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476159
Limits of sequences of sets I am starting to learn about the liminf and limsup of a sequence of sets. However, I do not understand the very basics, such as why [imath]\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k>n} A_k,[/imath] and why [imath]\lim\sup A_n=\bigcup_{n\in N}\bigcap_{k>n}A_k.[/imath] Can someone break these definitions down for me? Also, why is [imath]\lim\inf\subseteq\lim\sup[/imath]?
1787474
On the interpretation of the limsup of a sequence of events In the context of understanding Borel-Cantelli lemmas, I have come across the expression for a sequence of events [imath]\{E_n\}[/imath]: [imath]\bigcap_{n=1}^\infty \bigcup_{k\geq n}^\infty E_k[/imath] following Wikipedia notation. This is verbalized as: Limit supremum of the sequence of events when [imath]n\rightarrow\infty[/imath]: [imath]\limsup\limits_{\small n\rightarrow\infty}E_n[/imath]. The event that occurs "infinitely often": [imath]\{E_n \,\text{i.o.}\}[/imath] or the set that belongs to "infinitely many" [imath]E_n[/imath]'s. The explanation is that [imath]\displaystyle\bigcup_{k\geq n}^\infty E_k[/imath] is the event that at least one of the [imath]E_n,E_{n+1},E_{n+2,\cdots}[/imath] events occurs. The question is about the meaning of [imath]\displaystyle\bigcap_{n=1}^\infty [/imath] in front. The way I look at it is that it would play the role of excluding every [imath]\{E_n\}[/imath] imaginable with [imath]n\in\mathbb N[/imath]. For instance, when [imath]n=1[/imath], the union part of the expression would render: [imath]\{E_1,E_2,E_3,\cdots\}[/imath]; and when [imath]n=2[/imath], this same union operation would yield: [imath]\{E_2,E_3,\cdots\}[/imath]. So the intersection in front of the expression would eliminate [imath]\{E_1\}[/imath] and leave the rest. Next would come the intersection with [imath]\{E_3,E_4,\cdots\}[/imath], excluding [imath]\{E_2\}[/imath]. Doing this exercise at infinity would manage to exclude every single [imath]\{E_n\}[/imath]. Is this what the intersection/union operation is supposed to do: "clip" out every single set? My understanding is that this is not the case, and that we want to end up with a "tail" of events after point [imath]k[/imath]. What is the correct interpretation, and what is the mistake I am making? NOTE to self: It is not about the "clipped out" sets at the beginning, but rather about the tail events. The union part indicates that any of the events after event [imath]n[/imath] occurs, defining the "[imath]n[/imath]-th tail event." The intersection is defines the event that all the [imath]n[/imath]-th tail events occur.
1025789
Show that the Fourier Transform is differentiable This might be a silly question. For [imath]f[/imath] an integrable, complex-valued function, its Fourier transform is [imath] \hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-isx}f(x)\, \mathrm{d}x [/imath] I want to show that if [imath]\int xf(x) \, \mathrm{d}x[/imath] exists then [imath]\hat{f}[/imath] is differentiable, with [imath] (\hat{f})'(s) = - \frac{i}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} xe^{-isx}f(x)\, \mathrm{d}x [/imath] for every [imath]s[/imath]. I tried using the definition of derivative and got to [imath] (\hat{f})'(s) = \lim_{h \to 0}\frac{1}{h}\int_{-\infty}^{+\infty} f(x)e^{-isx}[e^{-ihx}-1] \, \mathrm{d}x [/imath] and I'm not really sure where to go. Can anyone point me in the right direction? Thanks.
1032137
Derivative of Fourier transform: [imath]F[f]'=F[-ixf(x)][/imath] Let us define the Fourier transform of the Lebesgue-summable function [imath]f\in L_1(\mathbb{R},\mu_x)[/imath] as [imath]F[f](\lambda)=\int_{\mathbb{R}}f(x) e^{-i\lambda x} d\mu_x[/imath], where [imath]\mu_x[/imath] is the Lebesgue linear measure. Kolmogorov-Fomin's (p. 430 here) states that, if [imath]x\mapsto xf(x)[/imath] is Lebesgue-summable too, then the Fourier transform is differentiable and[imath]\frac{d}{d\lambda}F[f](\lambda)=-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x[/imath] EDIT: I would be very grateful to the people having voted to close the question, and to anybody else, if they explained how to use the question linked as a duplicate to prove this equality, which I am not able to do. I am trying to teach myself and cannot afford university for the moment and have not got the sufficient mathematical ability to understand how that result implies this equality. I heartily thank you all!!! Previous trial of mine which may or may not be useful to prove the given equality: In order to prove the statement to myself I had thought that the fact that if the sequence [imath]\{\varphi_n\}\subset L_1(\mathbb{R},\mu_x)[/imath] converges to [imath]\varphi[/imath] with respect to the metric of [imath]L_1(\mathbb{R},\mu_x)[/imath], then [imath]F[f_n]\xrightarrow{n\to\infty} F[f][/imath], but I cannot apply it. In fact I see that the derivative [imath]F[f]'(\lambda)[/imath] can be seen, if we define [imath]\varphi_n(x)=f(x)\frac{e^{-ih_n x}-1}{h_n}[/imath], where [imath]\{h_n\}[/imath] converges to 0, as the limit [imath]\lim_n \int_{\mathbb{R}}f(x)\frac{e^{-i(\lambda+h_n)x}-e^{-i\lambda x}}{h_n}d\mu_x=\lim_n \int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x[/imath][imath]=\lim_n F[\varphi_n](\lambda)[/imath]but I cannot prove that [imath]\int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x[/imath] converges to [imath]-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x[/imath]. I thought about Lebesgue's dominated convergence theorem, but I am not convinced at all that we can majorate [imath]|\varphi_n(x)|[/imath] with a summable function and I am not sure this is the right path to prove the desired equality... P.S.: I assume that the equality holds for any [imath]f\in L_1(\mathbb{R},\mu_x)[/imath] because the book does not specify any other conditions, but it is perfectly in the style of the book to state theorems valid only under certain conditions without explicitly declare them. So, if you noticed that some other condition is needed...
251808
Periodic parametric curve on cylinder Given a cylinder surface [imath]S=\{(x,y,z):x^2+2y^2=C\}[/imath]. Let [imath]\gamma(t)=(x(t),y(t),z(t))[/imath] satisfy [imath]\gamma'(t)=(2y(t)(z(t)-1),-x(t)(z(t)-1),x(t)y(t))[/imath]. Could we guarante that [imath]\gamma[/imath] always on [imath]S[/imath] and periodic if [imath]\gamma(0)[/imath] on [imath]S[/imath]?
251567
Parametric curve on cylinder surface Let [imath]r(t)=(x(t),y(t),z(t)),t\geq0[/imath] be a parametric curve with [imath]r(0)[/imath] lies on cylinder surface [imath]x^2+2y^2=C[/imath]. Let the tangent vector of [imath]r[/imath] is [imath]r'(t)=\left( 2y(t)(z(t)-1), -x(t)(z(t)-1), x(t)y(t)\right)[/imath]. Would you help me to show that : (a) The curve always lies on ylinder surface [imath]x^2+2y^2=C[/imath]. (b) The curve [imath]r(t)[/imath] is periodic (we can find [imath]T_0\neq0[/imath] such that [imath]r(T_0)=r(0)[/imath]).If we make the C smaller then the parametric curve [imath]r(t)[/imath] more closer to the origin (We can make a Neighboorhood that contain this parametric curve) My effort: (a) Let [imath]V(x,y,z)=x^2+2y^2[/imath]. If [imath]V(x,y,z)=C[/imath] then [imath]\frac{d}{dt} V(x,y,z)=0[/imath]. Since [imath]\frac{d}{dt} V(x,y,z)=(2x,4y,0)\cdot (x'(t),y'(t),z'(t))=2x(2y(z-1))+4y(-x(z-1))=0[/imath], then [imath]r(0)[/imath] would be parpendicular with normal of cylinder surface. Hence the tangent vector must be on the tangent plane of cylinder. So [imath]r(t)[/imath] must lie on cylinder surface. (b) From [imath]z'=xy[/imath], I analyze the sign of [imath]z'[/imath] (in 1st quadrant z'>0 so the z component of [imath]r(t)[/imath] increasing and etc.) and conclude that if [imath]r(t)[/imath] never goes unbounded when move to another octan ( But I can't guarante [imath]r(t)[/imath] accros another octan.). I also consider the case when [imath](x=0, y>0), (x=0, y<0,z>1), (x>0, y=0,z>1[/imath]and so on) and draw the vector [imath]r'(t)[/imath]. Thank you so much of your help.
1157809
improper integral of exponential function I have a problem calculating improper integrals, this one for example, can you please help me solve it? [imath]\int_0^\infty t^3(e^{-t^2})dt[/imath] thanks in advance.
