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1117310 | A question about weak convergence in Lp space
Suppose [imath]1 \leq p<\infty[/imath], given [imath]f \in L^p (\mathbb{R})[/imath], define [imath]f_n (x)=n^{1/p} f(nx)[/imath] for n=1,2,3... Prove [imath]f_n[/imath] converges weakly to zero in [imath]L^p[/imath]. Now I can just know the that [imath] \|f_n\|_p[/imath]=[imath] \|f\|_p[/imath], but cannot prove for any function [imath]g \in L_q[/imath], where [imath]q[/imath] is the conjugate of [imath]p[/imath], the integral converges to zero. i.e. how to show [imath]\int f_ng[/imath] converges to zero. I'm thinking maybe I can use step functions or continuous functions since they are dense in Lp, so we can approximate, but not sure. | 805982 | Does scaling lead to weak convergence to the null function?
Let [imath]f\in L^p(\mathbb{R}^d)[/imath], with [imath]1<p<\infty[/imath]. Is it true that [imath]\lambda^{\frac{d}{p}}f(\lambda x ) \rightharpoonup 0\quad \text{ weakly in }L^p\text{ as }\lambda\to+\infty?[/imath] One has the easy case for [imath]f\in L^{p-\epsilon}\cap L^p[/imath]. This condition allows the use of Hölder's inequality as follows (here [imath]\frac{1}{p}+\frac{1}{p'}=1[/imath] and [imath]\phi[/imath] is a continuous function with compact support): [imath]\begin{split}\left\lvert \int_{\mathbb{R}^d} \lambda^{\frac{d}{p}} f(\lambda x)\phi(x)\, dx\right\rvert& \le \lambda^{\frac{d}{p}}\lVert f(\lambda x)\rVert_{L^{p-\epsilon}}\lVert \phi\rVert_{L^{(p-\epsilon)'}} \\ &= \lambda ^{\frac{d}{p}- \frac{d}{p-\epsilon} } \lVert f\rVert_{L^{p-\epsilon}}\lVert\phi\rVert_{L^{(p-\epsilon)'}}\to 0. \end{split} [/imath] But what happens if that condition is removed? I carried out an explicit check on the standard example of a function [imath]f[/imath] that belongs to [imath]L^2(\mathbb{R})[/imath] and does not belong to [imath]L^{2-\epsilon}(\mathbb{R})[/imath] for any [imath]\epsilon > 0[/imath], namely [imath]f(x)=\begin{cases} 0 , & x<2 \\ \frac{1}{\sqrt{x}\log x}, & x \ge 2\end{cases}.[/imath] Taking [imath]\phi=\chi_{[a, b]}[/imath] with [imath]2<a<b[/imath] one has [imath]\left \lvert \int_{\mathbb{R}} \lambda^{\frac{1}{2}}f(\lambda x)\phi(x)\, dx \right\rvert = \dfrac{ \int_{\lambda a }^{\lambda b} \frac{dy}{\sqrt{y}\log(y)} } {\lambda^{\frac{1}{2}}}, [/imath] and an application of l'Hôpital's rule shows that the right hand side tends to [imath]0[/imath] as [imath]\lambda \to \infty[/imath]. This implies weak convergence to [imath]0[/imath] by a standard density argument. This seems to point towards an affirmative answer to the question in the gray box. |
499009 | Number theory question on primes and divisibility
For which primes [imath]p[/imath] is [imath]2^p+1[/imath] divisible by [imath]p[/imath]? I am not quite sure how to approach this. | 499209 | Primes Number Theory
For which primes [imath]p[/imath] is [imath]2^p+1[/imath] divisible by [imath]p[/imath]? What I have been doing is: [imath]2^p+1\equiv 0\pmod p[/imath] [imath]2^p\equiv -1\pmod p[/imath] Then by Fermat's Theorem, we get [imath]2^p\equiv 2\pmod p[/imath] This shows [imath]p=1,3[/imath]. I feel like this is not right though... |
1117422 | Prove the convergence and find limit
Let [imath]x_n[/imath] be a sequence with general formula: [imath]x_{n+1}=\sqrt{2+x_n}[/imath] I am supposed to prove it's convergence and find the limit as [imath]n\to +\infty[/imath] I thought of proving this sequence to be monotone and bounded, which immediately means it's convergent. Proving that this sequence is monotone is quite trivial, but i can't seem to be able to find it's bounds and supremum. Is this a right way to solve this task? Can you provide a subtle hint. Taks like this always bug me because it's solution is only so logical and simple, yet i can't seem to reach it. | 333050 | How do I prove the sequence [imath]x_{n+1}=\sqrt{\alpha +x_n}[/imath], with [imath]x_0=\sqrt \alpha[/imath] converges? ( boundedness?)
I need to "study the limit behavior of the following sequence" and then compute the limit. I can compute the limit and prove the monotonicity, but I am having trouble proving boundedness. I tried to understand from other similar posts how this could be done, but since I didn't understand I'm asking it as a new question. The sequence is [imath]x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0[/imath]. It seems to be monotonically increasing. Proof: It can be shown by induction that the sequence is monotonically increasing. Claim that [imath]x_{n+1}\geq x_n[/imath] [imath]x_1=\sqrt{\alpha+\sqrt\alpha}>\sqrt\alpha=x_0[/imath] [imath]x_{n+1}\implies x_{n+2}[/imath] [imath]x_{n+1}=\sqrt{\alpha+x_n}<\sqrt{\alpha+x_{n+1}}=\sqrt{\alpha+\sqrt{\alpha+\sqrt\alpha}}=x_{n+2}[/imath] Boundedness: Because the sequence is monotonically increasing and [imath]x_o=\sqrt\alpha[/imath], it is bounded below by [imath]\sqrt\alpha[/imath]. Now comes the part where I have to prove that the sequence is bounded above. The problem is that I don't know how to start, so if anyone could give me a bit of a push I'd be grateful. The limit: I know that [imath]\lim\ x_n=\lim \ x_{n+1}[/imath], so letting [imath]\lim\ x_n=x[/imath] then [imath]x=\sqrt{\alpha+x}[/imath] and solving would give [imath]x=\frac{1+\sqrt{1+4\alpha}}{2}[/imath]. May somebody please confirm my proof thus far, and help me prove the upper bound? Thanks! |
1117872 | Calculate this infinite sum
[imath]s= \sum_{n=1}^\infty \frac{n+3}{(2^n)(n+1)(n+2)}[/imath] Any method to calculate this type of infinite sums? | 1116729 | Find the sum of the series [imath]\sum \limits_{n=1}^{\infty} \frac{2^{-n}(n+3)}{(n+1)(n+2)}[/imath]
Find the sum of [imath]\sum \limits_{n=1}^{\infty} \frac{2^{-n}(n+3)}{(n+1)(n+2)}[/imath] |
1117792 | Distance between a point and a line!
I have a big problem with geometry. How I do calculate the distance between the vectorial line [imath]r:(x,y,z)=(2,1,0)+\lambda(0,4,-3)[/imath] and the point [imath]A=(2,4,4)[/imath]? I tried to solve the problem but nothing... | 786813 | distance between a line and a point vector formula
Let [imath]a[/imath] and [imath]b[/imath] be the position vectors of two points [imath]A[/imath] and [imath]B[/imath], and let [imath]p[/imath] be the position vector of another point [imath]P[/imath]. Consider the line through the points [imath]A[/imath] and [imath]B[/imath], and show that the shortest distance between [imath]P[/imath] and this line is given by [imath] d = \frac{|(p-a) \times (p - b)|}{|b - a|} [/imath] let [imath]r = a + \lambda (b-a)[/imath] be the line connecting [imath]A[/imath] to [imath]B[/imath], my idea was to find a line connecting the point [imath]P[/imath] to [imath]r[/imath] i.e. [imath]r - p[/imath], then projecting this onto the normal of the line which I cannot find. How can I proceed? I noticed if I cross these [imath]r - p[/imath] and the normalised version of [imath]b-a[/imath] I get the answer. Why? |
1117712 | Prove that distinct Fermat Numbers are relatively prime
The Fermat numbers are defined by [imath]F_m = 2^{2^m} + 1[/imath]. Prove that for [imath]m \ne n[/imath] we have [imath](F_m, F_n) = 1[/imath]. I have to first prove that [imath]F_{m+1} = F_0F_1 \cdots F_m + 2[/imath] by representing [imath]F_{m+1}[/imath] in terms of [imath]F_m[/imath]. | 68653 | On the GCD of a Pair of Fermat Numbers
I've been working with the Fermat numbers recently but this problem has really tripped me up. If the Fermat theorem is set as [imath]f_a=2^{2^a}+1[/imath], then how can we say that for an integer [imath]b[/imath] less than [imath]a[/imath] that [imath]\gcd(f_b,f_a)=1[/imath]? |
538268 | Can you help me to solve the recurrence relation [imath]T(n) = T(\sqrt n) + 1 [/imath]?
I have this recurrence relation to solve : [imath]T(n) = T(\sqrt n) + 1 [/imath] I have tried to expand the recursion but I stopped here: \begin{align} T(n) &= T(n^{\frac12})+1\\ &= T(n^{\frac14})+1+1\\ &\text{after [imath]i[/imath] replacements I have}\\ &= T(n^{\frac1{2^i}}) + i\\ \end{align} I know that [imath]T(1) = 1[/imath] And now? How can I get to the solution? | 195037 | What is the solution to the following recurrence relation with square root: [imath]T(n)=T (\sqrt{n}) + 1[/imath]?
This looks like a question asked earlier, but it isn't. [imath]T(n)=\begin{cases} T (\sqrt{n}) + 1 \quad & \text{ if } n>1 \\ 1 & \text{ if }n=1\end{cases}[/imath] My professor gave this to me in class yesterday. This is where I'm stuck: [imath]T(n) = T(n^{1/2})+1 = T(n^{1/4})+2 = T(n^{1/8})+3 = \dots = T(n^{1/ 2 ^k})+k[/imath] Now this will continue till [imath]n^{1/ 2 ^k}=1[/imath] How do I proceed ahead? If I take log on both sides, then RHS becomes zero. How do I solve the relation? I don't want to use the Master theorem, I want to know where and why am I stuck. |
1107144 | Solutions to [imath]y^2 = x^3 + k[/imath]?
The equation [imath]y^2 = x^3 + k[/imath] for [imath]k = (4n-1)^3 - 4m^2[/imath], with [imath]m, n \in \mathbb{N}[/imath] and no prime number that p is congruent to 1 modulo 4 divids m, doesn't have any answer and its proof can be obtained by using quadratic reciprocity law. Do you know answers of this equation for two or three different values of [imath]k[/imath]? In addition, do you know any reference about that? | 1880650 | Proof that [imath]x^2-y^3=1[/imath] has only one solution in the domain of [imath]N^+[/imath]?
How can I prove that [imath]x^2-y^3=1[/imath] has only one solution in the domain of [imath]N^+[/imath]? Namely [imath]x=3, y=2[/imath]. My try: [imath]x^2-1=y^3 => (x-1)(x+1)-1=y^3[/imath]. [imath]gcd(x-1,x+1)[/imath] is 2 if [imath]x[/imath] is odd and [imath]1[/imath] if [imath]x[/imath] is even. In the case x-even the only possible solution would be if [imath]x=1[/imath] but [imath]y>0[/imath]. Therefore [imath]x[/imath] is odd. Let [imath]x=2p+1[/imath]. We can rewrite the equation as [imath]4p(p+1)=y^3[/imath]. [imath]gcd(p,p+1)=1[/imath], therefore [imath]p(p+1)=2[/imath]. Therefore [imath]x=3[/imath]. Is that correct ? |
1118216 | How to prove that [imath]f[/imath] is differentiable at every point beside [imath]x=-1[/imath]?
Consider [imath]f(x) = |x+1|[/imath]. I want to show that for every [imath]x_0\neq-1[/imath] , [imath]\lim_{h \rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}[/imath] is exist. So far i just wrote the definition, and looked on 2 options:When [imath]x>-1[/imath] and [imath]x<-1[/imath]. But, i got stuck when i go the definiton of differentiabily in those points, because what the meaning of [imath]\lim_{h \rightarrow 0}\frac{|x_0+h+1|- |x_0+1|}{h}[/imath]? tnx! | 1109557 | Why is [imath]|x|[/imath] not differentiable at [imath]x=0[/imath]?
The definition of a derivative is the slope of a function tangent to a point. It is also defined as [imath]\lim_{h\to0} \frac{f(x+h)-f(x)}{h}[/imath] If we apply this to [imath]f(x)= |x|[/imath], we get that it is [imath]\lim\limits_{h\to 0} \dfrac{|h|}{h}[/imath], which is undefined. However, if we look at the graph of [imath]|x|[/imath], we see that there can exist a tangent line at x=0, with slope 0. So why is the derivative undefined instead of 0? |
1110729 | How to prove that a ring is not flat over [imath]k[t,s][/imath]?
Let [imath]k[/imath] be a field, [imath]A=k[t,s][/imath], and [imath]C=A[z]/(tz-s)[/imath]. How can I prove, using the ideals [imath]tA[/imath] and [imath]sA[/imath], that [imath]C[/imath] is not flat over [imath]A[/imath]? (Liu, Algebraic Geometry and Arithmetic Curves, Exercise 2.6(c).) I know that if [imath]A[/imath] is a Dedekind domain then [imath]A[/imath]-module is flat if and only if it is not torsion-free over [imath]A[/imath]. But Dedekind domains are new structures for me so I'm not sure if [imath]k[t,s][/imath] is a Dedekind domain. Or do I have to show that not both of [imath]tA[/imath] and [imath]sA[/imath] can't be maximal or prime over [imath]A_{tA}[/imath] or [imath]A_{sA}[/imath]? | 110599 | Why isn't [imath]\mathbb{C}[x,y,z]/(xz-y)[/imath] a flat [imath]\mathbb{C}[x,y][/imath]-module
Why isn't [imath]M = \mathbb{C}[x,y,z]/(xz-y)[/imath] a flat [imath]R = \mathbb{C}[x,y][/imath]-module? The reason given on the book is "the surface defined by [imath]y-xz[/imath] doesn't lie flat on the [imath](x,y)[/imath]-plane". But I don't understand why this can be a reason. Since An [imath]R[/imath]-module [imath]M[/imath] is flat if and only if for any ideal [imath]S[/imath] of [imath]R[/imath], the map [imath]1_M \otimes i: M \otimes_RS \rightarrow M \otimes_RR[/imath] is injective. Here, [imath]i: S \rightarrow R[/imath] is the inclusion map. I tried to find an ideal [imath]S[/imath] in [imath]R[/imath] such that [imath]M \otimes_RS \rightarrow M \otimes_RR = M[/imath] is not injective, but I didn't succeed. Please give me some help. Thank you. |
1118420 | How do I evaluate [imath]\int \cot^2x[/imath]?
I have an integral with [imath]\frac{1}{\tan^2x}[/imath] needed to be evaluated. But instead of searching online for the antiderivative of [imath]\cot^2x[/imath], how would i find it from first principles? | 1073001 | How to evaluate [imath]\int \cot^2(x) \;\mathrm dx[/imath]?
How do you find the antiderivative of [imath]\cot^2x[/imath]? My steps to find it First [imath] \csc^2 x = \cot^2 x+ 1 [/imath] because of Pythagorean Identities, so [imath] \cot^2 x= \csc^2 x-1[/imath] so [imath] \int \cot^2 x\, \operatorname{d}x= \int \csc^2 x\, \operatorname{d}x - \int 1 \, \operatorname{d}x[/imath] simplying down to [imath]\int \cot^2 x = -\cot x- x + C[/imath] because I know the derivative of [imath]\cot x[/imath] is [imath]-\csc^2 x[/imath]. I want to know if this is the correct solution because I have many gaps in my logic and frequently make errors and I started learning calculus a couple weeks ago. If I got something wrong can you please tell me my mistakes. Thanks, Bot |
1108428 | Step-by-Step Solution for [imath]x^{1/x}=2^{1/2}[/imath]
I came across the equation [imath]x^{1/x}=2^{1/2}[/imath] where [imath]x\in\mathbb R[/imath]. One can immediately see that [imath]x=2[/imath] is a solution, but it is easy to miss that [imath]x=4[/imath] satisfies the equation as well. Verfiying that [imath]2,4[/imath] are solutions is not hard, but how would one go about formally solving this equation, i.e. how could one solve for [imath]x[/imath]? I am asking because if the equation was say [imath]x^{1/x}=2^{1/3}[/imath] then the two solutions would not be obvious. Basically, how to fill in the dots: [imath]x^{1/x}=2^{1/2} \iff \ldots\iff x=2 \text{ or } x=4[/imath] | 965836 | How to solve this equation [imath]x^{2}=2^{x}[/imath]?
