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1068123 | Joint density invariant under orthogonal transformations
I have a problem and I am totally stuck! I have to show that when the distribution of two random variables [imath]X[/imath] & [imath]Y[/imath] given by [imath]g(x,y)=f(x)f(y)[/imath] is invariant to orthogonal transformations, then it holds that [imath]f(x)f(y)=f(0)f\left(\sqrt{x^2+y^2}\right).[/imath] Any ideas on how to approach this? I have found that when the random variables are transformed by an orthogonal matrix like this [imath]\scriptstyle \begin{bmatrix}\tilde{X}\\\tilde{Y}\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\times \begin{bmatrix}X\\Y\end{bmatrix}[/imath], then the distribution of [imath](\tilde{X},\tilde{Y})[/imath] is given by [imath]\tilde{g}(x,y)=f\left(\frac{dx-by}{ad-bc}\right)f\left(\frac{ay-cx}{ad-bc}\right)[/imath], and my thought was that maybe it could be proven that [imath]f\left(\frac{dx-by}{ad-bc}\right)f\left(\frac{ay-cx}{ad-bc}\right)=f(0)f\left(\sqrt{\scriptstyle\left(\frac{dx-by}{ad-bc}\right)^2+\left(\frac{ay-cx}{ad-bc}\right)^2}\right)[/imath], but it doesn't get me anywhere. | 1067378 | Linear transformation of random variables
We have to stochastic variables X and Y, and we define [imath] \begin{pmatrix} \tilde{X} \\ \tilde{Y} \end{pmatrix}=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} [/imath] for [imath]a,b,c,d \in \mathbb{R}[/imath], i.e. [imath]\tilde{X}=aX+bY[/imath] and [imath]\tilde{Y}=cX+dY[/imath]. We know that both [imath]X[/imath] and Y, and [imath]\tilde{X}[/imath] and [imath]\tilde{Y}[/imath] are uncorrelated, and the distribution of (X,Y) has density [imath]g(x,y)=f(x)f(y)[/imath]. Here [imath]f[/imath] is a density with respect to the lebesgue measure for a probability measure on [imath]\mathbb{R}[/imath], and f is continous and strict positive. Furthermore, we say that the distribution of (X,Y) is invariant to orthogonal matrices [imath]\begin{pmatrix} a & b \\ c & d \end{pmatrix} [/imath] if [imath](X,Y)[/imath] and [imath](\tilde{X},\tilde{Y})[/imath], have equal distributions. In general, a matrix O is orthogonal if [imath]OO^T=O^TO=I[/imath], where I is the identity matrix. Now I have to show that if the distribution of [imath](X,Y)[/imath] is invariant to orthogonal transformations, then it holds that [imath]f(x)f(y)=f(0)f(\sqrt{x^2+y^2})[/imath] for alle [imath]x,y \in \mathbb{R}^2[/imath]. Can anyone help me with this problem? |
274815 | Entire function. Prove that [imath]f(\bar{z})=\overline{f(z)}, \forall z\in C[/imath]
Let [imath]f[/imath] a entire function: [imath]f(R)\subset R.\;[/imath] Prove that [imath]f(\bar{z})=\overline{f(z)}, \forall z\in C[/imath] | 1985230 | show the equality [imath]f(\bar{z})=\overline{f(z)}[/imath] for entire function [imath]f(z)[/imath]
i am struggling with this proof. Do you have some ideas? Let [imath]f(z)[/imath] be entire function. Show that, if [imath]f[/imath] maps [imath]\mathbb{R}[/imath] into [imath]\mathbb{R}[/imath], then equality [imath]f(\bar{z})=\overline{f(z)}[/imath] holds for all [imath]z \in \mathbb{C}.[/imath] Thanks for help. |
1067606 | [imath]f:\mathbb R\to\mathbb R[/imath] continuous function. Which of the following sets can not be image of [imath](0,1][/imath] under [imath]f[/imath]?
Let [imath]f:\mathbb R\to\mathbb R[/imath] continuous function. Which of the following sets can not be image of [imath](0,1][/imath] under [imath]f[/imath]? A. [imath]\{0\}[/imath]. B. [imath](0,1)[/imath]. C.[imath][0,1)[/imath]. D.[imath][0,1][/imath]. My effort: Continuous image of connected set connected. [imath](0,1][/imath] is connected and remove [imath]1[/imath] from the set left the set connected...but removing any point from [imath](0,1)[/imath] make it disconnected.....I am not sure though | 2308318 | If R is a continous function thn which cant be image of (0,1]
Let [imath]f:R\to R[/imath] be continous function. Then, which one of the following sets can't be the image of [imath](0,1][/imath] under [imath]f[/imath]. [imath]a) {0}[/imath] [imath]b) (0,1)[/imath] [imath]c) [0,1)[/imath] [imath]d) [0,1][/imath] the ans is [imath]b[/imath], but I don't know why, please explain it in detail. |
1069330 | complete metric space [imath]X[/imath] and nested sequence of closed sets [imath]A_m \subset X[/imath] where [imath]\bigcap_{n=1}^\infty = \emptyset[/imath]
What is an example of a complete metric space [imath]X[/imath] and a nested sequence of closed sets [imath]A_m \subset X[/imath] such that [imath]\bigcap_{n=1}^\infty A_m = \emptyset[/imath]? My analysis professor mentioned this in office hours as something interesting to think about, and I've thought about it for a while. But I haven't made much progress... | 116852 | An empty intersection of decreasing sequence of closed sets
What is an example of a family of closed subsets [imath]F_0 \supset F_1 \supset F_2 \supset \dots [/imath] of [imath]\mathbb{R}[/imath] so that [imath]F_n \neq \emptyset[/imath] for each [imath]n[/imath] and [imath]\bigcap_{i=1}^n F_i = \emptyset[/imath]? Thanks! |
1068644 | how to use Matlab ifft to calculate the following integral?
[imath]R(t)=\int_{-\infty}^\infty\dfrac{\omega e^{i\omega t}}{(3-\omega^2)^{2}+4\omega^2}\,d\omega[/imath] where t is a integer and [imath]t>0[/imath] I used to calculate this integral by numerical integral,but it seems that using ifft in matlab is more effcient. The ifft equation in Matlab is [imath]x(n)=\frac{1}{N}\sum_{k=1}^N X(k)e^{i2\pi\frac{k-1}{N}n-1}[/imath] where n=[imath]1,2...,N[/imath] I have no idea how to link the above Eq with the above-mentioned integral. Do someone have an idea? | 1036524 | How to evaluate this integral[imath]\int_{-\infty}^\infty\dfrac{\omega^\alpha e^{i\omega t}}{(\omega_0^2-\omega^2)^{2}+4(\zeta\omega_0\omega)^2}\,d\omega[/imath]
How to calculate the following integral? [imath]\int_{-\infty}^\infty\dfrac{\omega^\alpha e^{i\omega t}}{(\omega_0^2-\omega^2)^{2}+4(\zeta\omega_0\omega)^2}\,d\omega[/imath] where [imath]\alpha>0,t>0,\zeta>0,\omega_0>0[/imath]. When [imath]\alpha=0[/imath], I can figure it out by Residue theorem. However, I do not have any idea how to calculate this integral when [imath]\alpha>0[/imath]. Does this integral have an analytic solution? |
1069218 | Does there exist unique [imath]c \in (0,1)[/imath] such that [imath]f'(c)=f(c)[/imath]?
If [imath]f: [0,1] \to \mathbb R[/imath] be a continuous function differentiable in [imath](0,1)[/imath] such that [imath]f(0)=f(1)=0[/imath] then by Rolle's thorem for [imath]e^{-x}f(x)[/imath] , it is evident that [imath]f'(x)=f(x) [/imath] has a solution in [imath](0,1)[/imath] . Is this solution unique ? And does there exist a function for which the solution is unique ? | 1069000 | How many [imath]f(x)[/imath] are possible satisfying [imath]f(x)=f'(x)[/imath] and [imath]f(0)=f(1)=0[/imath].
Let [imath]f:[0,1]\to\Bbb{R}[/imath] be a fixed continuous function such that [imath]f[/imath] is differentiable on [imath](0,1)[/imath] and [imath]f(0)=f(1)=0[/imath]. Then the equation [imath]f(x)=f'(x)[/imath] admits how many solutions? The only solution that I am getting is [imath]y=0[/imath]. This is because [imath]y=y'[/imath] implies [imath]y=Ae^x[/imath], and [imath]A=0[/imath] when accounting for boundary conditions. However, I am not sure if this is the right answer. Most other examinees were saying that multiple such functions are possible. An explanation of this would be great. EDIT: The options were A. No solution [imath]x\in (0,1)[/imath] B. More than one solution [imath]x\in (0,1)[/imath] C. Exactly one solution [imath]x\in (0,1)[/imath] D. At least one solution [imath]x\in (0,1)[/imath] I feel that if C is right, then so is D! |
1069366 | Expected value of a Poisson variable conditioned on sum
Setting [imath]X_1 \overset{d}{\sim} \operatorname{Poisson}(\alpha_1)[/imath] [imath]X_2 \overset{d}{\sim} \operatorname{Poisson}(\alpha_2)[/imath] [imath]S = X_1 + X_2[/imath] Find [imath]E[X_1 | S =n][/imath] My argument is that since [imath]X_1 + X_2 = n[/imath], [imath]X_1[/imath] range from [imath]0[/imath] to [imath]n[/imath] only, thus the expected value is [imath]E[X_1 | S= n] = \sum_{k=1}^n k \Pr\{X_1 =k | S = n\}[/imath] where [imath]\Pr\{X_1=k | S = n\} = \sum_{k=0}^{n} \frac{e^{-\alpha_1} \alpha_1^k}{k!}[/imath] Please argue for or against it? Additionally, how would you evaluate such a series on the condition that it's correct? | 151272 | Proof that Conditional of Poisson distribution is Binomial
The classic example... [imath]X \sim Po\left (\lambda\right ), Y \sim Po\left (\mu\right)[/imath], X and Y are independent. Show that the conditional distribution of X is binomially distributed. Or in other words, [imath]$P(X=k\mid X+Y = n) = P (\tilde{X} = k), \tilde{X} \sim B\left (n ,\frac{\lambda}{\lambda + \mu}\right )$[/imath]. I've so far managed to reach to this step, and have been stuck since. Just somehow gotta get a [imath]\frac{1}{n!}[/imath] in the denominator, that would then complete the proof..or at least I think.. [imath]P(X=k\mid X+Y=n) = \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{P(X+Y = n)}= \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{\sum_{i=1}^{n} \frac{\lambda^{i}\mu^{n-i}}{i!(n-i)!}}[/imath] Thanks for the help! |
416547 | Hilbert subspace
Let be [imath]H[/imath] Hilbert space and [imath]M\subset H[/imath]. [imath]M=M^{\perp\perp}[/imath] if and only if [imath]M[/imath] subspace of [imath]H[/imath]. Does anyone know to prove this? | 1051514 | A subspace [imath]X[/imath] is closed iff [imath]X =( X^\perp)^\perp[/imath]
Let [imath]X[/imath] be a subspace of a Hilbert space [imath]H[/imath]. prove: [imath]X[/imath] is closed iff [imath]X =( X^\perp)^\perp[/imath] I do not know how to proceed. any hints would be appreciated. thanks. [imath]X^\perp = [/imath]{ [imath]y \in H[/imath]:[imath]\langle x,y\rangle = 0 \forall x \in X\}[/imath]. |
1069609 | What is the indefinite integral of zero?
From the definition of indefinite integral I might say: Since the derivative of a constant is zero, thus the indefinite integral of zero is a constant. Therefore: [imath] \frac{dc}{dx} = 0 \quad\iff\quad \int 0dx = c, \quad\forall c\in\mathbb{R} [/imath] However... we know that [imath]0\in\mathbb{R}[/imath], and since zero is a constant, I can pull it out the integral: [imath] \int 0dx = 0\cdot\int 1dx = 0\cdot(x+c) = 0 [/imath] And then we end up that integral of zero is zero, not an arbitrary constant. Where is wrong here? | 287076 | What is the integral of 0?
I am trying to convince my friend that the integral of [imath]0[/imath] is [imath]C[/imath], where [imath]C[/imath] is an arbitrary constant. He can't seem to grasp this concept. Can you guys help me out here? He keeps saying it is [imath]0[/imath]. |
1070134 | Help evaluating [imath]\int e^x \sqrt{1+e^{2x}}dx[/imath]
[imath]\int e^x \sqrt{1+e^{2x}}dx[/imath] It's probably been answered somewhere, but I havent found it so far so I decided to post it as a question (if it has been answered point me in the right direction and I will delete the question. What I have tried: Put it into mathematica, got a relatively simple answer: [imath]\frac{1}{2} (e^x \sqrt{1+e^{2x}}+\sinh^{-1}e^x)[/imath] Dont know how to get there! Obviously I tried [imath]e^x = u[/imath] substitution, ended up with [imath]\int \sqrt{1+u^2}du[/imath]. From here I am not really sure what do to, I tried another substitution of [imath]u=\cosh t[/imath], but to be honest I am very unfamiliar with hyperbolic trig: [imath]u=\cosh t \implies du=\sinh t~dt[/imath] [imath]\int \sinh t\sqrt{1+\cosh^2t} ~ ~dt[/imath] And i believe this is the same as [imath]\int \sinh^2 t ~ ~dt[/imath]? Once again, I am not too sure about this because I don't know much about hyperbolic trig. From here I don't know where to go (don't even know if my working so far is right).. | 431143 | Evaluating [imath]\int \sqrt{1 + t^2} dt[/imath]?
How do we solve (i.e. get the closed form of) [imath]\int \sqrt{1 + t^2} dt[/imath] ? The Wolfram page shows the closed form of it but not the steps in solving it. I think I need some algebraic trick.. I thought of some hyperbolic functions and their derivatives and equalities: If I make [imath] t = \sinh x[/imath], then I can reduce the integral to [imath] \int \cosh ^2 x dx [/imath] though I'm not sure if this is valid. But from here, I cannot go on evaluating the integral... |
1070238 | Complex Analysis analytic function 1[imath]f(z)=z[/imath]
if[imath]\text{ } f:D(0,1)\longrightarrow D(0,1)[/imath] is analytic such that there exists [imath]a,b\in D(0,1)[/imath] and [imath]\text{ }[/imath][imath]f(a)=a[/imath] , [imath]f(b)=b[/imath] prove that [imath]f(z)=z[/imath] [imath]\forall[/imath] [imath]z\in D(0,1)[/imath] | 801570 | an analytic function from unit disk to unit disk with two fixed point
prove that if [imath]f:\mathbb{D}\rightarrow\mathbb{D}[/imath] is analytic with two distinct fixed point then [imath]f[/imath] is identity. i thought if one of the fixed points were zero by schwarz lemma this statement is easily proved. but what can i do if fixed points were nonzero? pls don't answer:ur question is duplicated .i've seen other question & answers but i'm still confused. help pls thnx |
1063584 | Existence of Holomorphic function (Application of Schwarz-Lemma)
Let, [imath]D=\{z\in \mathbb C:|z|<1\}[/imath]. Which are correct? there exists a holomorphic function [imath]f:D \to D[/imath] with [imath]f(0)=0[/imath] & [imath]f'(0)=2[/imath]. there exists a holomorphic function [imath]f:D \to D[/imath] with [imath]f\left(\dfrac{3}{4}\right)=\dfrac{3}{4}[/imath] & [imath]f'\left(\dfrac{2}{3}\right)=\dfrac{3}{4}[/imath]. there exists a holomorphic function [imath]f:D \to D[/imath] with [imath]f\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}[/imath] & [imath]f'\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}[/imath] there exists a holomorphic function [imath]f:D \to D[/imath] with [imath]f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}[/imath] & [imath]f'\left(\dfrac{1}{4}\right)=1[/imath]. With the help of Schwarz lemma & its applications, we find that [imath](1)[/imath] is false & [imath](3)[/imath] is true. But, I can not think about options [imath](2)[/imath] & [imath](4)[/imath]. | 1766403 | Existence/non-existence of Holomorphic Function
Let [imath]D=\{z\in \Bbb C:|z|<1\}[/imath]. Show that there exists a holomorphic function [imath]f:D\to D[/imath] such that [imath]f(\frac{3}{4})=-\frac{3}{4}[/imath] and [imath]f^{'}{(\frac{3}{4})}=-\frac{3}{4}[/imath] Show that there exists no holomorphic function [imath]f:D\to D[/imath] such that [imath]f(\frac{1}{2})=-\frac{1}{2}[/imath] and [imath]f^{'}{(\frac{1}{4})}=1[/imath] By Schwartz Lemma we can at best say that [imath]|f^{'}(z)|\le \dfrac{1-|f(z)|^2}{1-|z|^2}[/imath] If I substitute the values of the two given problems in the equation then I can say that [imath]|f^{'}(z)|\le 1[/imath] which is satisfied in every case. How can I show existence/non-existence in two cases of the problem? Please help. |
261145 | Find an abelian infinite group such that every proper subgroup is finite
I found this question in Arhangel'skii and Tkachenko's book Topological Groups and Related Structures. The first chapter of the book is devoted to algebraic preliminaries. The question actually reads: Give an example of an infinite abelian group all proper subgroups of which are finite. What I have done is: Every element of this group has finite order, else we could find an infinite proper subgroup, namely the group generated by [imath]x²[/imath] if [imath]x[/imath] has infinite order. I think this can be strengthened: every element should have a prime order. Although I haven't proved this. Intuitively this group cannot be and infinite product of smaller groups, because you could take the product of the even group factors and find an infinite proper subgroup. Well, this is it, a highly non-trivial problem. Thanks in advance. | 2602496 | Abelian group with all proper subgroup finite
Let [imath]G[/imath] be an abelian group with all proper subgroup finite, then is [imath]G[/imath] finite or at least finitely generated ? |
1070883 | Is [imath]\mathbb R-\mathbb Q[/imath] open or closed?
