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598087 | What is the inverse of the following function?
What is the inverse function of [imath] y=x^3+x[/imath]? thanks | 60907 | Inverse of [imath]y = x^3 + x [/imath]?
Can you help me find the inverse function for [imath]y = x^3 + x[/imath]? This question was posed at the beginning of AP Calculus, so we can't use any math beyond precalc. Thanks! |
597524 | Get one of the two random variables's distribution function from limitation
This is a very fundamental problem. In the Stochastic Processes textbook, it says that: The Continuity Theorem of Probability allows us to conclude that [imath]F_X(x)=\lim_{y \to \infty}F_{XY}(x,y)[/imath] I know it's right but don't know why, I'm confused, so..can someone gives me a proof mathematically? The problem posing here is that, actually the [imath](\{X\leq x\},\{Y\leq y\})[/imath] stands for the Cartesian product of [imath]\{X\leq x\}\times\{Y\leq y\}[/imath] on [imath]R^2[/imath]. We have no information about [imath]P_{xy}[/imath] even though we know everything about [imath]X[/imath] and [imath]Y[/imath]. Hence we cannot simply say [imath]P(X\leq x,S) = P(X\leq x)[/imath], then how can we get the above equation? | 480477 | Limit of the joint cdf
How to see that given a joint cdf [imath]F(x,y)[/imath], we have [imath]\lim_{x\to\infty}F(x,y)=F(y)[/imath] and [imath]\lim_{y\to\infty}=F(x)[/imath] where [imath]F(y)[/imath] and [imath]F(x)[/imath] are marginal cdfs. |
587454 | Kernel of the ring homomorphism [imath]\phi\colon\mathbb{C}[x, y, z]\to\mathbb{C}[t][/imath] defined by [imath]\phi(x) = t[/imath], [imath]\phi(y) = t^2[/imath], [imath]\phi(z) = t^3[/imath]
Describe the kernel of the ring homomorphism [imath]\phi\colon \mathbb{C}[x, y, z]\to \mathbb{C}[t][/imath] defined by [imath]\phi(x) = t[/imath], [imath]\phi(y) = t^2[/imath], [imath]\phi(z) = t^3[/imath]. Some hints? I know [imath]\ker\phi = \{ a \in\mathbb{C}[x, y, z] \mid \phi(a) = 0\}[/imath] is closed under addition and multiplication. Also, it is an ideal of [imath]\mathbb{C}[x, y, z][/imath]. Using egreg hint, I came up with [imath]\ker \phi[/imath] ={[imath] (x^2 -y)r_1 + (xy-z)r_2 + (x^3-z)r_3 | r_1,r_2, r_3 \in\mathbb{C}[x, y, z] [/imath]} ? I am not sure if this is right. | 74195 | Kernel of a substitution map
Suppose [imath]R=k[x,y,z][/imath] and [imath]S=k[t][/imath]. Consider the map [imath]f:R\to S[/imath] s.t. [imath]f(x)=t[/imath], [imath]f(y)=t^2[/imath] and [imath]f(z)=t^3[/imath]. I suspect the kernel of this map is the ideal [imath](y-x^2,z-x^3)R[/imath]. It's clearly contained in the kernel, but I am not sure how to prove the reverse inclusion. |
599150 | Use the Maclaurin series for [imath]\cos(x)[/imath]to compute the value of [imath]\cos(5^\circ)[/imath] correct to five decimal places.
So we have been given a table for the Maclaurin series of [imath]\cos(x)[/imath] but I'm not sure how to even start with this problem. Here is the problem: Use the Maclaurin series for [imath]\cos(x)[/imath]to compute the value of [imath]\cos(5^\circ)[/imath] correct to five decimal places. And we were giving the following hint: Hints: Remember that the Maclaurin series assumes that the angle is in radians, not degrees. Also, since the Maclaurin series for [imath]\cos(x)[/imath] is alternating and decreasing in magnitude for small [imath]x[/imath], the error in using a partial sum of the series for the sum is less than the first term in the series omitted. Can you please help me to get started on this? | 366774 | Compute cos(5Β°) to 5 decimal places with Maclaurin's Series
I'm working on a problem: Compute cos(5Β°) to 5 decimal places with Maclaurin's Series I know that that function cos(x) has a Mclaurin representation of: [imath] \sum_{n=0}^\infty \frac{(-1)^n (x)^{2n}}{(2n)!} [/imath] The only part I am unsure of is how to find out many terms n to choose. I'm not sure if I am supposed to use the alternating series estimation or taylor series estimation or what. |
599655 | Prove that for a polynomial function [imath]f[/imath] if [imath]f(x) \geq 0[/imath] for all [imath]x[/imath], then [imath]f(x) + f'(x) + \cdots + f^{(n)} (x) \geq 0[/imath]
Prove that for a polynomial function [imath]f[/imath] with degree [imath]n[/imath] if [imath]f(x) \geq 0[/imath] for all [imath]x[/imath], then [imath]f(x) + f'(x) + \cdots + f^{(n)} (x) \geq 0[/imath]. Give me some hints for this and please explain to me how you have come to those hints. | 397973 | sum of polynoms of given property
I have [imath]P(x)[/imath] a polynomial with degree [imath]n[/imath] ,[imath]P(x) \ge 0[/imath] for all [imath]x \in[/imath] real. I have to prove that: [imath]f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0[/imath] for all [imath]x[/imath]. I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed. |
599885 | The trace of a matrix is the sum of its eigenvalues
If [imath]A[/imath] is a complex square matrix, I need to prove that the trace of [imath]A[/imath] is the sum of its eigenvalues. I've already proved that, if [imath]p(x)[/imath] is the characteristic polynomial, then [imath]p(x)=det(xI-A)=(x-\lambda_1)...(x-\lambda_n)=x^n-(\lambda_1+...+\lambda_n)+...+(-1)^ndet(A)[/imath] But IΒ΄m stuck with the proof... Thank you! PS: I know that is a direct consequence of the identity [imath]p(x)=x^n-Tr(A)+...+(-1)^ndet(A)[/imath], but in that case, I need a proof of this identity... | 27136 | Can it be determined that the sum of the diagonal entries, of matrix A, equals the sum of eigenvalues of A
I have a question to ask down below, that I have been having some trouble with and would like some help and clarification on. Suppose A is an [imath]n \times n[/imath] matrix with (not necessarily distinct) eigenvalues [imath]\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}[/imath]. Can it be shown that: (a) The sum of the main diagonal entries of A, called the trace of A, equals the sum of the eigenvalues of A. (b) A [imath]- ~ k[/imath] I has the eigenvalues [imath]\lambda_{1}-k, \lambda_{2}-k, \ldots, \lambda_{n}-k[/imath] and the same eigenvectors as A. Thank You very much. |
600032 | Combinatorics - find [imath]n![/imath] using inclusion-exclusion
difficult question I need help with. We are asked to show that [imath]n! = \sum_{k=0}^n (-1)^k\binom{n}{k}(n-k)^n[/imath] There is also a hint "try to think of the number of permutations of n elements using inclusion-exclusion" | 591350 | Expressing a factorial as difference of powers: [imath]\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n![/imath]?
The successive difference of powers of integers leads to factorial of that power. Here's the formula: [imath]\sum_{r=0}^{n}\binom{n}{r}(-1)^r(n-r)^n=n![/imath] Can anyone give a proof of this result? Note: The original question was to prove the more general [imath]\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n![/imath] for an integer [imath]l[/imath], which some of the answers address. |
599797 | Proof of a theorem about oscillation
There is a theorem in page 100 of Arnold's Mathematical Methods of Classical Mechanics, which says that: If [imath]\cfrac{dx}{dt} = f(x) = Ax + R_2(x)[/imath], where [imath]A = \cfrac{\partial f}{\partial x}|_{x = 0}[/imath], [imath]R_2(x) = O(x^2)[/imath], and [imath]\cfrac{dy}{dt} = Ay[/imath], [imath]y(0) = x(0)[/imath], then for any [imath]\vphantom{\cfrac12} T>0[/imath] and for any [imath]\xi > 0[/imath] there exists [imath]\delta > 0[/imath] such that if [imath]|x(0)| < \delta[/imath], then [imath]\vphantom{\cfrac12}|x(t) - y(t)| < \xi \delta[/imath] for all [imath]t[/imath] in the interval [imath]0 < t < T[/imath]. How can I prove this theorem rigorously? Cross-posted at math.se here. | 597622 | A theorem about oscillation in Arnold's mathematical methods of classical mechanics
There is a theorem in page 100 of Arnold's Mathematical Methods of Classical Mechanics, which says that: If [imath]\cfrac{dx}{dt} = f(x) = Ax + R_2(x)[/imath], where [imath]A = \cfrac{\partial f}{\partial x}|_{x = 0}[/imath], [imath]R_2(x) = O(x^2)[/imath], and [imath]\cfrac{dy}{dt} = Ay[/imath], [imath]y(0) = x(0)[/imath], then for any [imath]\vphantom{\cfrac12} T>0[/imath] and for any [imath]\xi > 0[/imath] there exists [imath]\delta > 0[/imath] such that if [imath]|x(0)| < \delta[/imath], then [imath]\vphantom{\cfrac12}|x(t) - y(t)| < \xi \delta[/imath] for all [imath]t[/imath] in the interval [imath]0 < t < T[/imath]. How can I prove this theorem rigorously? Cross-posted at physics.se here. |
597743 | Diagonalizable matrices A and B with [imath]\mathrm{Tr}(A^k)=\mathrm{Tr}(B^k)[/imath] have the same characteristic polynomial?
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] matrices with entries in a field F. Suppose [imath]A[/imath] and [imath]B[/imath] are diagonalizable in some extension field E of F and that [imath]\mathrm{Tr}(A^k)=\mathrm{Tr}(B^k)[/imath] for all integers [imath]k>0[/imath]. Show that A and [imath]B[/imath] have the same characteristic polynomial. [imath]A[/imath] and [imath]B[/imath] are diagonalizable over the field E, so there are invertible matrices [imath]R[/imath] and [imath]S[/imath], and diagonal matrices [imath]D_A[/imath] and [imath]D_B[/imath] with entries in E such that [imath]RAR^{-1}=D_A[/imath] and [imath]SBS^{-1}=D_B[/imath] Similar Matrices have the same characteristic polynomial, so if [imath]P_X(\lambda)[/imath] denotes the characteristic polynomial of a matrix X in variable [imath]\lambda[/imath], [imath]P_A(\lambda)=P_{D_A}(\lambda)[/imath] and [imath]P_B(\lambda)=P_{D_B}(\lambda)[/imath]. We also have [imath]\mathrm{Tr}(D_A^k)=\mathrm{Tr}((RAR^{-1})^k)=\mathrm{Tr}(RA^kR^{-1})=\mathrm{Tr}(A^k)=\mathrm{Tr}(B^k)=\mathrm{Tr}(D_B^k)[/imath] for all integers [imath]k>0[/imath]. I tried using the Cayley-Hamilton theorem on [imath]P_{D_A}(\lambda)[/imath] and taking the trace, but couldn't get anything out of it. | 88424 | If [imath] \mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)[/imath] for all [imath]1\leq k \leq n[/imath], then how do we show the [imath]M[/imath] and [imath]N[/imath] have the same eigenvalues?
Let [imath]M,N[/imath] be [imath]n \times n[/imath] square matrices over an algebraically closed field with the properties that the trace of both matrices coincides along with all powers of the matrix. More specifically, suppose that [imath]\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)[/imath] for all [imath]1\leq k \leq n[/imath]. The following questions about eigenvalues is then natural and I was thinking it would be an application of Cayley-Hamilton but I am having trouble writing out a proof. How do we show that [imath]M[/imath] and [imath]N[/imath] have the same eigenvalues? Added (because this question is now target of many duplicates, it should state its hypotheses properly). Assume that all the mentioned values of [imath]k[/imath] are nonzero in the field considered; in other words either the field is of characteristic [imath]$0$[/imath], or else its prime characteristic [imath]$p$[/imath] satisfies [imath]p>n[/imath]. |
600365 | Just a basic limit of a function
Find this limit,and if you can, do it without derivates or integrals or l'Hospital , just using the fundamental limits of functions. [imath]\lim_{x\to1} \left( \frac{x}{x-1} - \frac{1}{\ln(x)} \right)[/imath] | 554629 | Find the limit: [imath]\lim \limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right)[/imath]
Find: [imath]\lim\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right) [/imath] Without using L'Hospital or Taylor approximations Thanks in advance |
600482 | Does this constitute sufficient proof?
Task I have the following function [imath]f(x)=x^2+1[/imath] I need to prove, according to the [imath]\epsilon - \delta[/imath] definition of a limit, that [imath]f(x)[/imath] is continuous at [imath]x = 2[/imath]. Step 1 [imath]\forall \epsilon > 0 \enspace \exists \delta > 0 : |f(x) - L| < \epsilon, \mbox{ when } 0 < |x - c| < \delta \Rightarrow \lim_{x \to c}f(x)=L[/imath]. If a function is continious at a given point, the value of the limit and the function must be equal, so: [imath]\lim_{x \to c}f(x)=f(c)[/imath]. And in this case - [imath]f(2) = 2^2 + 1 = 5[/imath]. Which means that [imath]\lim_{x \to c}f(x)=5[/imath]. Step 2 Try to define [imath]x[/imath] in terms of [imath]\delta[/imath]. [imath]0 < |x - 2| < \delta \\ 2 - \delta < x < \delta + 2[/imath] Step 3 Try to define [imath]x[/imath] in terms of [imath]\epsilon[/imath]. [imath] |x^2+1 - 5| < \epsilon \\ 4- \epsilon < x^2 < \epsilon + 4 \\ \sqrt{4- \epsilon} < x < [/imath] Step 4 Expressing [imath]\delta[/imath] in terms of [imath]\epsilon[/imath]. [imath] \sqrt{4 - \epsilon} \leqslant 2 - \delta \\ \sqrt{4 - \epsilon} -2 \leqslant - \delta \\ 2 - \sqrt{4 - \epsilon} \geqslant \delta [/imath] And [imath] 2 + \delta \leqslant \sqrt{\epsilon + 4} \\ \delta \leqslant \sqrt{\epsilon + 4} - 2 [/imath] Step 5 I can now say that [imath]\forall \epsilon[/imath] I have a [imath]\delta[/imath] (two deltas?) and you can find them in terms of epsilon: [imath] \delta \leqslant \sqrt{\epsilon + 4} - 2 \mbox{ and } \delta \leqslant 2 - \sqrt{4 - \epsilon} [/imath] Questions Is this sufficient proof? What does the double [imath]\delta[/imath] definition mean? How should I format my final proof statement in this case (double deltas)? | 600430 | Proof (epsilon delta) for the continuity of a function at a point
Task I have the following function [imath]f(x)=x^2+1[/imath] I need to prove, according to the [imath]\epsilon - \delta[/imath] definition of a limit, that [imath]f(x)[/imath] is continuous at [imath]x = 2[/imath]. Step 1 [imath]\forall \epsilon > 0 \enspace \exists \delta > 0 : |f(x) - L| < \epsilon, \mbox{ when } 0 < |x - c| < \delta \Rightarrow \lim_{x \to c}f(x)=L[/imath]. If a function is continious at a given point, the value of the limit and the function must be equal, so: [imath]\lim_{x \to c}f(x)=f(c)[/imath]. And in this case - [imath]f(2) = 2^2 + 1 = 5[/imath]. Which means that [imath]\lim_{x \to c}f(x)=5[/imath]. Step 2 Try to define [imath]x[/imath] in terms of [imath]\delta[/imath]. [imath]0 < |x - 2| < \delta \\ 0 < x < \delta + 2[/imath] Step 3 Try to define [imath]x[/imath] in terms of [imath]\epsilon[/imath]. [imath] |x^2+1 - 5| < \epsilon \\ 4- \epsilon < x^2 < \epsilon + 4 \\ \sqrt{4- \epsilon} < x < \sqrt{\epsilon + 4} [/imath] Step 4 What next? I'm a little weak on how should I proceed. |
600889 | Solution to [imath]x^2[/imath] mod [imath]23=7^2[/imath]
Recently, I stumbled upon this problem, Solve [imath]x^2[/imath] [imath]mod[/imath] [imath]23 = 7^2[/imath], both here at MSE and somewhere surfing the web. I tried to solve it but don't know how. Although I can't remember where I found it, I doremember the question saying to solve the equation for [imath]x[/imath] rather than to determine whether some sort of solution existed. So by that, I presume that there is a such solution that I am not seeing. I tried the following: [imath]1^2[/imath] mod [imath]23[/imath], [imath]2^2[/imath] mod [imath]23[/imath], [imath]3^2[/imath] mod [imath]23[/imath], [imath]...[/imath] , [imath]22^2[/imath] mod [imath]23[/imath]. Following this procedure did not yield an answer of [imath]49[/imath]. Can anyone show me how a solution is determined because it has been over 2 days and it is driving me crazy from my inability to see the solution. | 586423 | Solve [imath]x^2[/imath] [imath]mod[/imath] [imath]23 = 7^2[/imath]
What is the procedure to solving [imath]x^2[/imath] [imath]mod[/imath] [imath]23 = 7^2[/imath]? According to WolframAlpha, there is no integer solution but I am completely confused as to what steps was taken to determine that. Before asking the question, I did try to solve this using brute force by plugging in some arbitrary numbers to see whether the square of that number mod [imath]23[/imath] gave me a remainder of [imath]49[/imath] but the procedure was quite tedious. Hence the reason I turned to WolframAlpha. Going back to the question, I want to know how WoflramAlpha determined that there was no integer solution to solving for [imath]x[/imath]. |
600885 | How find this integral [imath]I=\int\frac{1}{\sin^5{x}+\cos^5{x}}dx[/imath]
Question: Find the integral [imath]I=\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx[/imath] my solution: since \begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\ &=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\ &=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x}) \end{align*} then let [imath]\sin{x}+\cos{x}=t\Longrightarrow \sin{x}\cos{x}=\dfrac{1}{2}(t^2-2)[/imath] then [imath]\sin^5{x}+\cos^5{x}=t[1-\dfrac{1}{2}(t^2-2)-\dfrac{1}{4}(t^2-2)^2]=-\dfrac{1}{4}t(t^4-2t^2+4)[/imath] [imath]x+\dfrac{\pi}{4}=\arcsin{\dfrac{t}{\sqrt{2}}}\Longrightarrow dx=\dfrac{1}{\sqrt{2-t^2}}dt[/imath] so [imath]I=-4\int\dfrac{dt}{t\sqrt{2-t^2}(t^4-2t^2+4)}=-2\int\dfrac{1}{u\sqrt{2-u^2}(u^2-2u+4)}du[/imath] where [imath]u=t^2[/imath] Then I can't, because this wolf can't http://www.wolframalpha.com/input/?i=1%2F%28xsqrt%282-x%5E2%29%28x%5E2-2x%2B4%29%29dx Thank you for you help | 595038 | calculation of [imath]\int\frac{1}{\sin^3 x+\cos^3 x}dx[/imath] and [imath]\int\frac{1}{\sin^5 x+\cos^5x}dx[/imath]
Solve the following indefinite integrals: [imath] \begin{align} &(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\ &(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx \end{align} [/imath] My Attempt for [imath](1)[/imath]: [imath] \begin{align} I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\ &= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\ &= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx \end{align} [/imath] Using the identities [imath] \sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} [/imath] we can transform the integral to [imath]I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx [/imath] The integral is easy to calculate from here. My Attempt for [imath](2)[/imath]: [imath] \begin{align} J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\ &= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\ &= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx \end{align} [/imath] How can I solve [imath](2)[/imath] from this point? |
599601 | Properties of a continuous map [imath]f : \mathbb{R}^2\rightarrow \mathbb{R}[/imath] with only finitely many zeroes
Let [imath]f : \mathbb{R}^2\rightarrow \mathbb{R}[/imath] be a continuous map such that [imath]f(x)=0[/imath] only for finitely many values of [imath]x[/imath]. Which of the following is true? Either [imath]f(x)\leq 0[/imath] for all [imath]x[/imath] or [imath]f(x)\geq 0[/imath] for all [imath]x[/imath]. the map [imath]f[/imath] is onto. the map [imath]f[/imath] is one one. None of the above. What I have done so far is : I would take polynomial in two variables.. This need not be like [imath]f(x)\geq 0[/imath] for all [imath]x[/imath] or [imath]f(x)\geq 0[/imath] for all [imath]x[/imath].So,first option is eliminated. The map is not one one assuming that [imath]f[/imath] has more than one zero. So, third option is wrong. I could not think of an example in which it is not onto.. Only examples i am getting in my mind are polynomials and they are onto.. So, I am having trouble with surjectiveness of the function. Please help me to clear this. Thank you.. | 603185 | [imath]\,f \colon \Bbb R^2 \to \Bbb R [/imath] be continuous map such that [imath]\,f(x)=0\,[/imath]
I am stuck on the following problem : Let [imath]\,f \colon \Bbb R^2 \to \Bbb R [/imath] be continuous map such that [imath]\,f(x)=0\,[/imath] for only finitely many values of [imath]x[/imath]. Then which of the following options is correct? either [imath]f(x)\le 0 \forall x \,\text{or}\, f(x) \ge 0 \forall x[/imath] the map [imath]f[/imath] is one-to-one the map [imath]\,f[/imath] is onto none of the above My Attempt: Since [imath]\,f(x)=0\,[/imath] for only finitely many values of [imath]x[/imath], [imath]\exists x_1=(a_1,b_1),x_2=(a_2,b_2) \in \Bbb R^2[/imath] such that [imath]f(x_1)=0=f(x_2)[/imath] whereas [imath]x_1 \neq x_2[/imath]. So, [imath]f\,[/imath] is not one-to-one. Also,it is a many-to-one mapping and so can not be onto. I am confused about options 1 and 4. Can someone point me in the right direction with some explanation? Thanks and regards to all. |
599453 | upper bound on third moment
If I know that [imath]EX^4=1[/imath] and [imath]EX\leq0[/imath], how do I get an upper bound for [imath]EX^3[/imath]? I know that by Jensen's inequality, [imath]EX^3\leq1[/imath], but I need the upper bound less than 1. | 599283 | Is there any direct relation among first, third, fourth order moments?