How to solve this equation [imath]x^{2}=2^{x}[/imath] where [imath]x \in \mathbb{R}[/imath]. Por tentativa erro consegui descobri que [imath]2[/imath] é uma solução, mas não encontrei um método pra isso. Alguma sugestão?(*) (Translation: By trying different values I've found that [imath]2[/imath] is a solution, but I couldn't find any method to this though. Any suggestions? ) |
1119142 | Factorial formula problem
Prove that [imath](n-r)!(r!)[/imath] divides [imath] n! [/imath] i know its a factorial formula and it might be easy but i stuck .I tried induction to [imath]n[/imath] or analyzing the factorials but im missing something | 411363 | Dividing factorials is always integer
Is there a simple way to show that [imath]n!\over r!(n-r)![/imath] is always an integer? |
1119275 | Discrete math - Prove that a tree with n nodes must have exactly n - 1 edges?
I'm new in discrete math. Can someone prove simply that a tree with [imath]n[/imath] nodes must have exactly [imath]n - 1[/imath] edges. I have researched the solution but I haven't founded yet. I know of course, a tree with n nodes must have exactly n - 1 edges. But, I can't prove it. Thank you. induction on [imath]n[/imath] | 454639 | How many edges does an undirected tree with [imath]n[/imath] nodes have?
How many edges does an undirected tree with [imath]n[/imath] nodes have? |
1116367 | Series proof [imath]\sum_1^\infty|a_n|<\infty[/imath] then show that [imath]\sum_1^\infty{a_n^2}<\infty[/imath]
Iam stuck with this proof. There seems to be no property to help. If [imath]\sum_1^\infty|a_n|<\infty[/imath] then show that [imath]\sum_1^\infty{a_n^2}<\infty[/imath] and that the reverse isnt true. | 1282109 | [imath]\sum a_n[/imath] converges [imath]\implies\ \sum a_n^2[/imath] converges?
If [imath]\sum a_n[/imath] with [imath]a_n>0[/imath] is convergent, then is [imath]\sum {a_n}^2[/imath] always convergent? Either prove it or give a counter example. Im trying in this way, Suppose [imath]a_n \in [0,1] \ \forall\ n.\ [/imath] Then [imath]{a_n}^2\leq a_n\ \forall\ n.[/imath] Therefore by comparison test [imath]\sum {a_n}^2[/imath] converges. So If [imath]a_n[/imath] has certain restrictions then the result is true. what about the general case? How to proceed further? Hints will be greatly appreciated. |
234930 | If a measure is semifinite, then there are sets of arbitrarily large but finite measure
I am trying to solve following exercise from Folland, If [imath]\mu[/imath] is a semifinite measure and [imath]\mu(E) = \infty[/imath], for any [imath]C > 0[/imath], [imath]\exists[/imath] [imath]F \subset E[/imath] with [imath]C < \mu(F) < \infty[/imath]. It seems to follow from definition of semifinite measures, which you can find here, but I couldn't prove it. | 1786190 | Real Analysis, Folland Problem 1.3.14
If [imath]\mu[/imath] is semifinite measure and [imath]\mu(E) = \infty[/imath], for any [imath]C > 0[/imath] there exists [imath]F\subset E[/imath] with [imath]C < \mu(F) < \infty[/imath]. Attempted proof - Suppose [imath]E\in M[/imath] with [imath]\mu(E) = \infty[/imath] then there exists [imath]F\in M[/imath] with [imath]F\subset E[/imath] and [imath]0 < \mu(F) < \infty[/imath], [imath]\mu[/imath] is called semifinite. This is the definition from Folland. I am guessing that we have to set [imath]C[/imath] equal to something where the definition still holds true but I am not sure how to do this. Any suggestions is greatly appreciated. |
1119920 | Simple homomorphism of groups question
Show that a homomorphism of groups also has the property that [imath]f(a^{-1})=f(a)^{-1}[/imath] for all [imath]a \in G[/imath]. | 495009 | Let [imath]\phi:G_1\to G_2[/imath] be a group homomorphism. Show [imath]\phi(g^{-1})=(\phi(g))^{-1}[/imath].
Let [imath]G_1[/imath] and [imath]G_2[/imath] be groups and [imath]\phi:G_1\to G_2[/imath] be a group homomorphism. Show [imath]\phi(g^{-1})=(\phi(g))^{-1}[/imath] [imath]\forall g_1\in G_1[/imath]. So far I have: [imath]\phi(e)=e[/imath] by definition, where [imath]e[/imath] is the identity element. Then [imath]\phi(g_1)\phi(g_1^{-1})=\phi(g_1g_1^{-1})=\phi(e)=e[/imath] since [imath]\phi[/imath] is a group homomorphism. I know I can get [imath]\phi(g^{-1})=(\phi(g))^{-1}[/imath] [imath]\forall g_1\in G_1[/imath] from that, but I don't know HOW. Any help/hints would be welcomed. ^_^ |
568209 | Matrices which commute with all the matrices commuting with a given matrix
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix with entries from an arbitrary field [imath]F[/imath] and let [imath]C(A)[/imath] denote the set of all matrices which commute with [imath]A[/imath]. Is it true that [imath]C(C(A))= \{ \alpha_1 + \alpha_2 A + \cdots + \alpha_{n-1}A^{n-1}\mid\alpha_i \in F \}[/imath] ? This problem is a generalized version of Problems 6.3.13, 6.3.14 of Herstein's Topics in Algebra which I am unable to do. Herstein asks to prove that this holds for the cases [imath]n=2,3[/imath]. Of course, one way to go about it is by brute force calculation (which I have not tried), but I guess there is a more conceptual way to do this which I am unable to find. Clearly all the matrices of the type in R.H.S. are in [imath]C(C(A))[/imath] so the part I am unable to do is that these are the only matrices. I can see that [imath]C(A)[/imath] and [imath]C(C(A))[/imath] are themselves vector spaces but what to do further ? | 497806 | Matrices [imath]B[/imath] that commute with every matrix commuting with [imath]A[/imath]
There have been many questions in the vein of this one, but I can't find one that answers it specifically. Suppose [imath]A,B\in M_n(\mathbb C)[/imath] are two matrices such that, for any other matrix [imath]C\in M_n(\mathbb C)[/imath], [imath]AC=CA\implies BC=CB.[/imath] Prove that [imath]B=p(A)[/imath] for a polynomial [imath]p\in\mathbb C[t][/imath]. Nothing is assumed about [imath]A[/imath] being diagonalisable or having distinct eigenvalues or being invertible. Edit: after further work, this might not actually be true. If it's not, a counterexample would be great. |
1119836 | Logic - how to write [imath]\exists !x[/imath] without the [imath]\exists ![/imath] symbol
What is [imath]\exists ![/imath] equivalent to? I need to write [imath]\exists !x \,P(x)[/imath] without using the [imath]\exists ![/imath] symbol; thus, I am wondering what the [imath]\exists ![/imath] symbol is equivalent to. | 394609 | Write ‘There is exactly 1 person…’ without the uniqueness quantifier
During a lecture today the prof. posed the question of how we could write "There is exactly one person whom everybody loves." without using the uniqueness quantifier. The first part we wrote as a logical expression was "There is one person whom everybody loves.", ignoring the 'exactly one' part of the question initially. From this he wrote [imath]L(x,y): x[/imath] loves [imath]y[/imath]; domain for [imath]x[/imath] and [imath]y[/imath]: [imath]\{\text{people}\}[/imath] [imath]\exists x\forall y: L(y,x)[/imath] Which I understand to mean 'There is a person [imath]x[/imath] such that for all [imath]y[/imath], [imath]x[/imath] is loved by [imath]y[/imath]' AKA 'There is a person who is loved by everyone'. I get that part. The part I don't get is how the expression of 'exactly one'. [imath]\forall z(\forall y(L(y,z))\to x =z)[/imath] which then creates the joint expression [imath]\exists x\forall y(L(y,x))\land \forall z(\forall y(L(y,z))\to x=z)[/imath] I just can't seem to understand how [imath]\forall z(\forall y(L(y,z))\to x =z)[/imath] means exactly one here. I suppose you can take [imath]\forall z[/imath] here to mean 'for any given person', which means the [imath]\forall z[/imath] is considering every person in the world. This would translate the second expression block to something like, "For any given person [imath]z[/imath], if everybody loves [imath]z[/imath] then [imath]z[/imath] is the same person as [imath]x[/imath]". To me though [imath]\forall z[/imath] generally means for every element in the domain which I see as meaning every person in the world simultaneously, as it seems to for [imath]y[/imath]. Is that just plain wrong? How can I tell when [imath]\forall[/imath] means 'all' and when it means 'for any (one)'? In the previous English translation the only reason I was able to translate it (if it's even right) is because I already knew what the statement was suppose to mean. Is it just that [imath]\forall z[/imath] means that this statement could be true for any element, and if so what's the difference between [imath]\forall z[/imath] and [imath]\exists z[/imath]? Someone told me that [imath]\exists z[/imath] would be redundant because the expression says [imath]x=z[/imath] but how do I know that [imath]x[/imath] and [imath]z[/imath] are automatically the same person if [imath]\exists[/imath] is used for both? Sorry if this is a bit long with too many questions. I just wanted to try to make the cause of my confusion as clear as possible so you can help me figure this out. |
1120202 | Matrices that commute with all matrices
Let [imath]Z_n[/imath] be the set of all [imath]n \times n[/imath] matrices that commute with all [imath]n \times n [/imath] matrices. Show that [imath]Z_n = \{\lambda I_n \ | \ \lambda \in \mathbb R\}[/imath] ([imath]I_n[/imath] is the [imath]n \times n[/imath] identity matrix) I don't know how to use [imath]E_{ij}[/imath] (matrix with [imath]1[/imath] in [imath](i,j)[/imath] and [imath]0[/imath] elsewhere) and the elementary matrix [imath]P_{ij}[/imath] to prove this question. Can anyone explain it please? | 1476896 | Why is this linear transformation a scalar map?
Let [imath]T: V \rightarrow V[/imath] be a linear transformation such that [imath]T \circ S = S \circ T[/imath] for all [imath]S \in hom(V,V)[/imath]. Why is it that [imath]T = \lambda x[/imath] where [imath]\lambda \in Field[/imath] and [imath]x \in V[/imath]? |
1118455 | Which prime gaps are known to exist
It is easily proved that prime gaps can be arbitrarily large by constructing the sequence of composites [imath](n+1)! + 2, (n+1)! + 3, \dots, (n+1)! + (n+1)[/imath], which are divisible by [imath]2, \dots, n+1[/imath] respectively. It is also easily proved that prime gaps of size [imath]n[/imath] do not exist generally: if [imath]n > 1[/imath] is odd, [imath]p_{i+1} - p_i[/imath] is the difference of two odd numbers and is thus even, so [imath]p_{i+1} - p_i \not = n[/imath]. My question is if it is known, generally, whether for even [imath]n \in \mathbb N[/imath], there exists an [imath]i[/imath] such that [imath]p_{i+1} - p_i = n[/imath], and if so how it may be proved. | 359772 | Possible values of prime gaps
The nth prime gap is defined as [imath]p_{n+1} - p_n [/imath], [sequence A001223 in OEIX] (http://oeis.org/A001223). What values can occur as a prime gap? Clearly with the exception of [imath]1 = 3 - 2[/imath], all the prime gaps must be even. We also know that this sequence must contain infinitely large numbers, since there are no primes between [imath]n!+2[/imath] and [imath]n! + n[/imath]. Is it true that every even number occurs as a prime gap? |
1119950 | How many possible isomorphisms do we have between G and H?
Let [imath]G=(Z_4,+)[/imath] and let [imath]H=(U_5,*)[/imath] where [imath]U_5 = \{[1],[2],[3],[4] \}[/imath] . I know that [imath][1][/imath] and [imath][3][/imath] are both generators for [imath]G[/imath]. I also know that [imath][2][/imath] and [imath][3][/imath] are both generators for [imath]H[/imath]. In order to construct isomorphism we have to map a generator to a generator. And so, [imath]\phi([1]^x_4) = [2]^x_5[/imath] is an isomorphism from [imath]G \longrightarrow H[/imath]. Also [imath]\phi([1]^x_4) = [3]^x_5[/imath] is another isomorphism from [imath]G \longrightarrow H[/imath]. Can we do the same thing for the other generator of [imath]G[/imath]. And so we would have two more isomorphisms, Namely [imath]\phi([3]^x_4) = [2]^x_5[/imath] and [imath]\phi([3]^x_4) = [3]^x_5[/imath] is that true ? and if it is true, then does that mean that if we have [imath]n[/imath] generators for a group [imath]G[/imath] and [imath]m[/imath] generators of a group [imath]H[/imath] then i would have [imath]nm[/imath] isomorphisms in total ? Thanks | 1119900 | Isomorphisms between [imath](\mathbb Z_4,+)[/imath] and [imath](U_5,*)[/imath]
So I am asked to find all the isomorphisms between [imath]G = (\mathbb Z_4,+)[/imath] and [imath]H = (U_5,*)[/imath]. I solved it as follows: we will have two isomorphism corresponding to the two generators of [imath]U_5[/imath]. The first isomorphism : [imath]\phi([x]_4)[/imath] = [imath][2]_5^x[/imath] The second isomorphism is: [imath]\phi([x]_4)[/imath] = [imath][3]_5^x[/imath] Notice that [2] and [3] are the only generators of [imath]U_5[/imath] one can check it systematically or use the fact that [imath]a^m[/imath] is a generator iff gcd(m,4) = 1. So I just wanted a verification of my work I showed that they are bijective, well defined, and is a homomorphism. |
458795 | What's are all the prime elements in Gaussian integers [imath]\mathbb{Z}[i][/imath]
So far, I know if [imath]p[/imath] is a rational prime, then [imath](1)[/imath] if [imath]p\equiv 3\mod4[/imath], then [imath]p[/imath] is prime in [imath]\mathbb{Z}[i][/imath]. [imath](2)[/imath] If [imath]p\equiv1\mod4[/imath] then [imath]p=π_1 π_2[/imath] where [imath]π_1 [/imath] and [imath]π_2[/imath] are conjugate, Then [imath]π_1 [/imath] and [imath]π_2[/imath] are primes in [imath]\mathbb{Z}[i][/imath]. [imath](3)[/imath] [imath]2=(1+i)(1-i)[/imath], then [imath](1+i)[/imath]and[imath](1-i)[/imath] are primes in [imath]\mathbb{Z}[i][/imath]. What's are all the prime elements in Gaussian integers [imath]\mathbb{Z}[i][/imath]? For example, [imath]-3[/imath] are prime in [imath]\mathbb{Z}[i][/imath], but not in the above [imath]3[/imath] cases. | 1259236 | Show that if [imath]p \equiv 3[/imath] mod [imath]4[/imath] is a prime in [imath]\mathbb{Z}[/imath], then [imath]p[/imath] is a prime in [imath]\mathbb{Z}[i][/imath]
Show that if [imath]p \equiv 3[/imath] mod [imath]4[/imath] is a prime in [imath]\mathbb{Z}[/imath], then [imath]p[/imath] is a prime in [imath]\mathbb{Z}[i][/imath] I think I probably have to use something related to norm to solve this problem but I can't seem to figure out anything. Can someone suggest how I can solve this problem? Thanks |
1120862 | Hamiltonian cycle and Euler Cycle.
When [imath]G = K_n, n \ge 3[/imath] and [imath]n[/imath] is odd, then from the edges of the [imath]G[/imath] can be built edge-disjoint Hamiltonian cycles. Is it true? | 194247 | Hamiltonian Cycle Problem
At the moment I'm trying to prove the statement: [imath]K_n[/imath] is an edge disjoint union of Hamiltonian cycles when [imath]n[/imath] is odd. ([imath]K_n[/imath] is the complete graph with [imath]n[/imath] vertices) So far, I think I've come up with a proof. We know the total number of edges in [imath]K_n[/imath] is [imath]n(n-1)/2[/imath] (or [imath]n \choose 2[/imath]) and we can split our graph into individual Hamiltonian cycles of degree 2. We also know that for n vertices all having degree 2, there must consequently be [imath]n[/imath] edges. Thus we write [imath]n(n-1)/2 = n + n + ... n[/imath] (here I'm just splitting [imath]K_n[/imath]'s edges into some number of distinct Hamiltonian paths) and the deduction that [imath]n[/imath] must be odd follows easily. However, the assumption I made - that we can always split [imath]K_n[/imath] into Hamiltonian paths of degree 2 if [imath]K_n[/imath] can be written as a disjoint union described above - I'm having trouble proving. I've only relied on trying different values for [imath]n[/imath] trials and it hasn't faltered yet. So, I'm asking: If it is true, how do you prove that if [imath]K_n[/imath] can be split into distinct Hamiltonian cycles, it can be split in such a way that each Hamiltonian cycle is of degree 2? |
1120975 | Can someone help me prove that [imath]\tau(n)[/imath] is odd iff [imath]n[/imath] is a perfect square.