The set [imath]\Bbb R-\Bbb Q[/imath] is open, closed or neither? And how would one prove it? I tried to prove that [imath]\Bbb Q[/imath] is open, so [imath]\Bbb R-\Bbb Q[/imath] would be closed, but I am not sure that [imath]\Bbb Q[/imath] is open. | 116721 | Are the rationals a closed or open set in [imath]\mathbb{R}[/imath]?
I have a feeling they are neither closed nor open as the [imath]\mathbb{R} \backslash \mathbb{Q}[/imath] cannot be open or closed either... |
1071372 | How to prove [imath]x^{\phi(m)+1}\equiv x\pmod{p}[/imath]
How do I prove that [imath]x^{\phi(m)+1}\equiv x\pmod{p}[/imath] when [imath]m=pq[/imath], two distinct primes? I kind of have an idea that it involves Euler's Theorem but it doesn't seem to be working as well as I wanted it to. | 1069591 | Prove Euler's Theorem when the integers are not relatively prime
How can I prove Euler's Theorem: [imath]x^{\phi(m)+1} \equiv x \pmod m[/imath] is still true when [imath]x[/imath] is not relatively prime to [imath]m[/imath]? Edit: when m=pq where p and q are distinct primes |
1071555 | How do I prove that there doesn't exist a set whose power set is countable?
I don't even know where to begin on this one. Let [imath]A[/imath] be a countable set. In other words, [imath]|A| = |\mathbb{N}|[/imath]. I'm trying to prove that there doesn't exist an [imath]x[/imath] such that [imath]\mathcal{P}x = A[/imath]. Maybe a proof by contradiction? Suppose there was a set [imath]x[/imath] whose power set was countable ... | 201923 | Existence in ZF of a set with countable power set
Is it consistent with ZF for there to exist a set [imath]S[/imath] such that the power set [imath]P(S)[/imath] is countable? If so, what is the weakest form of the axiom of choice needed to prove that no such set exists? |
1072038 | Infinite series [imath]\sum_{k=1}^\infty \frac{k}{2^k}[/imath] and [imath]\sum_{k=1}^\infty \frac{k^2}{2^k}[/imath]
How to evaluate those infinite series? How are they called? [imath] \sum_{k=1}^\infty \frac{k}{2^k} \quad \text{and} \quad \sum_{k=1}^\infty \frac{k^2}{2^k} [/imath] I'm really sorry for asking, but I can't figure out how to google such stuff when I even don't know the names/categories. | 1500335 | Sum up the following series: [imath]\sum^n_{k=1}\frac{k^2}{2^k}[/imath]
How to find an explicit formula for the term [imath]\sum^n_{k=1}\frac{k^2}{2^k}[/imath] Then I discovered that [imath]\sum^n_{k=1}\frac{k^2}{2^k}=1^22^n+2^22^{n-1}+3^22^{n-2}+\cdots+n^22^1-\frac{n(n+1)(2n+1)}{6}[/imath] But how to write [imath]1^22^n+2^22^{n-1}+3^22^{n-2}+\cdots+n^22^1[/imath] into explicit formula?? [imath]\\[/imath] A hint would be grateful. |
1072099 | I need to find a rational numbers series that converging to irrational number
I found a series that is [imath]a_{n+1}=\frac{a_n^2 + 2}{2a_n}[/imath] yet I'm not sure. can someone give me a more umm solid example? thanks. | 1021876 | Give a concrete sequence of rationals which converges to an irrational number and vice versa.
Give a concrete sequence of rationals which converges to an irrational number and vice versa.... My work I could give a sequence of irrationals which converges to a rational number... Let [imath]r\in \mathbb Q,[/imath] [imath]a_n=\frac {\sqrt 2} n+r[/imath] But I couldn't give a sequence of rationals which converges to an irrational.. Help me to work out..... |
12571 | When is the [imath]n[/imath]th term sufficient to guarantee convergence of a series
If [imath]K[/imath] is a field complete with respect to a non-archimedean absolute value, then the [imath]n[/imath]th term test (checking whether the [imath]n[/imath]th term of a series goes to zero) is sufficient to check convergence of a series in [imath]K[/imath]. I was wondering if we have any results that guarantee, under additional hypothesis, whether a series in [imath]\mathbb{R}[/imath] converges by just making sure that the [imath]n[/imath]th term goes to zero. An example would be the alternating series test. Are there others? | 1071975 | When does [imath]\lim_{n\rightarrow \infty} a_n = 0[/imath] imply [imath]\sum_{n = 0}^{\infty} a_n[/imath] converges?
This is just a general question. I know the opposite statement is always true. I'm just asking when the converse holds. I know that the converse is not true in some cases, such as for harmonic series. What additional restrictions are needed to make this statement true? |
1072202 | Hermitian Matrix M(x) is continuous on x. Is its eigenvalue also continuous on x?
Each element of matrix [imath]M(x)[/imath] is a continuous function of [imath]x[/imath]. Does this imply that all the eigenvalues are continuous function of [imath]x[/imath] too? | 1010822 | Continuity of eigenvalues and spectral radius for a general matrix
Given a general matrix [imath]A(t), t>0[/imath], with real entries, I would like to know if the eigenvalues of [imath]A(t)[/imath] are continuous functions of [imath]t[/imath]. These eigenvalues may be real or complex. What about the spectral radius? A classical result from complex analysis states that the roots of a polynomial vary continuously with the coefficients. Can we use the theorem directly to prove the above? or there are other cases where the eigenvalues are actually discontinuous? What I'm actually doing is trying to prove that there exists a [imath]t[/imath] for which the spectral radius of [imath]A(t)[/imath] is in [imath](0,1)[/imath], and I'm doing that by proving that the spectral radius is 1 if [imath]t\rightarrow 0[/imath] and 0 if [imath]t\rightarrow \infty[/imath] (which I already know). Then I would invoke the continuity of the spectral radius to say that there must exist a value of [imath]t[/imath] for which the spectral radius is in [imath](0,1)[/imath] Thank you. |
1070487 | maximal ideal problem
I want to solve this problem, but I have no idea how I can start: If [imath]K[/imath] is a field, [imath](a_1,...,a_n) \in K^n,[/imath] and [imath]I[/imath] the ideal [imath]I=\langle x_1-a_1,...,x_n-a_n\rangle[/imath], then how can we prove that [imath]I[/imath] is a maximal ideal? One example: Is [imath]\langle x^2+1 \rangle[/imath] a maximal ideal of [imath] \mathbb{R}[x][/imath]? | 550729 | [imath](x_1-a_1, x_2-a_2)[/imath] is a maximal ideal of [imath]K[x_1,x_2][/imath]
[imath]K[/imath] is field. [imath]a_1[/imath],[imath]a_2[/imath] elements of [imath]K[/imath]. Show that [imath](x_1-a_1,x_2-a_2)[/imath] is a maximal ideal of [imath]K[x_1,x_2][/imath]. [imath]K[x_1,x_2][/imath] is UFD so if [imath]K[x_1,x_2]/(x_1-a_1,x_2-a_2)[/imath] is field then [imath](x_1-a_1,x_2-a_2)[/imath] is maximal ideal. If I can show that [imath]K[x_1,x_2]/(x_1-a_1,x_2-a_2)[/imath] isomorphic to [imath]K[/imath], we can verify that [imath](x_1-a_1,x_2-a_2)[/imath] maximal ideal of [imath]K[x_1,x_2][/imath]. thanks for helps and comments. |
1072480 | solve [imath]y'=ay+b[/imath]
I have this differential equation which I want to solve [imath]\displaystyle\frac{dT_i}{dt}=\frac{1}{RC}(T_a-T_i)+\frac{1}{C} \Phi_h[/imath] I know it is in the form [imath]y'=ay+b[/imath] But how can I solve it ? | 847888 | Tricky Separable Differential Equation
Please guide me: [imath]y' + ay +b = 0[/imath] ([imath]a[/imath] not zero) is supposed to be separable and has solution [imath]y = ce^{-ax} - \frac ba[/imath] Here is my start to this problem: [imath]\frac{dy}{dx} + ay = -b[/imath] is as far as I can go with this. How should I go about separating [imath]x[/imath] and |
1072657 | Show that [imath]a[/imath] is minimum
If [imath](A,<)[/imath] totally ordered, show that if [imath]a[/imath] is a minimal element of [imath]A[/imath] then [imath]a[/imath] is minimum. Could you give me a hint how we could do this? Definitions: Let [imath](A, \leq)[/imath] be an ordered set. We say that [imath]a \in A[/imath] is: minimal, when it does not exist in [imath]A[/imath] an element that is previous of [imath]a[/imath] and different from it, i.e. [imath](\forall x \in A)(x \leq a \rightarrow x=a)[/imath] [imath]$$[/imath] minimum when $(\forall x \in A) a \leq x$ $$ | 572504 | How to prove a total order has a unique minimal element
Let [imath]R[/imath] be a total order on set [imath]S[/imath]. Prove that if [imath]S[/imath] has a minimal element, than the minimum element is unique. I have difficulties with proofs. I know any graph of a total order is a straight line, which clearly has a minimal element. How doI tell when something is mathematically proven? EDIT: this is how I would prove it Let [imath]x, y \in S[/imath] such that x and y are minimal. Since S is a total order [imath]x[/imath] and [imath]y[/imath] are comparable and either case 1: one comes after the other and therefore one isn't the minimal case 2: both are the same and [imath]x=y[/imath] I know this isn't a mathy argument and could use help making it more presentable. |
928735 | Find solution of equation [imath](z+1)^5=z^5[/imath]
I attempt to solve the equation [imath](z+1)^5=z^5[/imath]. My first approach is to expand the left hand side but ı get more complicated equation. So I couldn't go further. Secondly, I write equation as, since [imath]z\neq0[/imath], [imath](\frac{z+1}{z})^5=1[/imath], put [imath]\xi=\frac{z+1}{z}[/imath] and attempt to solve equivalent equation [imath]\xi^5=1[/imath]. But this time it requires more computation to find solutions [imath]z[/imath]. Can anyone suggest a simple way to solve this equation? Thanks in advance.. | 577847 | Solving [imath](z+1)^5 = z^5[/imath]
The question says to solve this equation: [imath](z+1)^5 = z^5[/imath] I did. Just want to find out if I did it properly and if my run-around logic makes sense. First I begin my writing the equations as: [imath] (z+1)^5 = z^5[/imath] [imath] \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} [/imath] So [imath] \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i [/imath] [imath] \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i [/imath] Now, because the natural logarithm is one-to-one, I write: [imath] \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|[/imath] So assign [imath] z = a +bi[/imath] So that [imath]|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 [/imath] So, [imath]z = -\frac12 + bi[/imath] and [imath]z+1 = \frac12 + bi[/imath] for some [imath]b \in \mathbb R[/imath] Now to find [imath]b[/imath] [imath] \mathbf{Arg}(z+1) = \mathbf{Arg}(z)[/imath] [imath] \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}[/imath] I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly? Ultimately, I get [imath] z = -\frac12[/imath] which upon inspection...is wrong... |
1072925 | What's [imath]\int \frac{1}{\sqrt{25-x^2}}[/imath]
What is [imath]\int \frac{1}{\sqrt{25-x^2}}[/imath] WolframAlpha says [imath]\sin^{-1}(\frac{x}{5})[/imath] while I got [imath]\frac{1}{5}\sin^{-1}(\frac{x}{5})[/imath]. What is correct? Thanks in advance. | 138413 | Simplify [imath]\int \frac{1}{\sqrt{2-x^2}}\, dx[/imath]
Somewhere in the provided answer: [imath]\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}[/imath] How did they get that? What I have: [imath]\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}[/imath] [imath]\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}{\frac{x}{\sqrt{2}}}[/imath] So I have an extra [imath]\frac{1}{\sqrt{2}}[/imath] ... I probably had some stupid mistakes? |
1073131 | Inverse of [imath]f(x)= x+\sin(x)[/imath]?
How to find the inverse of [imath]f(x) = x+\sin(x)[/imath], analytically? Well how should I proceed to find the inverse of [imath]f(x)[/imath]? Basically I have applied graphical approach to solve the equation, but I want to know the inverse equation by analytical method. | 652277 | Inverse of [imath]f(x)=\sin(x)+x[/imath]
What is the inverse of [imath]f(x)=\sin(x)+x.[/imath] I thought about it for a while but I couldn't figure it out and I couldn't find the answer on the internet. What about [imath]f(x)=\sin(a \cdot x)+x[/imath] where [imath]a[/imath] is a known real constant. Thank you for taking the time to read this question! Sorry if this has been asked before... |
1070994 | About diagonalization
"Let A = [imath]\begin{bmatrix}1 & 1 & 4\\0 & 3 & -4\\0&0&-1\end{bmatrix}[/imath]. Is the matrix A diagonalizable? If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition [imath]A = PΛP^{-1}[/imath]." My solution attempt: det(λI - A) = [imath]\begin{vmatrix}λ - 1 & -1 & -4\\0 & λ - 3 &4\\0&0&λ + 1\end{vmatrix}[/imath] = (λ -1) (λ -3) (λ + 1) = 0 [imath]λ_{1}[/imath] = 1, [imath]λ_{2}[/imath] = 3, [imath]λ_{3}[/imath] = -1 ([imath]λ_{1}I - A)x_{1}[/imath] = 0 [imath]\begin{bmatrix}0 & -1 & -4\\0 & -2 &4\\0&0&2\end{bmatrix}[/imath] [imath]\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}[/imath] = [imath]\begin{bmatrix}0\\0 \\0\end{bmatrix}[/imath] [imath]\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}[/imath] = [imath]x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}[/imath] It follows that [imath]\begin{bmatrix}1\\0 \\0\end{bmatrix}[/imath] is a basis for the eigenspace of A corresponding to [imath]λ_{1}[/imath] = 1. In a similar way, I found other two basis vectors as [imath]\begin{bmatrix}1/2\\1 \\0\end{bmatrix}[/imath] and [imath]\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}[/imath], and I got [imath]P = \begin{bmatrix}1 & 1/2 & -2/5\\0 & 1 & 1\\0&0&1\end{bmatrix}[/imath]. Then my [imath]P^{-1} = \begin{bmatrix}1 & -1/2 & 9/10\\0 & 1 & -1\\0&0&1\end{bmatrix}[/imath]. And I said A's left-multiplication by [imath]P^{-1}[/imath] and right-multiplication by [imath]P[/imath] [imath](P^{-1}AP)[/imath] gives the diagonal matrix Λ that is similar to A. I found Λ as [imath]\begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0&0&21/5\end{bmatrix}[/imath]. Then I said that if we multiply both sides of the equation [imath]P^{-1}AP = Λ[/imath] by [imath]P[/imath] from the left and by [imath]P^{-1}[/imath] from the right we get [imath]PP^{-1}APP^{-1} = PΛP^{-1} = A[/imath] and that this is the eigendecomposition of A. Here are my questions: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition [imath]A = PΛP^{-1}[/imath]."? What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean? | 1069429 | Two questions about diagonalization
Let A = [imath]\begin{bmatrix}1 & 1 & 4\\0 & 3 & -4\\0&0&-1\end{bmatrix}[/imath]. Is the matrix A diagonalizable? If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition [imath]A = PΛP^{-1}[/imath]. A = [imath]\begin{bmatrix}3 & 1 & 0\\1 & 2 &1\\0&1&3\end{bmatrix}[/imath]. Is the matrix A diagonalizable? If so find a matrix P that orthogonally diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? (Note that A is real symmetric so that you do not have to compute the inverse of P). My solution attempts: det(λI - A) = [imath]\begin{vmatrix}λ - 1 & -1 & -4\\0 & λ - 3 &4\\0&0&λ + 1\end{vmatrix}[/imath] = (λ -1) (λ -3) (λ + 1) = 0 [imath]λ_{1}[/imath] = 1, [imath]λ_{2}[/imath] = 3, [imath]λ_{3}[/imath] = -1 ([imath]λ_{1}I - A)x_{1}[/imath] = 0 [imath]\begin{bmatrix}0 & -1 & -4\\0 & -2 &4\\0&0&2\end{bmatrix}[/imath] [imath]\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}[/imath] = [imath]\begin{bmatrix}0\\0 \\0\end{bmatrix}[/imath] [imath]\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}[/imath] = [imath]x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}[/imath] It follows that [imath]\begin{bmatrix}1\\0 \\0\end{bmatrix}[/imath] is a basis for the eigenspace of A corresponding to [imath]λ_{1}[/imath] = 1. In a similar way, I found other two basis vectors as [imath]\begin{bmatrix}1/2\\1 \\0\end{bmatrix}[/imath] and [imath]\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}[/imath], and I got [imath]P = \begin{bmatrix}1 & 1/2 & -2/5\\0 & 1 & 1\\0&0&1\end{bmatrix}[/imath]. Then my [imath]P^{-1} = \begin{bmatrix}1 & -1/2 & 9/10\\0 & 1 & -1\\0&0&1\end{bmatrix}[/imath]. And I said A's left-multiplication by [imath]P^{-1}[/imath] and right-multiplication by [imath]P (P^{-1}AP)[/imath] gives the diagonal matrix Λ that is similar to A. I found Λ as [imath]\begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0&0&21/5\end{bmatrix}[/imath]. Then I said that if we multiply both sides of the equation [imath]P^{-1}AP = Λ[/imath] by [imath]P[/imath] from the left and by [imath]P^{-1}[/imath] from the right we get [imath]PP^{-1}APP^{-1} = PΛP^{-1} = A[/imath] and that this is the eigendecomposition of A. Here is my first question: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition [imath]A = PΛP^{-1}[/imath]."? I started in the same manner and I got three eigenvectors as [imath]\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}[/imath] = [imath]x_{12} \begin{bmatrix}0\\1 \\0\end{bmatrix}[/imath], [imath]\begin{bmatrix}x_{21}\\x_{22}\\x_{23}\end{bmatrix}[/imath] = [imath]x_{21} \begin{bmatrix}1\\1 \\1\end{bmatrix}[/imath] and [imath]\begin{bmatrix}x_{31}\\x_{32}\\x_{33}\end{bmatrix}[/imath] = [imath]x_{33} \begin{bmatrix}1\\-2 \\1\end{bmatrix}[/imath]. I said A is not diagonalizable since the eigenvectors are not linearly independent. Here is my second question: What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean? |
1073550 | Relation Proofs on finite set
I have this problem I can't figure out how to do it Suppose A and B are finite sets and [imath]f : A → B[/imath] is surjective. Is it true that the relation [imath]“|A| < |B|”[/imath] is a sufficient condition for claming that [imath]f[/imath] is a bijection? | 1073556 | Is it true that the relation |A| < |B| is a sufficient condition for claiming that [imath]f[/imath] is a bijection?