Suppose that [imath]X[/imath] is a real-valued random variable, with [imath]EX^4=1[/imath] and [imath]EX\leq0[/imath]. Find an explicit constant [imath]c<1[/imath] such that [imath]EX^3<c[/imath]. |
439596 | Number of [imath](0,1)[/imath] [imath]m\times n[/imath] matrices with no empty rows or columns
I am looking to calculate the number of [imath]m\times n[/imath] matrices which have no empty rows or columns (at least one [imath]1[/imath] in each row and column). I have looked at the answers to a few similar questions such as 35019 and 329932 but I am struggling to make these approaches work in this case. | 2627026 | What is the number of all [imath]n\times n[/imath] matrices with entries in [imath] \{0,1\}[/imath] that in every row and every column has at least one 1?
What is the number of all matrices [imath]n*n[/imath] with entries in [imath] \{0,1\}[/imath] that in every row and every column has at least one 1? I think maybe it is easier to count those matrices that has some null-row or null-columns. Any ideas? Thanks in advance. |
602213 | Use epsilon-delta definition to show [imath]f(x)=x^2[/imath] is continuous
Let [imath]f[/imath] be a function [imath]f:\mathbb{R}\to \mathbb{R}[/imath] and [imath]f(x)=x^2.[/imath] Prove that [imath]f[/imath] is continuous on all of [imath]\mathbb{R}[/imath]. I tried this: Let [imath]\epsilon >0[/imath] and assume [imath]\mid x-x_0\mid < \delta[/imath] for some [imath]\delta[/imath] and [imath]\forall x_0 \in \mathbb{R}.[/imath] Then [imath]\mid f(x)-f(x_0) \mid=\mid x^2-x_0^2 \mid=\mid x-x_0\mid \mid x+x_0\mid <\delta \mid x+x_0\mid[/imath] Since the last term depends on [imath]\delta,[/imath] I can reduce it to any [imath]\epsilon.[/imath] Hence, [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath]. Does this look right? I think it might fail because I don't think I can choose any old [imath]\delta.[/imath] Is there a way to find an exact [imath]\delta[/imath] that guarantee [imath]\mid f(x)-f(x_0) \mid < \epsilon[/imath]? | 541110 | Show that [imath]f(x)=x^2[/imath] is continuous at [imath]a=2[/imath] using the [imath]\delta-\epsilon[/imath] definition of continuity.
So we want to find a [imath]\delta>0[/imath] such that for all [imath]2-\delta<x<2+\delta[/imath] , we will have [imath]4-\epsilon<x^{2}<4+\epsilon[/imath] for all [imath]\epsilon>0[/imath] . If we can find a way to express [imath]\delta[/imath] as a function [imath]\delta (\epsilon)[/imath], [imath]\delta:\mathbb{R}_+ \to \mathbb{R}_+ [/imath] then we will have solved the problem. But I can't see how to relate [imath]\epsilon[/imath] to [imath]\delta[/imath] in this case. My initial reaction is write [imath](2-\delta)^2<x^2<(2+\delta)^2[/imath], this makes the two inequalities look related, but I am not sure where to go from here. |
602439 | There exists no continuous function [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] with [imath]f=\chi_{[0,1]}[/imath] almost everywhere
I am trying to solve the same problem on this page. One gave an hint defining an inclusion function. Does someone know what is meant there? Thanks | 596921 | Prove that, there exists no continuous function [imath]f:\mathbb R\rightarrow\mathbb R[/imath] with [imath]f=\chi_{[0,1]}[/imath] almost everywhere.
Prove that, there exists no continuous function [imath]f:\mathbb R\rightarrow\mathbb R[/imath] with [imath]f=\chi_{[0,1]}[/imath] almost everywhere.[imath]\textbf(Make\ sure\ that\ your\ proof\ is\ completely\ rigorous)[/imath]. I don't know, which property to use. (It is not allowed to show it with [imath]\epsilon-\delta-criterion[/imath], our last topics were: Lp-Spaces, Radon-Nikodym Theorem, Riesz Representation Theorem,Lipschitz-Functions, Product measures, Fubini Theorem) but i can't find anything to do with continuity, which of them could be useful ? |
602556 | If [imath]A[/imath] and [imath]B[/imath] are [imath]F[/imath]-algebras, then [imath]A\otimes _{F}B [/imath] is [imath]F[/imath]-algebra.
Suppose [imath]A[/imath] and [imath]B[/imath] are two [imath]F[/imath]-algebras ([imath]F[/imath] is a field). Prove that [imath]A\otimes _{F}B [/imath] is an [imath]F[/imath]-algebra with the multiplication: [imath](a\otimes b )({a}'\otimes {b}')=(a{a}')\otimes (b{b}').[/imath] With this multiplication [imath]A\otimes _{F}B[/imath] is an [imath]F[/imath]-module and also a ring, and it is obvious that it is commutative with all elements of [imath]F[/imath]. The important thing that I couldn't show is how this multiplication is well-defined. | 548551 | Why is [imath]B\otimes_A C[/imath] a ring, where [imath]B,C[/imath] are [imath]A[/imath]-algebra
Given [imath]B,C[/imath] be two [imath]A[/imath]-algebra, it is said that [imath]B\otimes_AC[/imath] has a ring structure, with the multiplication being defined as: [imath] (b_1\otimes c_1)\cdot(b_2\otimes c_2):=(b_1b_2)\otimes (c_1c_2). [/imath] However I don't see an easy way to check it is well-defined. For, given another pair of representatives: [imath] (b_1'\otimes c_1')\cdot(b_2'\otimes c_2'):=(b_1'b_2')\otimes (c_1'c_2') [/imath] where [imath] b_1\otimes c_1=b_1'\otimes c_1'\quad\text{and}\quad b_2\otimes c_2=b_2'\otimes c_2'. [/imath] How to verify that [imath] (b_1b_2)\otimes (c_1c_2)=(b_1'b_2')\otimes (c_1'c_2')? [/imath] |
602721 | Intermediate field of [imath]\Bbb Q(\alpha)[/imath] and [imath]\Bbb Q[/imath]
Let [imath]f[/imath] be an irreducible polynomial of degree 4 over [imath]\Bbb Q[/imath] and [imath]Gal(f)=S_4[/imath]. Prove that there isn't nontrivial intermediate field between [imath]\Bbb Q(\alpha)[/imath] and [imath]\Bbb Q[/imath] where [imath]\alpha[/imath] is a root of [imath]f[/imath]. Here is my attempt: Let [imath]E[/imath] be a splitting field of [imath]f[/imath]. Then, [imath][E:\Bbb Q]=24[/imath]. If there is such intermediate field [imath]K[/imath], then clearly [imath][E:K]=12[/imath], which means [imath]Gal(E/K)=A_4[/imath]. But it's impossible? Or should I do different way by using irreducibility of [imath]f[/imath]? | 354098 | [imath]f \in K[X][/imath] of degree [imath]n[/imath] with Galois group [imath]S_n[/imath], why are there no non-trivial intermediate fields of [imath]K \subset K(a)[/imath] with [imath]a[/imath] a root of [imath]f[/imath]?
Let [imath]K[/imath] be a field and [imath]f \in K[X][/imath] of degree [imath]n[/imath] with Galois group [imath]S_n[/imath]. Let [imath]a[/imath] be a root of [imath]f[/imath], [imath]L = K(a)[/imath], and let [imath]E[/imath] be a intermediate field of the extension [imath]K \subset L[/imath]. Prove that [imath]E = K[/imath] or [imath]E = L[/imath]. Any suggestions on where to start? Thanks. |
598164 | For a differentiable function [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] what does [imath]\lim_{x\rightarrow +\infty} f'(x)=1[/imath] imply? (TIFR GS [imath]2014[/imath])
Question is : For a differentiable function [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] what does [imath]\lim_{x\rightarrow +\infty} f'(x)=1[/imath] imply? Options: [imath]f[/imath] is bounded [imath]f[/imath] is increasing [imath]f[/imath] is unbounded [imath]f'[/imath] is bounded I could not find counter examples but then, I strongly feel [imath]\lim_{x\rightarrow +\infty} f'(x)=1[/imath] would just imply that [imath]f'(x)[/imath] is bounded. I do not have much idea why would other three are false. I would be thankful if someone can suggest me some hints. Thank You. | 601571 | [imath]f:\mathbb{R} \to \mathbb{R}[/imath] be differentiable and [imath]\lim\limits_{x\to\infty}f'(x)=1[/imath], is [imath]f(x)[/imath] unbounded?
Suppose [imath]f:\mathbb{R} \to \mathbb{R}[/imath] is a differentiable function such that [imath]\lim\limits_{x\to\infty}f'(x)=1[/imath],then is it true necessarily true that [imath]f(x)[/imath] unbounded? I think that it will always intersect [imath]y=c[/imath] for every [imath]c\in \mathbb{R}[/imath] and thus cannot be bounded! |
603328 | measurable sets
please could you help me with this exercise. Prove or a counterexample to the following statement: [imath]f:\mathbb{R} \to \mathbb{R} [/imath] such that [imath]\forall \alpha \in \mathbb{R}[/imath] the set [imath]\{ x \in \mathbb{R} : f(x)=\alpha \} [/imath] is measurable. then [imath]f[/imath] is measurable. thank you very much in advance. | 264668 | Must [imath]f[/imath] be measurable if each [imath]f^{-1}(c)[/imath] is?
Suppose [imath]f[/imath] is a real-valued function on [imath]\mathbb R[/imath] such that [imath]f^{β1}(c)[/imath] is measurable for each number [imath]c[/imath]. Is [imath]f[/imath] necessarily measurable? |
603449 | Use [imath]\alpha, \beta, \gamma [/imath] roots of a polynomial to construct another polynomial
Let [imath]\alpha, \beta, \gamma [/imath] be roots [imath]\in \mathbb{C}[/imath] of [imath]x^3-3x+1[/imath]. Determinate a monic polynomial, degree [imath]3[/imath], witch roots are [imath]1- \alpha^{-1},1-\beta^{-1},1-\gamma^{-1}[/imath] The catch is that i can't use any formula to calculate the roots of [imath]x^3-3x+1[/imath] unless using the quadratic formula or the gauss theorem (though it might not be necessary) | 601508 | Let [imath]\alpha \in \overline{\Bbb Q}[/imath] a root of [imath]X^3+X+1\in\Bbb Q[X][/imath]. Calculate the minimum polynomial of [imath]\alpha^{-1}[/imath] en [imath]\alpha -1[/imath].
Let [imath]\alpha \in \overline{\Bbb Q}[/imath] a root of [imath]X^3+X+1\in\Bbb Q[X][/imath]. Calculate the minimum polynomial of [imath]\alpha^{-1}[/imath] en [imath]\alpha -1[/imath]. I don't really understand how to get started here. I know that [imath]\overline{\Bbb Q}=\{x\in \Bbb C : \text{ where $x$ is algegbraic over $\Bbb Q$} \}[/imath] A hint to get me started or a complete solution would be both very muchappreciated :) |
603718 | [imath]f(x)=x^{\frac{2}{3}}\ln x[/imath] for [imath]x>0[/imath] , is it uniformly continuous
Let [imath]f[/imath] be the real valued function 0n [imath][0,\infty][/imath] defined by [imath]f(x)=x^{\frac{2}{3}}\ln x[/imath] for [imath]x>0[/imath] and [imath]0[/imath] for [imath]x=0[/imath]. Is it uniform continuous on [imath][0,\infty)[/imath]? I think that if we break [imath][0,\infty)[/imath] in [imath][0,1]\cup(1,\infty)[/imath] then being continuous on [imath][0,1][/imath] it will be uniform there and by showing that its derivative is bounded on [imath][1,\infty)[/imath] we can collectively say that it is uniform on [imath][0,\infty)[/imath] , Is it correct? | 602554 | Uniform continuity of [imath]f(x)=x^{2/3}\log x[/imath] on [imath][0, \infty)[/imath]
Let [imath]f[/imath] be a real valued function on [imath][0,\infty)[/imath] defined by [imath]f(x)= \begin{cases} x^{{2}/{3}}\log x & \text{ if } x>0 \\ 0 & \text{ if } x= 0 \end{cases}[/imath] Then which of the following is true? A. [imath]f[/imath] is discontinuous at [imath]x=0[/imath]. B. [imath]f[/imath] is continuous on [imath][0,\infty)[/imath], but not uniformly continuous on [imath][0,\infty)[/imath]. C. [imath]f[/imath] is uniformly continuous on [imath][0,\infty)[/imath]. D. [imath]f[/imath] is not uniformly continuous on [imath][0,\infty)[/imath], but uniformly continuous on [imath][1,\infty)[/imath]. I could see that [imath]f(x)=x^{\frac{2}{3}}\log x[/imath] is continous at [imath]x=0[/imath] But for uniform continuity I thought boundedness of derivative would help But I found [imath]f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\log x + x^{-\frac{1}{3}}[/imath] This [imath]f'(x)[/imath] is not bounded so i can not say any thing about uniform contunuity. Please help me to clear this. Thank you :) P.S : This was already asked before (Uniform continuity of [imath]f(x)=x^{\frac{2}{3}}\log x[/imath] some other person but it was not having enough details so it got closed. I tried to edit that question but someone said I should not edit OP to add more details. I was afraid that may be against the rules of MSE. So, I thought i should ask this separately with my own ideas. |
156600 | Multiple choice question: Let [imath]f[/imath] be an entire function such that [imath]\lim_{|z|\rightarrow\infty}|f(z)|[/imath] = [imath]\infty[/imath].
Let [imath]\displaystyle f[/imath] be an entire function such that [imath]\lim_{|z|\rightarrow \infty} |f(z)| = \infty .[/imath] Then, [imath]f(\frac {1}{z})[/imath] has an essential singularity at 0. [imath]f[/imath] cannot be a polynomial. [imath]f[/imath] has finitely many zeros. [imath]f(\frac {1}{z})[/imath] has a pole at 0. Please suggest which of the options seem correct. I am thinking that [imath]f[/imath] can be a polynomial and so option (2) does not hold. Further, if [imath]f(z) = \sin z [/imath] then it has infinitely many zeros... which rules out (3) while for [imath]f(z) = z[/imath] indicates that it has a simple pole at [imath]0[/imath] and option (4) seems correct. | 829533 | Let [imath]f[/imath] be an entire function such that [imath]Lim_{|z|\rightarrow \infty}|f(z)| = \infty[/imath]. Which of the following statement is true?