Can someone help me prove that [imath]\tau(n)[/imath] is odd if and only if [imath]n[/imath] is a perfect square. So basically I have to prove that [imath]\tau(n)[/imath] is odd iff [imath]n = k^2[/imath] for some integer [imath]k[/imath]. [imath]\tau(n)[/imath] is the function which gives the number of positive divisors of n, including n itself. | 906159 | A number is a perfect square if and only if it has odd number of positive divisors
I believe I have the solution to this problem but post it anyway to get feedback and alternate solutions/angles for it. For all [imath]n \in \mathrm {Z_+}[/imath] prove [imath]n[/imath] is a perfect square if and only if [imath]n[/imath] has odd # of positive divisors. Thought to use induction but the perfect squares don't increase by 1. [imath]\Rightarrow[/imath]: If [imath]n[/imath] is a perfect square it must consist of 1+ prime factors each to an even power. If [imath]n[/imath] consists of 1+ prime factors each to an even power it must have an odd # of positive divisors. [imath]\Leftarrow[/imath]: If [imath]n[/imath] has an odd number of positive divisors it must consist of 1+ prime factors each to an even power. If [imath]n[/imath] consists of 1+ prime factors each to an even power it must be a perfect square. Thanks. |
1121279 | How to find the equation of tangent line?
Q: find the equation of the tangent line to the graph of [imath]f[/imath] at the indicated point. Then verify your answer by sketching both the graph of [imath]f[/imath] and the tangent line. [PS: the point of tangency (x,y)] [imath]f(x)=x^2+1 ,\quad (2,5)[/imath] So I got the [imath]f'(x)=2x[/imath] But how can I find the tangent line? Thank you so much! | 496765 | Find the equation of the tangent line to [imath]y=x^4-4x^3-5x+7[/imath]
Find the equation of the tangent line to [imath]y=x^4-4x^3-5x+7[/imath] at the point (1,-1) I'm completely at a lost for this problem. Am i going to use [imath]\frac{f(x+h)-f(x)}{h}[/imath] formula? Please Help!!! |
1121459 | Show that: [imath]\lim_{n \to \infty}\prod_{k=1}^n \cos(k\sqrt{\frac{3}{n^3}} t) = e^{- \frac{t^2}{2}}[/imath]
Is it true that [imath]\lim_{n \to \infty}\prod_{k=1}^n \cos\left(k\sqrt{\frac{3}{n^3}} t\right) = e^{- \frac{t^2}{2}}[/imath] ? How to proceed ? | 216617 | Find the value of : [imath]\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)[/imath]
Find the limit (where a is a constant) [imath]\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)[/imath] I think the answer is [imath]1-a^2/6[/imath] |
1121641 | probability of X+Y which are two independent random variable & uniform distribution[0,1]
Two random variables X, Y are independent and both uniform-distributed in[0, 1]. How to calculate the probability density function Z=X+Y ? I tried below, [imath]f_X(x) = \begin{cases} \frac1{1-0} \\ 0 \end{cases} = \begin{cases} 1, & 0 \le x \le 1 \\ 0, & otherwise \end{cases} [/imath] Similarly, [imath] f_Y(y) = \begin{cases} 1, & 0 \le y \le 1 \\ 0, & otherwise \end{cases} [/imath] As [imath]f(x,y)= f_X(x)f_Y(y)[/imath], and [imath] f(z) = \int_{-\infty}^{+\infty} f(x, y)\, dx = \int_0^1 f(x, z-x)\, dx = ? [/imath] but have no idea how to continue to finish above evaluation. | 220201 | Sum of two uniform random variables
I am calculating the sum of two uniform random variables [imath]X[/imath] and [imath]Y[/imath], so that the sum is [imath]X+Y = Z[/imath]. Since the two are independent, their densities are [imath]f_X(x)=f_Y(x)=1[/imath] if [imath]0\leq x\leq1[/imath] and [imath]0[/imath] otherwise. The density of the sum becomes [imath]f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)dy=\int_0^1f_X(z-y)dy[/imath] by convolution. I am stuck at this stage. How do I proceed with my integral? I think a diagram make it easy but I dont know how to proceed. |
1121736 | Is there an obvious reason why [imath]4^n+n^4[/imath] cannot be prime for [imath]n\ge 2[/imath]?
I searched a prime of the form [imath]4^n+n^4[/imath] with [imath]n\ge 2[/imath] and did not find one with [imath]n\le 12\ 000[/imath]. If [imath]n[/imath] is even, then [imath]4^n+n^4[/imath] is even, so it cannot be prime. If [imath]n[/imath] is odd and not divisible by [imath]5[/imath] , then [imath]4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5[/imath]. So, [imath]n[/imath] must have the form [imath]10k+5[/imath]. For [imath]n=35[/imath] and [imath]n=55[/imath], the number [imath]4^n+n^4[/imath] splits into two primes with almost the same size. So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ? | 52837 | If [imath]m^4+4^n[/imath] is prime, then [imath]m=n=1[/imath] or [imath]m[/imath] is odd and [imath]n[/imath] even
I have been stuck on this one for months, really simple to state, really giving me trouble. Show that if [imath]m^4 + 4^n[/imath] is prime, [imath]m>0[/imath], [imath]n>0[/imath], then [imath]m[/imath] is odd and [imath]n[/imath] is even, except when [imath]m=n=1[/imath]. The case [imath]m=n=1[/imath] gives [imath]5[/imath], so that is why it is excluded. Clearly [imath]m[/imath] is odd, for if it was even, the number would be divisible by at least [imath]4[/imath], but I can't seem to get rid of the case where [imath]n[/imath] must be odd. The only thing I've managed to try is to write [imath]m[/imath] as an odd number, say [imath]m = 2k+1[/imath], and [imath]n = 2\ell + 1[/imath], then get [imath] m^4 + 4^n = (2k+1)^4 + 4^{2\ell+1} = 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4^{2 \ell + 1} [/imath] and this is congruent to [imath]0 \mod 5[/imath] unless [imath]k \equiv 2 \mod 5[/imath], and now it gives you an even more ugly polynomial if I write [imath]k = 5j+2[/imath], and it's not working. I've tried other inconclusive approaches, like trying to see if two numbers of some form generate numbers of the same form.. (for instance primes of the form 8k+1 and 8k+7 always generate primes of the same form when multiplied together, stuff like that). Any ideas? Even just ideas you haven't gave a thought... they're welcome! |
1121697 | Determinant of the inverse matrix
I'm seeking for a proof of the following: Let [imath]A[/imath] be an invertible matrix. Then the determinant of [imath]A^{-1}[/imath] equals: [imath]\left|A^{-1}\right|=|A|^{-1} [/imath] I don't know where to begin the proof. Any suggestions? | 661633 | Prove that the determinant of [imath] A^{-1} = \frac{1}{det(A)} [/imath]- Linear Algebra
If I have a single matrix A that is non-singular, how can I prove the determinant of its inverse = [imath]\frac{1}{\det(A)}[/imath]? Prove: [imath] \det(\mathbf{A^{-1}}) = \frac{1}{\mathbf{\det(A)}} [/imath] I know that [imath](A)(A^{-1}) = I[/imath], but I am not sure what to do with that knowledge. |
1122542 | If [imath]f_n \rightarrow f[/imath] in [imath]L^p[/imath] and [imath]g_n \rightarrow g[/imath] in [imath]L^q[/imath], where [imath]\frac{1}{p} + \frac{1}{q} = 1[/imath], show that [imath]f_n g_n \rightarrow fg[/imath] in [imath]L^1[/imath]
I know that this will have something to do with Holder's inequality but I am at a loss as to how the [imath]L^p[/imath] and [imath]L^q[/imath] convergence in [imath]f[/imath] and [imath]g[/imath] dictate the convergence in [imath]L^1[/imath]. Any help is appreciated. | 251817 | Sequence of a product of functions in [imath]L^p. L^q[/imath] with [imath]p,q[/imath] conjugate
The question is: if [imath]f_i[/imath] is a sequence of functions in [imath]L^p[/imath] converging to [imath]f[/imath] and [imath]g_i[/imath] a sequence in [imath]L^q[/imath] converging to [imath]g[/imath] show that [imath]f_ig_i[/imath] converges to [imath]fg[/imath] in [imath]L^1[/imath] for [imath]p,q[/imath] finite and [imath]\frac{1}{p}+\frac{1}{q}=1[/imath]. Does this result hold if [imath]p=1, q=\infty[/imath]? So I think I showed the first part. [imath]||f_ig_i-fg||_1=||f_ig_i-f_ig+f_ig-fg||_1\leq||f_ig_i-f_ig||_1+||f_ig-fg||_1 = ||f_i(g_i-g)||_1+||g(f_i-f)||_1\leq||f_i||_p||g_i-g||_q+||g||_q||f_i-f||_p[/imath] (by Holder's inequality) which goes to [imath]0[/imath] as [imath]i\rightarrow\infty[/imath]. However the second part of the question throws me off because I don't see why this proof doesn't work just as well for [imath]p=1,q=\infty[/imath]. Am I missing something? Does the second part also hold? |
1122904 | A finite subset of an ordered set contains an [imath]\inf[/imath] and [imath]\sup[/imath]
Let S be an ordered set. Let A ⊂ S be a nonempty finite subset. Then A is bounded. Furthermore, inf A exists and is in A and sup A exists and is in A. Hint: Use induction. How do I use induction to prove that the infinum and supremum exist? | 259893 | Every finite set contains its supremum: proof improvement.
Every finite subset of [imath]\mathbb R[/imath] contains its supremum (and its infimum) Proof Let [imath]A=\{a_1,...,a_n\}[/imath] be a finite subset of [imath]\mathbb{R}[/imath]. Since it is non-empty and it is bounded ([imath]\max A[/imath] is an upper bound), it has supremum, that is [imath]\exists \sup A[/imath] and by definition [imath]\forall a \in A \;\, a \leq \sup A[/imath]. Let's suppose that [imath]\sup A \not\in A[/imath] then, since [imath]\max A \in A[/imath] we have that [imath]\max A < \sup A[/imath]. But considering that [imath]\mathbb Q[/imath] is dense in [imath]\mathbb R[/imath] we can conclude that [imath]\exists r \in \mathbb Q[/imath] s.t. [imath]\max A < r < \sup A[/imath], but this is absurd since [imath]r[/imath] is an upper bound of [imath]A[/imath] and it is lower than the supremum. Necessarily, [imath]\sup A\in A[/imath]. Is there anything wrong? Is there any way to prove this without using density of [imath]\mathbb Q[/imath] or another property? Thanks in advance. |
532355 | Prove the limit below using definition
Prove the limit using [imath]\epsilon[/imath] and [imath]\delta[/imath] definition [imath]\lim_{(x,y)\to (1,1)} xy = 1[/imath] Thanks! | 214828 | Limits using epsilon delta definition [imath]f(x,y)=xy[/imath] for functions of two variables
Prove: using [imath]\epsilon[/imath]-[imath]\delta[/imath] definition, the limit of both [imath]f[/imath] and [imath]g[/imath] as [imath](x,y)\to (0,0)[/imath] is [imath]0[/imath]. [imath]f(x,y)=xy[/imath] [imath]g(x,y)=\frac{xy}{x^2 +y^2+1}[/imath] Also, for Q2 can I convert [imath]g(x,y)[/imath] to [imath]m(x,y)/n(x,y)=g(x,y)[/imath] using arithmetic of limits, then prove using [imath]\epsilon[/imath]-[imath]\delta[/imath] definition the limit of function [imath]m[/imath] and [imath]n[/imath] separately; then combine the two? Thanks :) I wonder if this is correct: [imath]|xy-0|<\epsilon[/imath] given [imath]|x-0|< \delta [/imath] and [imath]|y-0|< \delta [/imath] [imath]|xy-0|< |x-0||y-0|<\delta^2=\epsilon[/imath] therefore: [imath]\delta<\epsilon^{1/2}[/imath] |
465094 | Differentiability of the function [imath]f(z)=|z|^2[/imath].
I'm stuck on the following problem: The function [imath]f(z)=|z|^2[/imath] is: Differentiable only at the Origin Not differentiable anywhere I have to determine which of the aforementioned options is true. The answer key to the problem says option 2 is true whereas I think option 1 is correct. We see that: [imath] f(z)=|z|^2 \implies u(x,y)+iv(x,y)=x^2+y^2,\;\text{where}\:z=x+iy,\:\text{say}. [/imath] Then: [imath] u(x,y)=x^2+y^2,v(x,y)=0.\;\text{So at the Origin}\;u_x=u_y=v_x=v_y=0 [/imath] So, C-R equation is satisfied, and hence option 1 holds true. Am I going in the right direction? | 103938 | Is [imath]|z|^2[/imath] complex differentiable?
I think I am a bit confused about the definition of (complex) differentiability. Yes, I know that's stupid, but I am hoping that someone could clear it up for me. I know that the definition of (complex) differentiability is when [imath]\lim\limits_{h\to 0}{f(z+h)-f(z)\over h}[/imath] exists. So, is [imath]|z|^2[/imath] considered differentiable? What I think is it is only differentiable at [imath]z=0[/imath] since at any other point if we take [imath]f(z+h)-f(z)\over h[/imath] as [imath]h\to 0[/imath] along a contour line of [imath]|z|^2[/imath] then the limit is [imath]0[/imath] whereas if we take a path say perpendicular to the contour lines, the "gradient" wouldn't be [imath]0[/imath], right? But then if this is true then all complex functions that are "not flat" would not be differentiable, so I must be wrong. Could someone kindly explain to me what is going on? Sorry for my stupidity! |
1122713 | Find a vector equation and parametric equations for the line segment that joins [imath]P[/imath] to [imath]Q[/imath].
Find a vector equation and parametric equations for the line segment that joins [imath]P[/imath] to [imath]Q[/imath]. Here [imath]P(1,-1,7)[/imath] and [imath]Q(7,5,1)[/imath]. I have tried to find [imath]r(t)[/imath] by using the formula [imath]r(t)=p+t(p-q)[/imath] but it is wrong. I am making a stupid mistake. | 495904 | Vector equation and parametric equation for a line segment
Suppose that [imath]P = (1,1,7)[/imath] and [imath]Q = (8,6,1)[/imath]. Inside parenthesis are x-coordinate value, y-coordinate and z-coordinate. The question is to find vector and parametric equation for a line segment. Now, I used equation that goes like [imath](1-t)<1,1,7>+t<8,6,1>, where, 0\leq t\leq 1[/imath]. But, the answer says that I am wrong. So, I am wondering what is going on. Also, why do we write parametric equations? Are they really different from vector functions anyway? If vector function is [imath]<2t,3t,1>[/imath], according to my knowledge, parametric equations would be [imath]x(t)=2t, y(t)=3t, z(t)=1[/imath] .... |
521469 | Exercise on Great Common Divisor and Prime Number
Let [imath]p[/imath] be a prime number and let [imath]1 \leq n < p[/imath] be a non-negative integer number. Show that there exist [imath]x,y \in \mathbb{Z}[/imath] such that [imath]n x + p y = 1[/imath]. | 321061 | Proving that [imath] \gcd(a,b) = as + bt [/imath], i.e., [imath] \gcd [/imath] is a linear combination.
