Parts bash? The [imath]\mathrm{d}\theta[/imath] part is very trivial. I mean for the [imath]\rho[/imath] and [imath]\phi[/imath]. |
2875623 | Pell type equation about prime
Let [imath]p=4k+1[/imath] be a prime number such that [imath]p=a^2+b^2[/imath] , with [imath]a[/imath] an odd integer. Prove that the equation [imath]x^2-py^2=a[/imath] has at least a solution in [imath]\mathbb{Z}[/imath]. Only a little progress (maybe useless): for any prime divisor [imath]q[/imath] of [imath]a[/imath], we get [imath](\frac{q}{p})=1[/imath] through Guass' quadratic reciprocity law, which suggests [imath](\frac{a}{p})=1[/imath]. | 179277 | Pell type equation: [imath]x^2-py^2=a[/imath]
Let [imath]p=4k+1[/imath] be a prime number such that [imath]p=a^2+b^2[/imath], where [imath]a[/imath] is an odd integer.Prove that the equation [imath]x^2-py^2=a[/imath] has at least a solution in [imath]\mathbb{Z}[/imath]. |
2875861 | Is there a solution to the diophantine equation [imath]a^3 + b^3 = c^2[/imath] other than [imath]1^3 + 2^3 = 3^2[/imath] or scalings thereof?
My question concerns the diophantine equation [imath]a^3 + b^3 = c^2[/imath]. I know one solution: [imath]1^3 + 2^3 = 3^2[/imath]. But this is special in (at least) two ways: the [imath]a[/imath] and [imath]b[/imath] are not coprime; the solution is a special case of the identity: sum of the first [imath]n[/imath] cubes [imath]=[/imath] the square of the [imath]n[/imath]th triangular number. Are there other solutions? If not, is there an elementary proof of the fact? | 369846 | Integer solutions of [imath]x^3+y^3=z^2[/imath]
Is there any integer solution other than [imath](x,y,z)=(1,2,3)[/imath] for [imath]x^3+y^3=z^2[/imath]? |
2875862 | Colouring squares in a grid
It is proposed to colour the squares of a [imath]4 \times 4[/imath] board black and white, so that there are exactly two black squares and two white squares in each row or column. In how many ways can this be done? I've tried this finding out the number of ways column 1 and 2 can be done which is [imath]36 = 6 \cdot 6[/imath] as [imath]\binom{4}{2}[/imath] since the second column doesn't get in the way of the first. Then I hit a road block. Can anyone provide a hint? | 1712169 | Hints/ Solution in finding a formula [imath]\phi(n)[/imath] for a binary square matrix
I don't know if this is a famous problem but my professor assigned this for a research project and I am hitting a dead end( I don't know what to do next). Note: In [imath]\phi(n)[/imath] ; [imath]n=rows=columns[/imath] and this function throws out total combination of matrices that satisfies the following rule: Rule for this square binary matrix: Every Row and Column has two 1's. So, [imath]\phi(2)=1[/imath] \begin{bmatrix} 1 1 \\ 1 1 \\ \end{bmatrix} [imath]\phi(3)=6[/imath] I can't write out all the possible matrices for [imath]n=3[/imath] but this is one of them (btw i did this by hand): \begin{bmatrix} 0 1 1 \\ 1 0 1 \\ 1 1 0 \\ \end{bmatrix} According to a program made by my friend, he told me: [imath]\phi(4)=90[/imath] [imath]\phi(5)=2040[/imath] |
2875393 | An example of a field extension where every polynomial has a root but is not algebraically closed
Let [imath]F\subset E[/imath] be an algebraic extension of fields and suppose that every polynomial [imath]f(x) \in F[x][/imath] has a root in [imath]E[/imath]. Is it possible that [imath]E[/imath] is not algebraically closed? Note that such an example must necessarily be of prime characteristic since over char [imath]0[/imath], [imath]E[/imath] is necessarily algebraically closed. Proof: Let [imath]g(x) \in E[x][/imath] be a polynomial with a root [imath]\alpha[/imath] (after fixing an algebraic closure of [imath]E[/imath]). Let [imath]f(x)[/imath] be the minimal polynomial of [imath]\alpha[/imath] over [imath]F[/imath] and let [imath]L[/imath] be the splitting field of [imath]f[/imath]. By the primitive element theorem, [imath]L=F[\beta][/imath] and if [imath]h(x)[/imath] is the minimal polynomial for [imath]\beta[/imath], it has some root [imath]\gamma \in E[/imath]. Since [imath]F[\beta] = F[\gamma][/imath], this shows that [imath]L\subset E[/imath]. In particular, if we are looking for counterexamples, we would like [imath]L[/imath] to not be a primitive extension. | 721608 | If [imath]E/F[/imath] is algebraic and every [imath]f\in F[X][/imath] has a root in [imath]E[/imath], why is [imath]E[/imath] algebraically closed?
Suppose [imath]E/F[/imath] is an algebraic extension, where every polynomial over [imath]F[/imath] has a root in [imath]E[/imath]. It's not clear to me why [imath]E[/imath] is actually algebraically closed. I attempted the following, but I don't think it's correct: I let [imath]f[/imath] be an irreducible polynomial in [imath]E[X][/imath]. I let [imath]\alpha[/imath] be a root in some extension, so [imath]f=m_{\alpha,E}[/imath]. Since [imath]\alpha[/imath] is algebraic over [imath]E[/imath], it is also algebraic over [imath]F[/imath], let [imath]m_{\alpha,F}[/imath] be it's minimal polynomial. I now let [imath]K[/imath] be a splitting field of [imath]m_{\alpha,F}[/imath], which is a finite extension since each root has finite degree over [imath]F[/imath]. If [imath]m_{\alpha,F}[/imath] is separable, then [imath]K/F[/imath] is also separable, so as a finite, separable extension, we can write [imath]K=F(\beta)[/imath] for some primitive element [imath]\beta[/imath]. By assumption, [imath]m_{\alpha,F}[/imath] has a root in [imath]E[/imath], call it [imath]r[/imath]. Then we can embed [imath]F(\beta)[/imath] into [imath]r[/imath] by mapping [imath]\beta[/imath] to [imath]r[/imath]. It follows that [imath]m_{\alpha,F}[/imath] splits in [imath]E[/imath]. Since [imath]f\mid m_{\alpha,F}[/imath], we must also have the [imath]f[/imath] is split in [imath]E[/imath]. But what happens if [imath]m_{\alpha,F}[/imath] is not separable? In such case, [imath]F[/imath] must have characteristic [imath]p[/imath]. I know we can express [imath]m_{\alpha,F}=g(X^{p^k})[/imath] for some irreducible, separable polynomial [imath]g(X)\in F[X][/imath]. But I'm not sure what follows after that. NB: I say [imath]E[/imath] is algebraically closed if every nonconstant polynomial in [imath]E[X][/imath] has a root in [imath]E[/imath]. |
2876776 | How can I find the nth derivative of the function [imath]f(x)=e^{-1/x^2} \text{, if } x \neq 0 \text{ and } 0 \text{, if } x=0[/imath].
There is a function [imath]f(x):= \begin{cases} e^{-1/x^2}, & \text{if [/imath]x \neq 0[imath]} \\ 0, & \text{if [/imath]x=0[imath]} \end{cases} [/imath]. That function's [imath]n[/imath]-th derivative is [imath]0[/imath] : [imath]f^{(n)}(0) = 0[/imath]. How to derive that? | 332142 | Looking for help with a proof that n-th derivative of [imath]e^\frac{-1}{x^2} = 0[/imath] for [imath]x=0[/imath].
Given the function [imath] f(x) = \left\{\begin{array}{cc} e^{- \frac{1}{x^2}} & x \neq 0 \\ 0 & x = 0 \end{array}\right. [/imath] show that [imath]\forall_{n\in \Bbb N} f^{(n)}(0) = 0[/imath]. So I have to show that nth derivative is always equal to zero [imath]0[/imath]. Now I guess that it is about finding some dependencies between the previous and next differential but I have yet to notice one. Could you be so kind to help me with that? Thanks in advance! |
2876909 | Is a real-analytic function which vanishes on a set of positive measure identically zero?