1130988
How to calculate [imath]\int e^{\frac{x^2}{2}} \cdot x^{3} dx[/imath]? [imath]\int e^{\frac{x^2}{2}} \cdot x^{3} dx=?[/imath] I tried to do the substitution [imath]du = x^3dx[/imath], so that [imath]u=\frac{x^4}{4}[/imath]. Then, [imath]\int e^{\frac{x^2}{2}} \cdot x^{3} dx= \int e^{\sqrt{u}} du[/imath] However, I would have to do a second substitution ([imath]v = \sqrt{u}[/imath]) which would give a product again. How should I proceed?
1158315
Proving an identity using a combinatorial proof? Let [imath]n\ge k\ge m\ge 0[/imath] be integers. Consider the following formula: [imath]\binom{n}k\binom{k}m=\binom{n}m\binom{n-m}{k-m}[/imath] Give two different proofs. On proof should use the factorial formula for [imath]\binom{n}k[/imath]. The other proof should be combinatorial. Try to develop a question that both sides of the equation answer. So this question asked us to prove the formula below in two ways, and while I got it for the 1st way [using the factorial formula for [imath]\binom{n}k[/imath]], I still don't understand how I can prove this using a combinatorial proof. I'm having trouble constructing a counting argument. Can anyone help me out?
388530
Proving an identity with a combinatorial proof: [imath]\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r}[/imath] For any integers [imath]n[/imath], [imath]k[/imath], [imath]r[/imath] where [imath]n\geq k\geq r \geq 0[/imath], give a combinatorial proof of the following identity: [imath]\begin{align} \binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r} . \end{align}[/imath] The problem is that I can't come up with a good counting argument of what exactly the two sides are doing. The left hand side is quite mysterious, and the right hand side is apparently choosing [imath]r[/imath] items and then choosing [imath]k-r[/imath] items from the remaining, which should be equivalent to [imath]\binom{n}{k}[/imath] but somehow isn't. How should I approach this problem?
1158192
Show equivalence of statement [imath]\left(P\rightarrow Q\right) \wedge \left(Q\rightarrow R\right)[/imath] to ... Show that [imath]\left(P\rightarrow Q\right) \wedge \left(Q\rightarrow R\right)[/imath] is equivalent to [imath]\left(P\rightarrow R\right) \wedge \left[\left(P\leftrightarrow Q\right) \vee \left(R\leftrightarrow Q\right)\right][/imath] After one page of expanding and collapsing the right hand side, I get as a result [imath] \left(P\rightarrow Q\right) \wedge \left(Q\rightarrow P\right) \wedge \left(P\rightarrow R\right) [/imath] Can anybody help? Although this question has been marked as a duplicate, the linked page only provides a partial answer (one direction) and the other answer is in Polish notation which I am not familiar with.
502656
Show by using logical connectives laws that [imath](P\to Q) \land (Q \to R) [/imath] is equivalent to [imath](P \to R) \land [(P \iff Q) \lor (R \iff Q)][/imath] I am having trouble with a problem in the book I'm self-studying from. It says the following: Show that [imath](P\to Q) \land (Q \to R) [/imath] is equivalent to [imath](P \to R)[/imath] [imath]\land [(P \iff Q) \lor (R \iff Q)][/imath] by using logical connectives I have dedicated so far a hefty amount of time on this problem, and now I'm asking you guys advice/hints or solution as to how to solve this problem. Here is one of the methods I used. Point any flaws that I made. [imath](P\to Q) \land (Q \to R) [/imath] (Conditional Law) [imath](\neg P \lor Q) \land (\neg Q \lor R) \Rightarrow[/imath] (Distributive Law) [imath][(\neg P \land \neg Q)] \lor [Q \land (\neg Q \lor R)] \Rightarrow[/imath] (Distributive Law) [imath][(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [(Q \land \neg Q) \lor (R \land Q)] \Rightarrow[/imath] Contradiction [imath][(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [(Contradiction) \lor (R \land Q)] \Rightarrow[/imath] (Contradiction Law) [imath][(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [ (R \land Q)] \Rightarrow[/imath] Typically at around step five I get stuck or get confused because the problem gets messy. I know you could show it by using the truth-tables. However, the problem says use logical connectives. My questions are: Am I on the right track into solving this problem? Did I make any mistakes? What advice/hints would you give me in my path to solving this problem? Edit Some of you guys want me to list the laws. Here they are: DeMorgan's laws [imath]\neg(P \land Q) \equiv \neg P \lor \neg Q[/imath] [imath]\neg(P \lor Q) \equiv \neg P \land \neg Q[/imath] Commutative laws [imath]P \lor Q \equiv Q \lor P[/imath] [imath]Q\lor P \equiv P \lor Q[/imath] Associative Laws [imath]P \land (Q \land R) \equiv (P \land Q) \land R [/imath] [imath](P \land Q) \land R \equiv P \land (Q \land R) [/imath] Idempotent Laws [imath]P \land P \equiv P[/imath] [imath]P \lor P \equiv P[/imath] Distributive Laws [imath]P \land (Q \lor R) \equiv (P \land Q ) \lor (P \land R)[/imath] [imath]P \lor (Q \land R) \equiv (P \lor Q) \land (P \lor R)[/imath] Absorption Laws [imath]P \lor (P \land Q) \equiv P[/imath] [imath]P \land (P \lor Q) \equiv P[/imath] Tautology Laws [imath]P \land (tautology) \equiv P[/imath] [imath]P \lor (tautology) \equiv (tautology)[/imath] Contradiction Laws [imath]P \land (contradiction) \equiv (contradiction)[/imath] [imath]P \lor (contradiction) \equiv P[/imath] Conditional laws [imath]P \to Q \equiv \neg P \lor Q[/imath] [imath]P \to Q \equiv \neg (P \land \neg Q)[/imath]
1158687
[imath]T^2 \setminus \{p\} \simeq S^1 \vee S^1[/imath] Why is the torus with a point missing homotopy equivalent to figure-8-space? [imath]T^2 \setminus \{p\} \simeq S^1 \vee S^1[/imath]
661219
Homotopy equivalence from torus minus a point to a figure-eight. Let [imath]I[/imath] be the unit interval, and define the torus using the usual identifications on [imath]I \times I[/imath]. I've shown that [imath]I \times I - \{x\}[/imath] (where [imath]x[/imath] is a point not on the boundary of [imath]I \times I[/imath]) is homotopy equivalent to its boundary (and that it even retracts onto its boundary). Let [imath]p: I \times I \rightarrow I \times I /\sim[/imath] be the canonical quotient map. Using the retraction above, call it [imath]r[/imath], is there a way I can explicitly show that [imath]p(I \times I - \{x\})[/imath] is homotopy equivalent to [imath]p(\text{boundary of } I \times I)[/imath], i.e., that torus minus a point to a figure-eight? More generally, if [imath]A \subset X[/imath], and [imath]X[/imath] is homotopy equivalent to [imath]A[/imath] (or [imath]X[/imath] retracts to [imath]A[/imath]), for which maps [imath]p[/imath] can I say that [imath]p[X][/imath] is homotopy equivalent to [imath]p[A][/imath]? Hints appreciated!
1158962
[imath]-1[/imath] isn't a quadratic residue [imath]\pmod{p}, p=4k+3[/imath]. I remember coming across this fact a while ago in a pdf somewhere, but I haven't been able to find it again. Can someone show me how to prove it? I would appreciate easier proofs. EDIT: I'm very sorry, I forgot to actually state it. Prove that [imath]-1[/imath] is not a quadratic residue modulo primes of the form [imath]4k+3[/imath]. Thanks!
735400
If p [imath]\equiv[/imath] 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p. If p [imath]\equiv[/imath] 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p. I suppose this would not be true if p [imath]\equiv[/imath] 1 (modulo 4). To prove something is a non-square I find to be tricky. It's difficult to see any straightforward way to do this using only the definition of congruence for example.
878962
Finite additivity in outer measure Let [imath]\{E_i\}_{i=1}^n[/imath] be finitely many disjoint sets of real numbers (not necessarily Lebesgue measurable) and [imath]E[/imath] be the union of all these sets. Is it always true that [imath] m^\star (E)=\sum_{i=1}^N m^\star(E_n) [/imath] where [imath]m^\star[/imath] denotes the Lebesgue outer measure? If not, please give a counterexample. The Vitali set is a counterexample in the countable case, but I am not sure whether it is false in finite case.
2517293
Lebesgue Outer Measure is not Finitely Additive. We know that Lebesgue Outer Measure is countably sub-additive. It is not even finitely additive. I am trying to find two sets of real numbers which are disjoint and yet [imath]\mu ^*(A\cup B) = \mu^*(A) +\mu^*(B)[/imath] doesn't hold.