For any nonzero integers [imath] a [/imath] and [imath] b [/imath], there exist integers [imath] s [/imath] and [imath] t [/imath] such that [imath] \gcd(a,b) = as + bt [/imath]. Moreover, [imath] \gcd(a,b) [/imath] is the smallest positive integer of the form [imath] as + bt [/imath]. I know of one proof of this question, in which we consider [imath] S = \{ am + bn ~|~ \text{$ m,n \in \mathbb{Z} $ and $ am + bn > 0 $} \}, [/imath] but I didn’t manage to get the proof. Can someone please explain this theorem? |
1123568 | Prove that [imath]\mathbb{Z}[/imath] is a closed subset of [imath]\mathbb{R}[/imath]
Let [imath]\mathbb{Z}[/imath] and [imath]\mathbb{Q}[/imath] represent the integers and the rationals, respectively. (a) Prove that [imath]\mathbb{Z}[/imath] is a closed subset of [imath]\mathbb{R}[/imath]. I am stuck on this topology problem from my homework, if anyone would be of some help it would be greatly appreciated. | 715979 | Topological Property of the Set of Integers
Is the set of integers [imath] \mathbb{Z}[/imath] closed in [imath]\mathbb{R}[/imath] equipped with the usual topology? |
1123587 | Proof that the formula [imath]((p\to q)\land (q\to r)\land p)\to r[/imath] is a tautology
Write down the assumptions in a form of clauses and give a resolution proof that the formula is a tautology. [imath]((p\to q)\land (q\to r)\land p)\to r[/imath] I got information that i need to use here Resolution calculus...Someone can help me? | 1119010 | Can anyone help me with a solution?
Write down the assumptions in a form of clauses and give a resolution proof that the proposition [imath]\Big((p \rightarrow q) \land ( q \rightarrow r) \land p \Big) \rightarrow r[/imath] is a tautology. |
1123438 | If [imath]p^q - 1[/imath] is a prime, then [imath]p=2[/imath] and [imath]q[/imath] is a prime
I was working my way through some number theoretic proofs and being a newbie am stuck on this problem : If [imath]p[/imath] and [imath]q[/imath] are positive integers ([imath]\mathbb{Z}^+[/imath]) such that [imath]q \gt 1[/imath] and [imath](p^q - 1)[/imath] is a prime, then [imath]p[/imath] is [imath]2[/imath] and [imath]q[/imath] is a prime. My Question : I am unable to make any concrete progress . Even a decent hint would be acceptable so that I can build on that ... | 927678 | Foundational proof for Mersenne primes
I know how to prove that, if [imath]2^n-1[/imath] is prime and [imath]n>1[/imath], then [imath]n[/imath] is prime. But how do we prove that, if [imath]a^n-1[/imath] is prime and [imath]n>1[/imath], then [imath]a[/imath] must equal 2? |
1123899 | Is it possible to prove that [imath]\text{tr}(AB)=\text{tr}(BA)[/imath] without using matrices?
Is it possible to prove that [imath]\text{tr}(AB)=\text{tr}(BA)[/imath] without using matrices or having to choose a particular base ? Such a proof should probably use a non matricial definition of traces. One could say for example that [imath]\text{tr}(A)[/imath] is the linear coefficient of the characteristic polynomial [imath]\det(A-X\text{Id})[/imath]. In case the latter doesn't satisfy everyone, here is a definition of trace suggested in Halmos's Finite Dimensional Vector Spaces. Let [imath]E[/imath] be a [imath]n[/imath]-dimensional vector space over some field [imath]k[/imath]. We recall that the space [imath]W[/imath] of alternating [imath]n[/imath]-linear forms [imath]\omega: E^{n}\rightarrow k[/imath] is one dimensional over [imath]k[/imath]. Given an endomorphism [imath]f[/imath] of [imath]E[/imath] we can define the linear application \begin{align*} \bar{f}:\quad W &\rightarrow W \\ \omega(x_{1},x_{2},...) &\mapsto \sum_{i} \omega(x_{1},...,x_{i-1},f(x_{i}),x_{i+1},..) \end{align*} Since [imath]W[/imath] is one dimensional [imath]\bar{f}=\lambda Id[/imath]. We set [imath]\text{tr}(f)=\lambda[/imath]. The analog definition of the determinant easily yields [imath]\det(AB)=\det(A)\det(B)=\det(BA)[/imath], but this no longer works here. One could fix a base [imath]e_{1},e_{2},..,e_{n}[/imath] and do the calculations but this would kind of beat the point, since it is essentially the same as computing with matrices. | 311580 | Coordinate-free proof of [imath]\operatorname{Tr}(AB)=\operatorname{Tr}(BA)[/imath]?
I am searching for a short coordinate-free proof of [imath]\operatorname{Tr}(AB)=\operatorname{Tr}(BA)[/imath] for linear operators [imath]A[/imath], [imath]B[/imath] between finite dimensional vector spaces of the same dimension. The usual proof is to represent the operators as matrices and then use matrix multiplication. I want a coordinate-free proof. That is, one that does not make reference to an explicit matrix representation of the operator. I define trace as the sum of the eigenvalues of an operator. Ideally, the proof the should be shorter and require fewer preliminary lemmas than the one given in this blog post. I would be especially interested in a proof that generalizes to the trace class of operators on a Hilbert space. |
1123815 | How to apply modular division correctly?
As described on Wikipedia: [imath]\frac{a}{b} \bmod{n} = \left((a \bmod{n})(b^{-1} \bmod n)\right) \bmod n[/imath] When I apply this formula to the case [imath](1023/3) \bmod 7[/imath]: [imath]\begin{align*} (1023/3) \bmod 7 &= \left((1023 \bmod 7)((1/3) \bmod 7)\right) \bmod 7 \\ &= ( 1 \cdot (1/3)) \mod 7 \\ &= ( 1/3) \mod 7 \\ &= 1/3 \end{align*} [/imath] However, the real answer should be [imath](341) \bmod 7 = \mathbf{5}[/imath]. What am I missing? How do you find [imath](a/b) \bmod n[/imath] correctly? | 25390 | How to find the inverse modulo m?
For example: [imath]7x \equiv 1 \pmod{31} [/imath] In this example, the modular inverse of [imath]7[/imath] with respect to [imath]31[/imath] is [imath]9[/imath]. How can we find out that [imath]9[/imath]? What are the steps that I need to do? Update If I have a general modulo equation: [imath]5x + 1 \equiv 2 \pmod{6}[/imath] What is the fastest way to solve it? My initial thought was: [imath]5x + 1 \equiv 2 \pmod{6}[/imath] [imath]\Leftrightarrow 5x + 1 - 1\equiv 2 - 1 \pmod{6}[/imath] [imath]\Leftrightarrow 5x \equiv 1 \pmod{6}[/imath] Then solve for the inverse of [imath]5[/imath] modulo 6. Is it a right approach? Thanks, |
1123866 | How are standard basis of polynomials linearly independent?
Consider the set [imath]\{1,z,z^2,...z^m\}[/imath]. As this is the standard basis for a vector space of polynomials, the list should span the space and also be linearly independent. One problem I'm having though is considering the equation [imath]0 = a_{0}1 + a_{1}z + ... + a_{m}z^m[/imath] If I let z = 0, then all the polynomials with the exception of the first one are equal to 0, should I should be able to pick scalars, not all 0 that lead to a representation of the 0 vector in this basis that does not correspond to the trivial solution. So this would contradict the linear independence of the set. I'm not exactly sure where I'm getting confused. Is it because you are not supposed to let z be any particular value? I can see how the list would be linearly independent if the value of z was not fixed. | 1122344 | Linear dependence of [imath]\left\{x^{n}\,\colon\, n\in\mathbb{N}\right\}[/imath]
Consider the set [imath]S=\left\{x^{n}\,\colon\, n\in\mathbb{N}\right\}[/imath]. (Note that [imath]x\in\mathbb{R}[/imath]) Is this set linearly dependent? Well thinking about it we want to find some non-trivial values [imath]\lambda_{n}[/imath] such that [imath]\lambda_{1}x+\lambda_{2}x^{2} + \ldots + \lambda_{n}x^{n} + \ldots = 0.[/imath] In other words we want to find if [imath]\sum_{k=1}^{\infty}\lambda_{n}x^{n} = 0[/imath] for any [imath]\lambda_{n}\not=0[/imath]. However i'm not really sure how to proceed. |
361264 | In how many ways can [imath]2n[/imath] players be paired?
I have this simple looking question: "In tennis tournament, there are [imath]2n[/imath] participants. So, there must be [imath]n[/imath] pairs for the first round. In how many ways can such a pairing be arranged??". Now, shouldn't the answer be [imath]^{2n}C_2[/imath]? But the book says it is [imath]1 \cdot 3 \cdot 5 \cdots (2n - 1)[/imath]. Where am I going wrong? | 1971619 | How many ways to pair tennis players?
I've come across the following problem in my textbook I tried solving this by figuring out how many choices there are for each position within a round and came up with the following. [imath]{8 \choose 2} \cdot {6 \choose 2} \cdot {4 \choose 2} \cdot {2 \choose 2}[/imath] My idea was that that are [imath]{8 \choose 2}[/imath] ways to pick the first pair, [imath]{6 \choose 2}[/imath] ways to pick the second pair, etc. so multiplying them would yield all possible first round pairs, but this is not correct and I don't understand why. Question: Can someone please explain why the above reasoning is not correct |
1124223 | [imath]\omega_2[/imath] is not a countable union of countable sets
Without using axiom of choice, can we show that [imath]\omega_2[/imath] is not a countable union of countable sets? I know this cannot be done for [imath]\omega_1[/imath]. | 1044535 | [imath]\omega_2[/imath] is a not countable union of countable sets without AC
This is an exercise from Jech 3.13 I would like to show that [imath]\omega_2[/imath] is not a countable union of countable sets without AC. I'm given the following hint: I'm not sure how to define the mapping. We know that there are isomorphisms, [imath]f_n :S_n \to \alpha_n[/imath], so if we define [imath]F: \omega \times \alpha \to \omega_2[/imath] by [imath]F(n, \gamma) = f_n^{-1}(\gamma)[/imath] if [imath]\gamma \in \alpha_n[/imath] and [imath]F(n, \gamma) = 0[/imath] otherwise, does this work? Also, if it works, does it avoid the axiom of choice? |
1124707 | I need to identify [imath]G/H[/imath] upto isomorphism.
[imath]\varphi:G=(\mathbb{C}^*,.)\to (\mathbb{C}^*,.)[/imath] given by [imath]\varphi(z)=z^4[/imath] clearly Ker[imath]\varphi=H=\{z:z^4=1\}=\{1,-1,i,-i\}[/imath] I need to identify [imath]G/H[/imath] upto isomorphism. Thanks for helping. | 1121845 | Kernel and Image of a group homomorphism
let [imath]G[/imath] be a multiplicative group of non-zero complex analysis.consider the group homomorphism [imath]\phi:G\rightarrow G[/imath] defined by [imath]\phi(z)=z^4[/imath]. 1.Identify kernel of [imath]\phi=H[/imath]. 2.Identify [imath]G/H[/imath] My try: let [imath]z\in \ker \phi[/imath] then [imath]\phi(z)=1\implies z^4=1[/imath] let [imath]z=re^{i\theta}\implies r^4\cos 4\theta =1;r^4\sin 4\theta =0[/imath] then [imath]r=1[/imath] and [imath]\tan 4\theta=0\implies 4\theta=0\implies \theta=\frac{n\pi}{2}[/imath] Is it correct? I cant proceed in the 2nd problem. Any hints in this regard. |
1124907 | Linear Transformation from [imath]\mathbb R^n[/imath] to [imath]\mathbb R^m[/imath]: image of [imath]1[/imath] dimensional subspace has dimension [imath]1[/imath] or [imath]0[/imath]
I am struggling to comprehend the question below. Especially the meaning of 'the image of [imath]L[/imath] under [imath]F[/imath]'. Let [imath]F : \mathbb R^n \to \mathbb R^m[/imath] be a linear transformation. Prove that if [imath]L[/imath] is a [imath]1[/imath]-dimensional subspace of [imath]\mathbb R^n[/imath], then [imath]F(L)=\{y \in \mathbb R^m \mid y=F(x),x \in L\}[/imath] the image of [imath]L[/imath] under [imath]F[/imath] is a subspace of [imath]\mathbb R^m[/imath] with dimension [imath]1[/imath] or [imath]0[/imath]. Hence prove that if the kernel of [imath]F[/imath] is [imath]\{0\}[/imath], the image of [imath]L[/imath] is a [imath]1[/imath]- dimensional subspace of [imath]\mathbb R^m[/imath] (and so in this case, [imath]F[/imath] takes lines in [imath]\mathbb R^n[/imath] to lines in [imath]\mathbb R^m[/imath]). | 1120161 | Linear Transformations: Proving 1 dimensional subspace goes to 1 dimensional
I am having trouble understanding this whole question, and how to prove it. Let [imath]F:\mathbb{R}^n\to\mathbb{R}^m[/imath] be a linear transformation. Prove that if [imath]L[/imath] is a [imath]1[/imath]-dimensional subspace of [imath]\mathbb{R}^n[/imath], then [imath]F(L)=\{y\in\mathbb{R}^m: y=F(x)\text{ for some }x\in L\}[/imath] the image of [imath]L[/imath] under [imath]F[/imath] is a subspace of [imath]\mathbb{R}^m[/imath] with dimension [imath]1[/imath] or [imath]0[/imath]. Hence prove that if the kernel of [imath]F[/imath] is [imath]\{0\}[/imath], then the image of [imath]L[/imath] is a [imath]1[/imath]-dimensional subspace of [imath]\mathbb{R}^m[/imath] (and so, in this case, [imath]F[/imath] takes lines in [imath]\mathbb{R}^n[/imath] to lines in [imath]\mathbb{R}^m[/imath]). |
943327 | Show that [imath] \limsup x_n ≤ \limsup y_n[/imath] and [imath]\liminf x_n ≤ \liminf y_n[/imath]
Let [imath](x_n)[/imath] and [imath](y_n)[/imath] be bounded sequences such that [imath]x_n ≤ y_n[/imath] for all [imath]n \in \mathbb{N}[/imath]. Show that [imath]\limsup x_n ≤ \limsup y_n[/imath] and [imath]\liminf x_n ≤ \liminf y_n[/imath]. | 353642 | [imath]X_n\leq Y_n[/imath] implies [imath]\liminf X_n \leq \liminf Y_n[/imath] and [imath]\limsup X_n \leq \limsup Y_n[/imath]
Can anyone prove this question? I tried but I didn't get any I idea, so I hope someone can solve it. Let [imath]X_n\leq Y_n[/imath] for each [imath]n\in \Bbb N[/imath]. Show that [imath]\liminf X_n \leq \liminf Y_n[/imath] and [imath]\limsup X_n \leq \limsup Y_n[/imath]. Please prove this question - thanks. The definition I have: Let [imath]X_n[/imath] be a sequence in real number and let [imath]E=\{x\in \Bbb R^\sharp:(X_{n_k}) \rightarrow x \text{ for some subsequence }(X_{n_k})\text{ of }(X_n)\}[/imath] for all [imath]n \in \Bbb N[/imath] and [imath]k[/imath] from [imath]1[/imath] to [imath]\infty[/imath]. Then by definition [imath]\lim\sup X_n = \sup E[/imath] and [imath]\lim\inf X_n = \inf E[/imath]. |
402362 | Group equals union of three subgroups
Suppose [imath]G[/imath] is finite and [imath]G=H\cup K\cup L[/imath] for proper subgroups [imath]H,K,L[/imath]. Show that [imath]|G:H|=|G:K|=|G:L|=2[/imath]. What I did: so if some of [imath]H,K,L[/imath] is contained in another, then we have [imath]G[/imath] being a union of two proper subgroups, which is impossible due to another result. So none of [imath]H,K,L[/imath] is contained is each other. If some element [imath]h[/imath] belongs to only [imath]H[/imath] and [imath]k[/imath] belongs to only [imath]K[/imath], then its product [imath]hk[/imath] cannot be in [imath]H[/imath] or [imath]K[/imath], so must belong to only [imath]L[/imath]. | 1926819 | For subgroups [imath]H_{1,2,3} \lneq G[/imath] so that [imath]\bigcup\limits_{i=1}^3H_i = G[/imath] show: [imath][G : H_i]=2[/imath]
Let [imath]H_1,H_2,H_3 \lneq G[/imath] be groups such that [imath]\bigcup\limits_{i=1}^3H_i = G[/imath]. Show that [imath]\forall 1\leq i \leq 3\ [G : H_i]=2[/imath] For three different numbers [imath]1\leq i,j,k \leq 3[/imath], if for some [imath]H_i \leq H_j[/imath] then [imath]H_i \cup H_j \cup H_k = H_j\cup H_k = G[/imath]. The union of two subgroups is a subgroup only if one is a subgroup of the other, so we get [imath]H_k = G[/imath], while it's given that [imath]H_k \lneq G[/imath]. So no [imath]H_i[/imath] is a subgroup of another [imath]H_j[/imath]. If so, then for the union [imath](H_i \cup H_j) \cup H_k[/imath] to be a group, it's required that [imath]H_k=<(H_i \cup H_j)-(H_i\cap H_j)> (= <H_i \Delta H_j>)[/imath]. But then, if [imath]h_1 \in (H_i \cap H_j),\ h_2 \in (H_i \Delta H_j) \subset H_k[/imath], let's say [imath]h_2 \in H_j[/imath] (it belongs to either [imath]H_j[/imath] or [imath]H_i[/imath]), then also [imath]h_2^{-1}, h_2h_1 \in H_i \Delta H_j \subset H_k \Longrightarrow h_1 \in H_k[/imath]. So we got [imath]H_i,H_j \leq H_k[/imath], but we showed that this contradicts with G being their union. Is there a problem with the question, or did I miss something? |
1125728 | Closed vector space and a subspace of a vector space
What is a closed operation in a vector space? I don't see any difference between a closed operation in some vector Space R[imath]^n[/imath] and the open operation. What I mean by the closed operation is addition sign that looks like [imath]\oplus[/imath] Also, is a subspace an element of a vector space R[imath]^n[/imath] or no? Like I must first assume that there is a space R[imath]^n[/imath] which is closed first. Thank you for the help. (Thank you everybody) ありがとう 皆さん | 486476 | Direct Sum of vector subspaces
How is direct sum of two vector subspaces different from the sum of two vector subspaces i.e. how is [imath]X\oplus Y[/imath] different from [imath]X + Y[/imath], where [imath]X, Y[/imath] are subspaces. |
431148 | When a 0-1-matrix with exactly two 1’s on each column and on each row is non-degenerated? [1]
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix with entries in the set [imath]\{0,1\}[/imath] which has exactly two ones in each column and two ones in each row. Give necessary and sufficient conditions for the rank of [imath]A[/imath] to be [imath]n[/imath]. | 433272 | When a [imath]0-1[/imath]-matrix with exactly two [imath]1[/imath]’s on each column and on each row is non-degenerated? [2]
Let [imath]A[/imath] be a [imath]n \times n[/imath] matrix with entries on the set [imath]\{0,1\}[/imath], with exactly two ones on each column and two ones on each row. Give necessary and sufficient conditions for rank[imath](A)[/imath] to be [imath]n[/imath]. I found two solutions in these two articles: http://ge.tt/1x0JYek?c but I really don't understand them, I was wondering if there is an easier way to solve this problem, or if you could explain to me the main idea of this articles in a less advanced and easier way, and/or point out which exactly are the necessary and sufficient conditions. Thank you. (I know a matrix [imath]A[/imath] is non-signular if and only if rank[imath](A)=n[/imath], that's why I think these articles can help). |
1124959 | Zero variance Random variables with density
I found here the question: Can a random variable have a density function whose variance is [imath]0[/imath] ? I understood as a random variable which has a density. What is your opinion on what I understood and if that is the case on the answer that was given here | 786003 | Zero variance Random variables
I am a probability theory beginner. The expression for the variance of a random variable [imath]x[/imath] (of a random process is [imath]\sigma^2 = E(x^2) - (\mu_{x})^2[/imath] If [imath]E(x^2) = (\mu_{x})^2[/imath], then [imath]\sigma^2 = 0[/imath]. Can this happen ? Can a random variable have a density function whose variance (the second central moment alone) is [imath]0[/imath] (other than the dirac delta function). |
1125709 | Show that [imath]a^2 + b^2 + c^2 \geq ab + bc + ca[/imath] for all positive integers [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath]
Show that [imath]a^2 + b^2 + c^2 \geq ab + bc + ca[/imath] for all positive integers [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath]. I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether [imath]a[/imath],[imath]b[/imath],[imath]c[/imath] and odd/even or is there a more general solution? | 64868 | How to prove [imath]a^2 + b^2 + c^2 \ge ab + bc + ca[/imath]?