Suppose [imath]n>1[/imath], and let [imath]U \subseteq \mathbb{R}^n[/imath] be an open connected set. Let [imath]f[/imath] be a real-analytic function on [imath]U[/imath]. Suppose that [imath]f=0[/imath] on a subset of [imath]U[/imath] of positive measure. Is it true that [imath]f[/imath] vanishes identically? I know that if [imath]f=0[/imath] on an open subset of [imath]U[/imath], then it is identically zero. (This is the identity theorem for real-analytic functions). | 1322858 | Zeros of analytic function of several real variables
Suppose [imath]f[/imath] is a non-constant analytic function on [imath]R^n[/imath], what can we say about its zeros set [imath]Z(f)=\{x\in R^n|f(x)=0\}[/imath]? Does [imath]Z(f)[/imath] have measure 0? Is it true that [imath]Z(f)[/imath] is union of manifolds of dimension less than [imath]n[/imath]? |
2877054 | Alternative methods of evaluating the Gaussian integral [imath]\int_{-\infty}^{\infty}e^{-z^2}dz[/imath]
A long while ago (at university) I learned the following (fairly standard) method for evaluating [imath]I:=\int_{-\infty}^{\infty}e^{-z^2}dz[/imath] by squaring [imath]I[/imath] then using polar coordinates: [imath] \begin{align} I^2 &= \int_{-\infty}^{\infty}e^{-z^2}dz \int_{-\infty}^{\infty}e^{-z^2}dz \\ &= 4\int_{0}^{\infty}e^{-x^2}dx \int_{0}^{\infty}e^{-y^2}dy \\ &= 4\int_{0}^{\infty}\int_{0}^{\infty}e^{-x^2}e^{-y^2}dxdy \\ &= 4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^2+y^2)}dxdy \\ &= 4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-r^2}rdrd\theta \\ &= 2\pi\int_{0}^{\infty}re^{-r^2}dr \\ &= \pi \left[ -e^{-r^2} \right]_{0}^{\infty} \\ &= \pi \end{align} \\ \\ \implies I=\sqrt{\pi} [/imath] I remember being amazed at the ingenuity of that first line: squaring [imath]I[/imath] in order to turn the single integral into a double integral. I am curious whether any other methods are known for evaluating [imath]I[/imath]. Especially methods that do not begin by squaring [imath]I[/imath]. I have not been able to do anything worthwhile with it myself. My research has yielded this method that does not require polar coordinates but still begins by squaring [imath]I[/imath]. I have also considered complex integration using a clever contour but have not managed to complete a method. Does anybody know of any other methods? | 846671 | Calculate the Gauss integral without squaring it first
We know that the integral [imath]I = \int_{-\infty}^{\infty} \mathrm{d}x e^{-x^2}[/imath] can be calculated by first squaring it and then treat it as a [imath]2-[/imath]dimensional integral in the plane and integrate it in polar coordinates. Are there any other ways to calculate it? I know that we may use the relation [imath]\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin{\pi x}},[/imath] but this, in effect, is still taking the square. Well, after I write down the above text, I figure that maybe there is no way to calculate it without squaring, since, after all, the result contains a square root, and it seems no elementary function can "naturally" produce a square root of [imath]\pi[/imath] starting from natural numbers (though I don't know how to describe this more concretely; you are also welcome to comment on this point). Nevertheless I still post this question in case there are some other ideas. EDIT: the Fourier transformation method at Computing the Gaussian integral with Fourier methods? appears kind of cheat to me, since the very proof of the Fourier transformation formula actually makes use of the value of the Gauss integral (at least in this wiki page http://en.wikipedia.org/wiki/Fourier_inversion_theorem#Proof). Thank you. |
1968420 | Is there a multinomial version of the DeMoivre-Laplace theorem?
The DeMoivre-Laplace theorem says that in the limit for large [imath]n[/imath], the distribution tends to a Gaussian. Can we say the same for multinomials? Can someone give me a reference as well? | 2859095 | Approximating a multinomial as [imath]p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right)[/imath]
Question Suppose we have a multinomial distribution with [imath]N[/imath] possible outcomes, with probabilities [imath]p_1,\ldots,p_N[/imath]. We sample this [imath]n[/imath] times, and denote the observed frequency of the [imath]i[/imath]th outcome as [imath]\xi_i[/imath]. In [1] the author claims that the distribution of the [imath]\xi_i[/imath] in the limit of large [imath]n[/imath] is: [imath]p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)[/imath] We can see immediately that this must be an approximation, as it assigns nonzero probabilities for [imath]\xi_1+\cdots+\xi_N>1[/imath]. However we can see that these have vanishing probability in the limit [imath]n\rightarrow\infty[/imath]. My question is how do we derive (1) from the multinomial distribution, and show that they match in the [imath]n\rightarrow\infty[/imath] limit? My thoughts My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean [imath]\mu_i=p_i[/imath] and covariance matrix [imath]\Sigma_{ij}=\delta_{ij}p_i-p_ip_j[/imath], so we would expect this in the large [imath]n[/imath] limit to be described by a multivariate Gaussian with mean [imath]\mu[/imath] and covariance [imath]\frac{1}{n}\Sigma[/imath]. However, things are complicated by the fact that the multinomial covariance is singular (since [imath]\xi_N[/imath] is determined by the other [imath]\xi_i[/imath]s), and so the multivariate Gaussian is not defined. To address this, we may try and consider only the first [imath]\xi_1,\ldots,\xi_{N-1}[/imath], which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution [imath]N=2[/imath]. The frequency [imath]\xi_1[/imath], this has mean [imath]p_1[/imath] and variance [imath]p_1(1-p_1)[/imath], so this would be described the the Gaussian: [imath]\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)[/imath] The expression (1) gives: [imath]\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)[/imath] If we substitute [imath]\xi_2\rightarrow 1-\xi_1[/imath], [imath]p_2\rightarrow 1-p_1[/imath] into (3), we can verify that this gives the same answer as (2). I have verified that this also works for [imath]N=4[/imath]. I'm sure that if I just bashed out the algebra for general [imath]N[/imath] we would get agreement between the central limit theorem and (1) when we restrict the latter to [imath]\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1[/imath]. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as [imath]n\rightarrow\infty[/imath] when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as [imath]n\rightarrow\infty[/imath], so I don't know if that argument is good enough. [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357. |
2877800 | Let [imath]p[/imath] be a prime. Compute [imath]1^k + 2^k + \ldots + (p-1)^k \pmod{p}[/imath].
An exercise at the end of a chapter on primitive roots asked me to compute [imath]1^k + 2^k + \ldots + (p-1)^k \pmod{p}[/imath] for any positive integer [imath]k[/imath] and any prime [imath]p[/imath]. Clearly, if [imath]k[/imath] is odd, the expression is congruent to [imath]0[/imath] modulo [imath]p[/imath]. But how can I even approach this problem if [imath]k[/imath] is even? And how do primitive roots apply? Hints would be greatly appreciated. | 268926 | What is [imath]1^k+2^k+\cdots+ (p-1)^k[/imath] modulo [imath]p[/imath]? (From Ireland and Rosen).
I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks Prove that [imath]1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}[/imath] if [imath]p-1\nmid k[/imath], and [imath]-1\pmod{p}[/imath] if [imath]p-1\mid k[/imath]. My work is leading where I didn't expect to go. Here I assume [imath]p[/imath] id odd. First, if [imath]p-1\mid k[/imath], then we can wrike [imath]k=(p-1)j[/imath] for some [imath]j[/imath]. But then [imath] 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} [/imath] since [imath]p-1[/imath] is even. Secondly, I suppose [imath]p-1\nmid k[/imath]. WLOG, I can assume [imath]0<k<p-1[/imath], Since if [imath]k=(p-1)j+r[/imath] with [imath]0<r<p-1[/imath], then [imath] a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. [/imath] Then if [imath]g[/imath] is a primitive root, [imath] 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. [/imath] Since [imath]g[/imath] is primitive, [imath]p\nmid 1-g^k[/imath], but [imath]p\mid 1-g^{k(p-1)}[/imath], so the above sum is again congruent to [imath]0\pmod{p}[/imath]. So in all cases, I get the sum is congruent to [imath]0[/imath] regardless. Have I messed up, or is there a typo? Thanks. |
2877756 | How to prove convergence of [imath]\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}[/imath]
Assume that [imath]a_n >0[/imath], [imath]n\in\mathbb{N}[/imath], and that [imath]\sum\limits_{n=1}^\infty a_n [/imath] converges. Show that for [imath]\alpha>1/2[/imath] the series [imath]\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}[/imath] converges as well I think we can use Abel's convergence test but I'm not sure if that is the way to go | 2878963 | if [imath]b_j>0[/imath] and [imath]\sum b_j[/imath] converges . prove that [imath]\sum \frac{\sqrt{b_j}}{j^\alpha}[/imath] converge for [imath]\alpha>1/2[/imath].