[imath]f(\frac{1}{z})[/imath] has essential Sinularity at [imath]0[/imath]. [imath]f[/imath] can not be polynomial. [imath]f[/imath] has finiyely many zeroes [imath]f(\frac{1}{z})[/imath] has pole at [imath]0[/imath]. Take [imath]f(z) = z^2[/imath],then [imath]Lim_{|z|\rightarrow \infty}|f(z)| = \infty[/imath]. So (1) and (2) are not true. I think (3) and (4)are true, But I do not know ,how will prove. Please give me Idea How to Solve this Thank you. |
308422 | Why do we accept Kuratowski's definition of ordered pairs?
I've been struggling understanding Kuratowski's definition of ordered pairs. I understand what it means but I don't see why I should accept it. I've seen this question and this one, most importantly --- through reading the wiki page I've realised one thing. The only reason [imath](a,b)=\{\{a\},\{a,b\}\}[/imath] is accepted is because it satisfies [imath](a,b)=(c,d) \iff (a=c) \land (b=d)[/imath] Am I not misunderstanding this? If I can come with my own exotic definition which satisfies the above iff statement, would it be accepted? | 742675 | Why do some authors define the ordered pair as the set: [imath](a,b)=\{\{a\},\{a,b\}\}[/imath]?
I am using a textbook called foundations of mathematical analysis by johnsonbaugh and in it, he defines the ordered pair of elements [imath]a[/imath] and [imath]b[/imath], writen as [imath](a,b)[/imath] as the set: [imath](a,b)=\{\{a\},\{a,b\}\}[/imath] where [imath]a[/imath] is called the first element of [imath](a,b)[/imath] and [imath]b[/imath] is called the second element of [imath](a,b)[/imath]. This definition is a bit strange, would anyone know how I can interpret it? Thank you! |
604173 | Alternating sum of binomial coefficients [imath]\sum(-1)^k{n\choose k}\frac{1}{k+1}[/imath]
I would appreciate if somebody could help me with the following problem Q:Calculate the sum: [imath] \sum_{k=1}^n (-1)^k {n\choose k}\frac{1}{k+1} [/imath] | 38623 | How to prove [imath]\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}[/imath]?
Other than the general inductive method,how could we show that [imath]\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}[/imath] Apart from induction, I tried with Wolfram Alpha to check the validity, but I can't think of an easy (manual) alternative. Please suggest an intuitive/easy method. |
604804 | Proof by Induction problem. Unsure of sigma notation.
I want to use proof by induction to verify [imath]\sum_{j=n}^{2n-1} (2j+1) = 3n^2.[/imath] First I assume that the expression is valid for some number k, [imath]S(k) = 2k + 1 [/imath] Then k + 1 so that, [imath]S(k+1) = s(k) + a(k+1) = 3k^2 + (2(k+1)+1) = 3k^2 + 2k +3 [/imath] However, if I try to insert j = k + 1 into the sum-formula [imath] 3n^2 [/imath] it equals [imath]3(k+1)^2= 3k^2 + 6k +3 [/imath]. Which is not the same. What am I doing wrong? | 604846 | Induction proof that [imath]\sum_{j=n}^{2n-1} (2j + 1) = 3n^2[/imath] - what happened?
So I have the general summation formula that I was to prove using mathemathical induction on a Calculus level. For all [imath]n = {1,2,3,...}[/imath] we have: [imath]\sum_{j=n}^{2n-1} (2j + 1) = 3n^2[/imath] In the answer, the one completing the proof gets this, on the [imath]n=k+1[/imath]: [imath]\sum_{j=(k+1)}^{2(k+1)-1} (2j + 1) = \sum_{j=k}^{2k-1} (2j + 1) - (2k+1) + (4k+1) + (4k+3)[/imath] [imath]= 3k^2 + 6k + 3 = 3(k+1)^2[/imath] Can anyone explain how on earth this was done? When I tried, I was unable to factorize - I tried by adding the [imath](2(k+1)+1)[/imath] to the [imath]S_n[/imath] formula and thereby confirm it, but I couldn't do it... |
604961 | Existence of surjection implies existence of injection?
Let [imath]A[/imath] and [imath]B[/imath] be sets. If there exists a surjection [imath]f : A \to B[/imath] then there exists an injection [imath]g : B \to A[/imath]. Proof: given [imath]b \in B[/imath] select an element [imath]a \in f^{-1}(b)[/imath]. Denote this element by [imath]g(b)[/imath]. Then [imath]g(b) \in f^{-1}(b)[/imath] so that [imath]f(g(b)) = b[/imath]. Consequently [imath]g(b_1) = g(b_2)[/imath] implies [imath]f(g(b_1)) = f(g(b_2))[/imath] so that [imath]b_1 = b_2[/imath]. We conclude [imath]g[/imath] is an injection. QED Is the axiom of choice required to make this argument rigorous? | 192460 | There exists an injection from [imath]X[/imath] to [imath]Y[/imath] if and only if there exists a surjection from [imath]Y[/imath] to [imath]X[/imath].
Theorem. Let [imath]X[/imath] and [imath]Y[/imath] be sets with [imath]X[/imath] nonempty. Then (P) there exists an injection [imath]f:X\rightarrow Y[/imath] if and only if (Q) there exists a surjection [imath]g:Y\rightarrow X[/imath]. For the P [imath]\implies[/imath] Q part, I know you can get a surjection [imath]Y\to X[/imath] by mapping [imath]y[/imath] to [imath]x[/imath] if [imath]y=f(x)[/imath] for some [imath]x\in X[/imath] and mapping [imath]y[/imath] to some arbitrary [imath]\alpha\in X[/imath] if [imath]y\in Y\setminus f(X)[/imath]. But I don't know about the Q [imath]\implies[/imath] P part. Could someone give an elementary proof of the theorem? |
605423 | Limit of [imath]L^r[/imath] norm in lebesgue measure theory
Let [imath]f\in L^r[/imath] for some [imath]r>0[/imath] and [imath]\mu (X)=1[/imath]. Then, prove that [imath]\lim_{p\to 0}||f||_p=\exp(\int \log|f|d\mu)[/imath]. This is from chapter [imath]L^p[/imath] spaces, but I don't have any idea. How to make [imath]\log[/imath]? Please give hint or proof. | 282271 | "Scaled [imath]L^p[/imath] norm" and geometric mean
The [imath]L^p[/imath] norm in [imath]\mathbb{R}^n[/imath] is \begin{align} \|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}. \end{align} Playing around with WolframAlpha, I noticed that, if we define the "scaled" [imath]L^p[/imath] norm in [imath]\mathbb{R}^n[/imath] to be \begin{align} \| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p} \end{align} then \begin{align} \lim_{p \to 0} \|x\|_p &= \left( \prod_{j=1}^{n} |x_j| \right)^{1/n}, \end{align} which is the geometric mean of the coordinates' absolute values. This is interesting maybe because the [imath]L^p[/imath] norm doesn't have a nice limit at zero. My questions: How do I prove this? Is this definition of "scaled [imath]L^p[/imath] norm" interesting, or known by another name, or used anywhere? Is there any interesting reason to define the [imath]L^0[/imath] norm as the geometric mean, as above? Further reading? Thanks! |
601779 | Vector bundle on a quadric
The problem: Consider the smooth quadric [imath]Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}[/imath] and the line [imath]L=V(X_{0},X_{2},X_{3})[/imath] contained in it. Prove that there exists a vector bundle [imath]F[/imath] on [imath]Q[/imath] with a section vanishing exactly at [imath]L[/imath]. | 603200 | Vector bundle on a quadric [imath]Q[/imath]
The problem: Consider the smooth quadric [imath]Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}[/imath] and the line [imath]L=V(X_{0},X_{2},X_{4})[/imath] contained in it. Prove that there exists a vector bundle [imath]F[/imath] on [imath]Q[/imath] with a section vanishing exactly at [imath]L[/imath]. The hardest thing is to get the transition matrices. Any help is appreciated. |
605398 | [imath]\mathbb{Z} \times \mathbb{Z} [/imath] is a PID or not?
we know Z is a PID but there exists no ring isomorphism between ZxZ and Z. So based on this observation can we conclude that ZxZ is not a PID ? I dont think we can because if A and B are isomorphic rings then we can say with certainty that both A & B have the same algebraic properties but if they are not isomorphic then it dose not necessarily mean that there can not exist some common property between A and B where some other property/properties are bound to differ. Am I wrong here? | 582699 | [imath]\mathbb Z\times\mathbb Z[/imath] is principal but is not a PID
I need to find an example of a ring that is not a PID but every ideal is principal. I know that [imath]\mathbb Z\times\mathbb Z[/imath] is not an integral domain, so certainly is not a PID, but here every ideal is principal. I already proved that if R and S are ring every ideal in R x S is I x J with ideals in the original ring. But I cant follow from that that [imath]\mathbb Z\times\mathbb Z[/imath] has only principal ideals. Explicitly, if [imath]I[/imath] is an ideal [imath](a)\times(b)[/imath] which would be its generator [imath](c,d)[/imath] in [imath]\mathbb Z\times\mathbb Z[/imath]? Thanks. |
605470 | If [imath]p[/imath] is an odd prime of the form [imath]p=3\pmod{4}[/imath], then prove that [imath]((p-1)/2)!=1[/imath] or [imath]-1\pmod{p}[/imath].
If [imath]p[/imath] is an odd prime of the form [imath]p=3 \pmod{4}[/imath], then prove that [imath] \left(\frac{p-1}{2}\right)!=1 \text{ or } -1\pmod{p}. [/imath] | 222372 | For primes [imath]pβ‘3\pmod 4[/imath], prove that [imath][(pβ1)/2]!β‘Β±1\pmod p[/imath].
I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it. |
605540 | pull back and push out problem
could you please help with the proof? If a category [imath]K[/imath] has pullbacks. For composable morphisms [imath]f: A\to B[/imath] and [imath]g: B \to C,[/imath] if [imath]g[/imath] and [imath]f[/imath] are extremal epimorphisms, prove that [imath]gf[/imath] is an extremal epimorphism. The definition of extremal epimorphism I have been given is: A morphism [imath]e: A \to B[/imath] is an extremal epimorphism when for each commutative diagram [imath]e: A \to B,[/imath] [imath]f: A \to C[/imath], [imath]m: C \to B[/imath] ([imath]e= mf[/imath]) if [imath]m[/imath] is monic, then [imath]m[/imath] is an isomorphism. I have let [imath]f = mk[/imath] and [imath]g = nh[/imath], then I have drawn out the pullbacks of [imath]f[/imath] and [imath]m,[/imath] and of [imath]g[/imath] and [imath]n[/imath]. I don't know where to go from here. | 591581 | questions about extremal epimorphisms in category theory
let [imath]K[/imath] be a category with equalizers, show that every extremal epimorphism is epic. for composable morphisms [imath]f: A \rightarrow B [/imath] and [imath]g: B \rightarrow C[/imath] in [imath]K[/imath], show that if [imath]gf[/imath] is an extremal epimorphism, then [imath]g[/imath] is an extremal epimorphism. let [imath]K[/imath] be a category with pullbacks. For composable morphisms [imath]f[/imath] and [imath]g[/imath] as given in #2, show that if [imath]g[/imath] and [imath]f[/imath] are extremal epimorphisms, then [imath]gf[/imath] is an extremal epimorphism. show that if an extremal epimorphism is monic, then it is an isomorphism. here is what I know: In any category [imath]K[/imath], an arrow [imath]f: A \rightarrow B[/imath] is called an isomorphism if there is an arrow [imath]g: B \rightarrow A[/imath] in [imath]K[/imath] such that [imath] g \circ f = I_{A} [/imath] and [imath] f \circ g = I_{B}[/imath] is called a monomorphism, if given any [imath]g,h : C \rightarrow A, fg=fh[/imath] implies [imath]g=h[/imath] is called an epimorphism, if given any [imath]i,j: B \rightarrow D, if = jf[/imath] implies [imath]i=j[/imath] is an extremal epimorphism, if for each commutative diagram, [imath]f = mh[/imath] where if [imath]m[/imath] is monic, then [imath]m[/imath] is an isomorphism I have tried to answer #2: since [imath]gf[/imath] is an extremal epimorphism, then [imath]gf = mh[/imath] where [imath]m[/imath] is monic and an isomorphism. now let [imath]g = nk[/imath] where [imath]n[/imath] is monic. I now need to show that [imath]n[/imath] is an isomorphism. my problem is how do I do this? I'm trying to visualise things using diagrams, but I'm lost. Here is what I have so far for #1: let [imath]f: A \rightarrow B[/imath] be and extremal epimorphism. consider [imath] g,h: B \rightarrow C[/imath] such that [imath]gf=hf[/imath] so I need to prove [imath]g=h[/imath] let [imath]e: E \rightarrow B[/imath] be the equaliser of [imath]g, h[/imath] now I don't know where to go from here to prove [imath]g=h[/imath]. I know that [imath]f[/imath] factors through the equaliser. since [imath]e[/imath] is an equaliser, there is a unique morphism [imath]\varphi : A \rightarrow E[/imath] therefore I now have [imath]f=e \varphi[/imath] which means that [imath]e[/imath] is now isomorphic since [imath]f[/imath] is an extremal epimorphism. So can I now conclude [imath]g=h[/imath] ? |
577690 | [imath]\frac{1}{\infty}[/imath] - is this equal [imath]0[/imath]?
I've seen that wolfram alpha says: [imath]\frac{1}{\infty} = 0[/imath] Well, I'm sure that: [imath]\lim_{x\to \infty}\frac{1}{x} = 0[/imath] But does [imath]\frac{1}{\infty}[/imath] only makes sense when we calculate it's limit? Because for me, [imath]1[/imath] divided by any large amount of number will be always almost zero. | 44746 | One divided by Infinity?
Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me. I know that [imath]0.999 \cdots[/imath] (repeating) = [imath]1[/imath]. So wouldn't [imath]1 - \frac{1}{\infty} = 1[/imath] as well? Because [imath]\frac{1}{\infty} [/imath] would be infinitely close to [imath]0[/imath], perhaps as [imath]1^{-\infty}[/imath]? So [imath]1 - 1^{-\infty}[/imath], or [imath]\frac{1}{\infty}[/imath] would be equivalent to [imath]0.999 \cdots[/imath]? Or am I missing something? Is infinity something that can even be used in this sort of mathematics? |
605871 | Ultrafilter problem
could you help me with this problem, please? If [imath]U[/imath] is a principal ultrafilter on [imath]I[/imath] such that [imath]\{a\}\in U[/imath]. Show that [imath]Ult(\mathfrak{A}_x:x\in I)[/imath] is isomorphic to [imath]\mathfrak{A}_a[/imath] and [imath][f]=f(a)[/imath] for each [imath]f[/imath] and [imath]j[/imath], the embedding is the identity. | 605091 | Show that if U is a principal ultrafilter then the canonical inmersion is an isomorphism between [imath]\mathcal A[/imath] and [imath]Ult_{U}\mathcal A[/imath]
Show that: If [imath]U[/imath] is a principal ultrafilter, then the canonical inmersion [imath]j[/imath] is an isomorphism between [imath]\mathcal A[/imath] and [imath]Ult_{U}\mathcal A[/imath] |
600500 | How to visualize [imath]SO(4) \simeq SO(3)\bigotimes SO(3) / \mathbb{Z}_2 [/imath]
Is there a simple way to see that [imath]SO(4) \simeq SO(3)\bigotimes SO(3) /\mathbb{Z}_2[/imath] ? | 46131 | Why is [imath]SO(3)\times SO(3)[/imath] isomorphic to [imath]SO(4)[/imath]?
Why is [imath]$SO(3)\times SO(3)$[/imath] isomorphic to [imath]$SO(4)$[/imath]? |
293243 | Proof of Euler's general formula for a sum involving harmonic numbers
I have seen this formula, but how to prove this? [imath]2\sum\limits_{k=1}^\infty \frac{H_k}{\left( k+1 \right)^m} =m\zeta \left( m+1 \right)-\sum\limits_{k=1}^{m-2}{\zeta \left( m-k \right)\zeta \left( k+1 \right)}[/imath] | 469023 | Generalized Euler sum [imath]\sum_{n=1}^\infty \frac{H_n}{n^q}[/imath]
I found the following formula [imath]\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)[/imath] and it is cited that Euler proved the formula above , but how ? Do there exist other proofs ? Can we have a general formula for the alternating form [imath]\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}[/imath] |
606292 | Group ring of galois group
Suppose [imath]E/F[/imath] is Galois extension. What is it known about structure of [imath]F[Gal(E/F)][/imath]? I've learned only one fact in this direction - existence of normal basis in [imath]E/F.[/imath] But it's not truly about [imath]F[Gal(E/F)].[/imath] | 387005 | Galois representations and normal bases
I am not very familiar with the theory of Galois representations, but I do know a bit about both Galois theory and representation theory. Recently I learned about the notion of a normal basis for a Galois extension which led me to consider following straight-forward Galois representation: Let [imath]L/K[/imath] be a Galois extension and choose a [imath]K[/imath]-basis for [imath]L[/imath]. Then each element of the Galois group is a [imath]K[/imath]-linear map, hence may be represented as a matrix with respect to our chosen basis. Question 1: What is known about the decomposition of this representation? Since a [imath]K[/imath]-basis for [imath]L[/imath] contains at most 1 element of [imath]K[/imath], it seems to me that there is at most 1 trivial subrepresentation. Can every irreducible representation occur? If we chose a normal basis for [imath]L/K[/imath], then the representation described above will be the permutation representation, no? The problem of constructing a normal basis for a Galois extension made me wonder about the Galois invariant subspaces. For example, in [imath]\mathbb{Q}(\sqrt{2})/\mathbb{Q}[/imath], I can quickly spot two Galois invariant subspaces: [imath]\mathbb{Q}[/imath] and [imath]\sqrt{2}\mathbb{Q}[/imath], which I am fairly confident exhausts the supply. Question 2: Can we use representation theory to classify/specify all the Galois invariant subspaces of an extension [imath]L/K[/imath]? Yesterday I read a proof of the existence of normal bases, which was not trivial. However, thinking about my example above for [imath]\mathbb{Q}(\sqrt{2})/\mathbb{Q}[/imath] geometrically, it seems very rare that a randomly chosen basis would not be normal: if we pick any element not on the two lines listed above, its Galois orbit will be a basis, no? Question 3: Will this observation continue to hold in general, and can this intuition be made into a proof of the existence of normal bases? |
433461 | Without appealing to choice, can we prove that if [imath]X[/imath] is well-orderable, then so too is [imath]2^X[/imath]?