How can the following inequation be proven? [imath]a^2 + b^2 + c^2 \ge ab + bc + ca[/imath] |
1125258 | What is the derivative of this?
I have a function of the following form: [imath]J = \|W^TW-I\|_F^2[/imath] Where, [imath]W[/imath] is a matrix and [imath]F[/imath] is the Frobenius Norm. How can I find the derivative of [imath]\frac{\partial J}{\partial W}[/imath] ? | 1125499 | What is the derivative of this?
I have a function of the following form: [imath]J = \|W^TW-I\|_F^2[/imath] Where, [imath]W[/imath] is a matrix and [imath]F[/imath] is the Frobenius Norm. How can I find the derivative of [imath]\frac{\partial J}{\partial W}[/imath] ? |
1126549 | How to prove the folowing theorem in probablity?
Show that for any continuous random variable [imath]X[/imath] that takes only positive real values [imath]\int_{0}^{\infty}\text{Pr}(X\geq x)dx=\mu[/imath] where [imath]\mu[/imath] is the mean. | 64186 | Intuition behind using complementary CDF to compute expectation for nonnegative random variables
I've read the proof for why [imath]\int_0^\infty P(X >x)dx=E[X][/imath] for nonnegative random variables (located here) and understand its mechanics, but I'm having trouble understanding the intuition behind this formula or why it should be the case at all. Does anyone have any insight on this? I bet I'm missing something obvious. |
605311 | Find all the solutions of [imath]x^2+7=2^n[/imath].
Checking for some small natural numbers [imath]n[/imath], I found out that [imath]2^n-7[/imath] is a perfect square for [imath]n=3,4,5,7,15[/imath]. How can we find all of the numbers [imath]n[/imath] for which [imath]2^n-7[/imath] is a perfect square? What I tried: For [imath]n=1,2[/imath] there is no [imath]x[/imath] to satisfy the equality. For [imath]n=3[/imath] we have an answer [imath]x=1[/imath] so suppose [imath]n>3[/imath]. If [imath]n>3[/imath] then [imath]8|2^n[/imath] so [imath]2^n\equiv0\bmod8[/imath]. We also know that for every [imath]x[/imath], [imath]x^2\equiv0,1,4\bmod 8[/imath]. so [imath]x^2+7\equiv 7,0,3\bmod8[/imath] and we had [imath]x^2+7=2^n[/imath] so [imath]x^2+7\equiv0\bmod8[/imath] and this only holds when [imath]x\equiv 3,5\bmod8[/imath]. So the answers for [imath]n>3[/imath] should all be on the form of: [imath]x=8m\pm3[/imath]. We can also say: [imath]x^2+7=2^n[/imath] so [imath]x^2-1=2^n-8=8(2^{n-3}-1)[/imath] so [imath](x-1)(x+1)=8(2^{n-3}-1)[/imath]. I don't know where to go from here. I would appreciate any help. Edit: Apparently this equation is called the Ramanujan–Nagell equation. I found this pdf in another question but I can't understand it. I would appreciate a good elementary proof (if there is any). | 1126527 | When is [imath]2^n -7[/imath] a perfect square?
This came up while solving another ENT problem. I want to ask when is: [imath]2^n -7 \text{ where } n\geq 3[/imath] a perfect square? Specifically, I also wanted to know what would be the solutions when [imath]n[/imath] is odd? How should I solve this? I can check that [imath]n=3, 4, 7[/imath] are solutions but cannot find more. As it is of the form [imath]4k+1[/imath], it doesn't help as well. |
1127156 | Open set in [imath]\mathbb{R}[/imath] as countable union of open intervals and which version of Choice
In proving, every non-empty open set in [imath]\mathbb{R}[/imath] is union of a countable collection of disjoint open intervals in [imath]\mathbb{R}[/imath]. It seems to me this result is using some version of Choice(probably AC). Am I correct is it using some version of AC(maybe weaker) or not? | 98923 | Open Sets of [imath]\mathbb{R}^1[/imath] and axiom of choice
In the proof of 'Every open set in [imath]\mathbb{R}^1[/imath] is a countable union of disjoint open intervals', we need to pick one rational representative from each of the intervals hence establish the countability. This seems to depend on the axiom of choice. Thus I wonder does this property of [imath]\mathbb{R}[/imath] is equivalent to the axiom of choice, that is, is there any construction such that this property is not true once we abandon the axiom of choice? Thanks! |
1126641 | If [imath]R[/imath] is a commutative simple ring with identity , then is any matrix ring [imath]M_n(R)[/imath] also simple?
If [imath]R[/imath] is a commutative simple ring with identity , then is any matrix ring [imath]M_n(R)[/imath] over [imath]R[/imath] of matrices of size [imath]n[/imath] also simple ? | 654451 | for any ring [imath]A[/imath] the matrix ring [imath]M_n(A)[/imath] is simple if and only if [imath]A[/imath] is simple
Let integer [imath]n\geq 1[/imath]. I have obtained that for any field [imath]k[/imath], the matrix ring [imath]M_n(k)[/imath] is simple, i.e., [imath]M_n(k)[/imath] contains no nonzero proper two sided ideals. Now I want to prove that: for any ring [imath]A[/imath] (not necessarily commutative), the matrix ring [imath]M_n(A)[/imath] is simple if and only if [imath]A[/imath] is simple. How to prove? |
186832 | Is this a sufficient condition for a subset of a topological space to be closed?
Let [imath]X[/imath] be a topological space, and let [imath]\{U_i\}[/imath] be an open cover. If [imath]Y[/imath] is subset of [imath]X[/imath] such that [imath]Y\cap U_i[/imath] is closed in [imath]U_i[/imath] (for each [imath]i[/imath]), does this imply that [imath]Y[/imath] is closed in [imath]X[/imath]? | 3145 | Is whether a set is closed or not a local property?
If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically, Let [imath]X[/imath] is a topological space with a given open cover [imath]{ U_i }[/imath]. Suppose that [imath]Z \subset X[/imath] is a set such that [imath]Z \cap U_i[/imath] is closed in [imath]U_i[/imath] for all [imath]i[/imath]. Does it follow that [imath]Z[/imath] is closed in X? This is clearly true if there are finitely many [imath]{ U_i }[/imath]. At first thought, it seems unlikely to be true in the infinite case, but I'm having trouble coming up with a suitable counter-example. |
1126252 | Proof of Geometric mean [imath]\le[/imath] Arithmetic mean
This is a problem from Spivak's Calculus. If [imath]a_1, \dots, a_n \ge 0[/imath], then the "arithmetic mean" [imath] A_n= \frac{a_1+\cdots +a_n}{n}[/imath] and "geometric mean" [imath] G_n= \sqrt[n]{a_1 \dots a_n}[/imath] satisfy [imath]G_n \le A_n[/imath]. Suppose that [imath]a_1 \lt A_n[/imath]. Then some [imath]a_i \gt A_n[/imath]; for convenience, say [imath]a_2 \gt A_n[/imath]. Let [imath]\bar a_1=A_n[/imath] and let [imath]\bar a_2=a_1+a_2- \bar a_1[/imath]. Show that [imath]\bar a_1 \bar a_2 \ge a_1a_2[/imath].(This part is easily done). Why does repeating this process enough times eventually prove that [imath]G_n \le A_n[/imath]? (This is another place where it is a good exercise to provide a formal proof by induction, as well as an informal reason.) When does equality hold in the formula [imath]G_n \le A_n[/imath]? I don't see how I can generalize this process to achieve the result. I would appreciate any hint, suggestions, or solutions. | 27328 | Geometric mean never exceeds arithmetic mean
This was a mathematical induction question proposed in a textbook, and I've exhausted multiple approaches (proving RHS - LHS > 0, splitting the fraction, fractional exponents, etc.) The geometric mean of [imath]n[/imath] positive numbers [imath]a_1, a_2,\ldots,a_n[/imath] is [imath]\sqrt[n]{a_1a_2 \ldots a_n}[/imath] and their arithmetic mean is [imath]\frac{a_1+a_2+\ldots+a_n}{n}[/imath]. If [imath]a_1, a_2,\ldots,a_n[/imath] are [imath]n[/imath] positive real numbers, prove by induction that their geometric mean is always smaller than or equal to their arithmetic mean, i.e. [imath]\sqrt[n]{a_1a_2\ldots a_n} \leq \frac{a_1+a_2+\ldots+a_n}{n}[/imath] |
904231 | Differential of the inversion of Lie group
Let [imath]G[/imath] be a Lie group and [imath]\iota \colon G\to G[/imath] denote the inversion. If [imath]e[/imath] is the identity of [imath]G[/imath], prove that: [imath]\text d \iota _e = -\text {id} _{T_e G}.[/imath] I understand that the differential at [imath]e[/imath] should be an involution, since [imath]\iota \circ \iota = \text{id}_G[/imath] implies that [imath]\text d \iota _e \circ \text d \iota _e=\text d (\iota \circ \iota)_e=\text d(\text {id} _G)_e=\text {id} _{T_e G}.[/imath] However I don't know how to conclude the thesis from this. Any hint? | 209682 | Pushforward of Inverse Map around the identity?
Let [imath]G[/imath] be a Lie group and [imath]i:G \rightarrow G[/imath] denote the inversion map. (Notation: [imath]f_*[/imath] is the pushforward map [imath]F_*:T_pG \rightarrow T_{i(p)}G[/imath] which takes [imath](F_{*}X)(f)=X(f\circ F)[/imath] and [imath]X[/imath] is a tangent vector, [imath]X\in T_pG[/imath].) I wish to show that [imath]i_{*}:T_{e}G\rightarrow T_{e}G[/imath] is given by [imath]i_{*}(X)=-X[/imath] As a first step, it is trivial to prove that [imath]i_*[/imath] is an involution as [imath]\mbox{Id}_{*}=(i\circ i)_{*}=i_{*}\circ i_{*}[/imath] but I can't seem to make any further progress. Any help would be appreciated. |
1127869 | There does not exist a polynomial [imath]p(x)[/imath] with integer coefficients which gives a prime number [imath]\forall x\in \mathbb{Z}[/imath]
There does not exist a polynomial [imath]p(x)[/imath] with integer coefficients which gives a prime number [imath]\forall x\in \mathbb{Z}[/imath] My attempt: I defined a polynomial as [imath]p(x)=a_0(x-a_1)(x-a_2)\ldots(x-a_n)[/imath]. So, whenever [imath]x[/imath] takes an integer value, the polynomial cannot generate a prime number unless the polynomial is a linear one. But I have serious doubts about this solution, suppose any factor is of the form [imath](x^2+x+1)[/imath] which cannot be factorised into linear factors with integer coefficients. Please help. Thank you. | 2350061 | Can a polynomial return only prime numbers?
I recently learned about The polynomial [imath]p(n)=n^2-n+41[/imath], and how for [imath]1\le n\le40, n\in\mathbb Z[/imath], [imath]p(n)[/imath] is prime. I understand that primes are very difficult to find, so from that I can conclude that we haven't found a polynomial [imath]q(m)[/imath] where [imath]m\in\mathbb N[/imath] with no upper bound is prime, but is it proven that no such polynomial exists? |
1127921 | For which prime numbers [imath]p[/imath] does the congruence [imath]x^2 + x + 1 \equiv 0 \pmod{p}[/imath] have solutions?
For which prime numbers p does the congruence [imath]x^2 + x + 1 \equiv 0 \pmod{p}[/imath] have solutions? We've recently learnt about quadratic reciprocity in class, however I am not sure how to tackle this problem. I have tried starting with the [imath]b^2-4ac[/imath] (discriminant) but that hasn't really helped. | 1101612 | For what prime numbers [imath]p[/imath] is [imath]x^2+x+1[/imath] irreducible in [imath]\mathbb{F}_p[X][/imath]
I think it's enough to search for the prime numbers where [imath]x^2+x+1=0[/imath] is not solvable, but I am not sure where to start. Thank you |
1128198 | FP3 Integration help
[imath]I_{n}=\int x^n(1-x^2)^{\frac{1}{2}} dx[/imath] Show that [imath](n+2)I_{n}=(n-1)I_{n-2}-x^{n-1}(1-x^2)^{\frac{3}{2}}[/imath] So far I have done this: [imath]\int x^{n-1}(x)(1-x^2)^{\frac{1}{2}} dx[/imath] [imath]u=x^{n-1}[/imath] [imath]u'=(n-1)x^{n-2}[/imath] [imath]v'=x(1-x^2)^{\frac{1}{2}}[/imath] [imath]v=\frac{-1}{3}(1-x^2)^\frac{3}{2}[/imath] [imath]I_{n}=\frac{-x^{n-1}}{3}(1-x^2)^\frac{3}{2}+(\frac{n-1}{3})\int x^{n-2}(1-x^2)^\frac{3}{2}dx[/imath] [imath]3I_{n}=-x^{n-1}(1-x^2)^\frac{3}{2}+(n-1)\int x^{n-2}(1-x^2)^\frac{3}{2}dx[/imath] What do I do next? Have I done something wrong? | 1128115 | Reduction formulae
[imath]I_{n}=\int x^n(1-x^2)^{\frac{1}{2}} dx[/imath] Show that [imath](n+2)I_{n}=(n-1)I_{n-2}-x^{n-1}(1-x^2)^{\frac{3}{2}}[/imath] I can't seem to get this answer. Can someone please explain how to get to this? Thanks in advance. |
1128553 | How to find the value of [imath]\sum\limits_{n=0}^\infty r^n \sin(n\theta)[/imath]?