[imath]b_j>0[/imath] and [imath]\sum b_j[/imath] converges . prove that [imath]\sum \frac{\sqrt{b_j}}{j^\alpha}[/imath] converge for [imath]\alpha>1/2[/imath]. I know that [imath]\sum \sqrt{b_j}[/imath] and [imath]\sum \frac{1}{j^{2\alpha}}[/imath] converge. O just don't know how to proceed from there. |
2877578 | Prove that if [imath]A,B[/imath] are closed and disjoint in a metric space, then they are contained in disjoint neighborhoods
Yesterday, I asked the question: Prove that if [imath]A,B[/imath] are closed then, [imath] \exists\;U,V[/imath] open sets such that [imath]U\cap V= \emptyset[/imath]. Here is the correct question: prove that if [imath]A,B[/imath] are closed sets in a metric space such that [imath]A\cap B= \emptyset[/imath], there exists [imath]U,V[/imath] open sets such that [imath]A\subset U[/imath], [imath]B\subset V[/imath], and [imath]U\cap V= \emptyset[/imath]. I am thinking of going by contradiction, that is: [imath]\forall\; U,V[/imath] open sets such that [imath]A\subset U[/imath], [imath]B\subset V[/imath], and [imath]U\cap V\neq \emptyset[/imath]. Let [imath] U,V[/imath] open. Then, [imath]\exists\;r_1,r_2[/imath] such that [imath]B(x,r_1)\subset U[/imath] and [imath]B(x,r_2)\subset V.[/imath] I got stuck here! I'm thinking of using the properties of [imath]T^4-[/imath]space but I can't find out a proof! Any solution or reference related to metric spaces? | 1133049 | Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric.
Question: Let [imath](X, d)[/imath] be a metric space. Let [imath]A[/imath] and [imath]B[/imath] be disjoint subsets of [imath]X[/imath] that are closed in the topology induced by [imath]d[/imath]. Prove that there exist disjoint open sets [imath]U[/imath] and [imath]V[/imath] such that [imath]A\subset U[/imath] and [imath]B\subset V[/imath]. Is this proof correct? : By assumption, [imath]A\cap B=\emptyset[/imath], so there is no pair like [imath](x_A,x_B)[/imath] (i.e., [imath]x_A\in A[/imath] and [imath]x_B\in B[/imath]), such that [imath]x_A=x_B[/imath]. Because every metric space is Hausdorff, so for every pair [imath](x_A,x_B)[/imath] there are two disjoint open sets [imath]U_A\ni x_A[/imath] and [imath]V_B\ni x_B[/imath]; and thus union of all [imath]U_A[/imath]s (which is an open set containing A) are disjoint from union of all [imath]V_B[/imath]s (which is an open set containing A). Thanks a lot for any help. PS In special case of standard metric topology on [imath]\mathbb{R}[/imath], if [imath][a,b][/imath] and [imath][c,d][/imath] are disjoint (say [imath]c>b[/imath]), intuitively it is apparent that there is a 'space' between [imath]c[/imath] and [imath]b[/imath] so we can have two disjoint open sets, each of them containing [imath][a,b][/imath] or [imath][c,d][/imath]. |
2878126 | Check proof that if [imath]a_{n+1} = \frac{1}{k}(a_n+\frac{k}{a_n})[/imath] with [imath]k>1[/imath] and [imath]a_1>0[/imath], then [imath]a_n\to\sqrt {\frac{k}{k-1}}[/imath]
Let [imath]a_n[/imath] be a Sequence such that [imath]a_{n+1} = \frac{1}{k}(a_n+\frac{k}{a_n}), k>1, a_1>0[/imath], show that [imath]\{ a_n \}[/imath] converges to [imath] \sqrt {\frac{k}{k-1}}[/imath] My attempt case 1: [imath]a_{n+1}-a_n \leq 0 \\ \frac{1}{k}(a_n + \frac{k}{a_n})-a_n \leq 0 \\ a_n^2+k-ka_n^2 \leq 0 \\ a_n \geq \sqrt {\frac{k}{k-1}}[/imath] Here sequence is decreasing and bounded below by [imath]\sqrt {\frac{k}{k-1}}[/imath]. so [imath]a_n[/imath] is convergent Case 2: if [imath]a_{n+1} - a_n \geq 0,[/imath] i get [imath]a_k \leq \sqrt {\frac{k}{k-1}}[/imath] Here sequence is increasing and bounded above. SO given sequence is convergent. Let the convergence limit be l, then [imath]l=1/k (l+ k/l)[/imath] gives given limit. can you pls tell me is this correct approach? | 2419165 | Convergence of recursive sequence [imath]a_{n+1} =\frac{ 1}{k} \left(a_{n} + \frac{k}{a_{n}}\right)[/imath]
Let [imath] a_{n+1} = \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k>1, a_1>0 [/imath] The problem is to show that it converges. Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track. Please provide any hints. |
2668398 | Find minimum value of [imath]\sum \frac {\sqrt a}{\sqrt b +\sqrt c-\sqrt a}[/imath]
Find the minimum value of [imath]\sum \frac {\sqrt a}{\sqrt b +\sqrt c-\sqrt a}[/imath] Where [imath]a, b, c[/imath] represent the sides of a triangle My approach [imath]\sum \frac {\sqrt a}{\sqrt b +\sqrt c-\sqrt a}[/imath] [imath]=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}[/imath] By Titu's lemma [imath]=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}\ge \frac {9}{\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)-3}[/imath] Now by Cauchy Schwarz inequality [imath]\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)^2 \le \left (\sum \frac ab +\frac ba\right)(6)[/imath] Hence we have [imath]=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}\ge \frac {9}{\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)-3}\ge \frac {9}{\sqrt {\left (\sum \frac ab +\frac ba\right)(6)}-3}[/imath] Now how to proceed further. | 2587231 | Find minimum value of [imath]\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}[/imath]
If [imath]a,b,c[/imath] are sides of triangle Find Minimum value of [imath]S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}[/imath] My Try: Let [imath]P=\sqrt{a}+\sqrt{b}+\sqrt{c}[/imath] we have [imath]S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}[/imath] [imath]S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}[/imath] Let [imath]x=\frac{P}{\sqrt{a}}[/imath], [imath]y=\frac{P}{\sqrt{b}}[/imath],[imath]z=\frac{P}{\sqrt{c}}[/imath] Then we have [imath]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1[/imath] By [imath]AM \ge HM[/imath] [imath]\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}[/imath] Hence [imath]x+y+z \ge 9[/imath] Any way to proceed further? |
2878546 | How can I demonstrate this series are the same?
How can I know that they are equivalent? [imath]\sum_{n\in \mathbb Z} e^{-n^2\pi x} = \frac{1}{\sqrt{x}}\sum_{n\in \mathbb Z}e^{\frac{-n^2\pi}{x}} [/imath] I find this in wikipedia, in an article about the gamma function, where they used this property to find the analytic continuation of the zeta function. Thank you. | 764276 | How to prove the transformation formula for Jacobi classic theta function
How to prove the following transformation formula: [imath] \theta(x)=\frac{1}{\sqrt{x}} \theta\left(\frac{1}{x}\right), [/imath] where [imath]\theta[/imath] is the Jacobi theta function [imath]\theta(x)=\sum_{n\in \mathbb{Z}} e^{-\pi n^2 x}[/imath]? |
2878799 | Let [imath]T: \mathbb{R}^4 \to \mathbb{R}^4[/imath] be any linear transformation. Then how can I show that [imath]T[/imath] has a proper non zero invariant subspace.