Without appealing to the axiom of choice, it can be shown that (Proposition:) if [imath]X[/imath] is well-orderable, then [imath]2^X[/imath] is totally-orderable. Question: can we show the stronger result that if [imath]X[/imath] is well-orderable, then so too is [imath]2^X[/imath]? Proof of Proposition. Pick any well-ordering of [imath]X[/imath]. Then the lexicographic order totally orders [imath]2^X[/imath]. More explicitly: for any two [imath]f,g \in 2^X[/imath], define [imath]f < g[/imath] iff there exists [imath]x \in X[/imath] such that [imath]f(x) \neq g(x)[/imath], and if [imath]x \in X[/imath] is minimal such that [imath]f(x) \neq g(x)[/imath], then [imath]f(x)=0[/imath] and [imath]g(x)=1[/imath]. It can be shown that [imath]<[/imath] totally-orders [imath]2^X[/imath]. | 455020 | AC iff [imath]P(\delta)[/imath] can be well-ordered for all [imath]\delta\in{\bf On}[/imath]
I'm trying to do the following exercise: Exercise. Assume ZF. Then AC is equivalent to the statement that [imath]P(\delta)[/imath] can be well-ordered for all [imath]\delta\in{\bf On}[/imath]. I'm struggling with the limit case in the induction proof of the "[imath]\Leftarrow[/imath]" direction; i.e. Foundation implies [imath]{\bf V}={\bf WF}[/imath], so I'm showing that [imath]R(\alpha)[/imath] (also sometimes called [imath]V_\alpha[/imath]) is well-ordered for all [imath]\alpha\in{\bf On}[/imath]. The hint tells me to fix some limit ordinal [imath]\gamma[/imath], find an ordinal [imath]\kappa\npreceq R(\gamma)[/imath] by Hartogs' lemma and fix a well-order [imath]\sqsubset[/imath] on [imath]P(\kappa)[/imath], which is possible by assumption. Now I have to find a way to recursively define a well-order [imath]\vartriangleleft_\alpha[/imath] on each [imath]R(\alpha)[/imath], from [imath]\sqsubset[/imath]. What I've tried and accomplished so far: If [imath]x\in R(\alpha)[/imath] then [imath]\forall\gamma\geq\alpha: x\in R(\gamma)[/imath]. Since the ordinal [imath]\kappa[/imath] is found via Hartog's lemma, it satisfies being [imath](R(\gamma))^+[/imath], and thus a cardinal. Since [imath]\kappa[/imath] is a cardinal, it is true that [imath]\forall \alpha\leq\gamma: R(\alpha)\preceq\kappa[/imath], but since [imath]\kappa\npreceq R(\gamma)[/imath], we have instead that [imath]\forall\alpha\leq\gamma: R(\alpha)\prec\kappa[/imath]. Since [imath]\forall\alpha\leq\gamma: R(\alpha)\prec\kappa[/imath], we know there exist a surjection [imath]f_\alpha:\kappa\to R(\alpha)[/imath] for each [imath]\alpha\leq\gamma[/imath], but not a corresponding injection (since that requires AC). We cannot do anything with the mostowski collapse [imath]\pi:R(\alpha)\cong\beta[/imath] ([imath]\beta\in{\bf On}[/imath]) of the [imath]R(\alpha)[/imath]'s, since that requires a fixed well-ordering on the [imath]R(\alpha)[/imath]'s, which requires AC (to choose the well-orderings). [imath]\forall\alpha\in{\bf On}: \alpha\in R(\alpha+1)[/imath]. We can't define [imath]x\vartriangleleft_\alpha y[/imath] iff [imath]x\vartriangleleft_{\alpha+1}y[/imath], since this requires the [imath]\sqsubset[/imath] ordering as the "last" [imath]\vartriangleleft[/imath], but to be able to do this, I'd suppose we need an injection, which is not possible (requires AC, as mentioned above as well). Am I heading in the right direction? Because it seems like I've hit one dead end after each other. I'd love to find [imath]\vartriangleleft_\alpha[/imath]'s that "coincide" with [imath]\sqsubset[/imath] somehow, but this seems near impossible, when we're not allowed to use injections. Is there just something blatantly obvious that I'm missing completely? |
607080 | Why do we use "if" in the definitions instead of "if and only if"?
I often write my notes as logical statements and constantly wonder why people use only the "if" direction in the definitions instead of the "if and only if". Consider: "A homomorphism [imath]\phi[/imath] is said to be an isomorphism if it is an injection and a surjection." would translate to: [imath]homomorphism(\phi) \wedge injection(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)[/imath] By adding the following statement consistent with the previous one: [imath]homomorphism(\phi) \wedge \neg injection(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)[/imath] we could deduce [imath]homomorphism(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)[/imath] which is obviously not what we intended to express in the definition, however the intention does not imply the deed. From the strict logical perspective do you agree that all definitions of this form are "flawed" owing to the lack of the attention to details? Or is there some reason which I have not managed to observe that explains why "if" is used in lieu of "if and only if"? Would you accept if I write in papers definitions in the iff version? I.e. "A homomorphism [imath]\phi[/imath] is an isomorphism if and only if it is an injection and a surjection." | 566565 | Are "if" and "iff" interchangeable in definitions?
In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if"). I'd like to know if in mathematical literature in general "if" in definitions means "iff". For example I am reading "Essential topology" and the following definition is written: In a topological space [imath]T[/imath], a collection [imath]B[/imath] of open subsets of [imath]T[/imath] is said to form a basis for the topology on [imath]T[/imath] if every open subset of [imath]T[/imath] can be written as a union of sets in [imath]B[/imath]. Should I assume the converse in such a case? Should I assume that given a basis [imath]B[/imath] for a topological space, every open set can be written as a union of sets in [imath]B[/imath]? This is just an example, I am not asking specifically about this sentence. |
607456 | Find the sum of [imath]\sum_{k=0}^n{{n \choose k}2^{k}}[/imath]
I know [imath]\sum_{k=0}^n{{n \choose k}=2^n}[/imath] but I'm not sure how to solve this with the [imath]2^{k}[/imath] term. | 534417 | Simplify a combinatorial sum
Why is [imath]\sum_{k=0}^n (_k^n) 2^k[/imath] simplified as [imath]3^n[/imath] and not just [imath]2^n[/imath] ? The answer uses the binomial theorem as a solution: let [imath]a=1[/imath] and let [imath]b=2[/imath]. So [imath]1+2 = 3[/imath]. But since [imath]1+2 = 3[/imath] why do we not use [imath]1+1 = 2[/imath]? |
607239 | Integral domain characteristic: contradictory ideas?
In my book (Herstein) this is a statement: An integral domain [imath]D[/imath] is said to be of finite characteristic if there exists a positive integer [imath]m[/imath] such that [imath]ma=0[/imath] for all [imath]a[/imath] in [imath]D[/imath]. But by definition an integral domain can't have zero-divisors right? So how can an integral domain have finite characteristic because isn't the characteristic a zero-divisor for every a in D? Is it that m can't be in D? I'm confused, if anyone could clarify this for me I'd really appreciate it. | 456215 | Characteristic of an integral domain must be either [imath]0[/imath] or a prime number.
Proposition: Characteristic of an integral domain must be either [imath]0[/imath] or prime number. I'm confused by this proposition. I think the characteristic of an integral domain should be always [imath]0[/imath]. Suppose it has characteristic [imath]n[/imath]. Then [imath]n * a = 0[/imath] for all a of the integral domain. since n is not [imath]0[/imath] and, if [imath]c * d = 0[/imath] in integral domain, it means [imath]c=0[/imath] or [imath]d=0[/imath], a should be [imath]0[/imath]. Hence [imath]n * a[/imath] is not [imath]0[/imath] when [imath]a[/imath] is nonzero. Therefore, characteristic should be always [imath]0[/imath]. What's wrong with my thought? |
607531 | How can I find the maximal interval of existance for following equation?
How can I find the maximal interval of existence for the solution of [imath]\dfrac{dx}{dt}=4x^2[/imath] , [imath]x(0)=5[/imath] including [imath]t=0[/imath] . | 585398 | Uniqueness and Existence problem
I just need a bit of help with this question. If I know that [imath]dg/dx = g^2[/imath], and that [imath]g(0) = g_0[/imath], then I can solve: [imath] dg/dx = g^2\\ \frac{1}{g^2} dg = dx \\ -\frac{1}{g} = x + \hat c \\ -g = \frac {1}{x + \hat c} \\ g = \frac{1}{-x - \hat c}. \\ [/imath] Solving for the specific solution, [imath] g_0 = \frac{1}{0 - \hat c} \\ \hat c = -\frac{1}{g_0} [/imath] This means that [imath]g(x) = \frac {g_0}{1-g_0x}[/imath]. My question is: How do I determine the maximum interval on which the solution is defined? |
608012 | Handwriting of special symbol in maths
What are standard ways of writing the symbol [imath]{\mathbb R}[/imath] in handwriting? | 152052 | How to write special set notation by hand?
Does anyone know a good resource (preferably pictures) that illustrates a conventional way to write the special sets symbols, i.e. [imath]\mathbb{N,Z,Q,R,C}[/imath] etc., by hand? |
242873 | Homeomorphism of the real line-Topology
I need to show that any open interval is homeomorphic to the real line. I know that [imath]f(x)=a+e^x[/imath] will work for the mapping [imath]f:R \to (a,\infty)[/imath] and [imath]f(x)=b-e^{-x}[/imath] will work for the mapping [imath]f:R \to (-\infty,b).[/imath] Without using two functions, how can I prove the result in general? | 1438133 | Prove that an open interval [imath](a,b)[/imath] considered as a subspace of a real line is homeomorphic to the real line.
I understand that in [imath](\Bbb R ,d)[/imath], any open interval [imath](a,b)[/imath] is topologically equivalent to [imath]\Bbb R[/imath]. But here the topology is unspecific, so I have no idea how to proceed. Could anybody help me? Thanks! |
608487 | converging sequence of a sum and an integral
prove that the sequence converges [imath]a_n = \sum_{k=1}^n \frac1k \ -\ \int_1^n\frac{dx}x[/imath] Attempt: I dont even know where to start here. the integral gives you -ln(n) | 607039 | How to show [imath]\gamma[/imath] aka Euler's constant is convergent?
I'm trying to show different convergents, and this is the first one i'm having problems with. [imath]\dfrac11 + \dfrac 12 + \ldots + \dfrac 1n - \log n \rightarrow \gamma[/imath] Like it's the definition of euler's constant, but how to show that this expression is convergent( i don't mean convergent to specific number, just convergent ). I tried d'alambert's criterion. This is : if [imath]\left| \frac{a_{n+1}}{a_{n}}\right | < 1[/imath] then [imath]a_{n}[/imath] is convergent. But i end up with something like [imath]\frac{H_{n} + (n+1) - ln(n+1)}{H_{n} - \ln(n)}[/imath] where [imath]H_{n}[/imath] is n-th harmonic number. I have no clue how to work on it further. Would love to get some hints or solutions on this. Cheers |
125890 | The [imath]n^{th}[/imath] root of the geometric mean of binomial coefficients.
[imath]\{{C_k^n}\}_{k=0}^n[/imath] are binomial coefficients. [imath]G_n[/imath] is their geometrical mean. Prove [imath]\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}[/imath] | 2377116 | Limit of [imath]\prod_{k=0}^n {n \choose{k}} ^{\frac{1}{n(n+1)}}[/imath]
What is the limit of [imath]\prod_{k=0}^n {n \choose{k}} ^{\frac{1}{n(n+1)}}?[/imath] It should be [imath]\sqrt e[/imath]. I've tried several approaches, including Stirling's formula (which I'd rather avoid), though I came close, I can't prove it. |
608794 | Suppose that n > 1. Prove that n divides [imath] Ο (2^n - 1) [/imath] .
Suppose that n > 1. Prove that n divides [imath] Ο(2^n - 1) [/imath] . Hint: Show that 2 has order n mod [imath] 2^n - 1 [/imath] | 421151 | Prove [imath]n\mid \phi(2^n-1)[/imath]
If [imath]2^p-1[/imath] is a prime, (thus [imath]p[/imath] is a prime, too) then [imath]p\mid 2^p-2=\phi(2^p-1).[/imath] But I find [imath]n\mid \phi(2^n-1)[/imath] is always hold, no matter what [imath]n[/imath] is. Such as [imath]4\mid \phi(2^4-1)=8.[/imath] If we denote [imath]a_n=\dfrac{\phi(2^n-1)}{n}[/imath], then [imath]a_n[/imath] is A011260, but how to prove it is always integer? Thanks in advance! |
139549 | How to show path-connectedness of [imath]GL(n,\mathbb{C})[/imath]
Well, I am not getting any hint how to show [imath]GL_n(\mathbb{C})[/imath] is path connected. So far I have thought that let [imath]A[/imath] be any invertible complex matrix and [imath]I[/imath] be the idenity matrix, I was trying to show a path from [imath]A[/imath] to [imath]I[/imath] then define [imath]f(t)=At+(1-t)I[/imath] for [imath]t\in[0,1][/imath] which is possible continous except where the [imath]\operatorname{det}{f(t)}=0[/imath] i.e. which has [imath]n[/imath] roots and I can choose a path in [imath]\mathbb{C}\setminus\{t_1,\dots,t_n\}[/imath] where [imath]t_1,\dots,t_n[/imath] are roots of [imath]\operatorname{det}{f(t)}=0[/imath], is my thinking was correct? Could anyone tell me the solution? | 974825 | What's an easy way to show that [imath]GL(n,\mathbb C)[/imath] is connected?
I think I've to show it's path connected, but can't figure out the path functions explicitly. Can anyone give these path maps? |
609239 | Find outlet of [imath]f(x)[/imath]
Find outlet of [imath]f(x)=x^x[/imath] I know I need to use [imath](a^x)'= a^x\cdot lna[/imath] But then I get [imath]f(x)=x^x\cdot lnx[/imath] and the solution is [imath]f(x)=x^x\cdot (lnx+1)[/imath] How do I calculate this? Thank you in advance! | 72426 | Derivative of [imath]x^x[/imath]
I can't get any of these on my own and I am attempting to do [imath]x^x[/imath] I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into [imath]x(\ln x)[/imath] which should be equivalent to the original term. Then the differentiation should be easy for most people from here. [imath]1(\ln x)+x(\frac1x) [/imath] or [imath]\ln x+1[/imath] This is of course wrong but I do not know why. |
609389 | Why do some people place the differential at the beginning of their integral?
I've seen several times on this site people writing integrals like [imath] \int \! dxf(x) [/imath]. This seems confusing to me, especially in an iterated integral or next to a long integrand and it's nonstandard compared to my own personal experience. I'm wondering what are the reasons people use this notation swap and if it has broader support than I'm aware of. | 793166 | Strange Integral Notation?
When reading about an certain algorithm (about parameter estimation for Kalman Filtering page 7 eq 57) I found this notation: [imath]\int dx f(x)[/imath] which is normally written as [imath]\int f(x) dx[/imath]. I spent a significant time to figure this out. Now my question: Why would anybody use such a misleading notation? Is this a convention in certain areas? I still don't know where the integrand ends and up to now I am pretty convinced that this notation is very harmful, since [imath]\int dx[/imath] also has a meaning. |
609561 | Analysis - Uniform Continuity
I need to find an example of a function [imath]f:I[/imath] to [imath]R[/imath] such that [imath]f[/imath] is uniformly continuous [imath]f'[/imath] exists but [imath]f'[/imath] is not bounded? I'm fairly stuck with this - it's part of a review and I guess I never handled any of these types of problems very well. Any direction is appreciated. | 352321 | A uniformly continuous function such that the derivative is not bounded and is not defined on a compact?
Is there a uniformly continuous function values in [imath]f: \mathbb{R}{\to}\mathbb{R}[/imath] such that its first derivative is not bounded and is defined on a non-compact set? And if [imath]f: X_{1}{\to}\mathbb{R}[/imath]? (let [imath](X_{1},d{1})[/imath] metric space ). And if [imath]f: X_{1}{\to}X_{2}[/imath]? (let [imath](X_{2},d{2})[/imath] metric space ). I know that the first derivative is bounded f:I-->R (let I interval) iff f is a lipschitz function; If X1 is a compact thanks to the Heine-Cantor theorem f is uniformly continuous. |
609560 | Suppose [imath]p[/imath] and [imath]q[/imath] are odd primes and [imath]p = q + 4a[/imath] for some [imath]a[/imath]. Prove that [imath](\frac ap) = (\frac aq)[/imath] holds.
This problem also had me prove that [imath](\frac pq) = (\frac aq)[/imath], but I've already managed to do that. I've tried messing around with the Law of Quadratic Reciprocity but can't get anything. I've also tried using the first identity by proving that [imath](\frac pq) = (\frac ap)[/imath], but again I've got nothing. | 582815 | Quadratic Reciprocity
For any integer [imath]a[/imath] and any two primes [imath]p,q[/imath] with [imath](p,q)=1[/imath]. Prove that if [imath]p \equiv q\pmod {4a}[/imath], then [imath]\displaystyle\left(\frac ap\right)=\left(\frac aq\right)[/imath] I know I need to write [imath]a=(β1)^{e_0} 2^{e_2} p_1 p_2 \cdots p_r[/imath] with [imath]p_i[/imath] odd and possibly equal; and then compute the Legendre symbol and apply the reciprocity laws, I am just unfamiliar with this process and would appreciate some help. |
609727 | Example of a function [imath]f[/imath] which is nowhere continuous but [imath]|f|[/imath] should be continuous at all points
So I had an exam today and one of the questions were: Give an example of a function [imath]f[/imath] which is nowhere continuous but [imath]|f|[/imath] should be continuous at all points. At first I had no idea how to do it then I came up with this, even though I know it's wrong: [imath]f(x)= \begin {cases} \ \sqrt{x}&\text{if }x <0\\ \sqrt{-x}\ &\text{if }x > 0\\ \end {cases}[/imath] Like I said I know it's wrong..but can someone give me an example of this? Thanks. | 75425 | Which function ([imath]f[/imath]) is continuous nowhere but [imath]|f(x)|[/imath] is continuous everywhere?