1159134
How do I compute [imath]\int_{-\infty}^{\infty}x^2e^{-x^2}dx[/imath]? I arrive at the partial solution of [imath]((x*e^{-x^2})/2)- (\pi^{1/2}/2)[/imath] using double integration by parts. How do I resolve the first part of the solution with e. Is the solution even right?
1149907
[imath] \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx[/imath] What are some different methods to evaluate [imath] \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx[/imath] for [imath]a > 0[/imath]. This integral arises in a number of contexts in Physics and was the original motivation for my asking. It also arises naturally in statistics as a higher moment of the normal distribution. I have given a few methods of evaluation below. Anyone know of others?
1159120
Find [imath]\lim \limits_{s\to 16} \frac{4-\sqrt{s}}{s-16}[/imath] How do you find this limit using algebra? [imath] \lim \limits_{s\to 16} \frac{4-\sqrt{s}}{s-16} [/imath] By graphing I get 1/8.
165515
How to find [imath] \lim\limits_{ x\to 100 } \frac { 10-\sqrt { x } }{ x-100 }[/imath] Find [imath] \lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }[/imath] (without using a calculator and other machines...?)
427423
Solve: [imath]T(n) = T(n-1) +(1/n)[/imath] by iteration Use iteration method to solve: [imath]1.[/imath] [imath]T(n) = T(n-1) + \frac{1}{n},\,(T(0)=1)[/imath] [imath] 2.[/imath] [imath]T(n) = 3T\left(\dfrac{n}{3}\right) +1,\,(T(3)=1)[/imath]
158791
Recurrence telescoping [imath]T(n) = T(n-1) + 1/n[/imath] and [imath]T(n) = T(n-1) + \log n[/imath] I am trying to solve the following recurrence relations using telescoping. How would I go about doing it? [imath]T(n) = T(n-1) + 1/n[/imath] [imath]T(n) = T(n-1) + \log n[/imath] thanks
143715
Form or asymptotic behaviour of [imath]T(n) =2T(n-1)+n[/imath] [imath]T(n) =[/imath] if [imath]n=1[/imath], then time execution is [imath]1[/imath], if [imath]n \geq 2[/imath] then [imath]2T(n-1)+n[/imath] The options are: [imath]T(n) = 2^{n+1} - n - 2[/imath] [imath]T(n) = O(n2^n)[/imath] [imath]T(n) = \Omega(n)[/imath] [imath]T(n) = \theta(2^n)[/imath] Thanks.
239974
Solve the recurrence [imath]T(n) = 2T(n-1) + n[/imath] Solve the recurrence [imath]T(n) = 2T(n-1) + n[/imath] where [imath]T(1) = 1[/imath] and [imath]n\ge 2[/imath]. The final answer is [imath]2^{n+1}-n-2[/imath] Can anyone arrive at the solution?
194510
Asymptotic of [imath]T(n) = T(n-2) + \frac{1}{ \lg n}[/imath] Trying to determine asymptotic of [imath]T(n) = T(n-2) + \displaystyle\frac{1}{ \lg n}[/imath] [imath]\lg n = \log_{2}n [/imath] Last term [imath]\frac{1}{ \lg n}[/imath] give me a lot of trouble. Iterative method doesn't work. Tried change of variables, there is nothing to change.
167418
Time Complexity of [imath]T(n)=T(n-2)+\frac{1}{\log(n)}[/imath] Solve [imath]T(n)=T(n-2)+\frac{1}{\log(n)}[/imath] for [imath]T(n)[/imath]. I am getting the answer as [imath]O(n)[/imath] by treating [imath]1/\log(n)[/imath] as [imath]O(1)[/imath]. The recursive call tree of this is a lop-sided tree of height [imath]n[/imath]. Hence, considering the amount of work done in each step, the answer comes out to be [imath]O(n)[/imath]. Please verify my answer, and tell me if I am correct.
19485
Dominoes and induction, or how does induction work? I've never really understood why math induction is supposed to work. You have these 3 steps: Prove true for base case (n=0 or 1 or whatever) Assume true for n=k. Call this the induction hypothesis. Prove true for n=k+1, somewhere using the induction hypothesis in your proof. In my experience the proof is usually algebraic, and you just manipulate the problem until you get the induction hypothesis to appear. If you can do that and it works out, then you say the proof holds. Here's one I just worked out, Show [imath]\displaystyle\lim_{x\to\infty} \frac{(\ln x)^n}{x} = 0[/imath] So you go: Use L'Hospital's rule. [imath]\displaystyle\lim_{x\to\infty} \frac{\ln x}{x} = 0[/imath]. Since that's [imath]\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0[/imath]. Assume true for [imath]n=k[/imath]. [imath]\displaystyle\lim_{x\to\infty} \frac{(\ln x)^k}{x} = 0[/imath]. Prove true for [imath]n=k+1[/imath]. You get [imath]\displaystyle\lim_{x\to\infty} \frac{(\ln x)^{k+1}}{x} = 0.[/imath] Use L'Hospital again: [imath]\displaystyle\lim_{x\to\infty} \frac{(k+1)(\ln x)^{k}}{x} = 0[/imath]. Then you see the induction hypothesis appear, and you can say this is equal to [imath]0[/imath]. What I'm not comfortable with is this idea that you can just assume something to be true ([imath]n=k[/imath]), then based on that assumption, form a proof for [imath]n=k+1[/imath] case. I don't see how you can use something you've assumed to be true to prove something else to be true.
2836728
On Assumptions In Induction I'm currently learning mathematical induction from this site: https://www.mathsisfun.com/algebra/mathematical-induction.html What I'm confused about is how it presents mathematical induction. It says that there are 3 steps to induction: Show it true for [imath]n=1[/imath]. Assume it is true for [imath]n=k[/imath] Prove it is true for [imath]n=k+1[/imath] (we can use the [imath]n=k[/imath] case as a fact.) There are many things I am confused about here, about all [imath]3[/imath] steps. Starting from the first step, why is it necessary to prove it true for [imath]n=1[/imath]? I don't get why this step is needed. Second, why choose [imath]1[/imath] of all numbers; can't a number like [imath]2[/imath] be chosen? Moving on to the second step, why is it legitimate to assume it true for all [imath]n=k[/imath]? Is this assumption proved true by the third step, if so, how? On the final step, first, how can we prove it true for all [imath]n=k+1[/imath] ? Because this prove fundamentally assumes that it is true for [imath]n=k[/imath], but there is no way to verify this. Second, what happens if the set we're doing induction on has only a limited amount of numbers, let's say 100 numbers. So if we go up to the 100th number, then how can [imath]n=k+1[/imath] still be true? Because there is no 101st term for it to be true on; there are only 100 numbers! Please explain this as simply as possible; I'm still a beginner. I will not be able to understand complicated proof notation. DETAILS: This question is different from the question above since the question above uses [imath]\lim_{x\to\infty}[/imath] notation and other pieces of calculus knowledge. However, I have not learnt calculus, and nothing about this question suggest prior calculus knowledge. An answer where induction is explained without calculus would benefit me greatly.
666067
Time Complexity for [imath]T(n) = T(\sqrt{n}) + 1[/imath] [imath]T(n) = T(\sqrt{n}) + 1[/imath] I am trying to find the time complexity of the given equation. I tried everything that I know but I could not find the answer. What I've tried: [imath]k = lg(n)[/imath] and [imath]n=2^k[/imath] -> [imath]\sqrt{n} = 2^{k/2}[/imath] Thanks in advance.
1134398
Tight bound for [imath]T(n) = T(n^{1/2}) + 1[/imath] Can someone help me figure out the big-O for the recurrence relation [imath]T(n) = T(n^{1/2}) + 1[/imath]? I didn't think the master theorem would work since it requires [imath]T(n) = T(n/b)[/imath]... to have [imath]b[/imath] as a constant. I was wondering if anyone could shed some light on a method to approach this problem? (not looking for a solution)
1159402
A limit without using L'hoptial's rule Define the natural number [imath]e[/imath] by [imath]e=\lim_{x\to 0} (1+x)^{1/x}[/imath]. Then, I can prove [imath]\lim_{x\to 0} \frac{e^x-1}{x}=1[/imath]. Let [imath]z=e^x-1[/imath]. Then, [imath]x=\ln(z+1)[/imath] and [imath]\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{z\to 0} \frac{z}{\ln(z+1)}=\frac{1}{\ln e}=1\text{.}[/imath] Using a similar trick (without L'Hoptial's rule), can we prove [imath]\lim_{x \to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}[/imath]?
184053
How to find [imath]\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}[/imath] without using l'Hopital's rule nor any series expansion? Is it possible to determine the limit [imath]\lim_{x\to0}\frac{e^x-1-x}{x^2}[/imath] without using l'Hopital's rule nor any series expansion? For example, suppose you are a student that has not studied derivative yet (and so not even Taylor formula and Taylor series).