This is an exercise of an assignment I have: Suppose [imath]A[/imath] and [imath]B[/imath] are finite sets and [imath]f\colon A\to B[/imath] is surjective. Is it true that the relation “[imath]|A| < |B|[/imath]” is a sufficient condition for claiming that [imath]f[/imath] is a bijection? Justify your answer. And this is my answer: No. In fact, if [imath]|A| < |B|[/imath], then there should exist at least [imath]1[/imath] element of [imath]A[/imath] that points to more than [imath]1[/imath] element of [imath]B[/imath], since all elements of [imath]B[/imath] must be pointed (surjective), but this is not a function, because the same element of [imath]A[/imath] point to different elements of [imath]B[/imath]. Is my answer correct? I don't know if the question actually is asking for a proof or what, and if my answer is a proof, if correct. |
1073332 | How to compute the integral [imath]\int^{\pi/2}_0\ln(1+\tan\theta)d\theta[/imath]?
How to compute the integral [imath]\int^{\pi/2}_0\ln(1+\tan\theta)d\theta[/imath]. If we let [imath]t=\tan\theta[/imath], then the integral becomes to [imath]\int^{\pi/2}_0\ln(1+\tan\theta)d\theta=\int_0^\infty\frac{\ln(1+t)}{1+t^2}dt[/imath]. Can we calculate this integral explicitly? | 396170 | Evaluating [imath]\int_0^\infty \frac{\log (1+x)}{1+x^2}dx[/imath]
Can this integral be solved with contour integral or by some application of residue theorem? [imath]\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}[/imath] It has two poles at [imath]\pm i[/imath] and branch point of [imath]-1[/imath] while the integral is to be evaluated from [imath]0\to \infty[/imath]. How to get [imath]\text{Catalan Constant}[/imath]? Please give some hints. |
1056433 | Mapping Circles by Möbius Transformation
I need to find a Möbius transformation which takes the circles [imath]|z|=1[/imath] and [imath]|z-\frac{1}{4}|=\frac{1}{4}[/imath] to concentric circles. I can assume that the images are centered at [imath]0[/imath], and that the unit circle is mapped to itself and the point [imath]1[/imath] is fixed. So I'm looking for a function of the form [imath]Tz= \frac{z-a}{\overline{a}z-1}\frac{\overline{a}-1}{1-a}[/imath]. I tried using [imath]a=4[/imath] which maps [imath]\frac{1}{4}[/imath] to [imath]0[/imath], but this doesn't work. So I'm not sure how to choose [imath]a[/imath]. | 1030394 | Find the linear fractional transformation that maps the circles |z-1/4| = 1/4 and |z|=1 onto two concentric circles centered at w=0?
I am very close to the solution I think. Since the circles cross the real axis, I want to find mappings from [imath]z \to w[/imath] such that [imath]1/2 \to m[/imath] and [imath]0 \to -m[/imath] and [imath]-1 \to n[/imath] and [imath]1 \to -n[/imath]. Using [imath] w = \frac{az+b}{cz+d}[/imath] we get [imath] \frac{a+2b}{c+2d} = -\frac{b}{d}[/imath] and [imath] \frac{b-a}{d-c} = - \frac{a+b}{c+d}[/imath] but where do we go from here? I am a little lost in the simplifying algebra here to solve for the [imath]a,b,c,d[/imath] in the transformation. |
1075116 | Irreducible polynomial and primes
Let [imath]n[/imath] be a prime number. How can I show that the polynomial [imath]f(x) = x^{n-1} + x^{n-2} + x^{n-3}+ \cdots + x+ 1[/imath] is irreducible over any finite field? | 433779 | Factoring polynomials of the form [imath]1+x+\cdots +x^{p-1}[/imath] in finite field
Suppose [imath]p[/imath] and [imath]q[/imath] primes and [imath]p[/imath] is odd. Then, are there nice and clever ways to factorize polynomials of the form [imath]f(x)=1+x+\cdots +x^{p-1}[/imath] in the ring [imath]\mathbb{F}_q[x][/imath] ? What about the case when [imath]q=2[/imath] ? I know that there are factorization algorithms but they are too general. I want to know if there are clever ways to do this for these special type of polynomials. Like for example, in [imath]\mathbb{F}_2[/imath] one might add terms [imath]x^r+x^r[/imath] and rearrange them so that [imath]f(x)[/imath] gets factorized. In case there are no good methods to factorize, are there nice ways to check whether [imath]f(x)[/imath] is irreducible in [imath]\mathbb{F}_q[x][/imath] ? |
1071437 | Determinant of Character table as a matrix
I'm studying for finals and came across this problem in a book. Suppose [imath]G[/imath] is a finite group with conjugacy class representatives [imath]g_1,...,g_k[/imath] and character table [imath]Z[/imath]. Consider [imath]Z[/imath] as a matrix. Then I want to show that [imath]\det(Z)[/imath] is either real or purely imaginary and that [imath]|\det(Z)|^2=\prod_{i=1}^k |C_G(g_i)|,[/imath] where [imath]|C_G(g_i)|[/imath] is the centralizer of [imath]g_i[/imath]. I am very lost as to how to approach this problem. Any help would be greatly appreciated. | 156110 | Determinant of the character table of a finite group [imath]G[/imath]
This is an exercise from the book "Groups and Representations" by Alperin & Bell. This quantity is well defined upto a sign. By column orthogonality relations, its squared norm is [imath]\displaystyle\prod_{\substack{g}} |C_G(g)|[/imath], where g runs over the representatives of the conjugacy classes. If the group is cyclic, the determinant is just a Vandermonde determinant. I wonder if there is a nice explanation for an arbitrary group. |
959002 | Find [imath]\{x\in \mathbb R\colon\neg(3x>21\implies x\leq 5) \}[/imath]
For which [imath]x\in\mathbb{R}[/imath] does [imath]\neg(3x>21\implies x\leq 5)[/imath] hold? I believe it holds for all [imath]x>7[/imath], but I don't know how to formally write this down. Can someone help me out? Is there a systematic way of doing this? | 507707 | For what [imath]a \in \Bbb R[/imath] does [imath]\neg(3a>21\implies a \leq 5)[/imath] hold?
For what [imath]a\in\mathbb{R}[/imath] will the statement [imath]\neg(3a>21\implies a\leq 5)[/imath] hold? My gut feeling says [imath]a>7[/imath], but I do not know how to formally write it down or prove it. Can someone help me? |
422120 | Known bounds for the number of groups of a given order.
The number of nonisomorphic groups of order [imath]n[/imath] is usually called [imath]\nu(n)[/imath]. I found a very good survey about the values. [imath]\nu(n)[/imath] is completely known absolutely up to [imath]n=2047[/imath], and for many other values of [imath]n[/imath] too (for squarefree n, there is a formula). In general, however, [imath]\nu[/imath] is very hard to calculate. So, I would like to at least have easy to calculate lower and upper bounds for [imath]\nu[/imath]. Are such bounds known? | 1377573 | Number of all possible groups of given order
Suppose [imath]n=18[/imath], then all possible groups of order [imath]18[/imath] is [imath]5[/imath]. Among them [imath]2[/imath] are abelian and [imath]3[/imath] are non-abelian. Let [imath]n[/imath] be a natural number. How can I determine the number of all possible groups (abelian and non-abelian) of order [imath]n[/imath]? Is there any theorem or result that can determine number of all possible groups of given order [imath]n[/imath]? |
1076094 | How can [imath] i [/imath] be distinguished from [imath] - i [/imath]?
Mathematicians designate one solution to [imath]x^2 = -1[/imath] as [imath]i[/imath] and the other as [imath]-i[/imath]. Would anybody notice if we switched their identities? Any polynomial [imath]p(x)[/imath] with a complex root will also have its conjugate as a root. Is there any equation in real numbers with a complex solution that does not have its conjugate as another solution? | 672795 | Difference between i and -i
Consider the two imaginary numbers [imath]i[/imath] and [imath]-i[/imath]. Is there any fundamental difference between the two of them? If I take the field [imath]\mathbb{C}[/imath] and apply the map [imath]a + bi \mapsto a - bi[/imath] does the image end up meaningfully different from the field I started with? Or when we write out complex numbers are we arbitrarily choosing which of the non-real solutions to [imath]z^4 = 1[/imath] to call [imath]i[/imath] and which to call [imath]-i[/imath]? |
1076348 | Evaluation of [imath]\int \frac{\sqrt{1+x^4}}{1-x^4}dx[/imath]
Evaluation of [imath]\displaystyle \int \frac{\sqrt{1+x^4}}{1-x^4}dx[/imath] [imath]\bf{My\; Try::}[/imath] Given [imath]\displaystyle \int\frac{\sqrt{1+x^4}}{1-x^4}dx\;,[/imath] Then We can write the above Integral as [imath]\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1}{2}\int\left\{\frac{1}{1-x^2}+\frac{1}{1+x^2}\right\}\cdot \frac{(1+x^4)}{\sqrt{1+x^4}}dx[/imath] So Integral is [imath]\displaystyle = \frac{1}{2}\int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx+\frac{1}{2}\int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx[/imath] Now Let [imath]\displaystyle I = \int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx[/imath] and [imath]\displaystyle J = \int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx[/imath] Now How can I evaluate [imath]I[/imath] and [imath]J\;,[/imath] plz help me Thanks | 1041531 | How to evaluate these indefinite integrals with [imath]\sqrt{1+x^4}[/imath]?
These integrals are supposed to have an elementary closed form, but Mathematica only returns something in terms of elliptic integrals. I got them from the book Treatise on Integral Calculus by Edwards. How can we evaluate them? [imath] I = \int{\frac{\sqrt{1+x^4}}{1-x^4}dx}\\ J = \int{\frac{x^2}{(1-x^4)\sqrt{1+x^4}}} dx [/imath] |
257456 | Number of semi direct products: Right answer but unsure of reason
The question is to work out the number of semi-direct products for from Q to H, where [imath] H = C_{42} , Q = C_{3} [/imath] I did: [imath]Aut(C_{42}) = C_2 \times C_6 = C_{12}[/imath] 3 divides 12 telling us that they're are some semidirect products. If we let Q = (1, 2, 3). We can see that they're are 3 elements in Q (1, 2 and 3) that divide 12, thus there are 3 semi direct products. Is this correct reasoning? | 257436 | How to construct semi direct products between Q and H
There is a semi direct product if you go [imath]\theta: Q \rightarrow Aut(H)[/imath]. [imath]H = C_{17}, Q = C_2[/imath]. [imath]Aut(H)\cong C_{16}[/imath]. From here, how do I construct the semi direct products? I said, in [imath]C_{16}[/imath], there are two elements of order that divide [imath]2[/imath], which are [imath]1[/imath] and [imath]2[/imath]. Therefore there are two semi direct products. Also [imath]17 \times 2 = 34[/imath] tells me that in both semi direct products there will be [imath]34[/imath] elements. How do I construct the semi direct products from here though? |
1076618 | If [imath]|f|+|g|[/imath] is constant then each of [imath]f, g[/imath] is constant
Let [imath]f,g: U \rightarrow \mathbb{C}[/imath] be holomorphic on the open and connected subset [imath]U[/imath]. If [imath]|f| + |g|[/imath] is constant on [imath]U[/imath] show that [imath]f, g[/imath] are constant on [imath]U[/imath]. What can we say about finite or countable number of holomorphic functions? | 193808 | sum of holomorphic functions
Does anyone know how prove the following? Suppose that [imath]f,g[/imath] are holomorphic functions on a non-empty open connected set [imath]\Omega \subset \mathbb{C}[/imath] and that [imath]|f|^2+ |g|^2[/imath] is constant on [imath]\Omega[/imath]. Show that [imath]f[/imath] and [imath]g[/imath] are constant on [imath]\Omega[/imath]. |
1076682 | How to continue on proving that rank (A+B) ≤ Rank A + Rank B?