Which function ([imath]f[/imath]) is continuous nowhere but [imath]|f(x)|[/imath] is continuous everywhere? I found this question here, the question seems much interesting but for obvious reason it is closed there, I was wondering how to derive such a function? |
609858 | 2x2 matrix and convergence under iteration connected to eigenvectors and eigenvalues.
When working with a general matrix [imath]A=[/imath] \begin{bmatrix}a & b\\c & d\end{bmatrix}, I find that the eigenvalues are: [imath]\lambda = \frac{d+a \pm \sqrt{(d-a)^2+4bc}}{2}[/imath] I then find that eigenvectors, [imath]e=[/imath] \begin{bmatrix}-b\\a-\lambda\end{bmatrix}. I also know that under iteration with initial vector [imath]x_n[/imath] = \begin{bmatrix}x_0\\y_0\end{bmatrix}, we have [imath]x_{n+1} = Ax_n[/imath]. The matrix [imath]A[/imath] converges to a line with a slope [imath]m[/imath]. I am told that the slope of this line is in the same direction as the eigenvector associated with the largest eigenvalue. How do I prove this? | 608735 | Repeated Iteration of a 2x2 matrix
Suppose I am given a [imath]2[/imath]x[imath]2[/imath] matrix [imath]A=[/imath] \begin{pmatrix} a & b \\ c & d \end{pmatrix} And an initial vector [imath]x_n[/imath] = \begin{pmatrix} x_0 \\ y_0\end{pmatrix}. Under repeated iteration [imath]x_{n+1} = Ax_n[/imath], we find that [imath]x_{n+1}[/imath] converges to a line in the direction of the [imath]eigenvector[/imath] of [imath]A[/imath] of the form [imath]X[/imath]= [imath][u,mu][/imath] where [imath]m[/imath] is the slope of the line. I know that this process somehow also involves the largest [imath]eigenvalue, \lambda,[/imath] of [imath]A[/imath], though I am unsure how it connects. My question is: How does the largest eigenvalue connect to this repeated iteration, how do I find the slope of the line of convergence [imath]m[/imath], and what is meant by the "direction of the eigenvector"? |
604928 | finding estimate of a polynomial outside the unit disk if it is bounded in the unit disk
Question : Let [imath]p(z)[/imath] be a polynomial of degree n satisfying [imath]|p(z)|\leq 1[/imath] for all [imath]z, |z| \leq 1[/imath]. Show that then [imath]|p(z)| \leq |z|^n[/imath] for all [imath]z, |z| >1[/imath]. This is what I could get done :- Let [imath]p(z) = \sum_{k=0}^{n-1} c_k z^k + z^n [/imath]. Given [imath]|p(z)| \leq 1[/imath] when [imath]|z|=1[/imath]. Then by Cauchy's estimates , [imath]|c_k| = \frac{|p^{(k)}(0)|}{k!} \leq 1[/imath] for all [imath]k[/imath]. Therefore, [imath]|p(z)| \leq \sum_{k=0}^{n-1} |c_k| |z|^k + |z|^n \leq \sum_{k=0}^{n-1} |z|^k + |z|^n \leq \dfrac{|z|^{n+1}-1}{|z|-1}[/imath]. From this we can say that when [imath]|z| >> 1[/imath], then [imath]|p(z)| \leq |z|^n[/imath]. But how do I complete the argument for all [imath]z \in \{|z|>1 \} [/imath]? | 598135 | Complex polynomial of degree [imath]n[/imath]
I've got following problem. Let [imath]z[/imath] be a complex number. Let [imath]P(z)[/imath] be a polynomial of degree [imath]n[/imath]. Suppose that [imath]|P(z)| \le M[/imath] for [imath]|z| \le 1[/imath]. Prove that for [imath]|z| \ge 1[/imath] we have. [imath]|P(z)| \le M \cdot |z|^n[/imath]. Can somebody help? |
610418 | Please find the mistake
Well, I was studying laws of indices and the following steps came to my mind [imath](-1)^{\frac {1}{2}}[/imath] = [imath](-1)^{\frac {2}{4}}[/imath] = [imath]((-1)^{2})^{\frac {1}{4}}[/imath] = [imath]1^{\frac {1}{4}}[/imath] = [imath]1[/imath] This is very strange, but I cannot find the mistake. In fact, [imath]\frac {1}{2}[/imath] and [imath]\frac {2}{4}[/imath] represent the same rational number and so the first step is alright. My second step is also correct as that is the definition. The remaining steps are also true. I'm confused! | 608023 | What is [imath](-1)^{\frac{2}{3}}[/imath]?
Following from this question, I came up with another interesting question: What is [imath](-1)^{\frac{2}{3}}[/imath]? Wolfram alpha says it equals to some weird complex number (-0.5 +0.866... i), but when I try I do this: [imath](-1)^{\frac{2}{3}}={((-1)^2)}^{\frac{1}{3}}=1^{\frac{1}{3}}=1[/imath].If it has multiple "answers", should we even call it a "number"? Because if we don't, it would be a bit different from what we were taught in elementary school. I actually thought if it doesn't have a variable in it, it should be a number. I'm a bit confused. Which one is correct and why? I would appreciate any help. |
610846 | Legendre symbol [imath]\left(\frac 3p\right)[/imath]
Show that: [imath]\left(\frac 3p\right)= \begin{cases} p \equiv 1, 11 \pmod {12} \\ p\equiv 5, 7 \pmod {12} \end{cases}[/imath] where [imath]\left(\frac 3p\right)[/imath] is Legendre symbol I have done this for [imath]\left(\frac {-3}p\right)[/imath] and reach [imath]\left(\frac 3p\right) = 1[/imath] if [imath]p=1[/imath] and [imath]\left(\frac 3p\right)= -1[/imath] if [imath]p=2[/imath]. in the case where [imath]p=1[/imath] the next step was to see that [imath]p=3k+1=6l+1[/imath] so then [imath]p=12j+1[/imath] and [imath]p=12j+11[/imath] give [imath]\left(\frac {-3}p\right)=1[/imath]. I'm very confused about all of this so if someone could help it'd be greatly appreciated. it would probably be better to start from the beginning with [imath]\left(\frac 3p\right)[/imath] or to use the same logic used to find [imath]\left(\frac 2p\right)[/imath] and Gauss lemma | 608481 | Prove that, for [imath]p > 3[/imath], [imath](\frac 3p) = 1[/imath] when [imath]p \equiv 1,11 \pmod{12}[/imath] and [imath](\frac 3p) = -1[/imath] when [imath]p \equiv 5,7 \pmod{12}[/imath]
[imath](\frac 3p)[/imath] is the Legendre symbol here, not a fraction, if that wasn't clear. This is what I have so far: We know by Gauss's Lemma that [imath](\frac qp) = (-1)^v[/imath] where [imath]v = \#\left\{1 \le a \le \frac{p-1}{2} \mid aq \equiv k_a \pmod{12}),\space -p/2 < k_a < 0\right\}[/imath] And we also know that multiples of [imath](\dfrac{p-1}{2}) \times 3[/imath] must be between [imath]0[/imath] and [imath]p/2[/imath], between [imath]p/2[/imath] and [imath]p[/imath], or between [imath]p[/imath] and [imath]3p/2[/imath]. Representing [imath]p[/imath] as [imath]12k + r[/imath] seems like the next step, but I'm not sure where to go from there. |
471304 | [imath]\pi_2(G)[/imath] for [imath]G[/imath] a Lie group.
It is well known that [imath]\pi_2(G)[/imath] is trivial for any Lie-group [imath]G[/imath]. Is there an elementary proof of this, say, that can be understood with minimal homotopy theory? Also, who gave the first proof of this theorem? | 61386 | First and second homotopy groups of a connected Lie group
I try to understand why for a connected Lie group [imath]G[/imath] the first fundamental group [imath]\pi_1(G)[/imath] is abelian, and mainly why the second fundamental group is trivial [imath]\pi_2(G)=0[/imath]? Thanks for anyone who give me references for a 'simple proof' of these results |
611351 | Prove that this factor ring is a finite ring without zero divisors
Let [imath]R=\mathbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\mid a,b,d\in\mathbb{Z}\}[/imath] and let [imath]d[/imath] be square-free. Let [imath]P[/imath] be a non-zero prime ideal in [imath]R[/imath]. I need to prove that the factor ring [imath]R/P[/imath] has no zero divisors and is finite. How do I do this? Thank you. | 607324 | Prove that every nonzero prime ideal is maximal in [imath]\mathbb{Z}[\sqrt{d}][/imath]
[imath]d \in \mathbb{Z}[/imath] is a square-free integer ([imath]d \ne 1[/imath], and [imath]d[/imath] has no factors of the form [imath]c^2[/imath] except [imath]c = \pm 1[/imath]), and let [imath]R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}[/imath]. Prove that every nonzero prime ideal [imath]P \subset R[/imath] is a maximal ideal. I have a possible outline which I think is good enough to follow. I think that we need to first prove that every ideal [imath]I \subset R[/imath] is finitely generated. So if [imath]I[/imath] is non-zero, then [imath]I \cap \mathbb{Z}[/imath] is a non-zero ideal in [imath]\mathbb{Z}[/imath]. Then I need to find [imath]I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}[/imath] for some [imath]x \in \mathbb{Z}[/imath]. That way if I let [imath]J[/imath] be the set of all integers [imath]b[/imath] such that [imath]a+b\sqrt{d} \in I[/imath] for some [imath]a\in \mathbb{Z}[/imath], then if there exists a integer [imath]y[/imath] such that [imath]J=\{ yt \mid t\in \mathbb{Z} \}[/imath], then there must exist [imath]s \in \mathbb{Z}[/imath] such that [imath]s+y\sqrt{d} \in I[/imath]. Then all I need to show is that [imath]I = ( x,s+y\sqrt{d} )[/imath]. Now I need to derive that the factor ring [imath]R / P[/imath] is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal [imath]P \subset R[/imath] is a maximal ideal, then I'll be done. |
534107 | For all [imath]X \geq 1[/imath],[imath]\sum_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] = 1.[/imath]
I'm currently sitting with the following number theory problem: Prove that for all [imath]X \geq 1[/imath], [imath]\sum_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] = 1.[/imath] A few ideas I have tried: Using that [imath][x]=x - \{x\}[/imath] which I couldn't make out to bring anything. Switching around on the summation variables, like [imath]\sum_{d \leq N} \mu(d) f\left( \frac{n}{d} \right) = \sum_{d \leq N} \mu \left( \frac{d}{n} \right) f(d)[/imath] but that didn't seem to bring anything. I tried writing out the first few terms, which give [imath]\mu(1) [X] - \left[\frac{X}{2}\right] - \left[\frac{X}{3}\right] - \left[\frac{X}{5}\right] + \left[\frac{X}{6}\right] - \left[\frac{X}{7}\right] - \left[\frac{X}{8}\right][/imath] but I can't seem to find a system where it simply reduces to one! What do you think? | 718717 | MΓΆbius function
For any natural number [imath]x[/imath], determine the sum; [imath] \sum_{\substack{ n\leq x }} \mu(n)\left\lfloor \frac{x}{n} \right\rfloor.[/imath] (Hint: Use [imath]\lfloor x \rfloor=\sum_{\substack{ k\leq x }}1.[/imath]) Here is my start: [imath]=\sum_{\substack{ n\leq x }} \mu(n)\sum_{\substack{ k\leq \frac{x}{n} }}1[/imath] if we change the order of summations then [imath]=\sum_{\substack{ k\leq x }}1 \sum_{\substack{ n\leq \frac{x}{k} }}\mu(n)[/imath] But I couldn't go further |
611722 | Need help with proving that group is not finitely-generated
I need to prove that [imath](\mathbb{Q}^*, \times)[/imath] (i.e rationals, zero excluded, under multiplication) is not finitely generated. So, suppose that G is finitely-generated. That means there exist a finite set [imath]S=\left\{x_1, x_2,...,x_n\right\}\subset \mathbb{Q}^*[/imath] s.t. [imath]\langle S\rangle=\mathbb{Q}^*[/imath]. Now, all I need to do is to find a number [imath]y\in\mathbb{Q}^*[/imath] that cannot be generated by [imath]S[/imath]. Naively, I first thought of prime numbers, but that surely won't work. Please help me out here... how can I find a number [imath]y\in\mathbb{Q}^*[/imath] that cannot possibly be multiplication of [imath]x_i[/imath]-s from [imath]S[/imath]? Thanks in advanced! | 604229 | show that rational numbers with the multipiciation are not abelian finitely generated group
we need to show that [imath]( Q \ast , \bullet )[/imath] is not abelian finitely generated group for all finite subset [imath]S \subseteq Q \ast [/imath] [imath] Q\ast=Q \setminus \big\{0\big\} [/imath] |
604941 | Proof of Lebesgue Density Theorem
If E is a Borel set in [imath]\Bbb R^n[/imath], the density [imath]D_E (x)[/imath] of [imath]E[/imath] at [imath]x[/imath] is defined as [imath]D_E(x) = \lim_{r \to 0} \frac{m(E \cap B(r,x))}{m(B(r,x))}[/imath] Show that [imath]D_E(x) = 1[/imath] for almost everywhere [imath]x \in E[/imath] and [imath]D_E(x) = 0[/imath] for almost everywhere [imath]x \in E^C[/imath]. I am not sure what to do to prove this statement. As far as I can tell, this is the Lebesgue Density Theorem, but I can only find proofs of this in the one-dimensional case (which is by far the easiest case). | 596676 | Lebesgue density theorem in the line
Suppose [imath]A \subseteq \mathbb{R} [/imath], [imath]m(A) > 0 [/imath]. Then for almost all [imath]x \in A [/imath] we have [imath] \lim_{\epsilon \to 0^+ } \frac{ m(A \cap (x - \epsilon, x + \epsilon))}{2 \epsilon} = 1.[/imath] Can someone help me with this problem? [imath]m[/imath] is the Lebesgue measure thanks |
613410 | continuity of a function when restricted to closed sets
Suppose [imath]X = A \cup B[/imath], where [imath]A[/imath] and [imath]B[/imath] are closed subspaces of [imath]X[/imath]. Let [imath]f: X \to Y[/imath]. Suppose [imath] f|_A : A \to Y [/imath] [imath] f|_B : B \to Y [/imath] are continuous. Then can we conclude that [imath]f[/imath] must be continuous? Any help would be greatly appreciated. thanks in advance. | 476239 | Continuity based on restricted continuity of two subsets
Let [imath]X=A\cup B[/imath], where [imath]A,B[/imath] are subspaces of [imath]X[/imath]. Let [imath]f:X\rightarrow Y[/imath]; suppose that the restricted functions [imath]f\mid A:A\rightarrow Y[/imath] and [imath]f\mid B:B\rightarrow Y[/imath] are continuous. Show that if both [imath]A[/imath] and [imath]B[/imath] are closed in [imath]X[/imath], then [imath]f[/imath] is continuous. [imath]f\mid A[/imath] continuous means that for any open subset [imath]V[/imath] of [imath]Y[/imath], the subset [imath](f\mid A)^{-1}(V)=f^{-1}(V)\cap A[/imath] is open in [imath]A[/imath]. Similarly, [imath]f^{-1}(V)\cap B[/imath] is open in [imath]B[/imath]. The meaning of [imath](f\mid A)^{-1}(V)=f^{-1}(V)\cap A[/imath] open in [imath]A[/imath] is that [imath]f^{-1}(V)\cap A = C\cap A[/imath] for some [imath]C[/imath] open in [imath]X[/imath]. What then? |
606295 | Are most matrices invertible?
I asked myself this question, to which I think the answer is "yes". One reason would be that an invertible matrix has infinitely many options for its determinant (except [imath]0[/imath]), whereas a non-invertible must have [imath]0[/imath] as the determinant. Do you have another approach to this question, in terms of probability for example? | 226128 | Probability of having zero determinant
Given a matrix [imath]A_{n \times n}[/imath], which has elements [imath]a_{i,j} \sim \mathrm{unif} \left[a,b\right][/imath], what is the probablity of [imath]\det(A)[/imath] being zero? What if [imath]a_{i,j}[/imath] have any other distribution? Added: Let's assume an extension of the about problem; What is the probability of, [imath]\mathbb{P}(|\det(A)| < \epsilon ), \; s.t. \; \epsilon \in \mathbb{R} [/imath] ? Thanks! |
613650 | Writing [imath](a,b)[/imath] as a disjoint union of closed intervals
I've been thinking about the following question: Is it possible to write [imath](a,b)[/imath] as a disjoint union of closed intervals? My first guess was no, but then I figured the question might be hiding something subtle. I tried various things, some which I asked about on this site, and I can't seem to construct such a union. So I decided to go back to my gut feeling and prove that it cannot be done. Attempt: Suppose there is such a union. Then each closed interval contains a rational. Given that the sets are disjoint, any rational specifies a unique interval. Therefore the union must be a countable union, and we can list the closed intervals. Choose some listing [imath]I_1,I_2,\dotsc[/imath] and let: [imath](a,b)=\bigcup_{k=1}^{\infty} I_k.[/imath] Now define [imath]J_n=(a,b)\setminus \bigcup_{k=1}^n I_k.[/imath] Specifically, consider that [imath]J_2[/imath] contains a middle interval [imath](a',b')[/imath] where [imath]a<a'<b'<b[/imath]. Construct a sequence where each [imath]u_n[/imath] is arbitrarily chosen in [imath]J_n \displaystyle\bigcap\, (a',b')[/imath]. By the Bolzano-Weierstrass theorem, [imath]u_n[/imath] contains a subsequence [imath]v_n[/imath] such that [imath]v_n\to c[/imath]. We must have [imath]c \in (a,b)[/imath] so [imath]J_{\infty}\neq \varnothing[/imath] which is a contradiction. Put differently, no matter how the closed intervals are chosen, we can find a sequence in [imath](a,b)[/imath] that converges to a limit in [imath](a,b)[/imath] which is contained in none of the closed intervals. Is my proof correct? | 6314 | Is [imath][0,1][/imath] a countable disjoint union of closed sets?