1159514
Showing that [imath]f_n[/imath] is the number closest to [imath]\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n[/imath] I want to prove that the the [imath]n[/imath]th Fibonacci number [imath]f_n[/imath] is the integer closest to [imath]\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n[/imath]. What would be a rigorous way to go about this? I assume I'll have to use a recurrence relation and maybe Binet's Forumla. I just don't know how to go about tying it all together, so any advice on how to proceed would be much appreciated.
992811
Prove the [imath]n[/imath]th Fibonacci number is the integer closest to [imath]\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n[/imath] Prove that the [imath]n[/imath]th Fibonacci number [imath]f_n[/imath] is the integer that is closest to the number [imath]\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n.[/imath] Hi everyone, I don't really understand the problem. I have the following hint, but I don't know how to work it. Hint: Show that the absolute value of [imath]\frac{1}{\sqrt{5}}\left( \frac{1 - \sqrt{5}}{2} \right)^n[/imath] is less than [imath]1/2[/imath].
1159680
Show that [imath]\lim (\sqrt{n^2+1)}-n) = 0[/imath] Can't use limit rules as [imath]\sqrt{n^2+1}[/imath] and n are not convergent sequences
546543
Show that [imath]\sqrt{n^2+1}-n[/imath] converges to 0 I want to use the definition of the limit to show that [imath]\sqrt{n^2+1}-n[/imath] converges to 0. The definition is as follows: if [imath]\sqrt{n^2+1}-n[/imath] converges to 0, then [imath]\forall \epsilon>0[/imath], there exists an [imath]N>0[/imath] such that [imath]n\ge N \implies \mid\sqrt{n^2+1}-n\mid<\epsilon[/imath]. Now I want to start backwords in order to figure out how to pick N. I know: [imath]\mid\sqrt{n^2+1}-n\mid=\sqrt{n^2+1}-n[/imath] since [imath]n[/imath] is a natural number and [imath]\sqrt{n^2+1}>n[/imath]. So I need to pick an N such that [imath]\sqrt{n^2+1}-n<\epsilon[/imath]. I tried multiplying [imath]\sqrt{n^2+1}-n[/imath] by [imath]\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}[/imath] but that didn't really seem to help. Do you have any ideas on how to find this N? Thanks
313046
Recurrence relation using the master theorem [imath] T(n) = 4T(n/2) + n^2 \log n[/imath] I am trying to solve the following recurrence relation using the master theorem: [imath] T(n) = 4T(n/2) + n^2 \log n[/imath] So: [imath]a=4 ,b=2, f(n)=n^2\log n[/imath] , then [imath]n^{\log_2 4}=n^2 [/imath] Now i know that [imath]n^2 \log n [/imath] is larger . Can I apply the master theorem to this problem? Thanks.
601295
Find the asymptotic tight bound for [imath]T(n) = 4T(n/2) + n^{2}\log n[/imath] Find the asymptotic tight bound in [imath] T(n) = 4T\left(\frac{n}{2}\right) + n^{2}\log n. [/imath] where [imath] \log n= \log _{2}n [/imath] and [imath]T(1) = 1[/imath]. I should solve this using all three common methods: iteration, master theorem and substitution. It is highly likely that this kind of recurrence equation will be in my test in two days. Thank you very much in advance!
1159432
[imath]L^2[/imath] is meager in [imath]L^1[/imath] This question is out of Rudin Functional analysis, but only attention was really brought to part (b) on this site, my apologizes if this is considered a re-post. I was just hoping for some assistance. I added what I have so far in my proof. Let [imath]L^1[/imath] and [imath]L^2[/imath] be the usual Lebegue space on the unit interval. Prove that [imath]L^2[/imath] is of the first catagory in [imath]L^1[/imath], by the following. Show that [imath]\{f:\int |f|^2 \leq n\}[/imath] is closed in [imath]L^1[/imath] but has empty interior. proof: Let [imath]f \in L^2[/imath] be arbitrary. Then [imath]\int |f|^2 d \mu \leq n[/imath], for some finite [imath]n[/imath]. Observe that [imath]|f g| \leq \frac{|f^2 + g^2|}{2}[/imath]. Suppose that [imath]g = 1[/imath]. Then [imath] \int |fg| d\mu = \int |f| d \mu \leq \frac{1}{2}\int |f^2 + 1| d\mu \leq \frac{1}{2}\int |f|^2 d\mu + \frac{1}{2} \int d \mu \leq \frac{1}{2}(n + 1) [/imath] Hence [imath]f \in L^1[/imath] [imath]f[/imath] arbitary and closed in [imath]L^2[/imath] implies that [imath]L^2[/imath] is closed in [imath]L^1[/imath]. I'm really struggling with showing that the interior is empty, and not sure how to make the argument.
140096
[imath]L^2([0,1])[/imath] is a set of first category in [imath]L^1([0,1])[/imath]? How to show that [imath]L^2([0,1])[/imath] is a set of first category in [imath]L^1([0,1])[/imath]? Thank you.
1159791
Is there any regular, balanced, connected bipartite graph that does not contain any Hamiltonian cycle? In a set of balanced, connected bipartite graphs, all with regularity [imath]r \ge 2[/imath], is it possible that there exists a bipartite graph that does not contain a Hamiltonian cycle ? My argument: The bipartite graph is balanced, so both of the partition has the same number of vertices. It's regularity [imath]r \ge 2[/imath] and the graph is connected, that means we can switch between the partitions, so there is a Hamiltonian path. Lets say that we have followed the Hamiltonian path from vertex [imath]u[/imath] and reached to a vertex [imath]v[/imath] and exhausted all the vertices, now how do I prove that there must be an edge that will take me back to [imath]u[/imath] from [imath]v[/imath]? I am getting lost with this construction of the "Hamiltonian cycle". So does this mean that there can be such bipartite graph (connected, balanced, [imath]r \ge 2[/imath] regular) that does not contain a Hamiltonian cycle? If yes, can someone give me a concrete example? N.B.: [imath]r[/imath]-regular: every edge is [imath]r[/imath]-degree.
830904
Does every connected [imath]r[/imath]-regular bipartite graph contain a Hamiltonian cycle? Here's a quickie: Let [imath]r\ge2[/imath]. Does every connected [imath]r[/imath]-regular bipartite graph contain a Hamiltonian cycle? I've been playing around with this for almost an hour, but I can't prove it.
1153949
if [imath]a,b[/imath] are nilpotent elements of a commutative ring [imath]R[/imath], show that [imath]a+b[/imath] is also nilpotent if [imath]a,b[/imath] are nilpotent elements of a commutative ring [imath]R[/imath], show [imath]a+b[/imath] is also nilpotent So then [imath]a^n=0, b^m = 0, n,m \in \mathbb{Z}^+[/imath] I know this is solvable using the binomial theorem but I would much rather solve it another way if possible. The answer using the binomial theorem isn't making the most sense at the present moment.
656860
Nilpotent elements of a commutative ring Show that if [imath]a[/imath] and [imath]b[/imath] are nilpotent elements of a commutative ring then [imath]a+b[/imath] is also nilpotent. Let [imath]R[/imath] be a commutative ring, since [imath]a[/imath] and [imath]b[/imath] are nilpotent we know that [imath]a^n = 0[/imath] for some [imath]n \in \mathbb{Z}^+[/imath]. Same argument for [imath]b[/imath] with some [imath]m \in \mathbb{Z}^+[/imath]. I wanted to first consider the linear combination of them, [imath]a^n + b^m = 0[/imath] but i'm not sure if this could get me anywhere. I can't really connect multiplication and addition in this way i have it here. And I can't make anything from the statement [imath]a^n = -b^m[/imath] because we aren't given that R is a field. (not sure if i could do anything with that anyways)
1160013
Formula for reflection across a line in [imath]\mathbb{R}^2[/imath]? [imath]\newcommand{\Reals}{\mathbb{R}}[/imath]I have an equation of a line: [imath]4x - 3y = 0[/imath]. Let [imath]S : \Reals^2 \to \Reals^2[/imath] be reflection through that line, and let [imath]P : \Reals^2 \to \Reals^2[/imath] be projection onto that line. I want to find a vector that represents the reflection through that line and a vector that represents the projection onto that line. Are there any formulas for these?
1159992
Need help with linear transformations (with projection and reflection)? Let [imath]L[/imath] be the line given by the equation [imath]4x − 3y = 0[/imath]. Let [imath]S : \mathbb{R}^2 \rightarrow \mathbb{R}^2[/imath] be reflection through that line, and let [imath]P : \mathbb{R}^2 \rightarrow \mathbb{R}^2[/imath] be projection onto that line. Determine, geometrically without doing any computations, whether there exist non-zero vectors [imath]x[/imath] such that: [imath] a) \qquad S(x) = 1 · x [/imath] [imath] b)\qquad P(x) = 1 · x [/imath] Can someone tell me how to solve this? Does anyone know what kind of arithmatic this is? I wanna look it up on google but I don't know what it's called. Looks a bit like eigenvectors to me but I'm not entirely sure... All help is appreciated.