Theorem: [imath]rank(A+B) \leq rank (A) + rank(B)[/imath]. Proof: Let [imath]U = Im(A)[/imath] and [imath]W = Im(B)[/imath]. By dimension theorem, we know that: [imath]Dim(U+W) = Dim(U) + Dim(W) - Dim (U \cap W)[/imath]. By substituting [imath]U[/imath] and [imath]W[/imath] we get: [imath]Dim(Im(A)+Im(B))= Dim(Im(A)) + Dim(Im(B)) - Dim(Im(A) \cap Im(B))[/imath]. I am stuck here. I know that [imath]dim(Im) = Rank[/imath] but I cannot continue from here. Please can someone help me? Thank-you:) | 851596 | Show [imath]\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)[/imath]
Show [imath]\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)[/imath], where [imath]A,B \in M_{m\times n}(\mathbb{F})[/imath]. I'm trying to think in terms of linear transformations. We can define [imath]T_a, T_b:\mathbb{F}^n\rightarrow \mathbb{F}^m[/imath]. I know that [imath]\dim_{\mathbb F}\operatorname{Im} T_a, \dim_{\mathbb F}\operatorname{Im} T_b \le m[/imath]. What should I do next? |
1074582 | Evaluating sums using residues [imath](-1)^n/n^2[/imath]
I am an alien towards compelx analysis, with very little know I am posing a question, who someone may want to help with. Evaluate: [imath]\frac{1}{4}\cdot \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}[/imath] In disguise this is similar to [imath]\zeta(2)[/imath] but how can this be done using residues, and complex analysis? I need some help. I am just interested. The answer is [imath]\displaystyle \frac{\pi^2}{48}[/imath] | 130666 | Complex Analysis Solution to the Basel Problem ([imath]\sum_{k=1}^\infty \frac{1}{k^2}[/imath])
Most of us are aware of the famous "Basel Problem": [imath]\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}[/imath] I remember reading an elegant proof for this using complex numbers to help find the value of the sum. I tried finding it again to no avail. Does anyone know a complex number proof for the solution of the Basel Problem? |
799601 | Existence of specific unitary transformation
Given vectors [imath]v_{1}, v_{2},\dots,v_{k}[/imath] and [imath]w_{1}, w_{2},\dots, w_{k}[/imath] which satisfy condition [imath]\forall i, j \in \{1,\dots,k\}[/imath], [imath]v_{i}\cdot v_{j} = w_{i}\cdot w_{j}[/imath] (dot product), prove that there exist such unitary transformation [imath]U[/imath] such that that [imath]U(v_{i})=w_{i}[/imath]. I can pretty much see what's going on here. These two sets of vectors are just rotated, so this tells me that there must be some kind of isometry that rotates them, but this is just intuition. | 583397 | Unitary map between sets of vectors
Suppose I have two sets of vectors, [imath]E_1=\{v_i\}_{i=1}^{k}[/imath] and [imath]E_2=\{u_i\}_{i=1}^{k}[/imath], with each vector belonging to [imath]\mathbb{C}^k[/imath]. When is it possible to find a unitary matrix that maps [imath]E_1[/imath] to [imath]E_2[/imath]? Is enough for the pairwise inner products between vectors in the set to be equal? |
994310 | How to prove [imath]\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0[/imath]?
How can I prove [imath]\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0[/imath], where [imath]a<1[/imath]? Thanks. | 352046 | A question in Complex Analysis [imath]\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx[/imath]
My problem is to integrate this expression: [imath]\int_0^{2\pi}\log(1-2r\cos x +r^2)dx.[/imath] where [imath]r[/imath] is any constant in [imath][0,1][/imath]. I know the answer is zero. Can you explain you idea to me or just prove that? Maybe you will use the "Cauchy integral theorem ". |
387333 | Are all limits solvable without L'Hôpital Rule or Series Expansion
Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion? For example, [imath]\lim_{x\to0}\frac{\tan x-x}{x^3}[/imath] [imath]\lim_{x\to0}\frac{\sin x-x}{x^3}[/imath] [imath]\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}[/imath] [imath]\lim_{x\to0}\frac{e^x-x-1}{x^2}[/imath] [imath]\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}[/imath] [imath]\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}[/imath] | 1081306 | [imath]\lim_{x\to0}\frac{\sin x-x}{x^3}[/imath] without de l'Hospital's Rule?
I know, how to calculate [imath] \lim_{x\to0}\frac{\cos x-1}{x^2} [/imath] without differential calculus. Calculating [imath] \lim_{x\to0}\frac{\sin x-x}{x^3} [/imath] using de l'Hospital's rule or Taylor expansion is also easy. Is there a method to calculate the previous limit without de l'Hospital's rule or stronger tools? |
1077576 | How can I prove [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2[/imath]
How can I prove [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2[/imath] I don't know which method can be used for this? | 87870 | Are these solutions of [imath]2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}[/imath] correct?
Find [imath]x[/imath] in [imath] \Large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}[/imath] A trick to solve this is to see that [imath]\large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}} \quad\implies\quad 2 = x^{\Big(x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}\Big)} = x^2 \quad\implies\quad x = \pm \sqrt{2} [/imath] Are these solutions correct? If not, why? If yes, are there other solutions? PS: An extension of this discussion can be found in What we can say about [imath](-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}[/imath]? |
1072458 | Non-standard 3D rotation of a set of points
I want to create a 3D surface as shown in the figure below. Toward this, I thought if I rotate a set of points in [imath]xy[/imath]-plane on a elliptical arc I may be able to get such a surface. I was thinking of rotation while changing the radius of rotation to create this surface. I maybe wrong in using this approach but here is the description of what I think. These points are a segment of an ellipse in [imath]xy[/imath]-plane. I want to find a non-standard rotation matrix which can give the trajectory of point A, as shown in the figure, as an elliptical segment in [imath]xz[/imath]-plane to ultimately create a 3D ellipsoidal surface as shown in the figure. [imath]\begin{bmatrix} x^{\prime}\\y^{\prime}\\z^{\prime} \end{bmatrix} = f\left(\phi\right)\begin{bmatrix} x\\y\\0 \end{bmatrix}[/imath] The projection of [imath]\left(x^{\prime},z^{\prime}\right)[/imath] in [imath]xz[/imath]-plane is an ellipse with the following equation: [imath]\begin{cases} x^\prime = R_x\cos\left(\phi\right)\\ z^\prime = R_z\sin\left(\phi\right) \end{cases}[/imath] The set of points to be rotated in a non-standard way by [imath]f\left(\phi\right)[/imath] to potentially create the 3D ellipsoidal surface are: [imath]\begin{cases} x = R_x\cos\left(\theta\right)\\ y = R_y\sin\left(\theta\right) \end{cases}[/imath] Could someone kindly let me know how I should find such a rotation function, [imath]f\left(\phi\right)[/imath]? | 1071328 | Creating an ellipsoidal 3D surface
I am trying to find the equation of a 3D ellipsoidal surface. I have thought of two approaches which are schematically shown below: By revolving an elliptical arc over a 3D elliptical path: Or by scaling an elliptical arc while translating in [imath]z[/imath]-direction: I thought the first approach (revolving) is easier. Here is my description of what I have done. The elliptical arc segment, which lies in [imath]xy[/imath]-plane, that revolves around [imath]y[/imath]-axis on an elliptical path (non-circular) to create a 3D ellipsoidal surface. The revolving elliptical segment has the following equation ([imath]0<\theta\le\pi/2[/imath]): [imath]\frac{x^2}{R_x^2}+\frac{y^2}{R_y^2} = 1,\mathrm{\ \ \ or\ \ \ } \begin{cases} x = R_x\cos\left(\theta\right)\\ y = R_y\sin\left(\theta\right) \end{cases}[/imath] The revolving angle is called [imath]\phi[/imath] and revolves [imath]0<\phi\le\pi/2[/imath] in the [imath]xz[/imath]-plane according to the following elliptical path: [imath]\frac{x^2}{{R^\prime}_x^2}+\frac{z^2}{R_z^2} = 1, \mathrm{\ \ \ or\ \ \ } \begin{cases} x = {R^\prime}_x\cos\left(\phi\right)\\ z = R_z\sin\left(\phi\right) \end{cases}[/imath] While revolving I want to scale down [imath]R_y[/imath] according to [imath]\phi[/imath]: [imath]R_y = h\cos\left(\phi\right)[/imath] where [imath]h[/imath] is a constant. So, as the elliptical arc revolves by [imath]\phi[/imath], [imath]R_y[/imath] decreases and so as [imath]R_x[/imath]. In the [imath]\phi[/imath]-plane, the points on the revolved ellipse has the following equations: [imath]\begin{cases} x = {\left(\left[{{R^\prime}_x}\cos\left(\phi\right)\right]^2+\left[R_z\sin\left(\phi\right)\right]^2\right)}^\frac12\cos\left(\theta\right)\\ y = h\cos\left(\phi\right)\sin\left(\theta\right)\\ z = R_z\sin\left(\phi\right) \end{cases}[/imath] When I plot the above parametric equations in MATLAB I don't get the expected surface. Could someone kindly help me? % Here is the MATLAB code I used. Rx = 1; Rz = 1; h = 2; n = 100; [theta, phi] = meshgrid(linspace(0,pi/2,n), linspace(0,pi/2,n)); x = (Rx^2*cos(phi).^2+Rz^2*sin(phi).^2).^0.5.*cos(theta); y = h*cos(phi).*sin(theta); z = Rz*sin(phi); surf(x,y,z), shading interp, axis equal, xlabel('x'), ylabel('y'), zlabel('z') Here is the MATLAB result: |
1078981 | Is [imath]x\in\{\{x\}\}[/imath]
Is [imath]x\in\{\{x\}\}[/imath]. I understand that [imath]x\in\{x\}\in\{\{x\}\}[/imath] does this mean [imath]x\in\{\{x\}\}[/imath]? Very simple just unsure about the properties of [imath]\in[/imath], not looking for an extravagant answer, thanks in advance for the help. | 1064663 | [imath]x\in \{\{\{x\}\}\}[/imath] or not?
I wonder if we can we say [imath]x\in \{\{\{x\}\}\}[/imath]? In one viewpoint the only element of [imath]\{\{\{x\}\}\}[/imath] is [imath]\{\{x\}\}[/imath]. In the other viewpoint [imath]x[/imath] is in [imath]\{\{\{x\}\}\}[/imath], for example all people in Madrid are in Spain. |
1079174 | find sum of ln(1- [imath]\frac{1}{n^2})[/imath]. prove it converges
![enter image description here][1] [imath] \sum = \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 - \frac{1}{16}) = \ln(\frac{3}{4}) + \ln (\frac{8}{9}) + \ln ( \frac{15}{16}) = -.287 + -.117 + -.064 = -.50 [/imath] it converges to [imath]-\frac{1}{2}[/imath]. | 1078043 | Proving the convergence of [imath]\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)[/imath]
How to prove that the series [imath]\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)[/imath] is convergent? What about finding the sum? My attempt: [imath]\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)[/imath] The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum. |
1079244 | Show this holomorphic function is costant
I have a holomorphic function [imath]f[/imath] defined on a neighborhood of the closed unit square. Further [imath]f(z+i)=f(z)[/imath] on [imath][0,1][/imath] and [imath]f(z+1)=f(z)[/imath] on [imath][0,i][/imath]. How do I show that [imath]f[/imath] is constant? | 1079108 | Show this holomorphic function is constant
I have a holomorphic function [imath]f[/imath] defined on a neighborhood of the closed unit square. Further [imath]f(z+i)-f(z)[/imath] is real and not negative on [imath][0,1][/imath] and [imath]f(z+1)-f(z)[/imath] is real and not negative on [imath][0,i][/imath]. How do I show that [imath]f[/imath] is constant? |
1076717 | If [imath]k, m, n[/imath], are natural numbers and [imath]k \leq n[/imath] What is the final answer of this :
If [imath]k, m, n[/imath], are natural numbers and [imath]k \leq n[/imath] What is the final answer of this: [imath]\sum_{r=0}^{m}\frac{k\binom{m}{r}\binom{n}{k}}{(r+k)\binom{m+n}{r+k}}[/imath] | 1076664 | Binomial-coefficients if, k, m, n natural numbers and k \leq n the result of
If [imath]k, m, n[/imath], are natural numbers and [imath]k \leq n[/imath] What is: [imath]\sum_{r=0}^{m}\frac{k\binom{m}{r}\binom{n}{k}}{(r+k)\binom{m+n}{r+k}}[/imath] |
1079280 | Proving that [imath] \prod_{i=1}^{n} (1 + a_{i}) \ge \left(1 + \prod_{i=1}^{n} a_{i}^{1/n} \right)^{n} [/imath]
I'd like some help on proving the following inequality [imath] \prod_{i=1}^{n}\left(1+a_{i}\right) \ge \bigg(1 + \prod_{i=1}^{n} a_{i}^{1/n}\bigg)^{n}, [/imath] given that [imath] a_{i} > 0\,\, \forall\, i\in\{1,2,\cdots,n\} [/imath]. I tried to use AM-GM inequality on the right-hand side, but without success. Any help would be appreciated. | 1078267 | Prove [imath](x+r_1) \cdots (x+r_n) \geq (x+(r_1 \cdots r_n)^{1/n})^{n}[/imath].
I can show that for [imath]x > 0[/imath] and [imath]r_{i} > 0[/imath] we have [imath] \left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.[/imath] However, I can't do this using straight up induction, strong or weak. Can someone do this? You can show the inequality holds for [imath]n[/imath] if it holds for [imath]n+1[/imath] using [imath]r_{n+1} = (r_1 \cdots r_n)^{1/n}[/imath], so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for [imath]n=a[/imath] and [imath]n=b[/imath] then it holds for [imath]n=ab[/imath]. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal". You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of [imath]x^k[/imath] and use the AM-GM inequality. |
1079510 | A question about eliminating square roots
If [imath]\sqrt{x^2} = \pm x[/imath], then why does [imath]\sqrt{(x+2)^2} = x+2[/imath] and not [imath]\pm (x+2)[/imath]? This is driving me crazy, so feel free to elucidate. Thanks! ---EDIT--- I'm not sure how the other questions' answers would help answer my own question, but it doesn't really matter now that I've figured it out. After thinking a bit about some of what @Daniel Hast typed on a similar question I asked, I realized that the reason why [imath]\sqrt{(x+2)^2} = x+2[/imath] was that more generally, given that [imath]\sqrt{x} = r[/imath] such that [imath]r^2 = x[/imath], squaring [imath]r[/imath] in [imath]\sqrt{x} = r[/imath] produced [imath](\sqrt{x})^2 = r^2 = x[/imath] and subsequently [imath](\sqrt{x})^2 = x[/imath]. | 1078370 | Why does the square root of a square involve the plus-minus sign?
If [imath]\sqrt{x^2}[/imath] can be simplified as follows: [imath]\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{\frac{2}{1}\times\frac{1}{2}} =x^\frac{2}{2} = x^1 = x[/imath] Then why would [imath]\sqrt{x^2} = \pm x[/imath]? |
1074345 | Two square matrices with the same minimial polynomial are similar for [imath]n=5[/imath] or [imath]n=6[/imath]
Let [imath]\mathbb{F}[/imath] be a field, [imath]\lambda \in \mathbb{F}[/imath] and [imath]A,B \in M_n(\mathbb{F})[/imath] such that [imath]m_A(x)=m_B(x)=(x-\lambda)^k[/imath] and such that the geometric multiplicity of [imath]\lambda[/imath] in [imath]A[/imath] equals to the geometric multiplicity of [imath]\lambda[/imath] in [imath]B[/imath]. a. Prove that for [imath]n=5[/imath] or [imath]n=6[/imath]- [imath]A[/imath] and [imath]B[/imath] are similar b. Find an example for [imath]n=7[/imath] for which [imath]A[/imath] and [imath]B[/imath] are NOT similar My main question here is whether I should try proving it for [imath]n=5[/imath] and [imath]n=6[/imath] seperately? Or should I try proving it for any [imath]n[/imath] and then observe that it holds only for [imath]n=5[/imath] or [imath]n=6[/imath]? Plus, any hints or suggestion regarding my main question or the question above will be welcome | 1049677 | What are examples of two non-similar invertible matrices with same minimal and characteristic polynomial and same dimension of each eigenspace?
I'm trying with matrices over [imath]\mathbb F_2[/imath] and trying to have a look at the Jordan canonical forms of these matrices. If the size of the biggest Jordan block is the same with 1's in all diagonal entries, we do get non-similar invertible matrices with same minimal and characteristic polynomial. But what do I do for satisfying the last condition on eigenspace? Please provide examples if possible and also please explain why it works. |
1079633 | Planning to integrate [imath]\int_{0}^{\infty} \frac{\log^2(x)}{x^2 + 1} dx[/imath] using complex analysis
This is just a plan-out. I want to evaluate: [imath]\int_{0}^{\infty} \frac{\log^2(x)}{x^2 + 1} dx[/imath] Using a keyhole contour a semi-circle, with base at the x-axis. First I must pick a branch. Since the logarithm is multivalued in that, [imath]\log(e^{i2\pi}) = \log(e^{(2ki\pi) + i2\pi})[/imath] We consider: [imath]f(z) = \displaystyle \frac{\log^2(z)}{z^2 + 1}[/imath] But how should I pick a branch. (1) What is a "Branch?" (2) Should I pick a place where the [imath]\log[/imath] is continous. That is everywhere except [imath]z=0[/imath] So then my contour will be half-washer shape. with the opening in the middle: Half a Washer Thanks | 191736 | Help with integrating [imath]\int_0^{\infty} \frac{(\log x)^2}{x^2 + 1} \operatorname d\!x[/imath] - contour integration?