Let [imath]T: \mathbb{R}^4 \to \mathbb{R}^4[/imath] be any linear transformation. Then how can I show that [imath]T[/imath] has a proper non zero invariant subspace. If [imath]T[/imath] has an eigen value then it is clear but if not then I can't solve it. Please help me. Thanks. | 241890 | Invariant subspaces of specific dimension
Could any one just tell me a sketch of the proof of the followings? [imath]1[/imath]. If [imath]n>1[/imath], does every linear transformation on [imath]\mathbb{R}^n[/imath] have an invariant subspaces of dimension [imath]2[/imath]? [imath]2[/imath] Is there a vector space [imath]V[/imath] and a linear transformation [imath]T[/imath] on [imath]V[/imath] such that it has exactly [imath]3[/imath] invariant subspace? Thank you! |
2878823 | [imath]Y\sim\operatorname{Poisson}(\lambda_2 = 15)[/imath]. Let [imath]Z = X + Y[/imath] . Compute [imath]\operatorname{Corr}(X, Z)[/imath].
Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables such that [imath]X\sim \operatorname{Poisson}(\lambda_1 = 5)[/imath] and [imath]Y\sim \operatorname{Poisson}(\lambda_2 = 15)[/imath]. Let [imath]Z = X + Y[/imath] . Compute [imath]\operatorname{Corr}(X, Z)[/imath]. Answer: [imath]\operatorname{Var}(X) = 5[/imath] [imath]\operatorname{Var}(Y) = 15[/imath] [imath]\operatorname{Cov}(Z) = \operatorname{Cov}(X, X+Y) = \operatorname{Cov}(X, X) = \operatorname{Var}(X) = 5[/imath] [imath]\operatorname{Var}(Z) = \operatorname{Var}(X + Y) = V(X) + V(Y) = 5 + 15 = 20[/imath] [imath]\operatorname{Corr}(X, Z) = \frac{\operatorname{Cov}(X, Z)}{\sqrt{\operatorname{Var}(X)} \sqrt{\operatorname{Var}(Z)}} = \frac{5}{\sqrt{5 \cdot 20}} = 0.5[/imath] I'm just wondering if this statement is true for all covariance values or just if independent. [imath]\operatorname{Cov}(Z) = \operatorname{Cov}(X, X+Y) = \operatorname{Cov}(X, X) = \operatorname{Var}(X) = 5[/imath] Couldn't find it on wiki | 2875777 | [imath]X\sim\mathrm{Poisson}(\lambda_1 = 5)[/imath] and [imath]Y\sim\mathrm{Poisson}(\lambda_2 = 15)[/imath]. Let [imath]Z = X + Y[/imath] . Compute [imath]\mathrm{Corr}(X, Z)[/imath].
Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables such that [imath]X\sim\mathrm{Poisson}(\lambda_1 = 5)[/imath] and [imath]Y\sim\mathrm{Poisson}(\lambda_2 = 15)[/imath]. Let [imath]Z = X + Y[/imath] . Compute [imath]\mathrm{Corr}(X, Z)[/imath]. Answer: [imath]\mathrm{Var}(X) = \lambda_1 = 5[/imath] [imath]\mathrm{Var}(Y) = \lambda_2 = 15[/imath] [imath]\mathrm{Cov}(Z) = \mathrm{Cov}(X, X+Y) = \mathrm{Cov}(X, X) = V(X) = 5[/imath] [imath]\mathrm{Var}(Z) = \mathrm{Var}(X + Y) = V(X) + V(Y) = 5 + 15 = 20[/imath] [imath]\mathrm{Corr}(X, Z) = \dfrac{\mathrm{Cov}(X, Z)}{\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Z)}} = \dfrac{5}{\sqrt{5 \cdot 20}} = 0.5[/imath] I get most of the solution but I'm confused on how they got [imath]\mathrm{Cov}(Z)[/imath] [imath]\mathrm{Cov}(Z) = \mathrm{Cov}(X, X + Y) = \mathrm{Cov}(X + Y, X) = \mathrm{Cov}(X, X) + \mathrm{Cov}(X, Y) \neq \mathrm{Cov}(X, X)[/imath] Could someone explain how they got [imath]\mathrm{Cov}(Z)[/imath]? |
2879010 | Sum of weighted binomial coefficients
I am struggling with computing the following sums: [imath]\sum_{k=1}^{n}k\binom{n}{k}=\binom{n}{1}+2\binom{n}{2}+...+n\binom{n}{n}[/imath] and [imath]\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\dots+\frac{1}{n+1}\binom{n}{n}[/imath] At first, I tried just rewriting the general terms in a form where the Binomial Theorem could be applied, but could not do so. ([imath]k[/imath] should be in the exponent of something with [imath]n-k[/imath] in the exponent of something else.) Then, for the first sum, I re-wrote it in the following manner: \begin{align} S &=\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}+\binom{n}{1}\\ &+\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}\\ &+\binom{n}{n}+\dots+\binom{n}{3}\\ &\quad\vdots\\ &+\binom{n}{n} \end{align} Noting that: [imath]\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\dots+\binom{n}{n}=\sum_{k=0}^{n}\binom{n}{k}1^k1^{n-k}=(1+1)^n=2^n,[/imath] I can rewrite the first line in [imath]S[/imath] as [imath]2^n-1[/imath] (the [imath]-1[/imath] is there because I'm over-counting the [imath]\binom{n}{0}=1[/imath]). Then I tried to say that if all the "blanks" were filled (if all the lines were like the first line), since there are [imath]n[/imath] lines, then I have [imath]n(2^n-1)[/imath]. From this point, we should have: [imath]S+\text{blanks}=n(2^n-1)\Longrightarrow S=n(2^n-1)-\text{blanks}[/imath] However, the problem of finding the value of "blanks" seems as hard as the original problem and I can't get to the right answer of [imath]n2^{n-1}[/imath] (according to WA). In one of my attempts, I ended up with a geometric series but that still didn't work. I have not spent much time on the second sum as I feel I should be able to do the first one before even tackling the second one. Am I going in the right direction with this? Is there a more intelligent way of working this out? Thanks in advance for any help/hints. | 2180181 | Find the value of [imath]\sum_{k=1}^{n}k\binom{n}{k}[/imath]?
Find the value of [imath]\sum_{k=1}^{n}k\binom{n}{k}[/imath] ? I know that [imath]\sum_{k=0}^{n}\binom{n}{k}= 2^{n}[/imath] and so, [imath]\sum_{k=1}^{n}\binom{n}{k}= 2^{n}-1[/imath] but how to deal with [imath]k[/imath] ? |
2870009 | Help Solving Integration [imath]\int_{-\infty}^{\infty} {(\text{Standard Normal} \times \text{Sigmoid}) }[/imath]
Does anyone know how to analytically solve the below expression? [imath]\int_{-\infty}^\infty p(x) \times \sigma(ax + b) \, dx [/imath] where [imath] p(x) = N(0, 1) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}[/imath] and [imath]\sigma(z) = \frac{1}{1+e^{-z}}[/imath]. I know that the integration of standard Gaussian is [imath]\pi[/imath]. Moreover, the integration of sigmoid alone is, [imath]\int \sigma(x) \, df = \int \frac{e^x}{1+e^x} \, dx = x + c[/imath] But I cant figure out how to integrate the first equation), when they are multiplied together. Can we use integration by part and solve above equation? Any help will be appreciated. Thanks Edit 1: Sorry about the complicated notations. If we simplify the above expression, then we get following equation. [imath] \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-\frac{x^2}{2}}}{1+e^{-x}} dx [/imath] | 2874576 | How to compute [imath] \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx [/imath] when [imath]b \ne 0[/imath]
I'm trying to solve this expression when [imath]b \ne 0[/imath]: [imath] \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx [/imath] Using the answers provided to this question we can solve [imath]\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-x}}\,\mathrm dx=\frac12[/imath] However, when we change the power of exponential term in the denominator to [imath]-(ax+b)[/imath] from [imath]x[/imath], the expression can not be solved using the same methods used in the post. I tried solving by substituting [imath]z=ax+b[/imath] and it also did not work. |
1343320 | Recursive sequence of square roots of previous elements
Bruckner, Bruckner, Thompson - Elementary Real Analysis [imath]a_1 = 1[/imath] and [imath]a_{n+1} = \sqrt{a_1+ a_2 + .. + a_n}[/imath] Show that [imath]\lim_{n \to \infty}\ \frac{a_n}{n} = \frac12[/imath] I cannot untangle the square root of the sum, to show that it 'converges' to [imath]\frac{n}2[/imath]. Any help much appreciated. | 2069879 | Let [imath]a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}[/imath] .Prove that [imath] \lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}[/imath]
A sequence [imath]a_n[/imath] is defined as [imath]a_1=1[/imath] and [imath]a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}[/imath] .Prove that [imath] \lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}[/imath] I have no idea how to approach this. But I have a feeling that Cesaro's lemma may come in handy |
2857901 | Find the derived group of [imath]GL(n,\mathbb{R})[/imath]
Find the derived group of [imath]GL(n,\mathbb{R})[/imath] I believe that the derived group is [imath]SL(n,\mathbb{R})[/imath], as it is easy to see that [imath]GL(n,\mathbb{R})' \subset SL(n,\mathbb{R})[/imath], since for any [imath]A,B \in GL(n,\mathbb{R})[/imath], we get that [imath]det(ABA^{-1}B^{-1})=1[/imath], and therefore [imath]ABA^{-1}B^{-1} \in SL(n, \mathbb{R})[/imath]. However, I'm not sure how to prove that other side - [imath] SL(n,\mathbb{R}) \subset GL(n,\mathbb{R})'[/imath]? | 2311331 | [imath]SL_n(\mathbb R)[/imath] is a subgroup of kernel of [imath]f\in[/imath] Hom[imath](GL_n(\mathbb R), A)[/imath], [imath]A[/imath] is an abelian group.