Can you express [imath][0,1][/imath] as a countable disjoint union of closed sets, other than the trivial way of doing this? |
379664 | Finite fundamental group in the Euclidean space
Is there an example of a (path-connected) subspace of [imath]\mathbb{R}^3[/imath] which has a nontrivial finite fundamental group? If not, why? | 287009 | Torsion on [imath]\pi_1(X)[/imath], [imath]X[/imath] connected and open in [imath]\mathbb{R}^n[/imath]
Can the fundamental group of an open connected subset [imath]X[/imath] of [imath]\mathbb{R}^n[/imath] have a torsion element? |
482633 | Image of Intersection of sects not equal to intersection of images of sets
How to disprove, if [imath]f[/imath] is a function, [imath]f(A \cap B) != f(A) \cap f(B)?[/imath] | 225333 | Is this proof correct for : Does [imath]F(A)\cap F(B)\subseteq F(A\cap B) [/imath] for all functions [imath]F[/imath]?
Is this proof correct? To prove [imath]F(A)\cap F(B)\subseteq F(A\cap B) [/imath] for all functions [imath]F[/imath]. Let any number [imath]y\in F(A)\cap F(B)[/imath]. We want to show [imath]y\in F(A\cap B).[/imath] Therefore, [imath]y\in F(A)[/imath] and [imath]y\in F(B)[/imath], by definition of intersection. By definition of inverse, [imath]y=F(x)[/imath] for some [imath]x\in A[/imath] and [imath]x\in B[/imath] And so, [imath]y=F(x)[/imath] for some [imath]x\in A\cap B[/imath] And therefore, [imath]y\in F(A\cap B)[/imath] I have a gut feeling deep down that something is wrong. Can anyone help me pinpoint the mistake? I am not sure why am I having so much problems with functions. Am I not thinking in the right direction? Sources : 2nd Ed, P219 9.60 = 3rd Ed, P235 9.12, 9.29 - Mathematical Proofs, by Gary Chartrand, P214 Theorem 12.4 - Book of Proof, by Richard Hammack, P257-258 - How to Prove It, by D Velleman. |
613978 | What is the ring [imath]\Bbb Z[x]/(x^2-3,2x+4)[/imath]?
How to describe an element of the Ring [imath]\Bbb Z[x]/(x^2-3,2x+4)[/imath]? Or is it isomorphic to any well known ring? What is meant by quotienting [imath]\Bbb Z[x][/imath] by a kernel like that? | 472235 | Describe [imath]R=\mathbb{Z}[X]/(X^2-3,2X+4)[/imath]
I need to describe a ring [imath]R=\mathbb{Z}[X]/(X^2-3,2X+4)[/imath] I know that its element would be of the form [imath]\{f(x)+(X^2-3,2X+4)|f(x)\in \mathbb{Z}[X]\}[/imath] After that will i divide [imath]f(x)[/imath] by [imath]X^2-3[/imath] first and then by [imath]2X+4[/imath] to reduce the elements in [imath]R[/imath]? |
291777 | [imath]f^{-1} (S\cup T) = f^{-1} (S) \cup f^{-1} (T)[/imath]
[imath]f^{-1}(S\cup T) = f^{-1} (S)\cup f^{-1}(T)[/imath] How does one go about showing this equality? What operations can one apply on functions to show this? | 357725 | Show that [imath]f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)[/imath]
Show that [imath]f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)[/imath] but not necessarily [imath]f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)[/imath]. Let [imath]S=A\cup B[/imath] I know that [imath]f^{-1}(S)=\{x:f(x)\in S\}[/imath] assuming that that [imath]f[/imath] is one to one. Is this true [imath]\{x:f(x)\in S\}=\{x:f(x) \in A\}\cup\{x:f(x)\in B\}[/imath]? Why doesn't the intersection work? Sources : β¦ 2nd Ed, [imath]\;[/imath] P219 9.60(d), [imath]\;[/imath] Mathematical Proofs by Gary Chartrand, β¦ P214, [imath]\;[/imath] Theorem 12.4.#4, [imath]\;[/imath] Book of Proof by Richard Hammack, β¦ P257-258, [imath]\;[/imath] Theorem 5.4.2.#2(b), [imath]\;[/imath] How to Prove It by D Velleman. |
591663 | For a prime [imath]p \equiv 1[/imath] or [imath]3[/imath] (mod [imath]8[/imath]), show that the equation [imath]x^{2} + 2y^{2} = p[/imath] has a solution.
For a prime [imath]p \equiv 1[/imath] or [imath]3[/imath] (mod [imath]8[/imath]), show that the equation [imath]x^{2} + 2y^{2} = p[/imath] has a solution. I'd like to see an elementary proof. | 714440 | [imath]p[/imath] is an odd prime of the form [imath]p=x^2+2y^2[/imath] iff [imath]p\equiv_8[/imath] [imath]1[/imath] or [imath]3[/imath]
How would I prove the following: Show that an odd prime [imath]p[/imath] can be written on the form [imath]p=x^2+2y^2[/imath] for some [imath]x,y\in\mathbb Z[/imath] iff [imath]p\equiv_8 1, 3[/imath]. Hint: use the quadratic reciprocity and the fact that [imath]\mathbb Z[\sqrt{-2}][/imath] is a Euclidean domain. |
614246 | [imath]\log _{-3}\left( 9 \right)[/imath]?
Can you have a negative base for a logarithm? If the answer is no, why can you not? Isn't the answer to the above just [imath]2[/imath]? Similarly, is [imath]-3[/imath] also a solution to the equation [imath]\log _{x}\left( 9 \right)\; =\; 2[/imath] (I know one solution is obviously 3). | 359855 | Why aren't logarithms defined for negative [imath]x[/imath]?
Given a logarithm is true, if and only if, [imath]y = \log_b{x}[/imath] and [imath]b^y = x[/imath] (and [imath]x[/imath] and [imath]b[/imath] are positive, and [imath]b[/imath] is not equal to [imath]1[/imath])[1], are true, why aren't logarithms defined for negative numbers? Why can't [imath]b[/imath] be negative? Take [imath](-2)^3 = -8[/imath] for example. Turning that into a logarithm, we get [imath]3 = \log_{(-2)}{(-8)}[/imath] which is an invalid equation as [imath]b[/imath] and [imath]x[/imath] are not positive! Why is this? |
613900 | Assuming: [imath]\forall x \in [0,1]:f(x) > x[/imath] Prove: [imath]\forall x \in [0,1]:f(x) > x + \varepsilon [/imath]
Let [imath]f[/imath] a continous function defined in the interval [imath][0,1][/imath]. Assuming: [imath]\forall x \in [0,1]:f(x) > x[/imath] Prove: [imath]\forall x \in [0,1]:f(x) > x + \varepsilon [/imath] I tried to use HeineβCantor theorem and did some Algebra tricks but it didn't bring me to a safe shore :) What do you suggest? Thanks! | 615098 | Proving with epsilon for a continuous function
Let f be a continous function in the section [imath][0,1][/imath]. [imath]f(x)>x \ \ \forall x\in[0,1][/imath]. Prove that [imath]\exists\epsilon>0 \ s.t \ f(x)>x+\epsilon :\ \forall x\in[0,1][/imath] From [imath]f(x)>x[/imath] I know that f is monotone increasing. So [imath]\forall \delta>0 \ |x-y|<\delta[/imath] and [imath]\forall \epsilon>0 \ |f(x)-f(y)|<\epsilon[/imath]. Though I have a feeling that this doesn't involve delta. How should I continue from here ? Maybe this have something to do with intermediate value theorem ? |
614511 | For any [imath]n[/imath], there are at most two simple groups of order [imath]n[/imath]?
How do you prove that for any [imath]n[/imath] there are at most two simple groups of order [imath]n[/imath]? | 1423 | Number of finite simple groups of given order is at most [imath]2[/imath] - is a classification-free proof possible?
This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that: [imath]L_4(2)[/imath] and [imath]L_3(4)[/imath] both have order [imath]20160[/imath] [imath]O_{2n+1}(q)[/imath] and [imath]S_{2n}(q)[/imath] have the same order for [imath]q[/imath] odd, [imath]n > 2[/imath] I think this means that for each integer [imath]g[/imath], there are [imath]0[/imath], [imath]1[/imath] or [imath]2[/imath] simple groups of order [imath]g[/imath]. Do we need the full strength of the Classification of Finite Simple Groups to prove this, or is there a simpler way of proving it? |
337125 | Consider [imath]P(n)[/imath] as a number of [imath]n[/imath]-permutations, which each cycle have even length, and ...
Consider [imath]P(n)[/imath] as a number of [imath]n[/imath]-permutations, which each cycle have even length, and [imath]N(n)[/imath] as a number of [imath]n[/imath]-permutations, which each cycle have odd length. Calculate [imath]P(2n)-N(2n)[/imath] | 331504 | Number of permutations of a specific cycle decomposition
Let [imath]X(n)[/imath] denote number of all permutations of [imath]\left\{1,\ldots,n \right\}[/imath] that have only cycles of even length. Let [imath]Y(n)[/imath] denote number of all permutations of the same set that have only cycles of odd length. Count [imath]X(2n)-Y(2n)[/imath]. I don't know even how to start. I counted this sum for [imath]n=1,2,3[/imath] and it seems that it is always equal to [imath]0[/imath], but it doesn't help with calculating it in general. |
614063 | prove that the following equation has exactly 2 solutions
Let [imath] a_1,a_2,a_3 >0[/imath] [imath]\lambda_1\lt \lambda_2 \lt \lambda_3 \in \mathbb{R}[/imath] prove that the following equation has exactly 2 solutions [imath]\dfrac{a_1}{x-\lambda_1} + \dfrac{a_2}{x-\lambda_2} + \dfrac{a_3}{x-\lambda_3} = 0[/imath] this should be closely related the continuity and the intermediate value theorem, but i couldn't connect the dots on this one. | 399330 | Prove that equation has exactly 2 solutions
Question: Given [imath]\lambda_1<\lambda_2<\lambda_3 \in \Bbb R, a_1,a_2,a_3>0 [/imath] Show that this equation has exactly 2 solutions: [imath]\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}=0[/imath] What I did: Try 1: Since we are studying continuity, Weirstrass theorem and Mean value theorem at the moment, I think one of these theorems should somehow be involved in the proof... Assuming that x is not equal to any of the lambdas I multiplied by the denumerators and got a quadratic equation- From this I figure that there must be at most 2 solutions. I thought about using the discriminant to prove that that there are exactly 2, but the calculation is very long, and I don't think this is what the lecturer intended for us to do, considering what we are studying now. Try 2: Thought of using the Intermediate value theorem. So I defined a function [imath]f(x)=\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}[/imath] and I want to find when it equals 0. but Intermediate value theorem only involves two lambdas, and also I don't know if the lambdas are negative or positive (some/all of them)- Should I divide to cases? Hints are welcome at first :-) |
105956 | Preimage of Intersection of Two Sets = Intersection of Preimage of Each Set : [imath]f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)[/imath]
Prove If [imath]f[/imath] is a function , [imath]f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)[/imath]. Proof attempt I am guessing here [imath]A[/imath] and [imath]B[/imath] are sets in the range of [imath]f[/imath]. Let's assume [imath]x[/imath] belongs to both [imath]A[/imath] and [imath]B[/imath] and [imath]f^{-1}[/imath] exists for both [imath]A[/imath] and [imath]B[/imath]. Then there must exist a [imath]y[/imath] such that [imath]y = f^{-1}(x)[/imath]. Now by our assumptions [imath]x[/imath] is in intersection of [imath]A[/imath] and [imath]B[/imath] and since [imath]f^{-1}(A)[/imath] and [imath]f^{-1}(B)[/imath] exist then [imath]f^{-1}(A) \cap f^{-1}(B)[/imath] must also exist. Also since [imath]A[/imath] and [imath]B[/imath] exist and are not equal to null set so [imath]A \cap B[/imath] exists and [imath]f^{-1}(A\cap B)[/imath] also must exist and contain our [imath]y[/imath]?? Not sure about the ending in this attempt. Any help would be much appreciated. Sources : β¦ 2nd Ed [imath]\;[/imath] P219 9.60(e) [imath]\;[/imath] Mathematical Proofs by Gary Chartrand, β¦ P214 [imath]\;[/imath] Theorem 12.4.#3 [imath]\;[/imath] Book of Proof by Richard Hammack, β¦ P257-258 [imath]\;[/imath] Theorem 5.4.2.#2(a) [imath]\;[/imath] How to Prove It by D Velleman. | 516374 | how to prove [imath]f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)[/imath]
I am given this equation: [imath]f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)[/imath] I want to prove it: what i did is I take any [imath]a \in f^{-1}(B_1 \cap B_2)[/imath], then there is [imath]b \in (B_1 \cap B_2)[/imath] so that [imath]f(a)=b[/imath]. Because of [imath]b \in (B_1 \cap B_2)[/imath], it is true that [imath]b \in B_1[/imath] and [imath]b \in B_2[/imath], so [imath]a \in f^{-1}(B_1)[/imath] and [imath]a \in f^{-1}(B_2)[/imath]. this means [imath]f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)[/imath]. is it ok? |
614786 | Series [imath]\sum\limits_{n=1}^{\infty}\frac{n^k}{n!}[/imath]
Is there a formula for [imath]\displaystyle\sum_{n=1}^{\infty}\frac{n^k}{n!}[/imath] in terms of [imath]k[/imath]? What about [imath]\displaystyle\sum_{n=1}^{\infty}\frac{(n-1)^k}{n!}[/imath]? | 503451 | Infinite Series [imath]\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}[/imath]
How can I find the value of the sum [imath]\sum_{k=1}^{\infty}\frac{k^n}{k!}[/imath]? for example for [imath]n=6[/imath], we have [imath]\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.[/imath] |
615337 | remainder of the division [imath]2^{1990}/1990[/imath]
How do we find the remainder of the division [imath]2^{1990}/1990[/imath]? I actually tried it through Fermat's theorem but couldn't arrive at the answer directly. | 545759 | What is the remainder when [imath]2^{1990}[/imath] is divided by [imath]1990[/imath]?
I actually do not have the basic idea on how to approach these type of questions....so please tell me a generalized method about all this too. It came in RMO, and the question is: What is the remainder when [imath]2^{1990}[/imath] is divided by [imath]1990[/imath] ? |
534608 | Solve the Sturm-Louiville boundary value Problem [imath]u_t(x,t)=u_{xx}(x,t)-bu(x,t)+q(t)[/imath], where [imath]b[/imath] is a constant.
Solve the Sturm-Louiville boundary value Problem [imath]u_t(x,t)=u_{xx}(x,t)-bu(x,t)+q(t)[/imath], where [imath]b[/imath] is a constant. Boundary conditions are [imath]u(x,0)=0[/imath] and [imath]u(0,t)=0=u(\pi,t)[/imath], with [imath]0<x<\pi[/imath] and [imath]t>0[/imath]. Following the only example in the textbook, I get: Assume [imath]u(x,t)[/imath] is of the form [imath]u(x,t)=\sum\limits_{n=1}^\infty B_ne^{-n^2t}\sin(nx)[/imath]. But that's for a heat equation: [imath]u_t(x,t)=ku_{xx}(x,t)+q(t)[/imath], boundary conditions [imath]u(x,0)=f(x)[/imath], [imath]u(0,t)=0=u(\pi,t)[/imath]. I don't know what I'm doing and the book and professor are incomprehensible. I really need help. | 534097 | Solve the pde [imath]u_t(x,t)=u_{xx}(x,t)-bu(x,t)+q(t)[/imath] for [imath]u(x,t)[/imath]
I have the example pde [imath]u_t(x,t)=u_{xx}(x,t)-b(t)u(x,t)+q_0[/imath], where [imath]b(t)[/imath] is a function of only [imath]t[/imath] and [imath]q_0[/imath] is a constant, [imath]0<x<\pi[/imath], [imath]t>0[/imath]. The subscripts denote derivatives. I also have some boundary conditions: [imath]u(x,0)=0[/imath] and [imath]u(0,t)=0=u(\pi,t)[/imath]. The problem solution using variation of parameters is [imath]u(x,t)=\frac{4q_0}{\pi a(t)}\sum\limits_{n=1}^\infty \frac{\sin((2n-1)x)}{2n-1}\int\limits_0^t e^{-(2n-1)^2(t-\tau)}a(\tau)d\tau[/imath] where [imath]a(t)=e^{\int\limits_0^t b(\sigma)d\sigma}.[/imath] The problem I'm working on is slightly different. It gives the pde [imath]u_t(x,t)=u_{xx}(x,t)-bu(x,t)+q(t)[/imath] where [imath]b[/imath] is a constant and [imath]q(t)[/imath] is a function of only [imath]t[/imath], [imath]0<x<\pi[/imath], [imath]t>0[/imath]. The solution should be very similar to that of the example. I don't know where to start. I'm using the textbook "Fourier Series and Boundary Value Problems" by Churchill and Brown, 7th edition, if anyone's interested. The "example" is problem 10 page 117 of that book. Any help/hints would be greatly appreciated. Thanks! |
615387 | Show that [imath]\int_{-\infty}^{\infty} \frac{1}{x^4 + 1} \,dx = \frac{\pi}{\sqrt{2}}[/imath]
Show that the following integral: [imath] \int_{-\infty}^{\infty} \frac{1}{x^4 + 1} \,dx = \frac{\pi}{\sqrt{2}}. [/imath] I know this is [imath]\pi/\sqrt{2}[/imath] which I was found with Mathematica. I think I could utilise the substitution [imath]t = x^2 + 1[/imath] or [imath]t = 1/x[/imath], but I can't seem to make an approach because the integral is quite different from what I have been dealing with. Is there a better approach? | 426152 | Evaluating [imath]\int_0^\infty \frac{dx}{1+x^4}[/imath].