George Arfken's book: Mathematical Methods for Physicists has the following problem in a chapter on contour integration: [imath]\displaystyle \int_0^{\infty} \dfrac{(\log x)^2}{x^2 + 1} dx[/imath]. Their suggestion is to make the substitution [imath]x \rightarrow z=e^t[/imath]. I'm not sure what they meant by this, but I tried making the substitution [imath]x = e^t[/imath], which turns the integral into: [imath]\displaystyle\int_{-\infty}^{\infty} \dfrac{t^2 e^t}{1+e^{2t}} dt[/imath]. The hint is then to take the contour from -R to R, to R+[imath]\pi i[/imath], to -R + [imath]\pi i[/imath], to -R. Since this has a pole at [imath]t = \pi i/2[/imath], a Laurent series expansion about this point gives the residue as [imath]i \pi^2/8[/imath], so the contour integral equals [imath]-\pi^3/4[/imath]. I've been able to show that the integral along R to R + [imath]\pi i[/imath] and along -R + [imath]\pi i[/imath] to -R goes to zero by the ML inequality - the denominator grows exponentially but the numerator quadratically. But at this point, I'm a bit lost as to what to do with the integral over Im [imath]t = \pi[/imath]. Any help? The book gives the answer as [imath]\pi^3 /8[/imath]. |
1080081 | Produce all solutions for [imath]2a^2 + b^2 = c^2[/imath]
I'm trying to generate a stream of tuples [imath](a, b, c)[/imath] that satisfy the equation [imath]2a^2 + b^2 = c^2[/imath] The first [imath]10[/imath] results are: [imath]2, 1, 3[/imath] [imath]4, 2, 6[/imath] [imath]6, 3, 9[/imath] [imath]4, 7, 9[/imath] [imath]6, 7, 11[/imath] [imath]8, 4, 12[/imath] [imath]10, 5, 15[/imath] [imath]12, 1, 17[/imath] [imath]12, 6, 18[/imath] [imath]8, 14, 18[/imath] There is a solution for [imath]x_2+2y_2=z_2[/imath] which is exactly the same but I cannot understand how I can produce all these numbers as a function. For instance, it says [imath]x_3=(a_2+2b_2)t=3[/imath] which is the solution of [imath]c[/imath], in my case, but it includes [imath]x_3[/imath] because [imath]t = \dfrac{1}{2}(x_1+x_3)[/imath]. In addition, he gives the values [imath]\{x_1,x_2,x_3\} = \{1,2,3\}[/imath] but I'm not sure which are the next values. Any help please. Thank you in advance!!! | 367443 | Find all solutions: [imath]x^2 + 2y^2 = z^2[/imath]
I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem: Find all solutions in positive integers to the following Diophantine equation [imath]x^2 + 2y^2 = z^2[/imath] I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem |
556351 | Proving that any finite integral domain is a field.
Let [imath]R[/imath] be an integral domain, and let [imath]a \in R, a \neq 0[/imath]. Let [imath]f_a: R \rightarrow R[/imath] be defined by [imath]f_a(r)=ar[/imath]. Prove that [imath]f[/imath] is injective. Then prove that every finite integral domain is a field. My ideas and concept: Can someone help articulate my understandings into a proper solution? | 2706822 | Prove that every finite integral domain is a field.
I am a little confused about the proof for the following theorem: Every finite integral domain is a field. [imath]R[/imath] is a finite integral domain with [imath]R^{*}[/imath] being the set of nonzero elements of [imath]R[/imath]. The proof defines a map for all [imath]a\in R^{*}[/imath] to be [imath]\lambda_{a}(r)=ar[/imath] for [imath]r\in R^{*}[/imath]. It showed it was injective. Then this is the part I am confused about: Since [imath]R^{*}[/imath] is a subset of [imath]R[/imath], [imath]R^{*}[/imath] is finite hence must be surjective. Could someone please explain this implication? |
1080575 | Proof that sum of first [imath]n[/imath] cubes is always a perfect square
I know that [imath]1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2[/imath] What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first [imath]n[/imath] cubes always a perfect square. | 1928283 | Induction involving series with exponents [imath](1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3[/imath]
Does anyone have any hints for proving the following via mathematical induction? Obviously the base case is easy enough... I'm getting stuck on how these exponents may be manipulated to show [imath]S_k\Rightarrow S_{k+1}[/imath]. Here is the proposition: Prove that [imath](1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3[/imath] for every [imath]n\in\mathbb{N}[/imath]. |
1080451 | solving a third-order nonlinear ordinary differential
I would like to solve: [imath]\left (\frac{d^{2}y}{dx^{2}} \right)^{2}+\frac{d^{3}y}{dy^{3}} \frac{dy}{dx}=0[/imath] Thanks in advance. | 1080458 | solving third-order nonlinear ordinary differential equation
I would like to solve: [imath](\frac{d^{2}y}{dx^{2}})^2+\frac{d^{3}y}{dx^{3}} \frac{dy}{dx}=0[/imath] Thanks in advance. |
481035 | Can This Matrix Proof Be Done Without the Definition?
The following was taken from Stephen Friedberg's Linear Algebra, 2nd Edition. Let [imath]A[/imath] be a [imath]m \times n[/imath] matrix, [imath]B[/imath] and [imath]C[/imath] be [imath]n \times p[/imath] matrices. Show that [imath]A(B+C) = AB + BC[/imath] and, for any [imath]k \in F[/imath], [imath]k(AB) = (kA)B = A(kB)[/imath]. Proof: [imath][ A(B+C) ]_{ij} = \sum_{i=1}^n A_{ik}(B+C)_{kj}[/imath] [imath]= \sum_{i=1}^n A_{ik}B_{kj} + \sum_{i=1}^n A_{ik}C_{kj} = (AB)_{ij} + (BC)_{ij} = [AB + BC]_{ij}[/imath] I did the second part myself: We have [imath][k(AB)]_{ij} = k \sum_{i=1}^n A_{ik}B_{kj}[/imath] [imath] = \sum_{i=1}^n kA_{ik}B_{kj} = \sum_{i=1}^n (kA)_{ik}B_{kj}[/imath] [imath] = \sum_{i=1}^n A_{ik}kB_{kj} = \sum_{i=1}^n A_{ik}(kB)_{kj}[/imath] As a result, [imath]k(AB) = (kA)B = A(kB)[/imath]. I noticed that the proof used the definition of matrix addition and scalar multiplication. [imath](B+C)_{ij} = B_{ij} + C_{ij}[/imath] and [imath](kA)_{ij} = k(A)_{ij}[/imath]. Now is it possible to prove without the definitions? I became curious because Mr. Friedberg introduced the matrix as a representation of a linear transformation over an ordered basis. So, I tried proving via the linear transformation over an ordered basis and could not do it. For example: Let [imath]\alpha = \{x_1, x_2 ... x_n\}[/imath] be an ordered basis of [imath]V, \gamma = \{z_1, z_2 ... z_m\}[/imath] be an ordered basis of [imath]W[/imath], and [imath]T: V \rightarrow W[/imath] be a linear transformation. Elsewhere, Mr. Friedberg already defined that [imath](kT)(x_j) = kT(x_j)[/imath]. Suppose that [imath]A[/imath] is the matrix representation of [imath]T[/imath]. [imath]k(A) = k *[ \space T(x_1) \space T(x_2) \space ... \space T(x_n) \space ] = [\space kT(x_1) \space kT(x_2) \space ... \space kT(x_n) \space ] = [ \space (kT)(x_1) \space (kT)(x_2) \space ... \space (kT)(x_n) \space][/imath] But this is possible only because of the definition of scalar multiplication. | 464151 | Proving Distributivity of Matrix Multiplication
If [imath]A,B,C[/imath] are matrices I am thinking how to show that [imath] A(B + C) = AB + AC[/imath] Is possible to show without sums like [imath]\sum_i a_i, ..., \sum_j b_j[/imath]? It seems if I do the proof with many indexes then is tedious and I don't learn much from it. |
912468 | Comparison theorem for ODE
Here is something I'm trying to prove: Conjecture: Suppose [imath]f'(x) \leq \phi(f(x), x)[/imath] and [imath]f(a)=\alpha[/imath]. Suppose [imath]g'(x)=\phi(g(x),x)[/imath] and [imath]g(a)\geq \alpha[/imath]. Then [imath]f(x)\leq g(x)\,\,\forall x[/imath]. It definitely seems like it should be true, and I don't think we even need continuity of [imath]\phi[/imath]. (Edit: a user has correctly pointed out that we should require [imath]\phi[/imath] be locally Lipshitz in the first variable, uniformly with respect to the second variable. Let's add that assumption.) I can prove it in the case that the inequality is strict. I'll place my proof below the fold. How can I extend it to weak inequality? If it's wrong, I'd love to see a counterexample. A reference is fine; I have the book by Teschl, for example. A similar question with stronger assumptions was asked here. Proposition: Suppose Suppose [imath]f'(x) < \phi(f(x), x)[/imath] and [imath]f(a)=\alpha[/imath]. Suppose [imath]g'(x)=\phi(g(x),x)[/imath] and [imath]g(a)\geq \alpha[/imath]. Then [imath]f(x)\leq g(x)\,\,\forall x[/imath]. Proof: Suppose not. Suppose WLOG there is a [imath]b>a[/imath] such that [imath]f(b)>g(b)[/imath]. Let [imath]c:=\inf \{x>a:f(x)>g(x)\}[/imath]. By definition of the derivative we have [imath]f'(c)\geq g'(c)[/imath], a contradiction, since [imath]f'(c)<\phi(f(c),c)=\phi(g(c),c))=g'(c)[/imath]. | 2957645 | If [imath]x'(t) \leq f(x(t))[/imath], then [imath]x(t)\leq y(t)[/imath] for which [imath]y'(t)=f(y(t))[/imath]?
Let [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] be some Lipschitz function, and assume [imath]x(t)[/imath] satisfies [imath]x'(t) \leq f(x(t))\ {\rm with }\ x(0) = a[/imath] Let [imath]y(t)[/imath] be the unique solution to [imath]y'(t) = f(y(t))[/imath] with [imath]y(0)=a[/imath], does it then hold that [imath]x(t) \leq y(t)[/imath] for all [imath]t\geq 0[/imath]? |
1080949 | Evaluate [imath]\int_{0}^{1} \frac{\log(x)}{\sqrt{1 - x^2}}[/imath] complex integration
Evaluate: [imath]2\cdot\int_{0}^{1} \frac{\log(x)}{\sqrt{1 - x^2}} dx[/imath] Using Complex Integration. I want to do something with the unit circle, but I am not quite sure how to work-around with the unit circle and the integral. Will someone have an idea? Thanks =) | 1080710 | Integral of [imath]\log(\sin(x))[/imath] using contour integrals
I know the integral is possible with a simple fourier series expansion of [imath]-\log(\sin(x))[/imath] But I am interested in complex analysis, so I want to try this. [imath]I = \int_{0}^{\pi} \log(\sin(x)) dx[/imath] The substitution [imath]x = \arcsin(t)[/imath] first comes to mind. But that substitution isnt valid as, The upper and lower bounds would both be [imath]0[/imath] because [imath]\sin(\pi) = \sin(0) = 0[/imath] What is a workaround using inverse trig or some other way? Inverse sine is good, because that gives us a denominator from which we will be able to find poles. [imath]x = arctan(t)[/imath] Is also good, but it would hard to do. Idea? |
1080822 | Linear equations in 3D space
I need to search a line in a 3D space. I have a starting point (coordinates) of the line and the angle at which it is suppose to go (relative to each of the axis). I need to start from the starting coordinates and search every element (every coordinate contains an element) [imath]1[/imath] by [imath]1[/imath] until I find one that matches the condition. Imagine a bullet going through air to some target. Doesn't seem that hard, but I haven't managed to find the solution for this. In 2D space this would be something like [imath]y = x\cos(\mbox{angle}) + \mbox{startPosition}[/imath] I assume you'll need [imath]2[/imath] formulas for my problem. Any idea how to solve it? | 404440 | What is the equation for a 3D line?
Just like we have the equation [imath]y=mx+b[/imath] for [imath]\mathbb{R}^{2}[/imath], what would be a equation for [imath]\mathbb{R}^{3}[/imath]? Thanks. |
1081930 | Question about how to prove [imath]x^5\equiv x \pmod {10}[/imath]
I was trying to prove why [imath]x^5\equiv x \pmod {10}[/imath] for all natural numbers [imath]x[/imath]. I saw a proof where they applied Euler's theorem to show this. They said that the totient function for [imath]10[/imath] is [imath]4[/imath]. ([imath]φ(10) = 4[/imath]). Therefore, [imath]x^5 ≡ 1^4 \cdot x^1 ≡ x\ \pmod { 10}[/imath] However my question is: This isn't a correct proof is it? Because it requires [imath]x[/imath] to be relatively prime with [imath]10[/imath] for this to work. So it's a partial proof only (it would prove for [imath]1[/imath], [imath]3[/imath], [imath]7[/imath], and [imath]9[/imath] but not for the other [imath]x[/imath]'s)? If this is true, is there a way to generalize using Euler's Theorem to prove for all [imath]x[/imath]? Or is there another way to go about doing this proof? | 451927 | [imath]n[/imath] and [imath]n^5[/imath] have the same units digit?
Studying GCD, I got a question that begs to show that [imath]n[/imath] and [imath]n^5[/imath] has the same units digit ... What would be an idea to be able to initiate such a statement? testing [imath]0[/imath] and [imath]0^5=0[/imath] [imath]1[/imath] and [imath]1^5=1[/imath] [imath]2[/imath] and [imath]2^5=32[/imath] In my studies, I have not got "mod", please use other means, if possible of course. I demonstrated in a previous period that [imath]2|n^5-n[/imath]because [imath]n^5-n=(n+1)n(5n^4+5n+5)[/imath], and[imath]5|n^5-n[/imath]By Fermat's Little Theorem Only I do not understand what should happen to the units of the two numbers are equal ... What must occur? |
1082361 | Number of words that can be formed using the word PHILOSOPHY
In the word PHILOSOPHY how many words would have the letters H,I,S,Y together when words are formed by using all 10 letters? This is what I did: Consider HISY as one letter.Then no.of words that can be formed with HISY together=[imath]7!*4!\over{2!2!}[/imath] But this includes the cases where the other H is in the following places: HHISY HISYH ISYHH. These are counted twice in the original calculation.Therefore number of ways of HHISY can happen(considering HHISY as one letter)=[imath]6!*3!\over{2!*2!}[/imath]. As HISYH ISYHH can also happen in [imath]6!*3!\over{2!*2!}[/imath] ways the required answer is= $7!*4!\over{2!2!}[imath]-([/imath]6!*3!\over{2!*2!}$ [imath]*3[/imath][imath])=[/imath] [imath]27000[/imath]. Is this correct?I am not sure of [imath]6!*3!\over{2!*2!}[/imath] [imath]*3[/imath] part | 636478 | Different arrangements of the word PHILOSOPHY
I want to figure out the number of different arrangements using all the letters in PHILOSOPHY such that the letters H,I,S,Y always stick together. The way I solved this is given below ; Selecting a H from the 10 letters = [imath]^2C_1[/imath] Hence total arrangements = $^2C_1[imath]6![/imath]4!$/[imath](2!2!)[/imath] I am not sure whether I have done this correctly. Appreciate any assistance thanks |
1082413 | On an algebra [imath]F[x][/imath], defining operators [imath]T,D[/imath] as integration, differentiation, respectively, show that [imath]DT = I[/imath]
I am finding difficulty in constructing a proof as requested in the following: Let [imath]F[/imath] be a subfield of the complex numbers and let [imath]T[/imath], [imath]D[/imath] be the transformations on [imath]F[x][/imath] defined by [imath] T \left( \sum_{i=0}^n c_i x^i \right) = \sum_{i=0}^n \frac{c_i}{1+i} x^{i+1}[/imath] and [imath] D \left( \sum_{i=0}^n c_i x^i \right) = \sum_{i=1}^n i c_i x^{i-1}.[/imath] I am asked to show that [imath]DT = I[/imath], where [imath]I[/imath] is the identity. (Please note the indices on the above sums.) Clearly, [imath]T[/imath] and [imath]D[/imath] correspond to integration and differentiation, respectively. I felt that this proof should be straightforward, but I'm not able to get it quite right. Here is what I have: [imath] \begin{align} (DT)\left(\sum_{i=0}^n c_i x^i \right) &= D \left(\sum_{i=0}^n \frac{c_i}{1+i} x^{i+1} \right) \\ &= \sum_{i=1}^n c_i x^i \end{align} [/imath] This uses the definitions of the operators, as given. However, the lower index in the last sum is [imath]i= 1[/imath], which is one higher than the [imath]i=0[/imath] in the original sum that [imath]DT[/imath] was applied to; in loose wording, the constants are gone (i.e., we integrated and then differentiated but the constants in the original expression are now gone). The value of this lower bound makes sense because of the [imath]c^{i-1}[/imath] term; if [imath]i[/imath] can be zero, we would have [imath]x^{-1}[/imath], which is not defined in this contact. To me, resolving this seems impossible, based on the definition of the operator [imath]D[/imath] as given in the text; no matter what the train of logic is, I do not see a way for the lower bound of the final index to be zero, which seems to me to be necessary in order to conclude that [imath]DT =I[/imath]. I do not see how to get around this in order to complete this proof. Where is my error? A comment on my question suggests that we can simply index the aforementioned sum from [imath]i=0[/imath] since [imath]ic_ix^{i-1}[/imath] can be taken to have value zero for [imath]i = 0[/imath]. However, the second part of this exercise asks to show that [imath]TD \neq I[/imath], which does seem to rely on the indices of the sums in the given definitions. Where is the subtlety here? This is Exercise 9 in Section 3.7 of Linear Algebra by Hoffman and Kunze. | 1024291 | Problem involving Derivative and Antiderivative operators in Hoffman and Kunze's Linear Algebra
I have a homework problem from Hoffman and Kunze's Linear Algebra. Let F be a subfield of [imath]\mathbb{C}[/imath] and let [imath]T[/imath], [imath]D[/imath] be the transformations of [imath]F[x][/imath] defined by [imath]\begin{align} T\left(\sum_{i=0}^n{c_ix^i}\right) &=\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1} \\ D\left(\sum_{i=0}^n{c_ix^i}\right) &=\sum_{i=1}^nic_ix^{i-1} \end{align}[/imath] These look like the anti-derivative and derivative operators. I am to show that [imath]DT=I[/imath] but [imath]TD\neq I[/imath]. So, [imath]DT\left(\sum_{i=0}^n{c_ix^i}\right)=D\left(\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1}\right)=\sum_{i=1}^n{(i+1)}\frac{c_i}{i+1}x^{i+1-1}=\sum_{i=1}^n{c_ix^i}.[/imath] It is close, but my index is off by 1, and so they can't be equal since I am missing my constant term. What am I missing? Do I need to rewrite the argument in the derivative operator before I operate? |
470804 | Simple Ordinary Differential Equation
As simple as this should be, I can not seem to solve it. I can't classify its type and thus figure out how to solve it. It's the only ODE in my problem sheet that I can't solve (embarrassing). [imath]t\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)=x+\sqrt{t^2+x^2}[/imath] | 458539 | Initial value problem [imath]t\frac{dx}{dt}=x+\sqrt{t^2+x^2}[/imath]
Another question on ODEs, this time just wondering how I should start with this one. [imath]t\frac{dx}{dt}=x+\sqrt{t^2+x^2}, \qquad x(1)=0.[/imath] It looks like a linear ODE and so after playing around with it I eventually got [imath]\frac{dx}{dt}-\frac{2}{t}x=\frac{dt}{dx}[/imath] to which I multiplied the integrating factor [imath]I(t)=t^{-2}=\frac{1}{t^2}.[/imath] This in turn gave me [imath]\frac{x}{t^2}=\int{\frac{1}{t^2}t'dt},[/imath] and then I used integration by parts to get [imath]\frac{x}{t^2}=\frac{-1}{t}+C.[/imath] Applying the initial conditions gave me [imath]x=t(t-1).[/imath] The answer in the book is however [imath]x=\frac{1}{2}(t^2-1).[/imath] Where did I go wrong? What should I do to solve this problem? Thanks. |
1082590 | Maclaurin series of [imath]\arctan(x)[/imath] up to degree [imath]4[/imath]
How can I find the Maclaurin series up to degree 4 for: [imath]\arctan(x)[/imath] Calculating the derivatives becomes complex very quickly. Is there a special expansion for [imath]\arctan[/imath] like there is for [imath]\cos(x)[/imath] and [imath]\sin(x)[/imath]? | 29649 | Why is [imath]\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots[/imath]?