983815
Vandermonde identity corollary [imath]\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2[/imath] I am trying to prove this identity: [imath]\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2[/imath] I think this identity (corollary of Vandermonde identity): [imath]{n\choose 0}^2+{n\choose 1}^2+{n\choose 2}^2+\cdots+{n\choose n}^2={2n \choose n}[/imath] is applicable for solving it. Please give me some hints. thank you
1656032
Verify the identity [imath]\sum_{k=0}^n \frac{(2n)!}{(k!)^2((n-k)!)^2} = \dbinom{2n}{n}^2 [/imath] I am struggling to verify this identity. I've tried using the Principle of Mathematical Induction, combinatorial proofs, and a generalization of Vandermonde's Identity, all to no avail. It's the [imath](2n)![/imath] that's giving me problems.
1002268
Construct a compact set of real numbers whose limit points form a countable set? (Rudin Exercise) I think [imath]S = \{r \in \Bbb Q: 0 \leq r \leq 1\}[/imath] in [imath]\Bbb R[/imath] can be a set that satisfies the conditions. First, it is compact by the Heine-Borel theorem since it is closed and bounded. It is closed because it contains all its limit points. Every rational number is a limit point; if we take an open neighbourhood of radius s, we can find some rational number contained in the open ball. Finally, the rationals are countable. Is this correct? Thank you.
444591
Construct a compact set of real numbers whose limit points form a countable set. I searched and found out that the below is a compact set of real numbers whose limit points form a countable set. I know the set in real number is compact if and only if it is bounded and closed. It's obvious it is bounded since [imath]\,d(1/4, q) < 1\,[/imath] for all [imath]\,q \in E.[/imath] However, I'm not sure how this is closed. Is there any simpler set that satisfies the above condition? Thank you! [imath]E = \left\{\frac 1{2^m}\left(1 - \frac 1n\right) \mid m,n \in \mathbb N\right\}.[/imath]
1160744
Approximate the size of a set given random items from the set. I'd like to know how would it be possible to approximate the size of a set that has no duplicate elements. We can make a limited amount of requests. Each request gives us a random element from the set (Without removing it). How do we approximate the size of the set given the amount of duplicate elements gotten from the set when doing [imath]x[/imath] requests?
75758
Estimate the size of a set from which a sample has been equiprobably drawn? Here is the problem I'm trying to solve: In order to send spam, a spammer generates fake nicknames, by picking random girl names (and appending a random number to it). I suppose it randomly and fairly picks names from a fixed list, and I get to observe the outcome. Over time, names eventually start repeating. From the distribution of the repetition count of every distinct name, is there a way to estimate the size of the fixed list ? Intuitively, if there was no repetition among the names I have observed, I would have no information about the size of the list, but the minimum bound given by the actual distinct names observed. On the other hand, if the distribution of the repeat counts is far away from zero, I can be reasonably confident that the sample size is much larger than the word list, and that I probably have observed the full list already. The problem lies in the middle zone. So far the distribution of repeat counts looks like this: [imath] \begin{matrix} n & k \\ 114 & 1 \\ 66 & 2 \\ 30 & 3 \\ 4 & 4 \\ 2 & 5 \\ 1 & 6 \end{matrix} [/imath] Here [imath]n[/imath] is the count of distinct names appearing [imath]k[/imath] times in the sample. Is there a simple estimator for the size of the list?
331286
Why does [imath]\int_a^b fg\, dx = 0[/imath] imply that [imath]f = 0[/imath]? Assume that f is a continuous function on [imath][a,b][/imath] such that for any continuous function g on [imath][a,b][/imath] [imath]\int_a^b f(x)g(x)dx = 0[/imath], then how can I show that f(x) = 0 for all [imath]x\in [a,b][/imath]?
1128621
How can I show that [imath]f[/imath] must be zero if [imath]\int fg[/imath] is always zero? Let [imath]f(x)[/imath] be continuous on [imath][a,b][/imath] and suppose [imath]\int_a^b f(x)g(x)dx = 0[/imath] for every continuous function [imath]g[/imath] on [imath][a,b][/imath]. Prove that [imath]f(x)=0[/imath] on [imath][a,b][/imath]. I understand that [imath]f(x)[/imath] must be zero otherwise the integral can't possibly be zero for all possible [imath]g[/imath]. But, I am not sure how to approach this systematically. Any suggestions would be appreciated. Thanks!
1161857
Compact set and continuous function Let [imath](E,d), (E',d')[/imath] be two metric space, and [imath]f:E\rightarrow E'[/imath] an injective function such that the image of any compact set from [imath]E[/imath] is compact in [imath]E'[/imath]. How can I prove that [imath]f[/imath] is continuous? Thank you.
150297
From injective map to continuous map Let [imath]X[/imath] and [imath]Y[/imath] metric spaces, [imath]f[/imath] is an injective from [imath]X[/imath] to [imath]Y[/imath], and [imath]f[/imath] sets every compact set in [imath]X[/imath] to compact set in [imath]Y[/imath]. How to prove [imath]f[/imath] is continuous map? Any comments and advice will be appreciated.
1162252
How do you prove the inequality [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} < 2[/imath]? How do you prove: [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} < 2[/imath] I have some competitions in my country, so I have to prepare.
1000642
Proof that [imath]\sum_{1}^{\infty} \frac{1}{n^2} <2[/imath] I know how to prove that [imath]\sum_1^{\infty} \frac{1}{n^2}<2[/imath] because [imath]\sum_1^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}<2[/imath] But I wanted to prove it using only inequalities. Is there a way to do it? Can you think of an inequality such that you can calculate the limit of both sides, and the limit of the rigth side is [imath]2[/imath]? Is there a good book about inequalities that helps to prove that a sum is less than a given quantity? This is not a homework problem, its a self posed problem that I was thinking about :)
1162095
Reciprocal squares sum inequality What is the easiest (preferably inductional) way without approximation of the sum_ to prove the following inequality: [imath]\frac{1}{1^2}+\frac{1}{2^2} + \ldots +\frac{1}{n^2} \le 2 - \frac{1}{n}[/imath]
1150388
Summation inductional proof: [imath]\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2[/imath] Having the following inequality [imath]\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2[/imath] To prove it for all natural numbers is it enough to show that: [imath]\frac{1}{(n+1)^2}-\frac{1}{n^2} <2[/imath] or [imath]\frac{1}{(n+1)^2}<2-\frac{1}{n^2} [/imath]
1162294
Prove that there exists a monotone nondecreasing function [imath]f:[0,1] \rightarrow \mathbb{R}[/imath] discontinuous in rationals Prove that there exists a monotone nondecreasing function [imath]f:[0,1] \rightarrow \mathbb{R}[/imath] discontinuous in rationals of [imath][0,1][/imath] I tried to test functions of the type [imath] f(x)=\begin{cases} \frac{1}{q} & \text{if } x=\frac{p}{q} \mid p,q \in \mathbb{Z}^+ \,\,\, \text{and} \,\,\, \gcd(p,q)=1 \\ x & \text{if } x \not\in \mathbb{Q} \end{cases}[/imath] which is discontinuous in all rationals, but is not monotone.
296195
Example of an increasing, integrable function [imath]f:[0,1]\to\mathbb{R}[/imath] which is discontinuous at all rationals? I have really no idea about this: Problem: Show that there exists a function [imath]f:[0,1]\rightarrow\mathbb{R}[/imath] such that: [imath]f[/imath] is discontinuous in all [imath]x\in \mathbb Q[/imath]. [imath]f[/imath] is increasing in [imath][0,1][/imath]. [imath]f[/imath] is integrable. EDIT: Sorry, it is not discontinuous in all [imath]x\in \mathbb R \setminus \mathbb Q[/imath], just in [imath]\mathbb Q[/imath].
1162669
cardinality of well-ordered set of real numbers This question is related to cardinality of well-ordered set of a subset of real number. If [imath]E[/imath] is a subset of a real number, and the set [imath]\langle E,\prec \rangle[/imath] or [imath]\langle E,\succ\rangle[/imath] is well ordered in the ordering inherited from the real number, then : [imath]|E|\le \aleph_0[/imath]
44559
Formal proof for A subset of the real numbers, well ordered with the normal order of [imath]\mathbb R[/imath], is at most [imath]\aleph_0[/imath] I tried to write a formal proof for the theorem: [imath]A[/imath] subset of [imath]\mathbb R[/imath] well ordered by the normal order [imath]\implies A[/imath] is at most of cardinality [imath]\aleph_0[/imath]. Any suggestions? Thanks.
1151457
How do you square [imath]\sin θ\,[/imath]? Is [imath](\sinθ)^2=\sin^2θ[/imath] or [imath](\sinθ)^2=\sin(θ^2)[/imath] or [imath](\sinθ)^2=\sin^2(θ^2)[/imath] Can you explain your answer, regards Tom. Also, does your answer work for [imath]\cos[/imath] and [imath]\tan[/imath]?