Question is to find the value of [imath]\sum_{n=0}^\infty r^n \sin(n\theta)\text{ for }r=0.5\text{ and }\theta=\pi/3[/imath] I don't know any tools which can solve this question. | 548150 | Show that [imath]\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)}[/imath]
Show that [imath]\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)},[/imath] where [imath]0\leq r <1[/imath]. Using this, prove that [imath]\sum_{n=0}^\infty r^n \cos(n\theta)[/imath] and [imath]\sum_{n=0}^\infty r^n \sin(n\theta)[/imath] are convergent. |
1128903 | Can a real function [imath]f[/imath] on [imath][0,1]\cap\mathbb{Q}[/imath] be differentiable?
can a real function [imath]f[/imath] on [imath][0,1]\cap\mathbb{Q}[/imath] be differentiable? if so, and if the derivative of [imath]f[/imath] is zero, then is [imath]f[/imath] is a constant function? | 802534 | If [imath]f'(x) = 0[/imath] for all [imath]x \in \mathbb{Q}[/imath], is [imath]f[/imath] constant?
Let [imath]f[/imath] be a differentiable function on [imath]\mathbb{R}[/imath] such that [imath]f'(x)=0,\forall x\in \mathbb Q.[/imath] Prove or disprove that [imath]f(x)=c[/imath] for some constant [imath]c[/imath]. I've heard this problem is true, but I'm not sure. Can you prove it or provide a counterexample? Auxiliary question: I wonder the converse of it, if [imath]f′=0[/imath] for all irrational, can we say that [imath]f[/imath] is constant? |
736709 | Proving [imath]\liminf_{n\to\infty}(-a_n)=-\limsup_{n\to\infty}(a_n)[/imath]
Prove: [imath]\displaystyle\liminf_{n\to\infty}(-a_n)=-\limsup_{n\to\infty}(a_n)[/imath] My general idea was: if [imath]a[/imath] is partial limit (PL) of [imath]a_n[/imath], then [imath]-a[/imath] is a PL of [imath]-a_n[/imath] so it follows that [imath]s[/imath] is the maximal PL iff [imath]-s[/imath] is the minimal PL. But how do you show it rigorously ? | 334114 | Prove that [imath]\liminf x_n = -\limsup (-x_n)[/imath]
How can we prove that [imath]\liminf x_n = -\limsup (-x_n)[/imath]? The definitions we are using are [imath]\limsup x_n = \lim\limits_{n\to\infty} \sup\{x_k; k\ge n\}[/imath] [imath]\liminf x_n = \lim\limits_{n\to\infty} \inf\{x_k; k\ge n\}[/imath] |
1129381 | Injectivity and localisation in Rings
Let [imath]A, B[/imath] be commutative rings with identity elements and let [imath]\mathfrak{p} \subseteq B[/imath] be a prime ideal. Let [imath]\varphi: A \to B[/imath] be an injective ring homomorphism. I want to show that the induced homomorphism [imath]\varphi_{\mathfrak{p}} : A_{\varphi^{-1}(\mathfrak{p})} \to B_{\mathfrak{p}}[/imath] is also injective. Moreover: Why is the induced morphism of schemes [imath]f: \text{Spec}(B) \to \text{Spec}(A)[/imath] dominant? The motivation of the first question: To show this is sufficient for showing that the induced morphism of sheaves [imath]\mathcal{O}_{\text{Spec}(A)} \to f_{*}\mathcal{O}_{\text{Spec}(B)}[/imath] is injective. | 223895 | Injectivity of Homomorphism in Localization
Let [imath]\alpha:A\to B[/imath] be a ring homomorphism, [imath]Q\subset B[/imath] a prime ideal, [imath]P=\alpha^{-1}(Q)\subset A[/imath] a prime ideal. Consider the natural map [imath]\alpha_Q:A_P\to B_Q[/imath] defined by [imath]\alpha_Q(a/b)=\alpha(a)/\alpha(b)[/imath]. Suppose that [imath]\alpha[/imath] is injective. Then is [imath]\alpha_Q[/imath] always injective? I think so, but I'm clearly being too dense to prove it! My argument goes as follows. Let [imath]\alpha(a)/\alpha(b)=0[/imath]. Then [imath]\exists c \in B\setminus Q[/imath] s.t. [imath]c\alpha(a)=0[/imath]. If [imath]B[/imath] is a domain we are done. If not we must exhibit some [imath]d\in A\setminus P[/imath] s.t. [imath]da=0[/imath]. Obviously this is true if [imath]c =\alpha(d)[/imath]. But I don't see how I have any information to prove this! Am I wrong and this is actually false? If so could someone show me the trivial counterexample I must be missing? Many thanks! |
1129649 | if [imath]f:X\to Y[/imath] is a surjective, then there is an injective function [imath]g:Y\to X[/imath]
I'm trying to prove if [imath]f:X\to Y[/imath] is a surjective, then there is an injective function [imath]g:Y\to X[/imath]. I know this intuitively, since [imath]f[/imath] is surjective, for every [imath]y[/imath] in [imath]Y[/imath] we need just to choice an element [imath]x\in f^{-1}(y)[/imath] to be the image of [imath]y[/imath], i.e., [imath]g(y)=x[/imath], am I right? I think if [imath]Y[/imath] is infinite we have to use axiom of choice. Thanks | 581980 | What does the Axiom of Choice have to do with right inversibility?
I have encountered an exercise that asks to prove that, these two statements are equivalent: every surjective function has a right inverse. Axiom of choice. Definition: Given a function [imath]f[/imath], we say that [imath]f[/imath] has a right inverse if there exists [imath]g[/imath] such that [imath]f \circ g[/imath] is the identity function. I must admit, I can't see the connection... Any help? Thanks! Shir |
1129641 | Is it possible (or meaningful) to do analysis or calculus with rational numbers only?
Irrationals and transcendentals are interesting and useful numbers, to be sure, but I think the importance of rationals can be forgotten when we get caught up in the mystique of other numbers. They seem, in some sense, to be the only numbers with which we can do completely precise computations. To this end, I've been wondering if there is a way to restrict calculus to rational numbers and still obtain similar results. Immediately there are problems with the exponential and trigonometric functions (since [imath]e[/imath] and [imath]\pi[/imath] are certainly not rational). But theories of rational trigonometry exist to handle the latter point. Anyway, is this something people have looked into before? My apologies if my question doesn't make sense. | 387234 | How far can one get in analysis without leaving [imath]\mathbb{Q}[/imath]?
Suppose you're trying to teach analysis to a stubborn algebraist who refuses to acknowledge the existence of any characteristic [imath]0[/imath] field other than [imath]\mathbb{Q}[/imath]. How ugly are things going to get for him? The algebraist argues that the real numbers are a silly construction because any real number can be approximated to arbitrarily high precision by the rational numbers - i.e., given any real number [imath]r[/imath] and any [imath]\epsilon>0[/imath], the set [imath]\left\{x\in\mathbb{Q}:\left|x-r\right|<\epsilon\right\}[/imath] is nonempty, thus sating the mad gods of algebra. As @J.M. and @75064 pointed out to me in chat, we do start having some topology problems, for example that [imath]f(x)=x^2[/imath] and [imath]g(x)=2[/imath] are nonintersecting functions in [imath]\mathbb{Q}[/imath]. They do, however, come arbitrarily close to intersecting, i.e. given any [imath]\epsilon>0[/imath] there exist rational solutions to [imath]\left|2-x^2\right|<\epsilon[/imath]. The algebraist doesn't find this totally unsatisfying. Where is this guy really going to start running into trouble? Are there definitions in analysis which simply can't be reasonably formulated without leaving the rational numbers? Which concepts would be particularly difficult to understand without the rest of the reals? |
1077227 | Prove [imath]X\times Y[/imath] is an equivalence relation
(Relation between two sets) If [imath]X[/imath] and [imath]Y[/imath] are sets, a relation between [imath]X[/imath] and [imath]Y[/imath] is a subset [imath]R \subset X \times Y.[/imath] For a relation [imath]R \subset X\times Y[/imath] and [imath]a \in X[/imath] and [imath]b \in Y[/imath] if [imath](x,y) \in R,[/imath] we write [imath]xRy[/imath] and if [imath](x,y)\notin R,[/imath] we write [imath]x/R/y.[/imath] (let /R/ be a slash across the R similar to what I have with [imath]\notin[/imath].) My question: Prove that [imath]X \times Y[/imath] is an equivalence relation. We have that there is a relation [imath]R[/imath] on [imath]X[/imath] and another relation [imath]R'[/imath] on [imath]Y[/imath] and that both are equivalence relations. The goal is to prove that [imath]R \times R'[/imath] is an equivalence relation. My thoughts: I am trying to teach this to myself. Some friend of mine was talking about this and I did not know too much about these types of question. I decided to learn this by myself and saw this question. How would we do this? Can someone please help me with this? I would like to see how this proof is done. This is what Hugh Denoncourt was saying in his comment. | 1077251 | If [imath]X,Y[/imath] are equivalence relations, so is [imath]X \times Y[/imath]
If [imath]X,Y[/imath] are reflexive, symmetric, and transitive, then [imath]X \times Y[/imath] is an equivalence relation where [imath]{(a,b):a\in X, b\in Y}[/imath]. I am trying to self learn these topics. I do know what an equivalence relation is. I wanted to know how would someone prove that [imath]X\times Y[/imath] be an equivalence relation? |
1115482 | Let [imath]A, B[/imath] be sets. Show that [imath]\mathcal P(A ∩ B) = \mathcal P(A) ∩ \mathcal P(B)[/imath].
Let [imath]A, B[/imath] be sets. Show that [imath]\mathcal P(A \cap B) = \mathcal P(A) \cap \mathcal P(B)[/imath]. I understand what this question is asking. The power set of an intersection equals the intersection of two power sets. I just have no idea how to prove it. | 1110845 | Does [imath]\wp(A \cap B) = \wp(A) \cap \wp(B)[/imath] hold? How to prove it?
I'm currently working on some discrete mathematics work and I've encountered a question I'm not sure how to answer exactly. Precisely, I'm trying to prove that two power, intersected sets statements are equal to each other and my understanding of how to do that doesn't seem to be enough. The statements in question go as follows: Determine whether, for any sets [imath]A[/imath] and [imath]B[/imath], through proof, it is true that [imath]\wp(A \cap B) = \wp(A) \cap \wp(B)[/imath] where [imath]\wp[/imath] denotes a power set. Should I make some form of example set to better understand how these statements are equivalent? Or even a venn diagram? Any help is appreciated. |
462977 | linear trasformations [imath]A\colon f\to \phi*f[/imath]
Let [imath]\phi[/imath] be a misurable function in [0,1] such that the linear trasformation [imath]A\colon f\to \phi*f[/imath] maps [imath]L^2[0,1][/imath] in [imath]L^2[0,1][/imath].Prove that [imath]\phi\in L^{\infty}[0,1][/imath]. Calculate t e norm of the operator A. When A is surjectiv,invertible,the inverse of A is continous? | 293489 | If [imath]\psi h[/imath] is in [imath]L^2[/imath] for all [imath]h\in L^2[/imath], must [imath]\psi[/imath] be essentially bounded?
If [imath]\psi:[a,b]\to\mathbb C[/imath] is a (measurable) function such that [imath]\psi h[/imath] is in [imath]L^2[a,b][/imath] for all [imath]h\in L^2[a,b][/imath], then must [imath]\psi[/imath] be essentially bounded? The converse direction is clear: If [imath]\psi[/imath] is essentially bounded, then [imath]\|\psi h\|_2\leq \|\psi\|_\infty\|h\|_2[/imath] for all [imath]h\in L^2[/imath], and thus the multiplication operator [imath]M_\psi:h\mapsto \psi h[/imath] is defined and bounded. The question here is whether, without prior assumptions on [imath]\psi[/imath], simply having [imath]M_\psi(L^2)\subseteq L^2[/imath] implies that [imath]\psi[/imath] is bounded. |
1047080 | Prove [imath]1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}[/imath] by Induction
The Question Prove [imath]1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}[/imath] where [imath]n\ge2[/imath] and [imath]n[/imath] is an integer by Induction My Work Basis Step: 1 + [imath]\frac{1}{4} = \frac{5}{4}[/imath] [imath]2-\frac{1}{2} = \frac{3}{2}[/imath] [imath]\frac{5}{4}<\frac{3}{2}[/imath] Inductive Hypothesis: [imath]1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}[/imath] Induction Step We must show [imath]1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2} < 2 - \frac{1}{(k+1)}[/imath] [imath]1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}[/imath] [imath]2 - \frac{1}{k} + \frac{1}{(k+1)^2} = 2 - \frac{(k^2+2k+1)+k}{k(k+1)^2} = 2-\frac{k^2 + 3k + 1}{k^3 +2k^2+k}[/imath] My Problem I can't seem to complete the proof. Any hints on how to change my approach for success? | 718872 | Help with proof using induction: [imath]1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}[/imath]
I am having trouble with the following proof: For every positive integer [imath]n[/imath]: [imath]1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}[/imath] My work: I have tried to add [imath]\frac{1}{(k+1)^2}[/imath] to [imath]2-\frac{1}{k}[/imath] in the inductive step and reduce it down to [imath]2-\frac{1}{k+1}[/imath] but cannot do so. I am beginning to think that my entire approach is wrong. |
1130223 | if R is a commutative ring in which all the prime ideals are finitely generated then R is Noetherian
Prove that if [imath]R[/imath] is a commutative ring in which all the prime ideals are finitely generated, then [imath]R[/imath] is Noetherian. Here is what I been told to do: Suppose that [imath]R[/imath] is not Noetherian, and use Zorn’s lemma to obtain a maximal element [imath]I[/imath] in the collection of all ideals of [imath]R[/imath] that are not finitely generated. Then use the following proposition: Let [imath]I[/imath] be an ideal of a commutative ring [imath]R[/imath], and let [imath]r ∈ R[/imath]. If the ideals [imath]I +rR[/imath] and {[imath]s ∈ R : sr ∈ I[/imath]} are finitely generated, then [imath]I[/imath] is a finitely generated ideal. Can anyone help please. thanks a lot. | 146884 | An ideal that is maximal among non-finitely generated ideals is prime.