Can anyone give me a hint to evaluate this integral? [imath]\int_0^\infty \frac{dx}{1+x^4}[/imath] I know it will involve the gamma function, but how? |
615479 | Prove a space is Hilbert
I got stucked in this problem and get no clue to solve this. Can any one please help me? Thanks Suppose [imath]X[/imath] is an inner product space. If for every bounded linear function [imath]f[/imath], there exists [imath]z \in X[/imath] such that [imath]f(x) = <x, z>[/imath] for all [imath]x \in X[/imath]. Prove that [imath]X[/imath] is a Hilbert space. I can't find any way to relate the complete of space [imath]X[/imath] with the existence of [imath]z[/imath]. Please help me solve this. I really appreciate. | 245298 | Show that if the Riesz map is surjective on [imath]H[/imath], then [imath]H[/imath] is a Hilbert space
Let [imath]H[/imath] be a vector space equipped with an inner product [imath](\cdot, \cdot)[/imath] and [imath]f:H\to H',\ f(x)=(\cdot,x)[/imath] surjective. Now, why [imath]H[/imath] is a Hilbert space? The other direction is clear by Riesz' representation theorem but what about this? |
614900 | How prove this inequality [imath]f(a)\le f(b)[/imath]
Question: Suppose [imath]f(x)[/imath] is continous on [imath][a,b][/imath], and for any [imath]x\in(a,b)[/imath] have [imath]\varliminf_{h\to 0^{+}}\dfrac{f(x+h)-f(x-h)}{h}\ge 0[/imath] show that [imath]f(a)\le f(b)[/imath] My try: I have posted these two questions: How prove this inequality [imath]f(a)\le f(b)[/imath] and How prove this [imath]f(a)\le f(b)[/imath] This problem is different from the above two questions, How does one prove it. Thank you very much! | 517786 | How prove this [imath]f(a)\le f(b)[/imath]
Suppose [imath]f(x)[/imath] is continuous on [imath][a,b][/imath], and that for any [imath]x\in (a,b)[/imath], the limit [imath]\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{h}[/imath] exists and [imath]\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{h}\ge 0.[/imath] Show that [imath]f(a)\le f(b)[/imath] this problem is from http://wenku.baidu.com/view/a643e6c26137ee06eff91855.html |
616072 | Sequential sums [imath]1+2+\cdots+N[/imath] that are squares
While playing with sums [imath]S_n = 1+\cdots+n[/imath] of integers, I have just come across some "mathematical magic" I have no explanation and no proof for. Maybe you can give me some comments on this: I had the computer calculating which Sn are squares, and it came up with the following list: Table row [imath]N[/imath] sum([imath]1+\cdots+N[/imath]) M (square root of sum) r=1 N=1 sum=1 M=1 r=2 N=8 sum=36 M=6 r=3 N=49 sum=1225 M=35 r=4 N=288 sum=41616 M=204 r=5 N=1681 sum=1413721 M=1189 r=6 N=9800 sum=48024900 M=6930 Of course we have [imath]1+\cdots+N = \frac{N(N+1)}{2}[/imath], but this gives no indication for which N the sum [imath]1+\cdots+N[/imath] is a square. Can you guess how in this table we can calculate the entries in row 2 from the entries in row 1? Or the entries in row 3 from the entries in row 2? Or the entries in row 4 from the entries in row 3? Or the entries in row 5 from the entries in row 4? I looked at the above table and made some strange observations: The value of the next M can be easily calculated from the previous entries: Take the M from the previous row, multiply by 6 and subtract the M from two rows higher up. [imath]M(r) = 6*M(r-1)βM(r-2)[/imath] How is this possible? The S(r) we calculate as [imath]S(r) = M(r)^2[/imath]. Note that we do not know whether this newly constructed number [imath]S_r[/imath] is in fact of the type [imath]1+\cdots+k[/imath] for some [imath]k[/imath]. The value of the next N can be calculated as N(r) = Floor([imath]M(r)*\sqrt 2[/imath]), where Floor means βrounding down to the next lower integerβ. Somewhat surprising, [imath]S(r)[/imath] is the sum [imath]1+\cdots+N(r)[/imath] ! It looks as if outside the entries in the above table there are no other cases. With other words, the method [imath]M(r) = 6*M(r-1)βM(r-2)[/imath] seems to generate ALL solutions n where the sum [imath]1+\cdots+n[/imath] is a square. Problems: Is there a proof for any of the three observations? Do observations 1 and 2 really work for the infinite number of rows in this table? Is there an infinite number of rows in the first place? Puzzled, Karl | 76040 | General formula to obtain triangular-square numbers
I am trying to find a general formula for triangular square numbers. I have calculated some terms of the triangular-square sequence ([imath]TS_n[/imath]): [imath]TS_n=[/imath]1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056 1882672131025, 63955431761796, 2172602007770041, 73804512832419600 I have managed to find that: [imath]m^2 = \frac{n(n+1)}{2}[/imath] where [imath]m[/imath] is the [imath]m^{th}[/imath] term of the square number sequence, and [imath]n[/imath] is the [imath]n^{th}[/imath] term of the triangular numbers sequence. Would anyone be able to point out anything that could help me derive a formula for the square-triangular sequence? |
616295 | If we accept a false statement, can we prove anything?
I think that the question is contained in the title. Suppose we begin from something that is false for example [imath]1=0[/imath]. Is it possible using only [imath]\Rightarrow[/imath] (and of course [imath]\lnot ,\wedge,\lor[/imath]) to prove any possible statement? It has been ages since i studied logic at the university so i would prefer a simple explanation (if possible) | 1840746 | What's going on with this 5-line proof of Fermat's Last Theorem?
I'm reading a book on the Philosophy of Mathematics, and the author gave a "5-line proof" of Fermat's Last Theorem as a way to introduce the topic of inconsistency in set theory and logic. The author acknowledges that this is not a real proof of the theorem, but the way it was presented implies that it was supposed to look somewhat convincing. I, however, have absolutely no idea how the proof given even remotely relates to FLT, and would greatly appreciate it if someone could make the connection for me. Below is an almost verbatim excerpt from the book. Theorem: There are no positive integers [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath], and integer [imath]n>2[/imath], such that [imath]x^n + y^n = z^n[/imath]. Proof. Let [imath]R[/imath] stand for the Russell set, the set of all things that are not members of themselves: [imath]R= \{x : x \notin x\}[/imath]. It is straightforward to show that this set is both a member of itself and not a member of itself: [imath]R \in R[/imath] and [imath]R \notin R[/imath]. So since [imath]R \in R[/imath], it follows that [imath]R \in R[/imath] or FLT. But since [imath]R \notin R[/imath], by disjunctive syllogism, FLT. End. |
585067 | Turn monotone increasing function to continuous function
Let [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] be a monotone increasing function. We know from a general fact that the set of discontinuities of [imath]f[/imath] is countable. Denote the set of discontinuities by [imath]D[/imath]. I want to construct a continuous function [imath]g[/imath] such that there exists a function [imath]h[/imath] defined on [imath]D[/imath] with [imath]f(b)-f(a)=g(b)-g(a)+\sum_{x\in (a,b)\cap D}h(x)[/imath]. The problem is that there can be a countable number of discontinuities, and it is not necessary that we can order them as [imath]\cdots<x_{-2}<x_{-1}<x_0<x_1<x_2<\cdots[/imath]. How can we construct this function [imath]g[/imath]? | 585265 | Eliminating discontinuities for increasing function
Let [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] be a monotone increasing function. We know from a general fact that the set of discontinuities of [imath]f[/imath] is countable. Denote the set of discontinuities by [imath]D[/imath]. Define [imath]g:\mathbb{R}\rightarrow\mathbb{R}[/imath] as follows: [imath]g(x)=f(x^-)-\sum_{y\in D \text{ and }y<x}(f(y^+)-f(y^-))[/imath] It should be relatively clear that [imath]g[/imath] is monotone increasing. But is [imath]g[/imath] necessarily continuous? |
616807 | Derivative polynomial
Let [imath]f :\mathbb R\to\mathbb R[/imath] be an infinitely differentiable function. Assume that for every [imath]x \in \mathbb R[/imath], there exists an [imath]n_x\in \mathbb N[/imath], such that [imath]\,f^{(n_x)}(x)=0[/imath]. Prove that [imath]f[/imath] is polynomial. | 232665 | Baire: Show that [imath]f\colon \mathbb{R}\to\mathbb{R}[/imath] is a polynomial in an open bounded set
Let [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] be an infinitely differentiable function, and suppose that for each [imath]x\in\mathbb{R},[/imath] [imath]\exists\ n=n(x)\in\mathbb{N}[/imath] such that [imath]f^{(n)}(x)=0[/imath]. For each fixed [imath]n\in\mathbb{N}[/imath], consider the sets [imath]F_n=\{x\in\mathbb{R} : f^{(n)}(x)=0\}[/imath] and let [imath]I=[a,b][/imath] a finite closed interval. Show that [imath]I=\bigcup_{n=0}^\infty(I\cap F_n).[/imath] Show that there is a non empty open set [imath]A\subset I[/imath] where [imath]f[/imath] is a polynomial. My attempt: If [imath]x\in I[/imath] then by hypotesis there is [imath]n[/imath] such that [imath]f^{(n)}(x)=0[/imath] then [imath]x\in F_n[/imath] so then [imath]\bigcup F_n\supset I[/imath] then [imath]I=I\cap\bigcup_{n=0}^\infty F_n=\bigcup_{n=0}^\infty(I\cap F_n).[/imath] Let [imath]B_k=\bigcap_{n=k}^\infty F_n.[/imath] I want to show that [imath]I\cap B_k[/imath] has non-empty interior for some [imath]k\in\mathbb{N}[/imath]. Then there is [imath]A\subset B_k[/imath] and by Taylor's Theorem [imath]f[/imath] will be a polynomial. But I didn't know how to prove this. I tried using the following bigger set: [imath]B=I\cap\bigcup_{k=0}^\infty B_k=I\cap\bigcup_{k=0}^\infty \bigcap_{n=k}^\infty F_n.[/imath] I think that it suffices to show that [imath]B^o\neq\emptyset[/imath] maybe using the Baire's Theorem proving that [imath]B[/imath] is a not meagre set. More general Problem https://mathoverflow.net/questions/34059/ Let [imath]f[/imath] be an infinitely differentiable function on [imath][0,1][/imath] and suppose that for each [imath]x \in [0,1][/imath] there is an integer [imath]n \in \mathbb{N}[/imath] such that [imath]f^{(n)}(x)=0[/imath]. Then does [imath]f[/imath] coincide on [imath][0,1][/imath] with some polynomial? If yes then how. I thought of using Weierstrass approximation theorem, but couldn't succeed. Answer: The proof is by contradiction. Assume [imath]f[/imath] is not a polynomial. Consider the following closed sets: [imath] S_n = \{x: f^{(n)}(x) = 0\} [/imath] and [imath] X = \{x: \forall (a,b)\ni x: f\restriction_{(a,b)}\text{ is not a polynomial} \}. [/imath] It is clear that [imath]X[/imath] is a non-empty closed set without isolated points. Applying Baire category theorem to the covering [imath]\{X\cap S_n\}[/imath] of [imath]X[/imath] we get that there exists an interval [imath](a,b)[/imath] such that [imath](a,b)\cap X[/imath] is non-empty and [imath] (a,b)\cap X\subset S_n [/imath] for some [imath]n[/imath]. Since every [imath]x\in (a,b)\cap X[/imath] is an accumulation point we also have that [imath]x\in S_m[/imath] for all [imath]m\ge n[/imath] and [imath]x\in (a,b)\cap X[/imath]. Now consider any maximal interval [imath](c,e)\subset ((a,b)-X)[/imath]. Recall that [imath]f[/imath] is a polynomial of some degree [imath]d[/imath] on [imath](c,e)[/imath]. Therefore [imath]f^{(d)}=\mathrm{const}\neq 0[/imath] on [imath][c,e][/imath]. Hence [imath]d< n[/imath]. (Since either [imath]c[/imath] or [imath]e[/imath] is in [imath]X[/imath].) So we get that [imath]f^{(n)}=0[/imath] on [imath](a,b)[/imath] which is in contradiction with [imath](a,b)\cap X[/imath] being non-empty. |
616791 | When do we have [imath]m\otimes n = 0[/imath]
Let [imath]M[/imath] and [imath]N[/imath] be [imath]R[/imath]-modules ([imath]R[/imath] a commutative ring with identity). Let [imath]m \in M[/imath] and [imath]n \in N[/imath]. Is there any necessary and sufficient condition to have [imath]m\otimes n = 0[/imath] (as an equation in [imath]M\otimes_RN)[/imath]. | 288431 | Testing whether an element of a tensor product of modules is zero
Let [imath]A[/imath] be a commutative ring and let [imath]F[/imath], [imath]E[/imath] be [imath]A[/imath]-modules. Then [imath]E[/imath] is the direct limit of its finitely generated submodules. Serge Lang, in his Algebra, p. 604, remarks that a technique to test whether an element of [imath]F \otimes E[/imath] is zero, is to examine its image in [imath]F \otimes N[/imath], as [imath]N[/imath] varies over the finitely generated submodules of [imath]E[/imath]. Could somebody please explain why this is a valid technique? What is the image of an element of [imath]F \otimes E[/imath] in [imath]F \otimes N[/imath]? |
617008 | Range A = range A^tA
let [imath]A\in\mathbb{R}^{m\times n}[/imath] with [imath]m\geq n[/imath]. ΒΏIs it true that rank [imath]A[/imath] is equal to rank [imath]A^tA[/imath]?. How show this? Thanks! | 349738 | Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any [imath]A\in M_{m \times n}[/imath]
How can I prove [imath]\operatorname{rank}A^TA=\operatorname{rank}A[/imath] for any [imath]A\in M_{m \times n}[/imath]? This is an exercise in my textbook associated with orthogonal projections and Gram-Schmidt process, but I am unsure how they are relevant. |
520491 | Semigroups isomorphism
Does there exist an isomorphism between the semigroups [imath]S(4)[/imath] and ββββββ[imath]\mathbf Z_{256βββββββ}[/imath].β [imath]S(4)[/imath] is the set of all maps from the set [imath]X[/imath] to itself and [imath]X = \{1, 2, 3, 4\}[/imath]. [imath]S(4)[/imath] is a semigroup under the composition of mappings and βββββ[imath]\mathbf Z_{256} = {0, 1, 2, β¦ , 255}[/imath] is the semigroup under multiplication of integers modulo 256. | 520369 | Semigroup isomorphism
Does there exist an isomorphism between the semigroups [imath]S(4)[/imath] and [imath]\mathbb{βββββZ}_{β256}[/imath]?β [imath]S(4)[/imath] is the set of all maps from the set [imath]X[/imath] to itself and [imath]X =\{1, 2, 3, 4\}[/imath], [imath]S(4)[/imath] is a semigroup under the composition of mappings, and [imath]ββββββ\mathbb{Z}_{β256}=\{0, 1, 2, \dots , 255\}[/imath] is the semigroup under multiplication modulo 256. |
617632 | uniformly integrable family of functions equivalent statement
I am studying the fourth edition of Royden Real Analysis and I have tried to solve the following exercise: Let [imath]\mathcal{F}[/imath] be a family of function, each of which is integrable over E. Show that [imath]\mathcal{F}[/imath] is uniformly integrable over E if and only if for each [imath]\varepsilon>0[/imath], there is a [imath]\delta>0[/imath] such that for each [imath]f \in \mathcal{F}[/imath], [imath]\text{if}\; A \subset E\;\text{is measurable and }m(A)<\delta,\; \text{then}\; \Biggl|\int_Af\Biggr|<\varepsilon.[/imath] Solution: The straight direction is obvious. Let [imath]\epsilon>0[/imath]. Since [imath]\mathcal{F}[/imath] is uniformly integrable over E, there is a [imath]\delta>0[/imath]such that for each [imath]f \in \mathcal{F}[/imath], [imath]\text{if}\; A \subset E\;\text{is measurable and }m(A)<\delta,\; \text{then} \int_A|f|<\epsilon,[/imath] and therefore, [imath]\Biggl|\int_Af\Biggr|<\int_A|f|<\epsilon.[/imath] To show the opposite direction we suppose that for each [imath]\varepsilon>0[/imath], there is a [imath]\delta>0[/imath] such that for each [imath]f \in \mathcal{F}[/imath], [imath]\text{if}\; A \subset E\;\text{is measurable and }m(A)<\delta,\; \text{then}\; \Biggl|\int_Af\Biggr|<\varepsilon.[/imath] We also know that every [imath]f \in \mathcal{F}[/imath] is integrable so for fixed [imath]f \in \mathcal{F}[/imath] and fixed [imath]\epsilon[/imath], there is a [imath]\delta_f>0[/imath] such that, [imath]\text{if}\; A \subset E\;\text{is measurable and }m(A)<\delta_f,\; \text{then}\; \int_A|f|<\epsilon.[/imath] My problem is that I cannot find a common [imath]\delta[/imath] for all [imath]f[/imath] to show the desire. | 126215 | An equivalent definition of uniform integrability
Let [imath](X,\mathcal{M},\mu)[/imath] be a measure space and [imath]\{f\}[/imath] be a sequence of functions on [imath]X[/imath], each of which is integrable over [imath]X[/imath]. Show that [imath]\{f_n\}[/imath] is uniformly integrable if and only if for each [imath]\varepsilon \gt 0[/imath], there is a [imath]\delta \gt 0[/imath] such that for any natural number [imath]n[/imath] and measurable subset [imath]E[/imath] of [imath]X[/imath], if [imath]\mu(E) \lt \delta[/imath], [imath] \left|\int_E f_n~d\mu\right| \lt \epsilon.[/imath] I think one direction [imath](\Rightarrow)[/imath] is clear. Since [imath]f_n[/imath] being uniformly integrable imply that for every [imath]\varepsilon\gt 0,~\exists \delta \gt 0[/imath] such that for any [imath]E\in \mathcal{M},~\mu(E)\lt \delta[/imath], [imath]\int_E |f_n|~d\mu[/imath] for every natural [imath]n[/imath]. But then [imath] \left|\int_E f_n~d\mu\right|\le \int_E|f_n|~d\mu \lt \varepsilon.[/imath] Any suggestions for the other direction? Edit: Following Davide's suggestions, I have [imath] \begin{align*} \int_E |f_n|~d\mu & \le \int_{E\cap [f_n \ge 0]} f_n~d\mu + \int_{E\cap [f_n \lt 0]} (-f_n)~d\mu\\ & = \left| \int_{E\cap [f_n \ge 0]} f_n~d\mu \right| + \left| \int_{E\cap [f_n \lt 0]}f_n~d\mu \right| \\ & \lt \varepsilon + \varepsilon = 2\varepsilon. \end{align*} [/imath] |
617793 | How to prove that [imath]n^{-2}[x+g(x)+g\circ g(x)+\cdots +g^{\circ n}(x)][/imath] converges when [imath]n\to\infty[/imath]
Let [imath]f:\mathbb R\to\mathbb R[/imath] be a periodic function with period [imath]1[/imath]. We assume that [imath]f[/imath] is Lipschitz continuous, and in particular, we assume that there exists an [imath]L\in (0,1)[/imath], such that [imath] |f(x)-f(y)| \le L|x-y|, \quad \text{for all $x,y\in\mathbb R$.} [/imath] Let also [imath]g(x)=x+f(x).[/imath] Show that the limit [imath]\lim_{n\to +\infty}\frac1{n^2}[x+g(x)+g(g(x))+\cdots +g^{\circ n}(x)][/imath] exists and is independent of [imath]x[/imath], where, for every [imath]n\ge1[/imath], [imath]g^{\circ n}[/imath] is [imath]g[/imath] composed with itself [imath]n[/imath] times, thus [imath]g^{\circ 1}=g[/imath] and, for every [imath]n\ge1[/imath], [imath]g^{\circ n+1}=g\circ g^{\circ n}[/imath]. My attempt: Since [imath]f(x+1)=f(x),\forall x\in R[/imath] then [imath]g(g(x))=g(x+f(x))=x+f(x)+f(x+f(x))[/imath] [imath]g(g(g(x)))=g(x+f(x)+f(x+f(x)))=x+f(x)+f(x+f(x))+f(x+f(x)+f(x+f(x)))[/imath] [imath]\cdots\cdots [/imath] and I have [imath]|g(x)-g(y)|=|x-y+f(x)-f(y)|\le |x-y|+|f(x)-f(y)|<(L+1)|x-y|[/imath] where [imath]L+1>1[/imath]. So [imath]g[/imath] is also Lipschitz continuous, and that's all I can do. Thank you for you help. | 597225 | Show that [imath]\lim\limits_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}[/imath] exists and is independent of the choice of [imath]a[/imath]
Suppose [imath]f:\mathbb{R}\to\mathbb{R}[/imath] has period 1, and for some [imath]q\in(0,1)[/imath]: [imath]|f(x)-f(y)|\leq q|x-y|\quad \forall x,y[/imath] Let [imath]g(x)=x+f(x)[/imath], for any [imath]a\in\mathbb{R}[/imath],define the following sequence: [imath]a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)[/imath] Show that [imath]\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}[/imath] exists and does not depend on the choice of [imath]a[/imath]. I don't even know what is the possible approach, any hints? Thanks! |
614128 | How to show that [imath]a^ab^b+a^bb^a\leq 1[/imath] for [imath]a,b\in(0,1)[/imath] with [imath]a+b=1[/imath].