Try to prove the following proposition. [imath]SL_n(\mathbb R)[/imath] is a subgroup of kernel of [imath]f\in[/imath] Hom([imath]GL_n(\mathbb R), A)[/imath], where [imath]A[/imath] is an abelian group. My first attempt is to show that [imath]SL_n(\mathbb R)[/imath] is in form of [imath]ABA^{-1}B^{-1}[/imath], then it's clear that [imath]ABA^{-1}B^{-1}[/imath] is in [imath]ker(f)[/imath]. However, there exists an elementary row reduction matrix, i.e. [imath]E_{ij}(m)[/imath] adding m times [imath]j[/imath]-th row to [imath]i[/imath]-th row. Then [imath]E_{ij}(m)\in SL_n(\mathbb R)[/imath]. Now, it is possible that [imath]E_{ij}(m)[/imath] is not in [imath]ker(f)[/imath]. Then I try to find any relation between [imath]\mathbb R^{\times}[/imath] and [imath]A[/imath], but, except knowing they are both abelian groups, I cannot figure out what else connects those two group. Could you give me some hints? |
2879524 | Prove limit of a sum is [imath]0[/imath]
Prove that, if [imath]a_k \geq 0[/imath] for large [imath]k[/imath], and that [imath]\sum_{k=1}^\infty \frac{a_k}{k}[/imath] converges. Prove that [imath]\lim_{j \rightarrow \infty} \sum_{k=1}^\infty \frac{a_k}{j+k} = 0[/imath] I believe I cant write [imath]\sum_{k=1}^\infty \frac{a_k}{j+k} \leq \sum_{k=1}^\infty \frac{a_k}{k}.[/imath] However, taking limit does not give me the answer. | 1135077 | Prove that $\lim_{j \to\infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = 0$
Exercise: Suppose that [imath]a_k \geq 0[/imath] for [imath]k[/imath] large and that [imath]$\sum_{k = 1}^{\infty} \frac{a_k}{k}$[/imath] converges. Prove that [imath]$$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = 0$$[/imath] Attempt in proof: Suppose [imath]a_k \geq 0[/imath] for any large [imath]k[/imath] and that [imath]$\sum_{k = 1}^{\infty} \frac{a_k}{k}$[/imath] converges. . Then give [imath]\epsilon >0[/imath] there is [imath]N \in N[/imath] such that [imath]$\left|\sum_{k = n}^{\infty} \frac{a_k}{k}\right| < \epsilon$[/imath]. Let [imath]s_n = \frac{a_n}{j+k}[/imath] denote the partial sums. Then [imath]$\sum_{k = 1}^{\infty} \frac{a_k}{j+k} $[/imath] will converge to zero if and only if its partial sum converges to zero as n approaches infinity. Taking the limit of [imath]$$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = \lim_{j \to \infty}\frac{a_1}{j + k} + \cdots+ \frac{a_n}{j + k} + \cdots = \lim_{j \to \infty}\frac{a_1/j}{1 + k/j} + \cdots+ \frac{a_n/j}{1 + k/j} +\cdots $$[/imath] Can someone please help me finish? I don't know if this a right way. Any help/hint/suggestion will be really appreciate it. Thank you in advance. |
2878848 | How to evaluate non-integer exponent?
I wonder how can calculators or Wolfram evaluate such expressions as [imath]2^{1.35}[/imath] or [imath]3^\pi[/imath]. Do they use Taylor series for [imath]2^x[/imath] and [imath]3^x[/imath] or do they employ other means? | 462443 | Calculating Non-Integer Exponent
I just wanted to directly calculate the value of the number [imath]2^{3.1}[/imath] as I was wondering how a computer would do it. I've done some higher mathematics, but I'm very unsure of what I would do to solve this algorithmically, without a simple trial and error. I noted that [imath] 2^{3.1} = 2^{3} \times 2^{0.1} [/imath] So I've simplified the problem to an "integer part" (which is easy enough) : [imath]2^3 = 2\times 2\times 2[/imath], but I'm still very confused about the "decimal part". I also know that : [imath] 2^{0.1} = e^{0.1\log{2}} [/imath] But that still presents a similar problem, because you'd need to calculate another non-integer exponent for the natural exponential. As far as I can see, the only way to do this is to let: [imath]2^{0.1}=a [/imath] And then trial and error with some brute force approach (adjusting my guess for a as I go). Even Newton's method didn't seem to give me anything meaningful. Does anybody have any idea how we could calculate this with some working algorithm? |
2875162 | Prove that: [imath]\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}[/imath]
Given three positive numbers a,b,c satisfying [imath]a^2+b^2+c^2=1[/imath] Prove that: [imath]\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}[/imath] The things I have done so far: [imath]\sum \limits_{cyc}\frac{bc}{a^2+1}=\sum \limits_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq \sum \limits_{cyc}\frac{bc}{2ab+2ac}[/imath] [imath]=\sum \limits_{cyc}\frac{bc}{2a(b+c)}\leq \frac{1}{4}\sum \limits_{cyc}\frac{(b+c)^2}{2a(b+c)}=\frac{1}{4}\sum \limits_{cyc}\frac{b+c}{2a}[/imath] [imath]=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)}{abc}=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)+3abc}{abc}-\frac{3}{8}[/imath] [imath]=\frac{1}{8}.\frac{(a+b+c)(ab+bc+ca)}{abc}-\frac{3}{8}[/imath] [imath]\leq \frac{1}{8}.\frac{\sqrt{3(a^2+b^2+c^2)}(a^2+b^2+c^2)}{abc}-\frac{3}{8}[/imath] [imath]=\frac{\sqrt{3}}{8abc}-\frac{3}{8}[/imath] I don't know what to do anymore. | 912905 | Prove [imath]\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}[/imath] if [imath]a^2+b^2+c^2=1[/imath]
Ff [imath]a,b,c[/imath] are positive real numbers that [imath]a^2+b^2+c^2=1[/imath] ,Prove: [imath]\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}[/imath] Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them. Things I have done: I tried to change LHS to something more easy to work but I was not successful. For example [imath]\frac{ab}{1+c^2}=\frac{1}{2}\left(\frac{a^2+b^2+2ab}{1+c^2}-\frac{a^2+b^2}{1+c^2}\right)[/imath] that was not useful. Any hint for starting step is appreciated. |
2879679 | Convergence of [imath]\sum\ln(n\sin\frac{1}{n})[/imath]
How to prove that [imath]\sum\ln(n\sin\frac{1}{n})[/imath] converges rigorously? The idea is to expand [imath]\sin(t)[/imath] and [imath]\ln(1+t)[/imath]. But I'm not sure how to do it rigorously. We have [imath]n\sin(1/n)=n(1/n+O(1/n^3))=1+nO(1/n^3)[/imath] as [imath]n\to \infty[/imath]. Then I need to substitute it into [imath]\ln (1+t)=t-t^2/2+O(t^3)[/imath]. (Or should I use the little o in one of both cases?) This involves taking powers of [imath]nO(1/n^3)[/imath] (or of the little o) and multiplying it by functions, and I'm not very familiar how the o's behave under such operations (and not aware of any source to find that out). So how to do the prove in a rigorous way? | 1190404 | Does [imath]\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))[/imath] converge?