I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over. Let [imath]R[/imath] be a commutative ring with identity and let [imath]U[/imath] be an ideal that is maximal among non-finitely generated ideals of [imath]R[/imath]. I wish to show that [imath]U[/imath] is a prime ideal. Assume that [imath]U[/imath] is not prime. Let [imath]x, y\not\in U[/imath] be such that [imath]xy\in U[/imath]. [imath]U[/imath] is contained in a maximal ideal [imath]M[/imath] and [imath]xy\in M[/imath], so either [imath]x[/imath] or [imath]y[/imath] is in [imath]M[/imath]; assume [imath]x\in M[/imath]. The condition [imath]U\subset M[/imath] then implies that there is a ring homomorphism [imath]\varphi: R/M\to R/U[/imath] Since [imath]R/M[/imath] is a field, [imath]\varphi[/imath] is injective. Hence, [imath]\varphi(x)\in U[/imath]. This is a contradiction, so [imath]U[/imath] must be prime. The thing that worries me is that I never explicitly used the hypothesis that [imath]U[/imath] was not finitely generated or the result that [imath]M[/imath] must be finitely generated. |
1130534 | [imath]K[x,y]/\langle x^2-y^3\rangle \cong K[t^2,t^3][/imath]
I'm stuck with this (should be easy) computation. I started by considering the most natural map [imath]K[x,y] \to K[t^2,t^3][/imath] which is the one that sends [imath]x \mapsto t^3[/imath] and [imath]y \mapsto t^2[/imath], and then extend by linearity. the map is easily seen to be surjective and factors through the quotient. The point is that I cannot find a way to show that the induced map is injective.. If I try to write down a general polynomial in two variables which is sent to zero I lose myself in the computations. Are there some other methods to show such claim? | 622095 | Verifying that the ideal [imath](x^3-y^2)[/imath] is prime
How to prove that the ideal [imath]I=(x^3-y^2)[/imath] in [imath]k[x,y][/imath] is prime? I have constructed a map from [imath]k[x,y][/imath] to [imath]k[t][/imath], which maps [imath]x[/imath] to [imath]t^2[/imath], and [imath]y[/imath] to [imath]t^3[/imath]. Then, I want to show that the kernel is exactly [imath]I[/imath]. But the difficulty is that how to prove the kernel is contained in [imath]I[/imath]. Could you help me? Thanks a lot! |
1130033 | If f and g are irreducible polynomials in K[X,Y ] that are not associates, show that the zero set Z(f,g) is either empty or finite
If [imath]f[/imath] and [imath]g[/imath] are irreducible polynomials in [imath]K[X,Y][/imath] that are not associates, show that the zero set [imath]Z(f,g)[/imath] is either empty or finite. Here is what I been told to do: If [imath](f,g)≠K[X,Y][/imath], show [imath](f,g)[/imath] contains a nonzero polynomial in [imath]K[X][/imath] and similarly a nonzero polynomial in [imath]K[Y][/imath]. and what i should do before that is let [imath]R=K[X][/imath] and [imath]F=K(X)[/imath], and apply Gauss’s Lemma to show [imath]f[/imath] and [imath]g[/imath] are relatively prime in [imath]F[Y][/imath]. Can any one help please, thanks a lot. | 384740 | Irreducible polynomials and affine variety
Let [imath]k[/imath] be any field, and let [imath]f,g\in k[x,y][/imath] be two irreducible polynomials such that [imath]g[/imath] is not divisible by [imath]f[/imath]. Prove that [imath]V(f,g)\subseteq A_k^2[/imath] is finite. |
1131307 | Show that f is not differentiable at (0,0)
Let [imath]f:\mathbb{R}^2\to\mathbb{R}[/imath] be defined by [imath]f(x,y)=\sqrt{|xy|}[/imath]. Show that [imath]f[/imath] is not differentiable at [imath](0,0)[/imath]. If you could start me out on how to show this, that would help a lot. | 182255 | Showing that a function is not differentiable
I want to show that [imath]f(x,y) = \sqrt{|xy|}[/imath] is not differentiable at [imath]0[/imath]. So my idea is to show that [imath]g(x,y) = |xy|[/imath] is not differentiable, and then argue that if [imath]f[/imath] were differentiable, then so would [imath]g[/imath] which is the composition of differentiable functions [imath]\cdot^2[/imath] and [imath]g[/imath]. But I'm stuck as to how to do this. In the one variable case, to show that [imath]q(x) = |x|[/imath] is not differentiable, I can calculate the limit [imath]\frac{|x + h| - |x|}{h}[/imath] as [imath]h\to 0^+[/imath] and [imath]h\to 0^-[/imath], show that the two one-sided limits are distinct, and conclude that the limit [imath]\lim_{h\to 0}\frac{|x + h| - |x|}{h}[/imath] does not exist. The reason this is easier is that I do not have to have in mind the derivative of the function [imath]q[/imath] in order to calculate it. But in the case of [imath]g(x,y) = |xy|[/imath], to show that [imath]g[/imath] is not differentiable at [imath]0[/imath], I would need to show that there does not exist a linear transformation [imath]\lambda:\mathbb{R}^{2}\to\mathbb{R}[/imath] such that [imath]\lim_{(h,k)\to (0,0)}\frac{\left||hk| - \lambda(h,k)\right|}{|(h,k)|} = 0[/imath] I thought of assuming that I had such a [imath]\lambda[/imath], and letting [imath](h,k)\to (0,0)[/imath] along both [imath](\sqrt{t},\sqrt{t})[/imath] and [imath](-\sqrt{t},-\sqrt{t})[/imath] as [imath]t\to 0^{+}[/imath], but this didn't seem to go anywhere constructive. |
1058736 | Prove f is uniformly continuous
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be such that [imath]|f '(x)|\leq B[/imath] for some [imath]B\in\mathbb{R}[/imath]. Prove that [imath]f[/imath] is uniformly continuous. Not exactly sure how to prove this. | 255652 | Questions about [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] with bounded derivative
I came across a problem that says: Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] be a function. If [imath]|f'|[/imath] is bounded, then which of the following option(s) is/are true? (a) The function [imath]f[/imath] is bounded. (b) The limit [imath]\lim_{x\to\infty}f(x)[/imath] exists. (c) The function [imath]f[/imath] is uniformly continuous. (d) The set [imath]\{x \mid f(x)=0\}[/imath] is compact. I am stuck on this problem. Please help. Thanks in advance for your time. |
1131540 | Rewriting Factorial Expressions
If I have an equation, say, [imath] y = x! +1 [/imath] Is it possible to rewrite the equation in terms of [imath]x[/imath]? For example, a simple algebraic equation might be [imath]y = x^2[/imath] --- and to rewrite it in terms of [imath]x[/imath], one might say [imath]x = \sqrt{y}[/imath]. I don't mind if the function isn't continuous, but I am curious as to whether or not this is possible... Thank you very much! | 18362 | Is there an Inverse Gamma [imath]\Gamma^{-1} (z) [/imath] function?
Since [imath]\Gamma[/imath] is not one to one over the complex domain, Is it possible to define some principal values ( analogues to Principal Roots for the Root function ) so we can have a [imath]\Gamma^{-1} (z)[/imath] (inverse [imath]\Gamma[/imath] function)? |
1131796 | Proof that inverse of a matrix is unique
If B and C are both inverses of the matrix A,then B=C. Can't i prove it in following way ? Proof: AB=BA=I and AC=CA=I,then BA=CA=I By postmultiplication [imath]\Rightarrow (BA)(A^{-1})=(CA)(A^{-1})=(I)(A^{-1})\Rightarrow B=C=A^{-1}[/imath], or by premultiplication [imath]AB=AC=I\Rightarrow (A^{-1})(AB)=(A^{-1})(AC)=(A^{-1})(I)\Rightarrow B=C=A^{-1}[/imath]. | 884470 | Proof that the inverse of a square matrix is unique
From my textbook ... if a 2×2 matrix [imath]A[/imath] is invertible then its inverse is unique. I wonder, how can one prove this? Also can one extend this proof to larger square matrices of order [imath]n[/imath]? Thanks |
1131125 | [imath]x \in \mathbb R^d[/imath], [imath]B[/imath](closed)[imath] \subseteq \mathbb R^d[/imath]. Where [imath]\mathbb R^d[/imath] is an inner product space. Proof help
[imath]x \in \mathbb R^d[/imath], [imath]B[/imath](closed)[imath] \subseteq \mathbb R^d[/imath]. Where [imath]\mathbb R^d[/imath] is an inner product space. Show that there exists a point [imath]b_0 \in B[/imath] such that [imath]d(x,B) = \|x - b_0\|[/imath] My attempt: [imath]d(x,B) \le \|x - b\|[/imath] for some [imath]b \in B[/imath] because [imath]d(x,B) = \inf\{\|x-b\|\}[/imath] For a point [imath]b_0 \in B[/imath](closed), does this imply that we can find a point [imath]b_0[/imath] such that it is equal to [imath]\inf\{\|x-b\|\}[/imath]? I feel like I can draw a picture to show this but I can't figure out the right logic to prove it. Any help is greatly appreciated. | 1130872 | Let [imath]B\subset\mathbb{R}^d[/imath] be closed and [imath]z\in\mathbb{R}^d[/imath], show that there is [imath]b_0\in B[/imath] such that [imath]\delta(z,B)=\|z-b_0\|[/imath].
If there is a closed subset [imath]B[/imath] of [imath]\mathbb{R}^d[/imath] and [imath]z\in \mathbb{R}^d[/imath] then show there is a point [imath]b_0\in B[/imath] so that we have [imath]\delta(z,B) = \|z-b_0\|.[/imath] I'm not sure how to begin this exercise and what theorems are available to help me solve it |
1130905 | Show that the series [imath]\sum_{n=1}^\infty \frac{a_n}{S_n}[/imath] diverges
Given that the series [imath]\sum_{n=1}^\infty a_n[/imath] of positive terms diverges and that [imath]S_n = a_1 + a_2 + ... + a_n[/imath] prove that [imath]\sum_{n=1}^{\infty} \frac{a_n}{S_n}[/imath] also diverges. | 808964 | Divergence of [imath]\sum\limits_{k=1}^n \frac{a_n}{S_n}[/imath] where [imath]S_n = \sum\limits_{k=1}^n a_k[/imath], when [imath]S_n[/imath] diverges
Let [imath]a_n[/imath] be a positive sequence such that [imath]S_n = \sum\limits_{k=1}^n a_k[/imath] diverges. I'm trying to prove [imath]\sum\limits_{k=1}^n \frac{a_n}{S_n}[/imath] diverges. I tried summation by parts, limit comparison and Stolz theorem in many different combinations and am still stuck with nothing. |
1132423 | integral domain as an intersection of localizations
I am stuck on trying to prove that [imath]\bigcap_{m \text{ maximal}}A_m = A[/imath] where [imath]A[/imath] is an integral domain (commutative, ring with unity). I would appreciate any hint! Thanks! | 630752 | An integral domain [imath]A[/imath] is exactly the intersection of the localisations of [imath]A[/imath] at each maximal ideal
This result appears to be ubiquitous as an algebra exercise. How do you prove this result? Let [imath]A[/imath] be an integral domain with field of fractions [imath]K[/imath], and let [imath]A_{\mathfrak{m}}[/imath] denote the localisation of [imath]A[/imath] at a maximal ideal [imath]\mathfrak{m}[/imath] considered as a subring of [imath]K[/imath]. Prove that [imath]A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,[/imath] where the intersection is taken over all maximal ideals [imath]\mathfrak{m}[/imath] of [imath]A[/imath]. |
1132540 | Monotonicity and limit of sequence
For each [imath]c \in [0;2][/imath] I have to examine monotonicity and limit of the sequence [imath]a_1 = c[/imath], [imath]a_{n+1} = 1 + \frac{(a_n - 1)^2}{17}[/imath]. I only solved equation [imath]g = 1 + \frac{(g-1)^2}{17}[/imath] and [imath]g = 1[/imath] or [imath]g = 18[/imath]. Can you help me with the rest? | 1132342 | Determining a limit of parametrized, recursively defined sequence [imath]a_{n+1}=1+\frac{(a_n-1)^2}{17}[/imath]
For every [imath]c\in [0;2][/imath] determine whether the sequence [imath]\{a_n\}_{n\geq 1}[/imath] which is defined as follows: [imath]a_1=c[/imath], [imath]a_{n+1}=1+\frac{(a_n-1)^2}{17}[/imath] for [imath]n\geq 1[/imath] is monotonic for sufficiently large [imath]n[/imath], and determine whether its limit exits and if it exists, give its value. I have no idea what to do with this problem. I was able to see that [imath]a_{n+1}-a_n[/imath] is a quadratic function and I also found ot that limit, if exists, is equal to either [imath]1[/imath] or [imath]18[/imath] (that's beacause if [imath]a_n[/imath] id convergent to [imath]g[/imath] then every subsequence is also convergent to [imath]g[/imath]). So how to determine the limit for every [imath]c\in [0;2][/imath]? |
1132806 | Proving [imath]b^{33} = e.[/imath]
Let [imath](G, *)[/imath] be a group and [imath]a[/imath], [imath]b[/imath] belongs to [imath]G[/imath]. Suppose that [imath]a^2 = e[/imath] and [imath]a*b^4*a = b^7[/imath]. Prove that [imath]b^{33} = e[/imath]. I dont know how to proceed. | 337626 | Please help me, Group Theory. Prove [imath]b^{33}=e[/imath].
Let [imath]G[/imath] be a group and [imath]a,b \in G[/imath]. Prove that if [imath]a^{2}=e[/imath] and [imath]ab^{4}a=b^{7}[/imath], then [imath]b^{33}=e[/imath], where [imath]e[/imath] is the identity of a group [imath]G[/imath]. |
1131947 | Find the value of k, (if any), for which the system below has unique, infinite or no solution.
The system of equations are: [imath]\begin{cases}x+y+kz = 1\\x+ky+z=1\\kx+y+z=1\\ \end{cases}[/imath] I am looking to finding values of [imath]k[/imath], for which this system has either no solutions, infinite many solutions or a unique solution (if any). I've had a look around the web for some help, and know that this question has been asked, but in a way that I have not yet learnt (something to do with determinants I think). However, I have only been working with reducing the matrix to REF/RREF and thus I'm trying to find rows, such as [imath]( 0 \ 0 \ 0 \mid 2)[/imath] and [imath]( 0 \ 0 \ 0 \mid 0)[/imath] etc to show if we have no solutions or infinite many solutions, for instance. So far, I've got the following: [imath]\left[\begin{array}{ccc|c}1&1&k&1\\1&k&1&1\\k&1&1&1\end{array}\right][/imath] Then, $R_2[imath]\mapsto[/imath]R_2-R_1$ [imath]\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\k&1&1&1\end{array}\right][/imath] Then, [imath]R_2[/imath] [imath]\leftrightarrow[/imath] [imath]R_3[/imath] [imath]\left[\begin{array}{ccc|c}1&1&k&1\\k&1&1&1\\0&k-1&1-k&0\end{array}\right][/imath] This is where I got stuck, and I'm not sure how to continue. Thank you in advance, and I'm sorry if this question has been asked and solved in the same way as presented above. Cheers. | 930252 | System of Linear Equations - how many solutions?
For which real values of t does the following system of linear equations: [imath] \left\{ \begin{array}{c} tx_1 + x_2 + x_3 = 1 \\ x_1 + tx_2 + x_3 = 1 \\ x_1 + x_2 + tx_3 = 1 \end{array} \right. [/imath] Have: a) a unique solution? b) infinitely many solutions? c) no solutions? I haven't done linear algebra in almost a year, so I'm really rusty and could use some pushes in the right direction. |
1132774 | Absolutely convergent series and convergent sequence
I have to examine whether if the series [imath]\sum_{n=1}^{\infty}{a_n}[/imath] is absolutely convergent then the sequence [imath]K_n = \frac{1}{ln(n^2 + 1)}\sum_{k=1}^{n}{a_k \frac{3k^3 - 2k}{7 - k^3}\sin(k)}[/imath] is convergent. Can you help me? | 1131957 | Do absolute convergence of [imath]a_n[/imath] implies convergence of [imath]K_n=\frac{1}{\ln(n^2+1)}\sum_{k=1}^{+\infty}a_k\frac{3k^3-2k}{7-k^3}\sin k[/imath]?