Why is [imath]\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots[/imath]? Can someone point me to a proof, or explain if it's a simple answer? What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series. A lot of the information I find when looking for that seems to point back to arctan. |
1082626 | How to find the value of [imath]\int_0^1 \frac{x-1}{\log x}\,dx[/imath]?
I want to find the value of [imath]\int_0^1 \frac{x-1}{\log x}\,dx[/imath] where [imath]\log x[/imath] stands for the natrual logarithm. I put it in the wolfram alpha, and it saids it's [imath]\log 2[/imath]. (Refer to : http://www.wolframalpha.com/input/?i=int_0%5E1+%28x-1%29%2F%28lnx%29) But I can't figure out how it comes to that value. Help me to find out the procedure from that integration to get [imath]\log 2[/imath]. | 1030296 | Evaluate [imath]\int_0^1 \frac{x^k-1}{\ln x}dx [/imath] using high school techniques
Is there a way to compute this integral, [imath]\int_0^1 \frac{x^k-1}{\ln x}dx =\ln({k+1})[/imath] without using the derivation under the integral sign nor transforming it to a double integral and then interchanging the order of integration. High school techniques only if possible. |
1045477 | convoluted recurrence: [imath]f(2n)=f(n)+f(n+1)+n, f(2n+1)=f(n)+f(n-1)+1[/imath]
For the recurrence relation: [imath]f(0)=1[/imath] [imath]f(1)=1[/imath] [imath]f(2)=2[/imath] [imath]f(2n)=f(n)+f(n+1)+n\ \ \ (\forall n>1)[/imath] [imath]f(2n+1)=f(n)+f(n-1)+1\ \ \ (\forall n\ge 1)[/imath] (the first numbers of the sequence are: 1, 1, 2, 3, 7, 4, 13, 6, 15, 11, 22, 12, 25, 18, 28) Given some number [imath]S[/imath], I would like to find [imath]n[/imath] such that [imath]f(n) = S[/imath], using an approach that runs in [imath]O(\log n)[/imath] time. Any ideas or tips how to solve it? I tried substitution so far but didn't get far. | 1072191 | Relation between 2 recurrence equations.
I am stuck with the following recurrence relations (actually asked in a programming question).Given the following relationships, is it possible to find the index of the occurrence of any given number which is in the series? [imath]f(0) = 1[/imath] [imath]f(1) = 1[/imath] [imath]f(2) = 2[/imath] [imath]f(2n) = f(n) + f(n + 1) + n[/imath] (for [imath]n > 1[/imath]) [imath]f(2n + 1) = f(n - 1) + f(n) + 1[/imath] (for [imath]n >= 1[/imath]) So the series is like:(Index: 0,1,2,3...n) [imath]1,1,2,3,7,4,13,6,15,11,22,12,25,18,28,20,34...[/imath] Example: Given: [imath]f(n) = 8074[/imath] Result: [imath]n = 2441[/imath] Also, some other condition given are: A number may occur more than once - the resultant index should be the last time it occurs. An example: [imath]f(14812) = 65389[/imath] Also, [imath]f(16623) = 65389[/imath] So, the result is [imath]n = 16623[/imath] By programming in python, I was able to observe that any number that repeats occurs utmost twice. Also, since the input can be pretty large - simple recursion (with hashing) takes too long and mostly causes memory errors. So, I tried to reduce the first recurrence formula by substituting the second in it, but still I was not able to find any proper relation between [imath]f(n)[/imath] and [imath]f(2n)[/imath] or [imath]f(n-1)[/imath] since only [imath]f(n)[/imath] is given as input. Is it possible to solve this using any other relations? (Edit: I've also posted on StackOverflow here) |
1081039 | Integrating [imath]\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx[/imath]
I am working on [imath]\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx[/imath] First: I use integrating by part then get [imath] x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx [/imath] then I have [imath]\displaystyle -\int_0^\pi \arctan(\sin x)\,dx[/imath] because [imath]x\arctan(\sin x)\Big|_0^\pi[/imath] is equal to [imath]0[/imath] However, I don't know how to integrate [imath]\displaystyle -\int_0^\pi \arctan(\sin x)\,dx[/imath] Can someone give me a hint? Thanks | 323603 | Integrate [imath]\int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}[/imath]
Integrate [imath]\displaystyle \int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}[/imath] |
296548 | I need assistance in integrating [imath] \frac{x \sin x}{1+(\cos x)^2}[/imath]
Find the integral [imath] \int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}[/imath] | 1244092 | Bonus integration problem we got at class: Integrate [imath]\frac {x \sin x}{1+\cos^2x}[/imath] between [imath]0[/imath] and [imath]\pi[/imath]
Bonus integration problem we got at class: Integrate [imath]\frac {x \sin x}{1+\cos^2x}[/imath] between [imath]0[/imath] and [imath]\pi[/imath] So the lecturer gave this problem. I tried this really hard but couldn't succeed. It doesn't give me any bonus points at class, he just gave it as a nice challenge and it got me quite frustrated. Any help will be great! Thank you. |
1083068 | How can you prove [imath] \left ( 1 + \frac{1}{k} \right ) ^ {k} < \left ( 1 + \frac{1}{k+1} \right ) ^ {k+1} [/imath]
I'm re-reading my old math books, and froze on the following theorem: [imath] \lim_{x\to\infty} \left ( 1 + \frac{1}{x} \right ) ^ {x} = e [/imath] [imath] e = 2,71... [/imath] I started thinking about it, and went on to prove: [imath] \left ( 1 + \frac{1}{k} \right ) ^ {k} < \left ( 1 + \frac{1}{k+1} \right ) ^ {k+1} [/imath] ..but couldn't. Please help. | 1011414 | Proof [imath](1+1/n)^n[/imath] is an increasing sequence
I need help proving [imath]a_n=\left(\dfrac{n+1}{n}\right)^n[/imath] is increasing sequence on the positive integers. An exercise in the analysis book by Mattuck asks to prove [imath]a_n=\left(\dfrac{2^n+1}{2^n}\right)^{2^n}[/imath] is increasing. But this is easy since [imath]\left(\dfrac{2^n+1}{2^n}\right)^{2n}=\left(\left(\dfrac{2^n+1}{2^n}\right)^{2}\right)^{2^{n-1}} =\left(\dfrac{2^{2n}+2\left(2^n\right)+1}{2^{2n}}\right)^{2^{n-1}}=[/imath] [imath]\left(\dfrac{2^{n-1}+1+\frac{1}{2^{n+1}}}{2^{n-1}}\right)^{2^{n-1}}>\left(\dfrac{2^{n-1}+1}{2^{n-1}}\right)^{2^{n-1}}[/imath] as desired. I am having difficulty in the general case: I must prove [imath]\left(\frac{n+2}{n+1}\right)^{n+1}>\left(\frac{n+1}{n}\right)^n[/imath]. This is equivalent to [imath](n+2)^{n+1}n^n>(n+1)^{2n+1}[/imath] which is the same as proving [imath]\frac{n+2}{n+1}>\left(\frac{(n+2)(n)}{(n+1)^2}\right)^n[/imath] which is the same as [imath]\frac{n+2}{n+1}>\left(1-\frac{1}{(n+1)^2}\right)^n[/imath] I'm stumped. Thanks in advance. Oh, and I don't want to use the derivative. |
235114 | How to solve complex equation, all values?
[imath](-81)^{1/4}[/imath] The answer, I got was -3,3..but when I input it in the webwork, it does not work. What is the method to solve this? | 487739 | Determine the fourth roots of -16
Determine the fourth roots of -16 in the form [imath]x +iy[/imath] where [imath]x[/imath] and [imath]y[/imath] are not trigonometric functions. I do not even know what they really want from me in this question. My initial thought wass: [imath]\sqrt[4]{-16} = 0 + 2i = 2i[/imath] but that seems overly simplistic for a problem that counts as 8 points in an exam. |
621705 | finding slope of a curve using derivative formulas
Why the slope will be infinitive if the tangent of curve is verticle ? Suppose [imath]x^2+2ax+y^2=0[/imath] is a curve and its tangent is verticle to [imath]x[/imath]-axis. From what basis we call the slope of this axis will be infinitive? Actually I searched a lot but didn't find any specific answer. Thanks in advance | 844565 | Slope of the tangent line, Calculus
Find the slope of tangent line to the curve at the point [imath](1, \pi/2)[/imath] the equation is [imath]\sin(xy) = x[/imath] The right answer was = Slope is infinite My answer was taking the derivative and I end up in this equation [imath] (1-y\cos(xy))/ x\cos(xy) [/imath] I substitute with the points and the calculator showed an error, Does the error means infinite ? |
1081404 | exactly represent Dirac distribution as derivatives of continuous function
It seems that in Rudin's functional analysis (P168, Thm6.27) that the Dirac distribution [imath]\delta_0[/imath] on [imath]R^1[/imath] can write as [imath] \delta_0=f_0+f'_1+f''_2, [/imath] for [imath]\{f_i\}_{i=0}^2\in C(R^1)[/imath], there [imath]'[/imath] should be understand as derivative of distruibution of course. My problem is, can we write down [imath]f_i[/imath] exactly? | 1081414 | Represent Dirac distruibution as a combination of derivatives of continuous functions?
It seems from Rudin's functional analysis (P168, Thm6.27), the Dirac distribution [imath]\delta_0[/imath] on [imath]R^1[/imath] can be written as [imath] \delta_0=f_0+f'_1+f''_2, [/imath] for [imath]\{f_i\}_{i=0}^2\in C(R^1)[/imath], there [imath]'[/imath] should be understand as derivative of distruibution of course. My problem is, can we write down [imath]f_i[/imath] exactly? The Dirac distribution is defined as [imath] \delta_0(\phi)=\phi(0),\quad\phi\in C_0^\infty(R) [/imath] |
1075452 | For [imath]E \subset \mathbb{R}[/imath] and [imath]\epsilon >0[/imath], [imath]\exists[/imath] [imath](a,b)[/imath] s.t. [imath]\theta(E \cap (a,b)) \geq (1-\epsilon)|b-a|[/imath] ([imath]\theta[/imath] Lesbegue Outer Measure)
In my notes this statement is left unproven. I want to show that for any measurable set [imath]E \subset \mathbb{R}[/imath] with [imath]\theta(E)>0[/imath], there exists an interval [imath](a,b)[/imath] that covers [imath]E[/imath] arbitrarily closely. From the definition of the Lebesgue Outer Measure: [imath] \exists \bigcup_i(a_i,b_i) \supseteq E : \sum_i(b_i-a_i) \leq \frac{1}{1-\epsilon}\theta(E) [/imath] And then [imath]\theta(E)=\theta(\cup_i E\cap(a_i,b_i))[/imath], but I'm not sure this helps. Can anyone show me how to finish this off? | 103306 | For a set of positive measure there is an interval in which its density is high, [imath]\mu(E\cap I)> \rho \mu(I)[/imath]
In my self-study, I came across the following two interesting, related results: Let [imath]E[/imath] be Lebesgue measurable, with [imath]\mu(E)>0[/imath] (here [imath]\mu[/imath] denotes the Lebesgue measure). Then: for any [imath]0<\rho<1[/imath], there exists an open interval [imath]I[/imath] such that [imath]\mu(E \cap I)> \rho \cdot \mu(I)[/imath]. In turn we should be able to use (1) with [imath]\rho > 3/4[/imath] to prove that the set [imath]E-E = \{x-y : x, y \in E\}[/imath] contains an (open) interval centered at [imath]0[/imath] (in particular, if [imath]\rho > 3/4[/imath], the text I am using suggests that [imath](-\frac{1}{2} \mu(I), \frac{1}{2}\mu(I)) \subseteq E-E[/imath]). I would like to see if anyone visiting today would be up for proving (1) or (2) (inclusive-or). I have tried to work out (1) a few times, but none of my attempts have been satisfactory. |
1071898 | Evaluate [imath]\lim_{n\to1^+}\left({\zeta(n)-\frac{1}{n-1}}\right)[/imath]
Let [imath]x=\lim_{n\to1^+}\left({\zeta(n)-\dfrac{1}{n-1}}\right)[/imath] where [imath]\zeta[/imath] is Riemann zeta function. What is the value of [imath]x[/imath]? At [imath]n\to1^+[/imath], [imath]\zeta(n)\to\infty[/imath] and [imath]\dfrac{1}{n-1}\to\infty[/imath], so this is indeterminate form of type [imath]\infty-\infty[/imath]. I wrote this limit as [imath]x=\lim_{n\to1^+}\dfrac{\zeta(n)^2-\left({\dfrac{1}{n-1}}\right)^2}{\zeta(n)+\dfrac{1}{n-1}}[/imath] Then I tried to apply L'Hopital's rule, but this limit become more complicated. Also, derivative of zeta function is very complicated, so L'Hopital's rule cannot help evaluating this limit. What is the easiest way to solve it? | 100790 | Limit of Zeta function
I'm looking for a reference for (or an elementary proof of) [imath] \lim_{s \rightarrow 1} \left( \zeta(s) - \frac{1}{s-1} \right) = \gamma[/imath] Thanks for your help. |
1083399 | A problem on Ito integral
Let [imath]W[/imath] be a standard, one-dimensional Brownian motion. Let [imath]T\in(0,+\infty)[/imath]. Then [imath]\lim_{\beta\to+\infty}\sup_{0\le t\le T}\left|e^{-\beta t}\int_0^te^{\beta s}\mathrm{d}W_s\right|=0\quad\text{a.s.}[/imath] Could someone give some hints or proofs? | 279678 | Limit of a Wiener integral
How to show that [imath] \lim _{\alpha \rightarrow \infty } \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.} [/imath] where [imath]\left (B_s \right)_{s\geq 0}[/imath] is a real standard brownian motion starting from zero ? |
1083457 | If [imath]A[/imath] is a non-empty set, then [imath]A \notin A[/imath]
Exercise 3.2.2 Use the axiom of regularity and the singleton set axiom to show that if [imath]A[/imath] is a set then [imath]A \notin A[/imath]. Axiom of Regularity: If [imath]A[/imath] is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A. Singleton Set Axiom: If [imath]a[/imath] is an object, then there exists a set [imath]\{a\}[/imath] whose only element is [imath]a[/imath], we refer to [imath]\{a\}[/imath] as the singleton set whose element is [imath]a[/imath] Furthermore, if [imath]a[/imath] and [imath]b[/imath] are objects, then there exists a set [imath]\{a,b\}[/imath] whose only elements are [imath]a[/imath] and [imath]b[/imath] I think I can give the proof when A is the non-empty set which contains finite elements. However, can anyone give any hint for the general case? Suppose [imath]A \in A[/imath] Then eixsts [imath]x[/imath] such that either [imath]x[/imath] is not a set or [imath]x \cap A[/imath]=[imath]\emptyset[/imath] Then it is clear thar [imath]x \ne A[/imath] (if [imath]x = A[/imath], then [imath]x \cap A \ne \emptyset[/imath]) We observe that as an element, [imath]A[/imath] [imath]\ne[/imath] [imath]x[/imath], [imath]A[/imath] [imath]\in[/imath] [imath]A[/imath] Then [imath]A[/imath] [imath]\in[/imath] [imath]A-x[/imath] Define the new set [imath]A-x[/imath] as [imath]A_1[/imath] then exists [imath]x_1[/imath] such that [imath]A \in A_1-x_1[/imath] After finite steps we find that [imath]A \in \emptyset [/imath] which leads to contradiction. | 2162155 | Prove that if [imath]x[/imath] is a set, then [imath]x[/imath] is not an element of itself.