A little while ago I came across this question on Yahoo! Answers and it had no replies. I have spent a while trying to figure this out as well, but I couldn't find an algebraic proof. The assumptions are [imath]a+b=1[/imath] and [imath]a,b\in(0,1)[/imath]. We want to show that [imath]a^ab^b+a^bb^a\leq 1[/imath]. Or equivalently that [imath]a^a(1-a)^{1-a}+a^{1-a}(1-a)^a\leq 1\;\forall a\in(0,1)[/imath]. I tried see if [imath]a[/imath] and [imath]b[/imath] are an upper bound for [imath]a^ab^b[/imath] and [imath]a^bb^a[/imath] respectively, for their sum would be [imath]1[/imath] by assumption. However, I was unable to do so. I managed to show the following, assuming without loss of generality that [imath]b\leq a[/imath] (i.e. [imath]a\geq\frac{1}{2}[/imath]): [imath]\begin{aligned} a^ab^b&=a^{1-b}b^b\\ &=a\left(\frac{b}{a}\right)^b\\ &\leq a. \end{aligned}[/imath] However, applying this same trick to [imath]a^bb^a[/imath] does not yield the necessary inequality: [imath]\begin{aligned} a^bb^a&=a^bb^{1-b}\\ &=b\left(\frac{a}{b}\right)^b\\ &\geq b. \end{aligned}[/imath] Which means that the upperbound [imath]a^bb^b\leq a[/imath] is too weak. By similar means one can find the following bounds: [imath]b\leq a^ab^b\leq a,[/imath] [imath]b\leq a^bb^a\leq a.[/imath] Now define [imath]f(a):=a^a(1-a)^{1-a}=a^ab^b[/imath]. One can show that the following holds for the derivative of [imath]f[/imath]: [imath]f'(a)=f(a)\log\left(\frac{a}{1-a}\right).[/imath] Since [imath]f(a)=\exp(a\log(a)+(1-a)\log(1-a))>0[/imath], we have that [imath]f'(a)=0\iff a=\frac{1}{2}[/imath]. Since [imath]f'(a)>0[/imath] for all [imath]a>\frac{1}{2}[/imath] we have that [imath]f[/imath] has a minimum at [imath]a=\frac{1}{2}[/imath]. Also note that [imath]\lim_{a\to 1}f(a)=1[/imath], hence by symmetry [imath]\frac{1}{2}\leq f(a)\leq 1[/imath] for all [imath]a\in(0,1)[/imath]. We'll define [imath]g(a):=a^{1-a}(1-a)^a=a^bb^a[/imath]. One finds that [imath]g'(a)=g(a)\left(\frac{1-2a}{a(1-a)}+\log\left(\frac{1-a}{a}\right)\right).[/imath] Now [imath]g'(a)=0\iff\log\left(\frac{1-a}{a}\right)=\frac{2a-1}{a(1-a)}[/imath]. It's clear that [imath]a=\frac{1}{2}[/imath] is a solution. Note that [imath]\frac{d}{dx}\log\left(\frac{1-a}{a}\right)=\frac{1}{a(a-1)}<0[/imath]. Since [imath]\frac{2a-1}{a(1-a)}[/imath] is increasing, [imath]a=\frac{1}{2}[/imath] is the only solution. As [imath]\lim_{a\to 1}g(a)=0[/imath], we have that this is a maximum. Hence [imath]0\leq g(a)\leq\frac{1}{2}[/imath] for all [imath]a\in(0,1)[/imath]. Combining these results with what we already had yields the following inequalities: [imath]\frac{1}{2}\leq a^ab^b\leq a,[/imath] [imath]b\leq a^bb^a\leq\frac{1}{2}.[/imath] This gives us [imath]b+\frac{1}{2}\leq a^ab^b+a^bb^a\leq a+\frac{1}{2}.[/imath] Which is still not all too helpful... So then I got another idea. I construct two parabolas [imath]\phi,\psi[/imath] both with their top at [imath]\left(\frac{1}{2},\frac{1}{2}\right)[/imath]. One of them, [imath]\phi[/imath], through [imath](0,0)[/imath] and [imath](1,0)[/imath] and [imath]\psi[/imath] through [imath](0,1)[/imath] and [imath](1,1)[/imath]. Then I'd show that [imath]a^a(1-a)^{1-a}\leq\psi(a)[/imath] and [imath]a^{(1-a)}(1-a)^a\leq\phi[/imath]. Then, since [imath]\phi(a)+\psi(a)=1\;\forall a[/imath] the statement is proven. However, this created some difficulty that I am now stuck on. I'll first define [imath]\phi[/imath] and [imath]\psi[/imath]: [imath]\phi:(0,1)\to\left(0,\frac{1}{2}\right][/imath] [imath]x\mapsto-2x^2+2x[/imath] Then [imath]\psi(x)=1-\phi(x)[/imath]. We have [imath]\begin{aligned} \phi(a)&\geq a^{1-a}(1-a)^a\\ \iff-2a^2+a&\geq a^{1-a}(1-a)^a\\ \iff2a&\geq\left(\frac{a}{1-a}\right)^{1-a}. \end{aligned}[/imath] I cannot figure out how to prove this inequality though; let alone the other one. Hence now I am stuck! I'm hoping someone can help figure this out. I realize parts of what I've done may be superfluous, but this is just everything that I have tried put together. | 586465 | Proving [imath]a^ab^b + a^bb^a \le 1[/imath], given [imath]a + b = 1[/imath]
Given [imath]a + b = 1[/imath], Prove that [imath]a^ab^b + a^bb^a \le 1[/imath]; [imath]a[/imath] and [imath]b[/imath] are positive real numbers. |
615157 | Introducing ordered pairs in an axiomatic way
Suppose that in [imath]ZFC[/imath] we have introduced ordered pairs not in the usual way as [imath](a,b) = \{\{a\}, \{a,b\}\}[/imath] but axiomatically, by extending [imath]ZFC[/imath] by adding to [imath]ZFC[/imath] a new binary functional symbol [imath]g[/imath] and a corresponding axiom: [imath]\forall a,b,c,d( g(a,b) = g(c,d) \rightarrow a=c \wedge b=d)[/imath]. The main advantage of this approach is absence of so called "junk" theorems - see https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems. Let us write instead of [imath]g(x,y)[/imath] just [imath](x,y)[/imath] Question 1. Will be the formula [imath]\forall S,u((S,u) \notin S)[/imath] a theorem of the extended theory? Question 2. Are there textbooks where ordered pairs were introduced in a similar axiomatic way? | 617062 | Problems with introducing ordered pairs axiomatically
(See also Introducing ordered pairs in an axiomatic way). Many feel that the usual way to introduce ordered pairs in set theory following K.Kuratowski as [imath](a,b) = \{\{a\}, \{a,b\}\}[/imath] is rather unnatural (B.Russell called Kuratowski's definition a trick). But the main drawback of this definition is the emergence of so called "junk" theorems - see https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems So it seems reasonable to try to introduce ordered pairs in an axiomatic way, for example, to extend [imath]ZFC[/imath] by adding to [imath]ZFC[/imath] a new binary functional symbol [imath]g[/imath] and an obvious axiom: Axiom1 := [imath]\forall a,b,c,d( g(a,b) = g(c,d) \rightarrow a=c \wedge b=d)[/imath]. But for further development of set theory we need more axioms, for example Axiom2 := [imath]\forall a,b(g(a,b) \in P(P(a \cup b)))[/imath]. So my question is: Are necessary some other axioms for the symbol [imath]g[/imath]? |
618548 | Counterexample that [imath]a\in G[/imath], [imath]a^n\notin H[/imath], for [imath]H[/imath] a subgroup of finite index [imath]n[/imath] in [imath]G[/imath].
Let [imath]G[/imath] be a group and [imath]H[/imath] a subgroup of finite index [imath]n[/imath]. Give a counterexample that [imath]a\in G[/imath], [imath]a^n\notin H[/imath] (although I can prove that there exists [imath]k\in\{1,2,\dots,n\}[/imath] such that [imath]a^k\in H[/imath]). Really do not know how to construct the counterexample... Thanks. | 545417 | If [imath][G:H]=n[/imath], is it true that [imath]x^n\in H[/imath] for all [imath]x\in G[/imath]?
Let [imath]G[/imath] be a group and [imath]H[/imath] a subgroup with [imath][G:H]=n[/imath]. Is it true that [imath]x^n\in H[/imath] for all [imath]x\in G[/imath]? Remarks. The answer is positive whenever [imath]H[/imath] is normal, e.g., for [imath]n=2[/imath]. In general, by using the normal core of [imath]H[/imath], one can find an [imath]m\ge 1[/imath] such that [imath]x^m\in H[/imath] for all [imath]x\in G[/imath]. |
618889 | How to show that the ring [imath]R= \mathbb Z_3[x]/\langle x^6-1\rangle[/imath] is finite without using the concepts of vector space?
I have understood that since [imath]\mathbb Z_3[/imath] is a finite field so [imath]R[/imath] has to be a finite dimensional vector space but is there any way of proving this from purely ring theoretic concepts ? | 618606 | Ring [imath]R= \mathbb Z_3[x]/\langle x^6-1\rangle[/imath]
i) Check whether [imath]R[/imath] is a finite ring or not. ii) Check whether [imath]R[/imath] has zero divisors. iii) Check whether [imath]R[/imath] has nilpotent elements. i) The field [imath]Z_3[/imath]={0,1,2} has 3 elements but as repetitions are possible while forming a polynomial for infinite times and power of x can also have infinitely possible positive integers thus [imath]Z_3[x][/imath] is not a finite ring. Again as [imath]Z_3[/imath][x] is not finite so will not be R. Please help rectify if I am wrong here and also need help for part ii) & iii). |
192947 | Inequality.[imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3[/imath]
Let [imath]a,b,c \gt 0[/imath]. Prove that (Using Cauchy-Schwarz) : [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3[/imath] I tried to use Cauchy-Schwarz in the following form [imath]\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.[/imath] I wrote [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.[/imath] and then I applied on [imath](1)[/imath]: \begin{eqnarray} A &=& 2a(c+a) &\mbox{and}& x=a+b;\\ B &=& 2b(a+b) &\mbox{and}& y=b+c;\\ C &=& 2c(b+c) &\mbox{and}& z=c+a , \end{eqnarray} but I did not obtain anything. Thanks for your help. :) | 1481348 | Prove inequality [imath]\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3[/imath]
How to prove the following inequality : [imath] \sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3 [/imath] with [imath]a>0,\ b>0[/imath] and [imath]c>0[/imath]. |
619110 | Prove/Disprove [imath]f(x) = x + \frac{x}{{x + 1}}[/imath] is uniformly continuous at [imath]\forall x,y \in [0,\infty )[/imath]
Prove/Disprove [imath]f(x) = x + \frac{x}{{x + 1}}[/imath] is uniformly continuous at [imath]\forall x,y \in [0,\infty )[/imath] This is my trial: [imath]\forall \varepsilon > 0\exists \delta > 0.\forall x,y \in [0,\infty ):[/imath] Let [imath]\left| {x - y} \right| < \delta [/imath] [imath]\left| {x + \frac{x}{{x + 1}} - y - \frac{y}{{y + 1}}} \right|... \le \delta + \frac{\delta }{{(x + 1)(y + 1)}}[/imath] As you can see after some algebra I got an expreesion with denominator involving [imath]x,y[/imath]. I need to "get rid" of [imath]x,y[/imath] at the denominator. One way doing it is just multiplying by [imath](x+1)(y+1)[/imath], because then I left with just [imath]2\delta[/imath] which is great. what do you think? Am I doing it right? | 618174 | Is [imath]f(x)=x+\frac{x}{x+1}[/imath] uniformly continuous on [imath](0,\infty)[/imath]
Is [imath]f(x)=x+\frac{x}{x+1}[/imath] uniformly continuous on [imath](0,\infty)[/imath] Going from the epsilon delta definition we get: [imath]\forall x,y>1,\text{WLOG}:x>y \ ,\ \forall\epsilon>0,\exists\delta>0 \ s.t. \ |x-y|<\delta\rightarrow |x+\frac{x}{x+1}-y-\frac{y}{y+1}|<\epsilon[/imath] But I'm not really sure on how to continue from here. Alternatively: if [imath]\lim\limits_{x\to \infty}f(x)[/imath] exists on [imath][0,\infty)[/imath] then the function is uniformly continuous, well, it's easy to see here that [imath]\lim\limits_{x\to \infty}f(x)=\infty[/imath] but is that enough ? |
288400 | Is the group isomorphism [imath]\exp(\alpha x)[/imath] from the group [imath](\mathbb{R},+)[/imath] to [imath](\mathbb{R}_{>0},\times)[/imath] unique?
I'm having a problem trying to find the simplest way of proving this, which has most probably been solved a hundred of times but I am unable to find a good reference. I have two groups, [imath](\mathbb{R},+)[/imath] and [imath](\mathbb{R}_{>0},\times)[/imath]. I am trying to prove that the only class of isomorphisms between them is the class [imath]F = \{f: f(x) = \exp(\alpha x), [/imath] for all [imath]\alpha \in \mathbb{R}_{>0}\}[/imath]. Existence is easy to prove: what I'm having trouble with is a clean algebraic uniqueness proof. Does anyone know the proof or a reference containing this proof? Thanks in advance! | 302257 | Is [imath]e^x[/imath] the only isomorphism between the groups [imath](\mathbb{R},+)[/imath] and [imath](\mathbb{R}_{> 0},*)[/imath]?
If so, how might I be able to prove it? EDIT: OK, thanks to many answers especially spin's and Micah's explanations. All of the answers were extremely helpful in understanding -- I have accepted Micah's because it seems the most complete, but all answers provide helpful additions/perspectives! I have tried to summarize: [imath]\phi[/imath] is an isomorphism between the groups if and only if [imath]\phi(x) = e^{f(x)}[/imath] where [imath]f[/imath] is an isomorphism from [imath](\mathbb{R},+)[/imath] back to [imath](\mathbb{R},+)[/imath]. Of course there are lots of such [imath]f[/imath], especially when we take the Axiom of Choice. However, it seems from the answers and Micah's link (Cauchy functional equation) that the only "nice" solutions are [imath]f(x) = cx[/imath] for a constant [imath]c[/imath]. It seems that all others must be "highly pathological" (in fact [imath]\{(x,f(x))\}[/imath] must be dense in [imath]\mathbb{R}^2[/imath]). A remaining question is, how strong is the statement All such isomorphisms have the form [imath]e^{cx}[/imath] for some [imath]c \in \mathbb{R}[/imath] or its negation? Or what is required for each to hold? One answer seems to be that supposing the reals have the Baire property is sufficient to rule out other solutions (as is assuming every subset of the reals is measurable, assuming the Axiom of Determinacy, and it holds in Solovay's model). For more, see this question, this question, and this mathoverflow question. |
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