The problem asks us to decide whether the following statement is true: Let [imath]\{a_n\}_{n\geq1}[/imath] be any absolutely convergent sequence. Does that imply that the sequence: [imath]K_n=\frac{1}{\ln(n^2+1)}\sum_{k=1}^{n}a_k\frac{3k^3-2k}{7-k^3}\sin k[/imath] is convergent? The term [imath]\frac{1}{\ln(n^2+1)}[/imath] goes to [imath]0[/imath] so if we showed that the series is convergent we would finish this problem. The thing is it doesn't look like a convergent series - and even if it is I have no idea how to show it - this [imath]\sin[/imath] function inside means I can't use any convergence test (at least I think so). So how to reason about [imath]K_n[/imath]? UPDATE: by absolute convergence od [imath]a_n[/imath] I of course mean that [imath]\sum |a_n|[/imath] is convergent |
1097067 | Describe explicitly all inner products on [imath]\mathbb{R}[/imath] and [imath]\mathbb{C}[/imath]
I know this is an elementary question, however I am really lost as to where to start. Since both [imath]\mathbb{R}[/imath] and [imath]\mathbb{C}[/imath] are finite-dimensional I think the inner product will be completely determined by the basis [imath]\{1\}[/imath]. I am not sure where to go from here. | 1132826 | Inner Product on [imath]\mathbb{R}[/imath] and on [imath]\mathbb{C}[/imath]
This is a question of the book Linear Algebra of Kenneth Hoffman. Describe explicitly all inner products on [imath]\mathbb{R}[/imath] and on [imath]\mathbb{C}[/imath]. I think that if [imath]<,>[/imath] is one inner product on [imath]\mathbb{R}[/imath] (or [imath]\mathbb{C}[/imath]) then [imath]< ,>=K[,][/imath], where [imath][,][/imath] is the standard inner product. Is that correct? |
1132984 | Deducing a [imath]\cos (kx)[/imath] summation from the [imath]e^{ikx}[/imath] summation
I'm trying to solve So far I've done the first part, evaluating the summation ; where a is just n. I'm not sure where to go from here or what it even means deduce the second summation. I understand that the summation of simply [imath]\sum\limits_{k=0}^n\cos(kx)[/imath] is derived from looking at the real part of [imath]\sum_{k=0}^n{\rm e}^{ikx} [/imath]I'm guessing they want me to see how the second summation in the question is defined by 'playing around' with the original? | 1154920 | complex variables, finding the identity
[imath]\sin(\theta)+ \sin (2\theta)+\ldots+\sin(n\theta).[/imath] Find the identity? Set [imath]z=e^{i\theta}[/imath]. Can someone point me in the right direction as to where to start/go from. So far I have [imath]\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}.[/imath] And I am stuck as to where to go from here. |
1132960 | Does the series Converge or Diverge
[imath]s_n=\sum_{k=1}^{n}\cos\left(\frac{\pi}{2}k\right)\frac{k}{k+1000}\frac{1}{\sqrt{k}}, \ \ \ \ \ \ n=1,2,...[/imath] | 1128804 | Series convergence or divergence how to test
I have the following series defined. [imath]\displaystyle\sum_{k=1}^{n} \cos \left( {\frac{\pi}{2}} k \right) \frac{k}{k+1000} \frac{1}{\sqrt{k}}[/imath] where [imath]n = 1,2...[/imath] How to test whether this series converges or diverges ? |
1133571 | Proof of the limit using only elementary techniques
Today I saw this limit and I was baffled by it. [imath]e=\lim_{ n\to \infty} \sqrt[n]{\operatorname{lcm}(1,2,3,4,\ldots,n)}[/imath] Is there an elementary proof of the result? | 1111334 | Reason for LCM of all numbers from 1 .. n equals roughly [imath]e^n[/imath]
I computed the LCM for all natural numbers from 1 up to a limit [imath]n[/imath] and plotted the result over [imath]n[/imath]. Due to the fast-raising numbers, I plotted the logarithm of the result and was surprised to find a (more or less) identity curve ([imath]x=y[/imath]). In other words, [imath]LCM(1, 2, 3, ..., n)[/imath] appears to be roughly the value [imath]e^n[/imath]. Is a there a simple explanation on why this is so? [imath]LCM(a, b, c, …)[/imath] shall be defined as the least common multiple of all arguments [imath]a, b, c, …[/imath] ln(LCM(1, 2, 3, …, n)) over [imath]n[/imath]"> |
1133449 | Number of integer partitions of a natural number
Consider [imath]S \in \mathbb{N}[/imath] where [imath]\mathbb{N}[/imath] is the set of natural numbers. I want to obtain [imath]S[/imath] by summing [imath]k[/imath] natural numbers strictly positive. I want to find how many possibilities I have to obtain [imath]S[/imath] by summing [imath]k[/imath] numbers, considering that: 1) order does not matter, e.g. [imath]2+1+3[/imath] is the same as [imath]3+1+2[/imath] 2) repetitions are allowed | 54635 | Integer partition with fixed number of summands but without order
For a fixed [imath]n[/imath] and [imath]M[/imath], I am interested in the number of unordered non-negative integer solutions to [imath]\sum_{i = 1}^n a_i = M[/imath] Or, in other words, I am interested in the number of solutions with distinct numbers. For [imath]n = 2[/imath] and [imath]M = 5[/imath], I would consider solutions [imath](1,4)[/imath] and [imath](4,1)[/imath] equivalent, and choose the solution with [imath]a_1 \ge a_2 \ge ... \ge a_n \ge 0[/imath] as the representative of the class of equivalent solutions. I know how to obtain the number of total, ordered, solutions with the "stars and bars" method. But unfortunately, I cannot just divide the result by [imath]n![/imath] since that would only work if all the [imath]a_i[/imath] are distinct. |
1132869 | Differential Equation of the Form [imath]\frac{dy}{dx}=\sin(x+y)[/imath]
I have been attempting to solve the above differential equation for some time now, and I remain stuck on one step. After substituting [imath]u=x+y[/imath], separating the variables, and integrating both sides, I am left with [imath]\frac{\sin(u)-1}{\cos(u)}=x+c[/imath] I have to solve for [imath]y[/imath], and thus, for [imath]u[/imath], but I cannot think of any identity which helps me here. Any help would be appreciated. I followed the step in the solution, and set [imath]\frac{\tan(\frac{u}{2})-1}{\tan(\frac{u}{2})+1}=x+c[/imath] Then, I used the substitution [imath]z=\frac{1}{1+\tan(\frac{u}{2})}[/imath], which led to [imath]1-2z=c+x[/imath], which in turn, simplified to [imath]z=\frac{1}{2}-\frac{c}{2}-\frac{x}{2}[/imath]. Then, substituting back: [imath]u=-2\arctan\left(1-\frac{1}{\frac{1}{2}-\frac{c}{2}-\frac{x}{2}} \right)[/imath] Thus, for all [imath]n \in \mathbb Z[/imath], the solution of the differential equation is [imath]y=-x-2\arctan\left(\frac{c+x+1}{c+x-1}\right)+2\pi n[/imath] Is my process correct, and do I require the [imath]2\pi n[/imath] generalization in my final expression? Edit: The solution can also be expressed in indefinite form as [imath]\tan(x+y)-\sec(x+y)=x+c[/imath] | 751910 | Ordinary differential equation [imath]y'(t)=\sin(f(t,y))[/imath]
One whose solution never makes me happy is the following: [imath]y'(t)=\sin(y+t)\text{.}[/imath] I would start by substituting [imath]z(t)=y(t)+t[/imath] to get an ODE in [imath]z(t)[/imath], but then I'm not sure about how to substitute back my solution to check if it's correct or not[imath]\dots[/imath] |
1134819 | How to prove this? [imath] \lim_{x \to 0}\frac{e^x-1}{x}=1 [/imath]
Any idea how do I prove the following? [imath] \lim_{x \to 0}\frac{e^x-1}{x}=1 [/imath] Thanks | 77348 | How to prove that [imath]\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a[/imath]
In order to find the derivative of a exponential function, on its general form [imath]a^x[/imath] by the definition, I used limits. [imath]\begin{align*} \frac{d}{dx} a^x & = \lim_{h \to 0} \left [ \frac{a^{x+h}-a^x}{h} \right ]\\ \\ & =\lim_{h \to 0} \left [ \frac{a^x \cdot a^h-a^x}{h} \right ] \\ \\ &=\lim_{h \to 0} \left [ \frac{a^x \cdot (a^h-1)}{h} \right ] \\ \\ &=a^x \cdot \lim_{h \to 0} \left [\frac {a^h-1}{h} \right ] \end{align*}[/imath] I know that this last limit is equal to [imath]\ln(a)[/imath] but how can I prove it by using basic Algebra and Exponential and Logarithms properties? Thanks |
1134750 | Rings in which every maximal ideal is finitely generated
Suppose that [imath]R[/imath] is a commutative ring with unity in which every maximal ideal is finitely generated. Then is [imath]R[/imath] Noetherian? | 183199 | Commutative non Noetherian rings in which all maximal ideals are finitely generated
In commutative rings we have the following Theorem. [imath]R[/imath] is Noetherian if and only if each prime ideal of [imath]R[/imath] is finitely generated. From this Theorem I am looking for commutative rings [imath]R[/imath] in which every maximal ideal is finitely generated but [imath]R[/imath] is non Noetherian. Question: Is there a straightforward example of a commutative ring [imath]R[/imath] so that each maximal ideal is finitely generated, but [imath]R[/imath] is non Noetherian? Thank You |
1135287 | Stratonovich integral of Wienere process
I need an help with the following exercise. Let [imath](W_t)_{t\geq 0}[/imath] a Wiener process on [imath](\Omega, \mathcal E, \mathbb P)[/imath] and let [imath]I=[0,T][/imath] be an interval. We want to prove that the Stratonovich integral [imath]\int_0^T W_tdW_t\to W^2_T/2 \qquad \text{in } L^2(\Omega).[/imath] Denote with [imath]\Pi[/imath] a partition of the interval [imath]I[/imath], i.e. [imath]\Pi:=\{t_0=0,\dots, t_n=T\}[/imath]. Denote also [imath]\tau_i=(t_{i+1}+t_i) /2[/imath]. We want to prove that [imath]\lim_{n\to \infty} \sum_{i=0}^{n-1}W_{\tau_i}(W_{t_{i+1}}-W_{t_{i}})=W^2_T/2,[/imath] in the sense of [imath]L^2(\Omega)[/imath] norm, i.e. [imath]\mathbb E((\sum_{i=0}^{n-1}W_{\tau_i}(W_{t_{i+1}}-W_{t_{i}})-W^2_T/2)^2)\to 0.[/imath] And now? I don't know how to continue. | 861125 | Evaluating Stratonovich integral from definition
[imath]\bf 3.9.[/imath] Suppose [imath]f\in\mathcal V(0,T)[/imath] and that [imath]t\to f(t,\omega)[/imath] is continuous for a.a. [imath]\omega[/imath]. Then we have shown that [imath]\int\limits_0^T f(t,\omega)dB_t(\omega)=\lim_{\Delta t_j\to0}\sum_jf(t_j,\omega)\Delta B_j\qquad\text{ in }\, L^2(P).[/imath] Similarly we define the Stratonovich integral of [imath]f[/imath] by [imath]\int\limits_0^Tf(t,\omega)\circ dB_t(\omega)=\lim_{\Delta t_j\to0}\sum_j f(t_j^*,\omega)\Delta B_j,\quad\text{ where }\; t_j^*=\tfrac12(t_j+t_{j+1}),[/imath] whenever the limit exists in [imath]L^2(P)[/imath]. In general these integrals are different. For example, compute [imath]\int\limits_0^T B_t\circ dB_t[/imath] and compare with example [imath]3.1.9.[/imath] I am struggling to evaluate the integral [imath]\displaystyle \int^{T}_{0} B_t \circ dB_t [/imath] from definition. So far I have that [imath]\begin{align} \displaystyle \sum B_{\frac{t_j+t_{j+1}}{2}}(B_{t_{j+1}}-B_{t_{j}}) &= \\ \sum [ B_{\frac{t_j+t_{j+1}}{2}}(B_{t_{j+1}}-B_{\frac{t_j+t_{j+1}}{2}})+B_{t_{j}}(B_{\frac{t_j+t_{j+1}}{2}}-B_{t_j})]+ \sum (B_{\frac{t_j+t_{j+1}}{2}}-B_{t_j})^2 \end{align}[/imath] I can see that the first term goes to [imath]\int^{T}_{0} B_t dB_t[/imath] in [imath]L^2(P)[/imath]. Not sure where to go next nor with the general direction of where this solution is going. |
1135064 | Prove that [imath]4^n > n^4[/imath]by induction for [imath]n\ge 5[/imath]
That is a simple question and I can't start a simple desenveloment. Just [imath]k=5[/imath] we have [imath]4^5 = 1024 > 5^4 = 625[/imath] for [imath]k+1[/imath]: [imath]4^{k+1} > (k+1)^4\Rightarrow 4^k > (k+1)^4/4[/imath] And How can i proceed after that? Best, | 934089 | Proving [imath]4^n > n^4[/imath] holds for [imath]n\geq 5[/imath] via induction.
I know that it holds for [imath]n=5[/imath], so the first step is done. For the second step, my IH is: [imath]4^n > n^4[/imath], and I must show that [imath]4^{n+1} > (n+1)^4[/imath]. I did as follows: [imath]4^{n+1} = 4*4^n > 4n^4[/imath], and if I show that [imath]4n^4 > (n+1)^4[/imath] holds for [imath]n\geq 5[/imath] I'm done, so I expand the right hand side and arrive to: [imath]4n^4 > n^4 + 4n^3 + 6n^2 + 4n + 1[/imath] And this holds iff: [imath]3n^4 - 4n^3 - 6n^2 - 4n - 1 >0[/imath] And here I'm stuck, I don't really know how to show that the given inequality holds. Any hints? |
1135297 | Let [imath]G[/imath] be a group and let [imath]\Phi\colon G \to G[/imath] be an isomorphism. Define [imath]H = \{ a \in G\ |\ \Phi(a) = a^{-1} \}[/imath]
It asks to prove that if [imath]H[/imath] is a subgroup of [imath]G[/imath], then [imath]G[/imath] is abelian. Solution: I showed that for every [imath]a[/imath] and [imath]b[/imath] in [imath]H[/imath], [imath]a[/imath] and [imath]b[/imath] commute. But how do I generalize to elements in [imath]G[/imath] NOT in [imath]H[/imath]? | 1133476 | Groups and Symmetries
Let [imath]G[/imath] be a group and let [imath]f:G\to G[/imath] be an isomorphism. Let [imath]H = \{ a \in G \; | \; f(a)=a^{-1}\}[/imath] Prove that [imath]H[/imath] is a subgroup of [imath]G[/imath] if and only if [imath]G[/imath] is abelian. My attempt: I already know that [imath]H[/imath] is a subset of [imath]G[/imath] since [imath]a[/imath] is in [imath]G[/imath]. And that [imath]H[/imath] is [imath]a[/imath] group if and only if for every [imath]a[/imath] and [imath]b[/imath] in [imath]H[/imath], [imath]ab^-1[/imath] is in [imath]H[/imath]. That is, if [imath]f(a)=a^{-1}[/imath] and [imath]f(b)=b^{-1}[/imath] then [imath]f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a[/imath]. But I don't know how to derive this nor the converse based on the abelian property of [imath]G[/imath] that for every [imath]a[/imath] and [imath]b[/imath] in [imath]G[/imath], [imath]ab=ba[/imath]. Please help some. Thanks! |
1136003 | Use differentiation to find a power series representation for [imath]f(x)=\frac{1}{(1+x)^2}[/imath]
Need help on all steps. Use differentiation to find a power series representation for [imath] f(x)=\frac{1}{(1+x)^2} [/imath] | 369889 | What's the Maclaurin Series of [imath]f(x)=\frac{1}{(1-x)^2}[/imath]?
This function seemed to be pretty much straight forward, but my solution is incorrect. I have two questions: 1. Where did I make a mistake? 2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here? [imath]f(x)=\frac{1}{(1-x)^2} \qquad f(0)=1[/imath] [imath]f'(x)=\frac{2}{(1-x)^3} \qquad f'(0)=2[/imath] [imath]f''(x)=\frac{6}{(1-x)^4} \qquad f''(0)=6[/imath] [imath]f'''(x)=\frac{24}{(1-x)^5} \qquad f'''(0)=24[/imath] [imath]f''''(x)=\frac{120}{(1-x)^6} \qquad f''''(0)=120[/imath] The series is then [imath]1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}[/imath] and when simplified is [imath]1+x+x^2+x^3+x^4[/imath] But the correct answer is [imath]1+2x+3x^2+4x^3+5x^4 [/imath] |
1136264 | Show that [imath]y_1[/imath] and [imath]y_2[/imath] are not linearly independent part II
How can I show that if [imath]y_1(x)[/imath] and [imath]y_2(x)[/imath] attain a max or min at the same point then they are not linearly independent? Thanks a lot. | 1136259 | Show that [imath]y_1[/imath] and [imath]y_2[/imath] are not Linearly Independent
Suppose that [imath]y_1(x)[/imath] and [imath]y_2(x)[/imath] are solutions of the differential equation [imath]y''+py'+qy=0[/imath] on [imath]I[/imath]. How can I show that if [imath]y_1[/imath] and [imath]y_2[/imath] vanish at the same point then they are not linearly independent? Here is my attempt to prove the problem. Since [imath]y_1(x)[/imath] and [imath]y_2(x)[/imath] are solutions of the differential equation [imath]y''+py'+qy=0[/imath] on [imath]I[/imath], it is enough for me to show that the Wronskian of [imath]y_1[/imath] and [imath]y_2[/imath] denoted by [imath]W(y_1,y_2)[/imath] is zero. Now since [imath]y_1[/imath] and [imath]y_2[/imath] vanish at the same point say [imath]p[/imath] we have [imath]y_1(p)=0[/imath] and also [imath]y_2(p)=0[/imath]. Solving for [imath]W(y_1,y_2)(p)[/imath] we have: [imath]y_1(p)y_2'(p)-y_2(p)y_1'(p)=0[/imath] since [imath]y_1(p)=0[/imath] and also [imath]y_2(p)=0[/imath]. Am I correct? Thanks |
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