I am having troubles with proving the following statement: if [imath]x[/imath] is a set, then [imath]x \notin x[/imath]. I am only allowed to use the axioms from the Zermelo-Fraenkel Set Theory. Thanks so much if someone could give me a hand! |
585918 | Prove that : [imath]x^n+4[/imath] is irreducible in [imath]\mathbb{Z}[x][/imath].
Prove that : [imath]x^n+4[/imath] is irreducible on [imath]\mathbb{Z}[x][/imath] if only if [imath]n\neq 4k[/imath] with [imath]k\in\mathbb{N}[/imath]. | 133581 | When is [imath]X^n-a[/imath] is irreducible over F?
Let [imath]F[/imath] be a field, let [imath]\omega[/imath] be a primitive [imath]n[/imath]th root of unity in an algebraic closure of [imath]F[/imath]. If [imath]a[/imath] in [imath]F[/imath] is not an [imath]m[/imath]th power in [imath]F(\omega)[/imath] for any [imath]m\gt 1[/imath] that divides [imath]n[/imath], how to show that [imath]x^n -a[/imath] is irreducible over [imath]F[/imath]? |
542162 | Prove the absolute value function of a continuous function is continuous
Suppose that [imath]f[/imath] is a continuous function defined on an interval [imath]I[/imath]. Prove that [imath]|f|[/imath] is continuous on [imath]I[/imath]. Our definition of continuity: Let [imath]I[/imath] be an interval, let [imath]f:I\rightarrow\Bbb{R}[/imath], and let [imath]c\in I[/imath]. The function [imath]f[/imath] is continuous at [imath]c[/imath] if for each [imath]\epsilon>0[/imath] there exists [imath]\delta>0[/imath] such that [imath]|f(x)-f(c)| < \epsilon[/imath] for all [imath]x\in I[/imath] that satisfy [imath]|x-c|<\delta[/imath]. The function [imath]f[/imath] is continuous on [imath]I[/imath] if [imath]f[/imath] is continuous at each point of [imath]I[/imath]. | 660138 | Continuity of absolute value of a function
Let [imath]f(x)[/imath] be a continuous function. Prove that [imath]\left|f(x)\right|[/imath] is also continuous. Is it correct to say that, by the reverse triangle inequality, [imath]\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|[/imath] in all cases, so we will always have that for all [imath]\left|x-c\right|<\delta[/imath] implies [imath]\left|f(x)-f(c)\right|\leq \epsilon[/imath]? I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution! |
1083795 | [imath]C([0,1], \mathbb R)[/imath] with supremum norm is not sigma compact
Assume the banach space [imath]C([0,1], \mathbb R)[/imath] with supremum norm. I guess that this banach space is not sigma compact, in the other word [imath]C([0,1], \mathbb R)[/imath] is not union of countable compact subsets of it. My strategy : I assume it is sigma compact the because [imath]C([0,1], \mathbb R)[/imath] is a complete metric space then I want to use baire theorem to reach contradiction. but I am confused to complete it. | 295552 | Show that [imath]C_0([a, b], \mathbb{R})[/imath] is not [imath]\sigma[/imath]-compact
[imath]C_0([a, b], \mathbb{R})[/imath] is the space of real-valued continuous function on [imath][a, b][/imath]. The hint says think Baire. So I assume that [imath]C_0([a, b], \mathbb R)[/imath] is [imath]\sigma[/imath]-compact. Then it is the countable union of compact, and hence closed, subsets [imath]M_i[/imath]. Since [imath]C_0([a, b], \mathbb R)[/imath] is a complete metric space, by Baire's theorem at least one of the [imath]M_i[/imath]'s has a non-empty interior. But I got stuck here. I think I can derive a contradiction from this, but I'm not sure how. Any hints on how to solve this (using contradiction or not) would be much appreciated. |
1083066 | Closed form for series [imath]\sum_{m=1}^{N}m^n\binom{N}{m}[/imath]
How can we calculate the series [imath] I_N(n)=\sum_{m=1}^{N}m^n\binom{N}{m}? [/imath] with [imath]n,N[/imath] are integers. The first three ones are [imath] I_N(1)=N2^{N-1}; I_N(2)=N(N+1)2^{N-2}; I_N(3)=N^2(N+3)2^{N-3} [/imath] | 714194 | Evaluate [imath]\sum_{k = 0}^{n} {n\choose k} k^m[/imath]
So, I wonder what is the evaluation of [imath]\sum_{k = 0}^{n} {n\choose k} k^m\text{,}\qquad (*)[/imath] where [imath]m,n\in \mathbb{N}[/imath]. One of my tries: knowing that [imath]k^m = \sum_{j = 0}^{m}\text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{,}[/imath] where [imath]\text{S}(m,k)[/imath] are Stirling numbers of the second kind, and [imath]\sum_{k = 0}^n {n\choose k} k(k-1)\cdots(k-j+1) =2^{n-j}\cdot n(n-1)\cdots(n-j+1)\text{,}[/imath] I have rewritten the upper sum into [imath]\sum_{k = 0}^{n}\sum_{j = 0}^m {n\choose k} \text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{.}[/imath] Changing the order of summation, we get [imath]\sum_{j = 0}^{m}\text{S}(m,j) \cdot 2^{n-j}\cdot n(n-1)\cdots (n-j+1)\text{,}[/imath] and here it stops. |
1083881 | solve integral with residue theorem
I want to show that for positive [imath]a[/imath] [imath]\int_{-\infty}^{\infty}{\frac{\cos(x)}{x^2+a^2}} dx = \frac{\pi e^{-a}}{a}[/imath] I'm not even sure how to define a smart contour… I guess it can't be a half circle, since [imath]\cos(z)[/imath] is unbounded for big imaginary parts. If I take a rectangle, then the vertical lines will have no impact in the limit since [imath]\cos(z)[/imath] is bounded there and [imath]\frac{1}{z^2}[/imath] decreases rapidly, but for the "way back" i can't find a good choice since the nominator isn't periodic… :( | 140580 | Computing [imath]\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx[/imath] using residue calculus
I need to find [imath]\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx[/imath] where [imath]a > 0[/imath]. To do this, I set [imath]f(z) = \displaystyle\frac{\cos z}{z^{2} + a^{2}}[/imath] and integrate along the semi-circle of radius [imath]R[/imath]. For the residue at [imath]ia[/imath] I get [imath]\displaystyle\frac{\cos(ia)}{2ia}[/imath]. Then letting [imath]R \rightarrow \infty[/imath], the integral over the arc is zero, so I get [imath]\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx = 2 \pi i \frac{\cos(ia)}{2ia} = \frac{\pi \cos(ia)}{a}[/imath]. But this is supposed to be [imath]\displaystyle\frac{\pi e^{-a}}{a}[/imath], so I am doing something wrong. In a similar problem, I have to evaluate [imath]\displaystyle\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2} + a^{2}}\ dx[/imath] and get [imath]\pi i \sin(ia)[/imath] whereas this is supposed to be [imath]\pi e^{-a}[/imath]. I think I am getting some detail wrong in both cases. Can anyone enlighten me? |
1084191 | Enumerating function of [imath]a_n[/imath] i.e. the function [imath]\sum_{n=0}^{n=\infty}a_n x^n.[/imath]
Consider the Fibonacci Series {[imath] a_n[/imath]} defined by [imath]a_0=0,a_1=1,a_{n+1}=a_{n-1}+a_n [/imath] for [imath]n\ge1.[/imath] Then what will be the enumerating function of [imath]a_n[/imath] i.e. the function [imath]\sum_{n=0}^{n=\infty}a_n x^n.[/imath] | 338740 | The generating function for the Fibonacci numbers
Prove that [imath]1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}[/imath] The coefficients are Fibonacci numbers, i.e., the sequence [imath]\left\{1,1,2,3,5,8,13,21,...\right\}[/imath]. |
367911 | Lebesgue integral of a bounded measurable function over a measurable subset of a measurable set of finite measure.
Let [imath]f[/imath] be a bounded measurable function on a set [imath]E[/imath] of finite measure. For a measurable subset [imath]A[/imath] of [imath]E[/imath], show that [imath]\int_A f=\int_E f\cdot\chi_A[/imath], where [imath]\chi_A[/imath] is a characteristic function on [imath]A[/imath] and [imath]\int_A f[/imath] is the Lebesgue integral of [imath]f[/imath] on [imath]A[/imath]. How am I going to establish this? Help please. Thanks | 285655 | Bounded measurable function and integral with charcteristic function
I have been struggling with the following for quite some time now. If anyone can give me some help, it will be much appreciated: Let [imath]f[/imath] bounded, measurable and [imath]E[/imath] be a set of finite measure. Let [imath]A \subset E[/imath] be measurable. Prove that: [imath]\displaystyle \int_{A} f = \int_{E} f \chi_A [/imath] |
936092 | If [imath]f(x+a)[/imath] is irreducible over [imath]F[/imath] then [imath]f(x)[/imath] is irreducible over [imath]F[/imath]
Actually we can prove the statement in the reverse way. Here [imath]a[/imath] is non zero element, [imath]F[/imath] is a field. Suppose [imath]f(x)[/imath] is reducible then [imath]f(x) = g(x)h(x)[/imath], then [imath]f(x+a)= g(x+a) h(x+a)[/imath]. But I can't understand what it exactly wants to convey. It's an exercise problem from Joseph Gallion Algebra under Factorization of Polynomials. | 707537 | Show [imath]p(X)[/imath] (over a field) is irreducible iff [imath]p(X+a)[/imath] is irreducible
Let [imath]A[/imath] be a field and let [imath]p(X)[/imath] be a polynomial over [imath]A[/imath]. Let [imath]a\in A[/imath]. Want to show: [imath]p(X)[/imath] is irreducible if and only if [imath]p(X+a)[/imath] is irreducible. I suspect that I should use the substitution principle somehow, but that's as far as I've come. Completely stumped. |
935653 | What does it mean that the product of two vectors produces real number?
I am going over inner product space. I know that linear space has an inner product as long as it satisfies [imath]4[/imath] conditions. And, the book says that for [imath]x,y[/imath] in [imath]V[/imath], there is a real number corresponding [imath](x,y)[/imath] What does that mean? For [imath]\mathbb R^2[/imath], where [imath]x=(x_1,x_2)[/imath], [imath]y=(y_1,y_2)[/imath], what is inner product of [imath](x,y)[/imath], would it be [imath]x_1y_1+x_2y_2[/imath]? | 934575 | Can anybody explain about real linear space and complex linear space?
This is definition "A real linear [imath]V[/imath] is said to have an inner product if for each pair of elements [imath]x[/imath] and [imath]y[/imath] in [imath]V[/imath], there corresponds a unique real number [imath](x.y)[/imath] satisfying what we know as axioms for inner product." What does it mean? [imath](x.y)=xy[/imath] and [imath]xy[/imath] is a real number? When [imath](x.y)[/imath] is complex, it is called complex space, does that mean that [imath](x.y)=xy=[/imath] complex? I don't understand what inner product is. |
1084287 | Limit-Fundamental Concept
Actually,my question is here. Limit-Fundamental Concept? This part is taken from a calculus book of I.A Maron. According to this concept,[imath]\lim\limits_{x\to a}(-2)^2[/imath] cannot be defined because [imath]f(x)[/imath] is negative. Can anyone kindly explain this to me? | 1082338 | Limit-Fundamental Concept?