I must determine whether the following series converges: [imath]\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)[/imath] I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process: For [imath]n[/imath] large, we have that [imath]\lim_{n\to \infty}n\sin\frac1n=1[/imath], and so, [imath]\ln(1)=0[/imath]. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated. |
2879027 | Dividing a Set into Subsets & The Pigeonhole Principle
I need help with a question. I don't know how to start, but I do know that it focuses on the pigeonhole principle. I have no problem understanding the principle, but I want to know how to find the pigeons and the pigeonholes. A set [imath]M = \{1, 2, ..., 100\}[/imath] is divided into seven subsets with no number in two or more subsets. Show that at least one subset either contains four numbers [imath]a, b, c[/imath] and [imath]d[/imath] such that [imath]a + b = c + d[/imath] or three numbers [imath]p, q[/imath] and [imath]r[/imath] such that [imath]p + q = 2r[/imath]. | 2865068 | Any partition of [imath]\{1,2,\ldots,100\}[/imath] into seven subsets yields a subset with numbers [imath]a,b,c,d[/imath] such that [imath]a+b=c+d[/imath].
A set [imath]M = \{1,2,\ldots,100\}[/imath] is divided into seven subsets with no number in [imath]2[/imath] or more subsets. How do you prove that one subset either contains four numbers [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath] such that [imath]a + b = c + d[/imath] or three numbers [imath]p[/imath], [imath]q[/imath], and [imath]r[/imath] such that [imath]p + q = 2r\,?[/imath] I am having some issues with this question whether I don't fully understand the question or my arithmetic is wrong. Grateful for any help. |
2875519 | non-linear ordinary differential equation (Jacobi's?)
I am working on some firm optimization problem and got to the following differential equation. [imath]\frac{dy}{dx} = \frac{(y-x)(1-y)}{(c-x)(1-2y+x)}[/imath] with [imath]x,y \in [0, 1][/imath] and c is a constant with [imath]0<c\leq1[/imath]. I know the obvious solution [imath]y=1[/imath] but I was wondering if there is any other solutions. Also I am looking for a generic solution that depends on any [imath]c[/imath] within the specified interval. Thanks. | 2873565 | Non-linear differential equation example
I'm dealing with the following differential equation [imath]\frac{dy}{dx} = \frac{(y-x)(a-by)}{(1-\frac{b}{a}x)(a-2by+bx)}[/imath] with [imath]x,y \in [0, \frac{a}{b}][/imath] Can anyone help me with the non trivial solution? Thanks. |
2874548 | Quadratic equation sum from Russian book
Solve [imath]\sqrt{5-x}=5-x^2[/imath] for [imath]x[/imath]. This is what I have done so far. Method 1: \begin{align} \sqrt{5 - x} & = 5 - x^2 \\ 5 - x & = (5 - x^2)^2 \\ 5 - x & = 25 - 10x^2 + x^4 \\ 0 & = x^4 - 10x^2 + x + 20 \end{align} Method 2: \begin{align} \sqrt{5 - x} & = (\sqrt5)^2 - (x)^2 \\ \sqrt{5 - x} & = (\sqrt5 - x)(\sqrt5 + x) \\ 1 & = \sqrt5 + x\\ x & = 1 - \sqrt5 \end{align} | 1393046 | real values of [imath]x[/imath] in [imath]\sqrt{5-x} = 5-x^2[/imath].
Calculate the real solutions [imath]x\in\mathbb{R}[/imath] to [imath] \sqrt{5-x} = 5-x^2 [/imath] My Attempt: We know that [imath]5-x\geq 0[/imath] and thus [imath]x\leq 5[/imath] and [imath] \begin{align} 5-x^2&\geq 0\\ x^2-\left(\sqrt{5}\right)^2&\leq 0 \end{align} [/imath] which implies that [imath]-\sqrt{5}\leq x \leq \sqrt{5}[/imath]. Now let [imath]y=\sqrt{5-x}[/imath]. Then [imath] \tag1 y^2=5-x [/imath] and the equation converts into [imath] \begin{align} y &= 5-x^2\\ x^2 &= 5-y\\ y^2-x^2 &= 5-x-(5-y)\\ y^2-x^2 &= y-x\\ (y-x)(y+x)-(y-x) &= 0\\ (y-x)(y+x-1) &= 0 \end{align} [/imath] So either [imath]y=x[/imath] or [imath]x+y=1[/imath]. Case 1 ([imath]y=x[/imath]): We can plug this into [imath](1)[/imath] to get [imath] \begin{align} y^2 &= 5-x\\ x^2 &= 5-x\\ x^2+x-5 &= 0\\ x &= \frac{-1\pm \sqrt{1+20}}{2} \end{align} [/imath] Since [imath]-\sqrt{5}\leq x\leq \sqrt{5}[/imath], the only solution is [imath] x = \frac{-1+\sqrt{21}}{2} [/imath] Case 2 ([imath]y=1-x[/imath]): We can plug this into [imath](1)[/imath] to get [imath] \begin{align} y^2 &= 5-x\\ (1-x)^2 &= 5-x\\ 1+x^2-2x &= 5-x\\ x^2-x-4 &= 0\\ x &= \frac{1\pm\sqrt{17}}{2} \end{align} [/imath] Since [imath]-\sqrt{5}\leq x\leq \sqrt{5}[/imath], the only solution is [imath] x = \frac{1-\sqrt{17}}{2} [/imath] So final solution is [imath] x \in \left\{\displaystyle \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\} [/imath] Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola? |
2879633 | Are there functions satisfying [imath]\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx[/imath]
Are there functions satisfying [imath]\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx[/imath] I came up with one which is [imath]f(x)=e^{x \sec ^2 \alpha}[/imath] and [imath]g(x)=e^{x \csc^2 \alpha}[/imath] [imath]\int f(x)g(x)dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}[/imath] Also [imath]\int e^{x\sec^2 \alpha}dx \times \int e^{x \csc^2 \alpha} dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha \csc ^2 \alpha}=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}[/imath] | 804163 | [imath]f,g[/imath] such that [imath]\int fg = \int f \int g[/imath]
Suppose [imath]f,g[/imath] are real valued on [imath]\mathbb{R}[/imath] (and no further restrictions apart from the obvious requirement that the integrals exist), then when does [imath]\displaystyle\int f(x)g(x)\,dx = \int f(x) \, dx \int g(x) \, dx[/imath]? (all integrals indefinite) This question was posed on another site, and I was wondering whether there are any large classes of functions [imath]f,g[/imath] that work. Aside from the trivial solution, things like [imath]f(x) = e^{nx}, g(x) = e^{\frac{n}{n-1}x}[/imath] [imath](n\not= 1)[/imath] work by inspection. I've tried re-casting the problem as: [imath]\int F^2 - \left(\int F\right)^2 = \int G^2 - \left(\int G\right)^2[/imath] With [imath]F=f+g, G=f-g[/imath], to introduce some symmetry. This gives rise to similarities with the variance formulae but nothing more than that. Using power series [imath]F(x) = \displaystyle\sum_{n=0}^{\infty} a_nx^n, G(x) = \sum_{n=0}^{\infty} b_nx^n[/imath], and the cauchy product, this equation boils down to: [imath]\sum_{m=0}^n \frac{(n-m)(a_ma_{n-m} - b_mb_{n-m})}{(n+1)(n-m+1)}= 0\qquad \forall n\geq 0[/imath] But again, progress is limited. |
2879887 | Solve [imath]i^{i^{i^\ldots}}[/imath]
How to find [imath]i^{i^{i^\ldots}} \quad :\quad i=\sqrt{-1}[/imath] I'm able to find the solution for the finite powers using [imath]i=e^{i(2k\pi+\frac{\pi}{2})}\quad:\quad k\in\mathbb{Z}[/imath] [imath]i^{i}=e^{-(2k\pi+\frac{\pi}{2})}[/imath] [imath]i^{i^{i}}=e^{-i(2\pi k+\frac{\pi}{2})}=-i[/imath] [imath]i^{i^{i^i}}=e^{(2\pi k+\frac{\pi}{2}) }[/imath] [imath]\text{and so on}[/imath] but what should be the approach to solve for infitie powers[imath]\space[/imath]? | 1976018 | What is the value of [imath]i^{i^{i^\ldots}}[/imath]?