932903
Ambiguity of notation: [imath]\sin(x)^2[/imath] Several people have told me that [imath]\sin(x)^2 = \sin(x^2)[/imath]. However, on several computing platforms, such as the TI-84 and Wolfram|Alpha, [imath]\sin(x)^2 = \sin^2(x)[/imath]. Can I safely conclude that the notation [imath]\sin(x)^2[/imath] is ambiguous and should always be avoided in favor of [imath]\sin^2(x)[/imath] or [imath]\sin^2 x[/imath]? I am having trouble finding any reference through Google or in textbooks (which, I presume, avoid notation like this).
1162462
Borel functions and continuous functions Suppose we have a set [imath]A\subset\mathbb {R}[/imath] and let [imath]f\in\mathcal{B}(A)[/imath] and [imath]g\in\mathcal{B}_b(A)[/imath] (Borel function on [imath]A[/imath] and bounded Borel function on [imath]A[/imath], resp.) Is it possible to approximate [imath]f[/imath] and/or [imath]g[/imath] by continuous functions on [imath]A[/imath] in a certain sense (pointwise, uniform, etc.)? So yes, how to prove that, or what are the references? So no, what is a counterexample? Thanks a lot.
549135
Construction of a function which is not the pointwise limit of a sequence of continuous functions This is somewhat linked to a prior question of mine which was looking to see if a proof of mine regarding the Dirichlet function was correct (it wasn't). I now have an answer to the question which can be answered without using directly using the Baire category theorem or the likes; as it is sufficiently different to my original approach, I feel that a new question would be the best way of going about this. The question is to Construct a function [imath]f:[0,1] \rightarrow \mathbb{R}[/imath] which is not the pointwise limit of any sequence [imath](f_n)[/imath] of continuous functions to which I will have an answer below (while I didn't come up with much of this myself, I feel it's an interesting result to discuss). Finally, as a warning to those who are seeing this as a result of searching for answers to their example sheet/homework questions, please have a think about the question before reading the answer below.
913890
Alternative proof: Matrix [imath]A[/imath] is similar to [imath]B[/imath] iff [imath]\lambda I - A[/imath] is equivalent to [imath]\lambda I - B[/imath] We have this theorem for square matrices [imath]A,B[/imath] over a field [imath]K[/imath], and [imath]\lambda[/imath] an indeterminate: If [imath]\lambda I - A[/imath] is equivalent to [imath]\lambda I - B[/imath], then [imath]A[/imath] is similar to [imath]B[/imath]. ([imath]A[/imath], [imath]B[/imath] are matrices in [imath]K^{n\times n}[/imath], and [imath]\lambda I - A[/imath] is the characteristic matrix of [imath]A[/imath]). (Note The word "equivalent" above means that you can get one from the other by doing row operations and column operations: a sequence of elementary operations on rows or on columns: add multiple of one row to another, swap two rows, or multiply one by an invertible scalar, or similar operations for columns. In this case the equivalence allows polynomial scalars in [imath]K[\lambda][/imath]). The proof which I'm aware of is very ... "calculation-style": It first proved a lemma, (something like [imath]M(\lambda) = P(\lambda)(\lambda I - A) + R[/imath]), then show you the result is correct by doing some calculations. (I think there is good chance that you know which proof I am talking about). After reading the proof, I still have no idea why it is true. Knowing a proof usually helps you understanding something, but for this one, it confuses me. So, my question is: Is there an alternative proof for this theorem, (undergraduate level), that helps me to understand what's going on.
633582
If [imath](A-\lambda{I})[/imath] is [imath]\lambda[/imath]-equivalent to [imath](B-\lambda{I})[/imath] then [imath]A[/imath] is similar to [imath]B[/imath] When reading the topic about primary and rational canonical form of matrices I stuck myself on this theorem: The matrices [imath]A,B\in K^{n\times n}[/imath] are similar if and only if their characteristic [imath]\lambda[/imath]-matrices [imath](A-\lambda{I})[/imath] and [imath](B-\lambda{I})[/imath] are [imath]\lambda[/imath]-equivalent; more precisely if [imath]P(\lambda), Q(\lambda) \in K[\lambda]^{n\times n}[/imath] are invertible such that [imath](B-\lambda{I})=Q(\lambda)(A-\lambda{I})P(\lambda)[/imath] then [imath]P(\lambda)=P[/imath] and [imath]Q(\lambda)=Q[/imath] are constant matrices satisfying [imath]Q=P^{-1}[/imath] and [imath]B=Q A P[/imath]. Theorem 21.5.1 on page 448 (in Slovak). The main point in the proof of this theorem seems to be the assertion: If [imath]P(\lambda), Q(\lambda)[/imath] are invertible and [imath](B-\lambda{I})=Q(\lambda)(A-\lambda{I})P(\lambda)[/imath] then [imath]P(\lambda)=P[/imath] and [imath]Q(\lambda)=Q[/imath] are constant matrices. What is the proof of this assertion?
1162924
Commutative ring where [imath]r[/imath], [imath]s[/imath] are associates but [imath]r \neq us[/imath] for any [imath]u[/imath] unit. First of all I think it's important to note that the definition of associates that [imath]r[/imath] divides [imath]s[/imath] and [imath]s[/imath] divides [imath]r[/imath]. Secondly, I know that my ring [imath]R[/imath] has to have zero divisors since if [imath]R[/imath] is an integral domain these two definitions of associate are equivalent. I figured I could probably find a pretty simple example of this with [imath]\mathbb{Z}_c[/imath] where [imath]c[/imath] is some composite number. I started with [imath]\mathbb{Z}_6[/imath], then [imath]\mathbb{Z}_8[/imath], then [imath]\mathbb{Z}_{10}[/imath] and finally [imath]\mathbb{Z}_{30}[/imath] before I became pretty convinced that I was unlikely to find a counterexample this simple. Unfortunately I don't have a particularly deep pool of examples of commutative rings with no zero divisors. Any help would be much appreciated. Edit: I'm also comfortable with the equivalent definition of associates that [imath](r) = (s)[/imath]. I'm assuming [imath]R[/imath] is a commutative ring with unity.
14270
Are associates unit multiples in a commutative ring with [imath]1[/imath]? Recall the following relevant definitions. We say that [imath]b[/imath] is divisible by [imath]a[/imath] in [imath]R[/imath], or [imath]a\mid b[/imath] in [imath]R[/imath], if [imath]b = r a[/imath] for some [imath]r\in R[/imath]. [imath]a[/imath] and [imath]b[/imath] are associates in [imath]R[/imath] if [imath]a\mid b[/imath] and [imath]b\mid a[/imath], (or, equivalently, if [imath]aR = bR[/imath]). [imath]u\in R[/imath] is a unit if it has a multiplicative inverse (a [imath]v\in R[/imath] such that [imath]uv=vu=1[/imath]). [imath]a[/imath] and [imath]b[/imath] are unit multiples in [imath]R[/imath] if [imath]a = ub[/imath] for some unit [imath]u\in R[/imath]. Given these definitions, my question is, If [imath]R[/imath] is a commutative ring with unity and [imath]a,b\in R[/imath] are associates in [imath]R[/imath], are [imath]a[/imath] and [imath]b[/imath] unit multiples in [imath]R[/imath]? I was told that this not always true. But I encountered some difficulties in finding a counterexample.
1163250
Solving recurrence equation with generating indices of positive indices I don't know how to solve recurrence equation with positive indices like [imath]a_{n+2} + 4a_{n+1}+ 4a_n = 7[/imath] by generating functions. How to solve such kind of problems.
357338
Recurrence Relations: general process for solving first order So I had asked a question prior to this one about recurrence relations, but apparently it was a bad one to ask. So I'm trying again to understand how to solve these babies... Here it is: [imath] 3a_{n+1}-4a_n=0, n>=0, a_1=5 [/imath] What is the general process for solving a relation like this?
1163781
Evaluate the following sum using a combinatorial argument Evaluate the following sum using a combinatorial argument: [imath] \sum\limits_{k=0}^n {n \choose k} {m \choose k} [/imath] Can someone push me in the right direction with this? I thought for combinatorial proofs there has to be a left side and a right side where one side can be used to form a question? (if that makes sense? haha) Is there a difference with combinatorial arguments? Any help would be greatly appreciated.
855538
Evaluate [imath]\sum_{k=0}^{n} {n \choose k}{m \choose k}[/imath] for a given [imath]n[/imath] and [imath]m[/imath]. How do I evaluate [imath]\sum_{k=0}^{n} {n \choose k}{m \choose k}[/imath] for a given [imath]n[/imath] and [imath]m[/imath]. I have tried to use binomial expansion and combine factorials, but I have gotten nowhere. I don't really know how to start this problem. The answer is [imath]{n+m \choose n}[/imath]. Any help is greatly appreciated. EDIT: I'm looking for a proof of this identity.