Can anyone clear me this fundamental concept about this. I am confused for many months over this. It is said that in [imath]\lim_{x\to a} f(x)^{g(x)}[/imath]. [imath]f(x)[/imath] should be greater then [imath]0[/imath]. Can anyone explain the reason? If [imath]f(x)[/imath] is [imath]-2[/imath] and [imath]g(x)[/imath] is [imath]\frac{1}{2}[/imath],then it is not possible. But if [imath]f(x)[/imath] is [imath]-2[/imath] and [imath]g(x)[/imath] is [imath]2[/imath],then why it is not possible? (It is true as [imath](-2)^2 = 4[/imath]) |
1065015 | Proper Bernoulli Function Generating Function
Consider the function [imath]\frac{t}{e^t - 1} = \sum_{i=0}^{\infty}\frac{B_i}{i!}t^i[/imath] This has been one of the famous generating functions for the bernoulli numbers. What about the function associated with [imath]\sum_{i=0}^{\infty}B_it^i[/imath] What function could this be? | 370985 | Ordinary generating function for Bernoulli polynomial
I know the exponential generating function for the Bernoulli polynomial [imath]B_n(x)[/imath]:[imath]\frac{te^{tx}}{e^t-1}=\sum_{n=0}^\infty B_n(x)\frac{t^n}{n!}.[/imath] But is there an ordinary generating function? i.e a function [imath]f[/imath] such that [imath]f(t,x)=\sum_{n=0}^\infty B_n(x)t^n[/imath] valid for some open interval of [imath]x[/imath] and some open interval of [imath]t[/imath]. |
531772 | Prove that [imath]\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k[/imath]
Prove that [imath]\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k[/imath] What should I do for this equation? Should I focus on proving [imath]\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}{k}2^k[/imath]? | 461762 | Alternative combinatorial proof for [imath]\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}=\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r[/imath]
I have a question with regards to combinatorics. I am supposed to show the following combinatorial identity: [imath]\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}=\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r[/imath]. The algebraic way is to consider the coefficient of [imath]t^m[/imath] in the expansion of [imath](1+t)^n(1-t)^{-(n+1)}[/imath]. For [imath]r\in\{0,1,\cdots,n\}[/imath], we have the coefficient of [imath]t^{n-r}[/imath] in the expansion of [imath](1+t)^n[/imath] to be equal to [imath]\binom{n}{r}[/imath], and the coefficient of [imath]t^{m+r-n}[/imath] in the expansion of [imath](1-t)^{-(n+1)}[/imath] to be equal to [imath]\binom{m+r}{n}[/imath]. By the addition and multiplication principles this will show that the coefficient of [imath]t^m[/imath] in the expansion of [imath](1+t)^n(1-t)^{-(n+1)}[/imath] will be equal to [imath]\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}[/imath]. On the other hand, we note that [imath](1+t)^n(1-t)^{-(n+1)}=(1+\frac{2t}{1-t})^n(1-t)^{-1}=\sum\limits_{r=0}^n\binom{n}{r}2^rt^r(1-t)^{-(r+1)}[/imath]. As the coefficient of [imath]t^{m-r}[/imath] is equal to [imath]\binom{m}{r}[/imath], this implies that the coefficient of [imath]t^m[/imath] in the expansion of [imath](1+t)^n(1-t)^{-(n+1)}[/imath] is equal to [imath]\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r[/imath]. What I am thinking of, is to think of a way to prove this identity without the use of the generalized binomial series. I was trying to think of a combinatorial method to solve this problem. In particular, I was trying to prove the combinatorial identity as follows: Firstly, let us be given a class of [imath]n[/imath] boys and a class of [imath]m[/imath] girls, where [imath]n\leq m[/imath]. Choose a team [imath]A[/imath] of [imath]r[/imath] boys, and choose a team [imath]B[/imath] of [imath]r[/imath] girls, and finally choose any subset (may be empty) of boys from [imath]A[/imath] to clean the classroom, where [imath]r\in\{0,1,\cdots,n\}[/imath]. For each value of [imath]r[/imath], we see that there are [imath]\binom{n}{r}[/imath] choices for [imath]A[/imath], [imath]\binom{m}{r}[/imath] choices for [imath]B[/imath], and subsequently there are [imath]2^r[/imath] ways to choose a subset from [imath]A[/imath]. In this way we get the number of ways to do so to be equal to the expression on the RHS of the identity. What I am struggling though, is to be able to count the number of desired ways in a way that will give me the expression on the LHS of the identity. I tried firstly to choose the boys to clean the classroom before choosing the sets [imath]A[/imath] and [imath]B[/imath], but I wound up with some messy expression that does not seem close to matching the expression on the LHS. I would like to know if there is another way of looking at this method (or I could have been missing out on something), or if there is any other combinatorial method (or any other algebraic method that does not involve the use of generalized binomial series) that one could think of to show the desired identity. |
1084327 | Solving the integral [imath]\int_0^{\frac\pi2}\frac{dx}{1+\sin^2(\tan x)}[/imath]
[imath]\int_0^{\frac\pi2}\frac{dx}{1+\sin^2(\tan x)}[/imath] First, I tried to set [imath]t=\tan x[/imath] Then I got [imath]\int_0^\infty\frac{dt}{(1+t^2)(1+\sin^2t)}[/imath] applied a trig identity, [imath]1+ \sin^2t=\frac{1}{2}(3-\cos2 t)[/imath] I got [imath]\int_0^\infty\frac{dt}{(1+t^2)\left(\frac12(3-\cos 2t)\right)}[/imath] I don't know how to keep going to solve it. can some one give me a hint? Thanks | 730876 | Prove [imath]\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)[/imath]
Prove the following integral [imath]I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)[/imath] This integral result was calculated using Mathematica and I like this integral. But I can't solve it. My idea: Let [imath]\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt[/imath] so [imath]I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}[/imath] then I can't proceed. Can you help me? Thank you. |
1084018 | If two continuous maps into a Hausdorff space agree on a dense subset, they are identically equal
Let [imath]f, g : X \to Y[/imath] be continuous functions. Assume that [imath]Y[/imath] is Hausdorff and that there exists a dense subset [imath]D[/imath] of [imath]X[/imath] such that [imath]f(x) = g(x)[/imath] for all [imath]x \in D[/imath]. Prove that [imath]f(x) = g(x)[/imath] for all [imath]x \in X[/imath]. Here is what I have so far, Proof: Let [imath]f : X \to Y[/imath] and [imath]g : X \to Y[/imath] be continuous and suppose that [imath]f(x)=g(x)[/imath] for some dense [imath]D\subset X[/imath]. Let [imath]x \in X[/imath], since [imath]D[/imath] is a dense subset of [imath]X[/imath], [imath]x \in \mathrm{Cl}(D)[/imath]. I'm unsure how to proceed from here. | 543962 | [imath]f,g[/imath] continuous from [imath]X[/imath] to [imath]Y[/imath]. if they are agree on a dense set [imath]A[/imath] of [imath]X[/imath] then they agree on [imath]X[/imath]
Problem: Suppose [imath]f[/imath] and [imath]g[/imath] are two continuous functions such that [imath]f: X \to Y [/imath] and [imath]g : X \to Y [/imath]. [imath]Y[/imath] is a a Hausdorff space. Suppose [imath]f(x) = g(x) [/imath] for all [imath]x \in A \subseteq X [/imath] where [imath]A[/imath] is dense in [imath]X[/imath], then [imath]f(x) = g(x) [/imath] for all [imath]x \in X [/imath]. Attempt at a solution: Put [imath]h(x) = f - g [/imath]. Therefore, [imath]h: X \to Y [/imath] is continuous and [imath]Y[/imath] is Hausdorff by hypothesis. Also we know [imath]h(x) = 0 [/imath] for all [imath]x \in A [/imath] such that [imath]A[/imath] is dense in [imath]X[/imath]. I want to show that [imath]h(x)[/imath] vanishes everywhere in [imath]X[/imath]. We can show [imath]h(x) = 0 [/imath] for all [imath]x \in X \setminus A [/imath]. Suppose [imath]h(x) > 0 [/imath] on [imath]X \setminus A[/imath]. Pick points [imath]y_1,y_2 \in Y [/imath]. Since [imath]Y[/imath] is Hausdorff, can find open set [imath]O_1, O_2 \subseteq Y [/imath] which are disjoint such that [imath]y_1 \in O_1[/imath] and [imath]y_2 \in O_2[/imath]. By continuity, [imath]f^{-1}(O_1), f^{-1}(O_2)[/imath] are open in [imath]X[/imath]. I know that if I can show that one of the [imath]f^{-1}(O_i)[/imath] lies in [imath]X \setminus A [/imath], then we would have a contradiction since we have non-empty open set in [imath]X \setminus A[/imath] and this implies [imath]A[/imath] cannot be dense in [imath]X[/imath]. But this is the part I am stuck. Any help would greatly be appreciated. Also, Would be be possible to prove this without using the Hausdorff condition on [imath]Y[/imath]? |
1084994 | Spectrum of [imath]\mathbb R[X,Y][/imath]
Let [imath]A=\mathbb R[X,Y][/imath]. Is it easy to classify the [imath]\operatorname{Spec}A[/imath]? I guess it contains at least [imath](0)[/imath] and [imath](p)[/imath] for primes [imath]p\in A[/imath] but maybe some else sets. Is it easy to classify those? Edit. Yes. I forgot maximal ideals. I think those are given in http://mathhelpforum.com/advanced-algebra/68149-maximal-ideal-r-x-y.html I'm not sure but I guess those can be written as [imath](p,q)[/imath] where [imath]p=X-a[/imath] and [imath]q=Y-b[/imath] for [imath]a,b\in \mathbb R[/imath] or [imath]p[/imath] is linear and [imath]q[/imath] is [imath]X^2+xX+d[/imath] or [imath]Y^2+cY+d[/imath] where [imath]c,d\in\mathbb R[/imath] and [imath]c^2<4d[/imath]. | 56916 | What do prime ideals in [imath]k[x,y][/imath] look like?
Suppose that [imath]k[/imath] is an algebraically closed field. Then what do the prime ideals in the polynomial ring [imath]k[x,y][/imath] look like? As far as I know, the maximal ideals of [imath]k[x,y][/imath] are of the form [imath](x-a,y-b)[/imath] where [imath]a,b\in k[/imath]. What can we say about the prime ideals? Are there similar results? And what about [imath]k[x,y,z], k[x,y,z,w][/imath] and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much! |
1021121 | Why is the antiderivative of [imath]\frac{1}{1+x^2}=\tan^{-1}(x)[/imath]?
My textbook says the antiderivative of [imath]\frac{1}{1+x^2}[/imath] is [imath]\tan^{-1}(x)[/imath]. To confirm this to myself I took the derivative of [imath]\tan^{-1}(x)[/imath] expecting to get [imath]\frac{1}{1+x^2}[/imath] , but instead I ended up with [imath]-\frac{1}{\sin^2(x)}[/imath]. So why is [imath]\tan^{-1}(x)[/imath] the antiderivative of [imath]\frac{1}{1+x^2}[/imath] if the derivative of [imath]\tan^{-1}(x)[/imath] is not [imath]\frac{1}{1+x^2}[/imath]? Shouldn't the derivative of the antiderivative of a function give you the original function? | 1009871 | Prove that [imath]\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx[/imath]
Prove that [imath]\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}[/imath] |
1085524 | How find x if [imath]\frac{1}{x}+\frac{2}{x^3}+\frac{6}{x^5}+\cdots+\frac{\binom{2n}{n}}{x^{2n+1}}+\cdots=1[/imath]
Let [imath]x>2[/imath] and [imath]\frac{1}{x}+\frac{2}{x^3}+\frac{6}{x^5}+\cdots+\frac{\binom{2n}{n}}{x^{2n+1}}+\cdots=1[/imath]. How find [imath]x[/imath]? | 69270 | Show [imath]\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}[/imath]
How do you prove that [imath]\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}[/imath]? I tried to identify the sum as a binomial series, but the [imath]4[/imath] and the [imath]-1/2[/imath] puzzle me. (This series arises in studying the first passage time of a simple random walk.) |
1085733 | Polynomial division [imath]p(x) = x^4-(2m + 4)x^2 + (m-2)^2[/imath]
For which values of [imath]m[/imath] can the polynomial [imath]p(x) = x^4-(2m + 4)x^2 + (m -2)^2[/imath] be factored into two non-constant polynomials whose coefficients are integers? | 1081652 | Solve [imath]p_4(x) = x^4 −(2m + 4)x^2 + (m−2)^2 [/imath]such that [imath]p_4[/imath] is a product of two non-constant integer-coeficient polynomials
I'm having trouble getting the starting idea for a problem I've been presented with: I need to find values for m (integer) such that the following polynomial [imath]p_4(x) = x^4 −(2m + 4)x^2 + (m−2)^2[/imath]; [imath]m[/imath] is integer can be described as the product of two non-constant integer-coeficient polynomials. Any help/pointers/tips on how to go about this are greatly appreciated! Edit: I meant polynomials with integer coeficients, but for some reason typed positive. |
816389 | Minimal polynomial and diagonalization of a block diagonal matrix.
Let [imath]A \in \mathbb C^{m\times m}[/imath] and [imath]B \in \mathbb C^{n\times n}[/imath], and let [imath]C=\begin{pmatrix} A & 0 \\ 0 & B\\ \end{pmatrix} \in \mathbb C^{(m+n)\times (m+n)}[/imath]. Calculate the minimal polynomial of [imath]C[/imath] based on the minimal of [imath]A[/imath] and the minimal of [imath]B[/imath]. Prove that [imath]C[/imath] is diagonalizable if and only if [imath]A[/imath] and [imath]B[/imath] are. The attempt at a solution I have no idea how to prove 1). For 2) I got stuck in a lot of parts: [imath]\Leftarrow[/imath] If [imath]C[/imath] is diagonalizable, then [imath]C=P^{-1}DP[/imath] where [imath]D[/imath] is a diagonal matrix. Somehow, I must construct from D two diagonal matrices [imath]D_1 \in \mathbb C^{m\times m}[/imath] and [imath]D_2 \in \mathbb C^{n \times n}[/imath] and two invertible matrices [imath]Q[/imath] and [imath]S[/imath] so that [imath]A=Q^{-1}D_1Q[/imath] and [imath]B=S^{-1}D_2S[/imath], I don't know how to construct all these matrices. [imath]\Rightarrow[/imath] Suppose [imath]A[/imath] and [imath]B[/imath] are diagonalizable, so [imath]A=Q^{-1}D_1Q[/imath] and [imath]B=S^{-1}D_2S[/imath], with both [imath]D_1[/imath] and [imath]D_2[/imath] diagonal matrices. My guess is [imath]C[/imath] can be written as [imath]\pmatrix{Q^{-1}&0\\ 0&S^{-1}}\pmatrix{D_1&0\\ 0&D_2}\pmatrix{Q&0\\ 0&S}[/imath]. Now, I would have to prove that [imath]\pmatrix{Q^{-1}&0\\ 0&S^{-1}}\pmatrix{Q&0\\ 0&S}=Id_{m+n}[/imath] and that [imath]C=\pmatrix{Q^{-1}&0\\ 0&S^{-1}}\pmatrix{D_1&0\\ 0&D_2}\pmatrix{Q&0\\ 0&S}[/imath] in order to show [imath]C[/imath] is diagonalizable. I would appreciate help in all these points where I am stuck and any suggestion or hint with regard to 1. | 350028 | Block Diagonal Matrix Diagonalizable
I am trying to prove that: The matrix [imath]C = \left(\begin{smallmatrix}A& 0\\0 & B\end{smallmatrix}\right)[/imath] is diagonalizable, if only if [imath]A[/imath] and [imath]B[/imath] are diagonalizable. If [imath]A\in\mathbb{C}^n[/imath] and [imath]B\in\mathbb{C}^m[/imath] are diagonalizable, then is easy to check the [imath]C\in\mathbb{C}^{n+m}[/imath] is diagonalizable. But if I suppose that [imath]C[/imath] is diagonalizable, then exists [imath]S = [S_1, S_2, \ldots, S_{n+m}][/imath], [imath]S_i\in\mathbb{C}^{m+n}[/imath], such that [imath]S^{-1}CS = \mbox{diag}(\lambda_i)[/imath] . Now [imath]CS_i = \lambda_iS_i[/imath], and if [imath]S_i = \left(\begin{smallmatrix}x_i\\y_i\end{smallmatrix}\right)[/imath], [imath]x_i\in\mathbb{C}^n[/imath] and [imath]y_i\in\mathbb{C}^m[/imath], then [imath]Ax_i = \lambda_ix_i\quad\mbox{ and }\quad By_i = \lambda_iy_i.[/imath] So, if I can justify that [imath]\{x_1,\ldots,x_{n+m}\}[/imath] have exactly [imath]n[/imath] linear independent vectors and [imath]\{y_1,\ldots,y_{n+m}\}[/imath] have [imath]m[/imath] linear independent vectors, I will prove that [imath]A[/imath] and [imath]B[/imath] are diagonalizables, but I don't know how to prove that? Please, anyone have an idea? Thanks in advance. |
1086048 | Let [imath]Y = X^2[/imath]. Find the pdf of Y when the distribution is [imath]N(0,1)[/imath].
I've performed a change of variable: [imath]X = \sqrt{y}[/imath] [imath]X'=\frac{1}{2}Y^{-\frac{1}{2}}[/imath] Thus: [imath]f(\sqrt{y})*X'=f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}[/imath] However the book gives: [imath]f(y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}[/imath] Where did I go wrong? | 1016986 | Characteristic function of a square of normally distributed RV
Let's assume that [imath] X \sim \mathcal{N}(0,1) [/imath]. I'm supposed to compute the characteristic function of [imath] X^2 [/imath]. As far as I got is that the density of [imath] X^2 [/imath] is [imath] g(y) = \frac{1}{2\sqrt{2\pi}} e^{-y/2} \frac{1}{\sqrt{y}} \textbf{1}_{[0,\infty)}[/imath] and therefore [imath] \psi_{X^2}(t) = \frac{1}{2\sqrt{2\pi}}\int\limits_{0}^{\infty} e^{itx} e^{-x/2}\frac{1}{\sqrt{x}} dx[/imath] The problem is I'm not acquainted with computing comples integrals and don't know how to continue from this point. Is there a way around / an easy non-residue method of calculating that? |
1077619 | Calculate the limit : [imath]\lim_{x \to 0}\frac{x-\sin{x}}{x^3}[/imath] WITHOUT using L'Hopital's rule
I was given a task to find [imath]\lim_{x\to0}\frac{x-\sin{x}}{x^3}[/imath] at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks. | 1396309 | Limit problem without l'Hopital's rule
The problem goes : Solve the following limit without using l'hopital's rule : [imath]\lim\limits_{x\to 0} \frac{x-\sin(x)}{x^3}.[/imath] I've tried multiplying with conjugate "[imath]x+\sin(x)[/imath]", I've tried extracting [imath]x[/imath] from the numerator, and I always end up in a dead end. Pleas help. |
1086108 | How to show the Riemann Zeta function converges for [imath]s>1[/imath] where [imath]s[/imath] is Real.
How to show the Riemann Zeta function converges for [imath]s>1[/imath] where [imath]s[/imath] is Real. I know it's elementary, I just can't find any proof online. How do you prove: [imath]\sum_{n=1}^\infty \frac{1}{n^x} \mbox{ converges for } x>1[/imath] | 101131 | p-series convergence
Show that if [imath]p>1[/imath], [imath]\sum\frac{1}{n^{p}}[/imath] converges and if [imath]p<1[/imath] it diverges for [imath]p\in\mathbb{R}^{+}[/imath]. Is there any way to show another series converges or diverges and then use the Comparison Test to prove this? |
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