What is the value of [imath]i^{i^{i^\ldots}}[/imath]? My effort is the following: If [imath]z, \alpha \in \mathbb{C}[/imath] with [imath]z \neq 0[/imath] then we can write [imath]z^{\alpha}=e^{\alpha \log z} = e^{\alpha [ \log |z|+i \text{ Arg z} + 2 \pi i m]}[/imath] where [imath] m=0, \pm 1, \pm 2, \ldots[/imath] and Arg z is the principle argument of the complex number [imath]z[/imath]. But in above expression we don't have the finite no of exponents, so I am not sure how should I use this formula to the solve above problem. Thanks and Regards! |
2825960 | If [imath]|f'_n(x)|\leq \frac{1}{\sqrt x}[/imath] and [imath]\int_0^1f_n(x)dx =0[/imath], then exists a subsequence which converges uniformly
Let [imath]f_n:[0,1]\rightarrow \mathbb{R}[/imath] be continuous functions such that: [imath]|f'_n(x)|\leq \frac{1}{\sqrt x}[/imath] and [imath]\int_0^1f_n(x)dx =0[/imath] for every [imath]n\in N[/imath]. Prove that exists a subsequence of [imath](f_n)[/imath] which converges uniformly. I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence [imath](f_n)[/imath] is uniformly bounded and that it is equicontinuous. What I have noted/done so far: Since [imath]|f'_n(x)|\leq \frac{1}{\sqrt x}[/imath], if [imath]x,y\in [0,1],x\neq y[/imath], then [imath]|f_n(y)-f_n(x)|=|\int_x^yf'_n(t)dt |\leq\int_x^y|f'_n(t)|dt\leq\int_x^y\frac{1}{\sqrt t}dt\leq\int_0^1\frac{1}{\sqrt t}dt = K[/imath] but I can't use this to solve the problem. Any help? Thanks in advance | 270556 | Prove [imath]f_n[/imath] has a uniformly convergent subsequence on [imath][0,1][/imath].
Let [imath]f_n[/imath]: [imath][0,1]\to\mathbb{R}[/imath] be differentiable satisfying both: [imath]\begin{align} (1)& \int_0^1f_n(x)dx=0,\quad \forall n\in\mathbb{N},\\ (2)& |f'_n(x)|\le \frac{1}{\sqrt{x}},\quad\forall x\in(0,1]. \end{align}[/imath] Prove that [imath]f_n[/imath] has a uniformly convergent subsequence on [imath][0,1][/imath]. I want to apply Arzela-Ascoli Theorem, so I try to show that [imath]f_n[/imath] is uniformly bounded and equicontinuous. But I only get that [imath]f_n[/imath] is pointwise bounded. And I have no idea how to use the first condition. |
2879791 | Rearrangement of Sequence [imath]\left\{x_n\right\}[/imath]
We say that a sequence [imath]\left\{y_n\right\}[/imath] is a rearrangement of a sequence [imath]\left\{x_n\right\}[/imath] if there is a 1-1 correspondence [imath]f : \mathbb{N} \rightarrow \mathbb{N}[/imath] such that [imath]\forall[/imath] [imath]n \in \mathbb{N}[/imath], [imath]y_n = x_{f(n)}[/imath]. Suppose [imath]\left\{y_n\right\}[/imath] is a rearrangement of [imath]\left\{x_n\right\}[/imath]. Prove that [imath]\left\{y_n\right\} \rightarrow L[/imath] iff [imath]\left\{x_n\right\} \rightarrow L[/imath]. 1) How do I prove this ? | 2129797 | Proof that injective rearrangements of a convergent sequence preserves its limit
Is this statement true? i.e. If [imath]f(n)[/imath] is injective and [imath]p_n \rightarrow L[/imath], does [imath]p_{f(n)} \rightarrow L[/imath]?? (I realize that it's technically only a rearrangement if [imath]f(n)[/imath] is bijective.) My attempt at a proof: Since [imath]p_n \rightarrow L[/imath], we have that, for all [imath]n[/imath] except [imath]n \le N[/imath], [imath]d(p_n, L) < \epsilon[/imath]. Let [imath]S = \{n \ | \ f(n)\le N\}[/imath], let [imath]n_0[/imath] be the largest [imath]n \in S[/imath], we know there is such a largest [imath]n[/imath] because [imath]f(n)[/imath] is injective. Now we have that [imath]\forall \ n > n_0 \ f(n) > N[/imath] which implies that [imath]p_{f(n)} \rightarrow L[/imath], as required. Is this correct? Does my proof work? |
2880252 | Find Polynomials with Integer Coefficients with Particular Roots
Is there any simple way to find a polynomial with integer coefficients so that ([imath]x=\sqrt{2} +\sqrt{3}[/imath]) is one of its roots? I know one way is to get rid of all the square roots in the equation to be left with rational numbers, but is there any other simple way? | 1494018 | Minimal polynomial of [imath]\sqrt{2}+\sqrt{3}[/imath] over [imath]\mathbb Q[/imath]
I had an example in the book given as follows: Find the minimal polynomial of [imath]\sqrt{2}+\sqrt{3}[/imath] over [imath]\mathbb Q[/imath] . Solution: [imath]~~~~~[/imath][imath](\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6[/imath] [imath](\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6[/imath] Then [imath](\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.[/imath] Thus [imath]a=\sqrt{2}+\sqrt{3}[/imath] satisfies [imath]f(x)=x^4-10x^2+1[/imath] over [imath]\mathbb Q[/imath]. Let [imath]p(x)[/imath] be the minimal polynomial of [imath]\sqrt{2}+\sqrt{3}[/imath] over [imath]\mathbb Q[/imath].Then [imath](\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})[/imath] are also roots of [imath]p(x).[/imath] So degree of [imath]p(x)[/imath] is atleast [imath]4[/imath]. But [imath]f(a)=0 [/imath] and [imath]f(x) \in \mathbb Q[x] \implies p(x)[/imath] divides [imath]f(x)[/imath] . So [imath]f(x)[/imath] is minimal polynomial of [imath]\sqrt{2}+\sqrt{3}.[/imath] But I can't get the step how are [imath](\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})[/imath] also roots of [imath]p(x).[/imath] Kindly help with this. |
2879251 | How to verify large Mersenne Primes
As of December 2017, the largest known prime number was the Mersenne prime [imath]2^{77232917} – 1[/imath]. For such a large Mersenne prime, what are the techniques available for one to verify that it is in fact a prime? Is it simply a matter of brute force checking, or are there other more elegant techniques available? | 483298 | How do you prove a number is prime?