1163665
How can I prove that [imath]\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1[/imath]? How can I prove that [imath]\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?[/imath] I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
1354643
Summation of [imath]\frac{1}{k^2 - k}[/imath] from [imath]k=2[/imath] to [imath]\infty[/imath]. I couldn't get an idea how to get this summation?Can you help me please!!
1164202
Elementary tensors I need to determine whether the following function is tensor on [imath]\Bbb R^4[/imath] and express it in terms of elementary tensors. Can someone please help me with it? I do not know what elementary tensor means either. [imath]f(x,y)=3x_1y_2+5x_2 x_3[/imath] Thanks in advance!
243998
Showing that a function is a tensor I am trying to solve question 4 in Munkres Analysis on Manifolds section 26. The question is determine if the following is a tensor on [imath]\mathbb{R}^4[/imath] and express those that are in terms of the elementary tensors on [imath]\mathbb{R}^4[/imath]. [imath]f(x,y) = 3 x_1 y_2 + 5 x_2 x_3[/imath] I am trying to follow the definition of a tensor (which I don't think I fully understand) to show this. I know I must find a function [imath]T: \mathbb{R}^4 \rightarrow \mathbb{R}[/imath] that is linear on the [imath]i^{th}[/imath] variable. Since there are only 2, then I believe I have to somehow write [imath]f[/imath] as a projection. I'm just not sure where to go from here. Any hints would be great, thanks!
578459
Minimal polynomial is invariant under field extensions I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet. My question is as follows: Suppose [imath]p \in F[x][/imath] is the minimal polynomial of some matrix [imath]A \in M_n(F)[/imath] and [imath]F \subseteq K[/imath] I need to show that [imath]p[/imath] is also the minimal polynomial of [imath]A[/imath] over the field [imath]K[/imath]. In other words: Show that the minimal polynomial remains the same when we move to a bigger field. What I thought of: Obviously [imath]p[/imath] is irreducible in [imath]F[/imath], suppose it is reducible in [imath]K[/imath], which means there are [imath]q,t \in K, q,t \notin F[/imath] such that [imath]p=qt[/imath], now we just need to show that [imath]q(A) \neq 0[/imath] and [imath]t(A) \neq 0[/imath]...But why is that so? Edit: Another way of presenting this question is "Show that if [imath]p[/imath] is the minimal polynomial of [imath]A[/imath] over the field [imath]K[/imath] while the entries of[imath]~A[/imath] are contained in a subfield [imath]F \subseteq K[/imath], then all the coefficients of [imath]p[/imath] are elements of [imath]F[/imath]".
66443
A square matrix has the same minimal polynomial over its base field as it has over an extension field I think I have heard that the following is true before, but I don't know how to prove it: Let [imath]A[/imath] be a matrix with real entries. Then the minimal polynomial of [imath]A[/imath] over [imath]\mathbb{C}[/imath] is the same as the minimal polynomial of [imath]A[/imath] over [imath]\mathbb{R}[/imath]. Is this true? Would anyone be willing to provide a proof? Attempt at a proof: Let [imath]M(t)[/imath] be the minimal polynomial over the reals, and [imath]P(t)[/imath] over the complex numbers. We can look at [imath]M[/imath] as a polynomial over [imath]\Bbb C[/imath], in which case it will fulfil [imath]M(A)=0[/imath], and therefore [imath]P(t)[/imath] divides it. In addition, we can look at [imath]P(t)[/imath] as the sum of two polynomials: [imath]R(t)+iK(t)[/imath]. Plugging [imath]A[/imath] we get that [imath]R(A)+iK(A)=P(A)=0[/imath], but this forces both [imath]R(A)=0[/imath] and [imath]K(A)=0[/imath]. Looking at both [imath]K[/imath] and [imath]R[/imath] as real polynomials, we get that [imath]M(t)[/imath] divides them both, and therefore divides [imath]R+iK=P[/imath]. Now [imath]M[/imath] and [imath]P[/imath] are monic polynomials, and they divide each other, therefore [imath]M=P[/imath]. Does this look to be correct? More generally, one might prove the following Let [imath]A[/imath] be any square matrix with entries in a field[imath]~K[/imath], and let [imath]F[/imath] be an extension field of[imath]~K[/imath]. Then the minimal polynomial of[imath]~A[/imath] over[imath]~F[/imath] is the same as the minimal polynomial of [imath]A[/imath] over[imath]~K[/imath].
1164711
Fallacious proof with induction my teacher gave the following as an example of a fallacious proof: We'll prove that a group of [imath]n[/imath] people are either all male or all female. For [imath]n = 1[/imath] The claim says that a group containing 1 person is a group containing either only males or only females, which is correct. Inductive Step: Let's assume P(n) is correct. Let [imath]A = \{{a_1,a_2,...a_n,a_{n+1}}\}[/imath] be a group of [imath]n+1[/imath] people. We'll define [imath]A' = \{{a_1,a_2,...a_n}\}[/imath] and [imath]A'' = \{{a_2,...a_n,a_{n+1}}\}[/imath]. There are two options, either [imath]a_2[/imath] is male or female. Let's assume [imath]a_2[/imath] is a male. Since A' and A'' are n-sized groups, they contain all females or all males, and since [imath]a_2[/imath] is a member of both, they both contain all males. But [imath]A = A'\cup A'' [/imath], so A contains only males. A similiar approach is used if [imath]a_2[/imath] was a female. What is the fallacy in this proof?
898992
Flaw in proof that in a classroom with [imath]n[/imath] students, if there is a girl, then all students are girls Problem: Prove that: In a classroom with n student, if there is a girl student, all students of this class are girl. Solving: Let f(n) is the clause: In the class, if there is 1 girl student, all of n students are girl. First: with n=1 ---> the clause f(1) is right (of course) Second: with n=k, assume that the clause f(k) is right we will prove that the clause f(k+1) is right. Consider a set of (k+1) students {a1, a2, a3,....a(k),a(k+1)} (with a1 is the girl student) with first k student A= {a1, a2, a3,....ak}: using the inductive supposition above, we have all student in A are girl. Now we have B= {a1, a2, a3,....a(k-1),a(k+1)}: also using the inductive suppositon, we have have all student in B are girl. So the f(k+1) clause is right. Conclusion: In the class, if there is a girl student, all student are girl !!!!!!!!!!! I can't find the wrong in this proof. Please help me. thanks
1164762
Show a matrix is invertible How to show that [imath]A=\begin{pmatrix}1233&2344&1324&3456\\ 2342&11233&1432&13256\\234132&32432&1234567&43254\\423412&42354&452356&13245\end{pmatrix}[/imath] is invertible ? I could compute the determinant, but it's very very long...
175561
Is the following matrix invertible? [imath]\begin{bmatrix} 1235 &2344 &1234 &1990\\ 2124 & 4123& 1990& 3026 \\ 1230 &1234 &9095 &1230\\ 1262 &2312& 2324 &3907 \end{bmatrix}[/imath] Clearly, its determinant is not zero and, hence, the matrix is invertible. Is there a more elegant way to do this? Is there a pattern among these entries?
1164781
[imath]P,Q,R[/imath] be subspaces of a vector space [imath]V[/imath] such that [imath]V=P \cup Q \cup R[/imath] , then must one of [imath]P,Q,R[/imath] be equal to [imath]V[/imath]? Let [imath]P,Q,R[/imath] be subspaces of a vector space [imath]V[/imath] such that [imath]V=P \cup Q \cup R[/imath] , then is it true that one of [imath]P,Q,R[/imath] must be equal to [imath]V[/imath] ? I know the result about subspaces that tells that if for subspaces [imath]A,B[/imath] , [imath]A \cup B[/imath] is also a subspace then [imath]A \subseteq B[/imath] or [imath]B \subseteq A[/imath] , but for this three union case I cannot make any headway , Please help . Thanks in advance
858576
Prove that the union of three subspaces of [imath]V[/imath] is a subspace iff one of the subspaces contains the other two. Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two. I can do this problem when I am working in only two subspaces of [imath]V[/imath] but I don't know how to do it with three. What I tried is: If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??). If the union of three subspaces is a subspace..... How do I prove that one of the subspaces must contain the other two from here? *When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?
1164937
Concise proof of [imath]\operatorname{dim} V =\operatorname{dim\,ker}T+\operatorname{dim \, range}T[/imath] Can I have a concise proof of the following: [imath]\operatorname{dim} V =\operatorname{dim\,ker}T+\operatorname{dim \, range}T[/imath] I have read a few proofs of this, and they are all so long, I always forget how it is done within a few days. I could 'see' why it was true prior to seeing the proofs, so it isn't a lack of intuition.