I am a software engineer but try to keep up with maths as I really enjoyed the subject at school. I just saw a great TED talk: Why I fell in love with monster prime numbers The talk states that the current largest known prime is [imath]2^{57885161}-1[/imath] This got me thinking - how does one actually prove whether a number this large is prime or not? Is it just done by brute force via a computer algorithm or are there any well known mathematical techniques for doing so? I would be really grateful for any pointers as I am eager to learn about these techniques (if they exist). |
2683423 | Find [imath]\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx[/imath]
How do you evaluate this integral? [imath]\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx[/imath] where [imath]f(x)=1[/imath] and [imath]g(x)=\sqrt{R^2-x^2}[/imath]. Wolfram tells me I exceeded my computational limit. Mathematica gives me a long answer which is very difficult to read (for me). EDIT: I was told to consider instead [imath]x=f(y)[/imath], so: [imath]\int_0^{1}\sqrt{R^2-y^2}\,\mathrm dy[/imath] | 2683356 | Area of intersection using integration
Suppose, for the sake of illustration, that [imath]AF=2[/imath], where [imath]F\mathbb{'s}[/imath] coordinates are [imath](2,0)[/imath] and [imath]A[/imath] is the point of origin; [imath]E=(0,1)[/imath] and circle [imath]A[/imath] has [imath]R=1.5[/imath] How do I solve the area of the intersection of the circle and rectangle by integration? If we let [imath]f(x)=1[/imath], then the area of the rectangle is defined by: [imath]\int_0^Ff(x)dx[/imath] If we let [imath]g(x)=\sqrt{R^2-x^2}[/imath], then the area of the circle is defined by: [imath]4\int_0^Rg(x)dx[/imath] Now my best attempt to find the area is through: [imath]\int_0^Rg(x)dx-\int_0^{\sqrt{R^2-1}}(g(x)-1)dx[/imath] The first integral gives me a quarter of the area circle [imath]R[/imath], the second gives me the area of the circle that is outside the rectangle, because the circle and the top line intersects at [imath]x=\sqrt{R^2-1}[/imath] Is my solution correct? |
1325122 | Derivative Problem: [imath]e^{-x}f(x)=2+\int{\sqrt{t^{4}+1}}dt[/imath]
Let [imath]f(x)[/imath] be a real valued function defined on [imath](-1,1)[/imath] such that [imath]e^{-x}f(x)=2+\int_0^x{\sqrt{t^{4}+1}}\;\mathrm{d}t[/imath] for all [imath]x[/imath] belonging to [imath](-1,1)[/imath] and [imath]g(x)[/imath] be its inverse function. How to find the value of [imath]g'(2)[/imath] (the derivative of [imath]g(x)[/imath] at [imath]x=2[/imath])? I tried writing [imath]f(g(x))=x[/imath] and then took derivative on both sides of original equation. But it's not working. I think I'm missing something important. And please don't use hyper-geometric functions. | 2879394 | [imath]e^{-x}f(x)=2+ \int_0^x\sqrt{t^4+1}dt[/imath] function for finding the inverse
Let f be the real-valued function defined on the interval [imath](-1,1)[/imath] such that [imath]e^{-x}f(x)=2+ \int_0^x\sqrt{t^4+1}dt[/imath] for all [imath]x\in(-1,1)[/imath] and let [imath]f^{-1}[/imath] be the inverse function of [imath]f[/imath]. Then find the value of [imath](f^{-1})'(2)[/imath]. This is my step 1-[imath]f(x)=y[/imath] 2-[imath]f^{-1}(y)=x[/imath] 3-[imath]f^{-1}(x)=y[/imath] 4-[imath]f^{-1}(x)'=\frac{dy}{dx}[/imath] 5- [imath]y=f(x)=2e^{x}+e^{x} \int_0^x\sqrt{t^4+1}dt[/imath] 6- [imath]y'=2e^{x}+e^{x} \int_0^x\sqrt(t^4+1)dt+e^x\sqrt{x^4+1}[/imath] I am lost after this step, please help me as I have jolted down the step. |
2799398 | Transcendental Equation.
I was able to solve [imath] \ln x = - x [/imath] using Lambert's function. I was wondering how does one solve [imath] \ln x = x[/imath] Does the solution even exist for this equation? | 668901 | The problem of x = ln(x)
I am trying to find x values for points along the normal distribution curve, and I ended up with a problem that goes back to the method of solving [imath]x = \ln x[/imath]. Right now, I have [imath]\ln(a \mu) - \ln(10) = \frac {-1} {2a^2} [/imath]. This is not something I know how to solve. I tried Googling this type of problem and nothing came up. How does one solve for a variable both inside and outside of a natural logorithm, specifically in the example problem above? Could anyone explain how this could be done? |
2880421 | Rational points on the elliptic curve [imath]y^2=x^3 -x[/imath]
I've been trying to find rational points on the elliptic curve [imath]y^2=x^3 -x[/imath] but I can't find anything else apart from [imath](-1,0), (0,0), (1,0)[/imath]. I have like an 'edgy' proof that it may not be possible to find rational points on this curve(apart from the ones already mentioned), but I'm unsure about it. Can anyone help? | 1786448 | Rational solutions of [imath]y^2 = x^3 - x[/imath]
I believe that the only rational solutions of [imath]y^2 = x^3 - x[/imath] are the obvious ones [imath](-1,0)[/imath], [imath](0,0)[/imath], [imath](1,0)[/imath], and that this was proved by Fermat using the method of descent. Can anyone outline a proof, either Fermat’s proof or a different one; or, failing that, point to a suitable reference, preferably available online? |
2836197 | Computing the electric flux
Let [imath]B_r=\{x\in \mathbb R^3 : |x|\le r\}[/imath] and let [imath]dS_r[/imath] denote the area element on [imath]\partial B_r[/imath]. Set [imath]E(x)=C\int_{\partial B_R}\nabla_x |x-y|^{-1} dS_y[/imath] Show that for [imath]|x|< R[/imath], [imath]E[/imath] is zero and for [imath]r<R[/imath], the flux [imath]\int_{\partial B_r} E(x)\cdot \nu \ dS_x[/imath] is zero. First of all, what does [imath]\nabla_x[/imath] stand for? The [imath]\nabla[/imath] without subscripts usually stands for the gradient, but I'm not sure about [imath]\nabla_x[/imath]. For the flux integral, is it better to use the divergence theorem or use the definition? In the former case, how do I find the divergence (i.e., how to differentiate a line integral)? | 1992071 | Using Divergence Theorem
Let [imath]B_r[/imath] denote the ball [imath]|x|\le r[/imath] in [imath]\mathbb{R}^3[/imath], and write [imath]dS_r[/imath] for the area element on its boundary [imath]\partial B_r[/imath]. The electric field associated with a uniform charge distribution on [imath]\partial B_R[/imath] may be expressed as [imath]E(x)=C\int\limits_{\partial B_R}\nabla_x|x-y|^{-1}dS_y,[/imath] a) Show that for any [imath]r<R[/imath], the electric flux [imath]\int\limits_{\partial B_r}E(x).\nu dS_x[/imath] through [imath]\partial B_r[/imath] equals zero. **b)**Show that [imath]E(x)\equiv 0[/imath] for [imath]|x|<R[/imath]("a conducting spherical shell shields its interior from outside electrical effects"). For part a, I tried to to using divergence theorem as follows: [imath]\int\limits_{\partial B_r}E(x).\nu dS_x=\int\limits_{B_r}divE(x)dx= C\int\limits_{B_r}\int\limits_{B_R}div\nabla_x|x-y|^{-1}dS_ydx,[/imath] However, I don't know how to proceed next steps to prove the integrand is zero. By the way, could you please suggest me a method for part b)? Thank you very much for your help. |
2880989 | Natural number that has a remainder of [imath]1, 2, 3, 4[/imath] respectively after dividing...
A number when divided by 2 leaves a remainder of 1. When it is divided by 3 leaves a remainder 2. When it is divided by 4 it leaves a remainder of 3. And when it is divided by 5 it leaves remainder of 4. What should be the number ? Note : I already formulated some formula but I think it might not work: x = 2a + 1 x = 3b + 2 x = 4c + 3 x = 5d + 4 | 1544093 | Find an integer having the remainders [imath]2,3,4,5[/imath] when divided by [imath]3,4,5,6[/imath], respectively.
Find an integer having the remainders [imath]2,3,4,5[/imath] when divided by [imath]3,4,5,6[/imath], respectively. My work: We consider the congruences [imath]x \equiv 2 \pmod 3[/imath], [imath]x \equiv 3 \pmod 4[/imath], [imath]x \equiv 4 \pmod 5[/imath], [imath]x \equiv 5 \pmod 6[/imath]. We can reduce this further to [imath]x \equiv 2 \pmod 3[/imath], [imath]x \equiv 3 \pmod 4[/imath], [imath]x \equiv 4 \pmod 5[/imath]. We have [imath]N_1 = 4 \cdot 5 = 20 \implies 20 x_1 \equiv 1 \pmod{3} \implies 2x_1 \equiv 1 \pmod{3} \implies x_1 \equiv 2 \pmod {3}[/imath] [imath]N_2 = 3 \cdot 5 = 15 \implies 15x_2 \equiv 1 \pmod{4} \implies -x_2 \equiv 1 \pmod{4} \implies x_2 \equiv 3 \pmod {4}[/imath] [imath]N_3 = 3 \cdot 4 = 12 \implies 12 x_3 \equiv 1 \pmod{5} \implies 2x_3 \equiv 1 \pmod{5} \implies x_3 \equiv 3 \pmod {5}[/imath] Now, \begin{align*} \bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 \\ &= 3 \cdot 20 \cdot 2 + 4 \cdot 15 \cdot 3 + 5 \cdot 12 \cdot 3 \\ &= 480 \equiv 0 \pmod {3 \cdot 4 \cdot 5} \end{align*} Is this correct, or is something wrong in my work? I don't like how I have [imath]0[/imath] remainder. |
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