208378
Intuitive explanation of why [imath]\dim\operatorname{Im} T + \dim\operatorname{Ker} T = \dim V[/imath] I'm having a hard time truly understanding the meaning of [imath]\dim\operatorname{Im} T + \dim\operatorname{Ker} T = \dim V[/imath] where [imath]V[/imath] is the domain of a linear transformation [imath]T:V\to W[/imath]. I've used this equation several times in many problems, and I've gone over the proof and I believe that I fully understand it, but I don't understand the intuitive reasoning behind it. I'd appreciate an intuitive explanation of it. Just to be clear, I do understand the equation itself, I am able to use it, and I know how to prove it; my question is what is the meaning of this equation from a linear algebra perspective.
1164891
Zeros of discontinuous functional is dense. Let [imath]X[/imath] be a normed linear space, then the zeros of a discontinuous linear functional is dense in [imath]X[/imath]. If [imath]Z_f[/imath] is set of zeros then [imath]Z_f[/imath] will not be closed i.e. [imath]Z_f\subsetneq \overline{Z_f}[/imath], but don't know what to do after that.
123282
Linear functional on a Banach space is discontinuous then its nullspace is dense. I need to prove that: If a nonzero linear functional [imath]f[/imath] on a Banach Space [imath]X[/imath] is discontinuous then the nullspace [imath]N_f[/imath] is dense in [imath]X[/imath]. To prove that [imath]N_f[/imath] is dense, it suffices to show that [imath]\overline N_f = X[/imath] which is equivalent to [imath](X \setminus N_f)^o=\emptyset[/imath]. (the interior of complement of [imath]N_f[/imath] is null set.) Since [imath]f[/imath] is a linear functional and is discontinuous, it has to be unbounded. I don't know exactly how to utilize these observations. Also on a related topic, I'm a little confused about how to exploit the a Linear Functional [imath]f:X \to R[/imath] or a Linear Operator [imath]T:X \to Y[/imath] being unbounded. Can I say that if a linear operator is unbounded then exists a sequence [imath]<x_n>[/imath] in [imath]X[/imath] s.t. [imath]||Tx_n|| > n^2||x_n||[/imath] for each [imath]n[/imath] or [imath]||Tx_n|| > n||x_n||[/imath] ?
488022
How to prove [imath]\inf(S)=-\sup(-S)[/imath]? Let [imath]S\subset \mathbb{R}[/imath] which is bounded below. Let [imath]x\in S[/imath] such that [imath]-x\in -S[/imath]. Prove [imath]\inf(S)=-\sup(-S).[/imath] By definition a lower bound on [imath]S\subset \mathbb{R}[/imath] is a number [imath]a\in \mathbb{R}[/imath] such that [imath]s\ge a[/imath] for each [imath]s\in S[/imath]. Since [imath]S[/imath] is non-empty and is bounded below then there exists an infimum. How can I continue this proof?
1947589
Proving [imath]\inf F = - \sup E[/imath] Assume [imath] E \subset \mathbb{R}[/imath] is nonempty and bounded above. Define [imath]F:\{-x:x\in E\}[/imath]. Prove that F is nonempty and bounded below and that [imath]\inf F = - \sup E[/imath]. Here's my rough proof: F is nonempty because [imath]-1 \in F[/imath]. I know for bounded below we need [imath] \exists \beta \in E[/imath] such that [imath] x \geq \beta \ \forall X \in F[/imath]. Other than this I'm not sure how to proceed. Any help would be much appreciated! I understand there have been similar arguments made for proving inf(A)=-sup(-A), but I'm very confused now because it seems like I need to prove inf(-A)=-sup(A)
1165079
Simplifying help: If [imath]{n \choose 3} + {n+3-1 \choose 3} = (n)_3[/imath], compute [imath]n[/imath]. If [imath]{n \choose 3} + {n+3-1 \choose 3} = (n)_3[/imath], compute [imath]n[/imath]. So far, I have: [imath]\frac{n!}{3! (n-3)!} + \frac{(n+2)!}{3! (n-1)!}[/imath] Then I simplified to: [imath]\frac{1}{6} (n-2)(n-1)(n) + \frac{1}{6} (n)(n+1)(n+2)[/imath] Is this correct, and if so, what is the next move? It needs to equal [imath](n)_3[/imath]. Thanks!
1163917
If [imath]{n \choose 3} + {n+3-1 \choose 3} = (n)_3[/imath], compute [imath]n[/imath]. If [imath]{n \choose 3} + {n+3-1 \choose 3} = (n)_3[/imath], compute [imath]n[/imath]. Just making sure I am on the right track... Do I expand each piece out and solve algebraically?
990698
How to give a good guess to the recurrence relation problem I have been trying to solve the following recurrence relation [imath]T(n)=2T(\frac{n}{2}) + nlgn[/imath] by using substitution method. I started to compute [imath]T(1)[/imath] ,[imath]T(4)[/imath],[imath]T(8)[/imath],[imath]T(16)[/imath] to guess a solution as below: [imath]T(1)=1[/imath] [imath]T(2)=2+2lg2=4[/imath] [imath]T(4)=8+8lg2=16[/imath] [imath]T(8)=32+24lg2=56[/imath] [imath]T(16)=112+64lg2=176[/imath] My guess is that [imath]T(n)[/imath] should be something like [imath]nlg^2n[/imath] but I couldn't get a correct term. Can anyone improve my guess?
159720
How to solve this recurrence [imath]T(n) = 2T(n/2) + n\log n[/imath] How can I solve the recurrence relation [imath]T(n) = 2T(n/2) + n\log n[/imath]? It almost matches the Master Theorem except for the [imath]n\log n[/imath] part.
1165783
Closed Linear spaces and the direct sum Prove that if M and N are closed linear spaces and [imath]M \perp N[/imath], then [imath]M \oplus N[/imath] is a closed linear space. I'm having trouble starting this one. Do I need to consider cauchy sequences in each M and N?
435588
Direct sum of orthogonal subspaces I'm working on the following problem set. Let [imath]\mathcal{H}[/imath] be a Hilbert space and [imath]A[/imath] and [imath]B[/imath] orthogonal subspaces of [imath]\mathcal{H}[/imath]. Prove or disprove: 1) [imath]A \oplus B[/imath] is closed, then [imath]A[/imath] and [imath]B[/imath] are closed. 2) [imath]A[/imath] and [imath]B[/imath] are closed, then [imath]A \oplus B[/imath] is closed. I could prove 1) and if my proof is correct, it even holds if [imath]\mathcal{H}[/imath] is just an inner product space. Unfortunately, I can't manage to prove 2). Since [imath]\mathcal{H}[/imath] is by assumption a Hilbert space and I didn't use that fact to prove 1), I should probably use it here. It means that [imath]A[/imath] and [imath]B[/imath] are also complete. Given some convergent sequence in [imath]A \oplus B[/imath], I want to show that the limit is also in [imath]A \oplus B[/imath]. Here I'm stuck. I want to use the completeness of [imath]A[/imath] and [imath]B[/imath], but I don't see how to obtain suitable Cauchy sequences. Can anyone drop me a hint? Or is my approach all wrong?
1165951
Prove that the norms ||.||2 and ||.||infinity on a finite dimensional space are equivalent I have to prove that the norms [imath]||\cdot ||_2[/imath] and [imath]||\cdot ||_\infty[/imath] on a finite dimensional space are equivalent. I know how to do this with two integer values for the norms where for example [imath]\displaystyle ||x||_p= \left(\sum_{i=1}^n|a)i|^p \right)^\frac{1}{p}[/imath] and [imath]p[/imath] is a constant but not with infinity since [imath]||x||_\infty=\max(|x_1|,...,|x_n|)[/imath]. Someone please help!
733595
proving topological equivalence How would I show that [imath]d_E = \sqrt{\sum_{i=1}^n(x_i-y_i)^2}[/imath] and [imath]d_\infty= \sum_{i=1}^n|x_i-y_i|[/imath] and [imath]\sup\{|x_1-y_1|,|x_2-y_2|,...,|x_n-y_n|\}[/imath] are topologically equivalent on [imath]\mathbb{R}^n[/imath]? I can do it for [imath]\mathbb{R}^2[/imath], but I'm not sure how a generalization looks like.
1165858
Show that [imath]\|A\|_\infty \leq \sqrt{n}\|A\|_2[/imath]. Show that [imath]\|A\|_\infty \leq \sqrt{n}\|A\|_2[/imath] for any [imath]m \times n[/imath] matrices. I know [imath]\|A\|_\infty[/imath] is the largest row sum of the matrix, but I'm not sure how to compare with [imath]\|A\|_2[/imath] especially since matrix 2-norm is induced by some other vector. All hints will be appreciated!
957909
Norm equivalence (Frobenius and infinity) I am trying to prove the matrix norm equivalence for norms 1, 2, [imath]\infty[/imath] and Frobenius. I have managed to solve find the constants for [imath]||.||_{1}[/imath] and [imath]||.||_{2}[/imath] but I cannot see how to continue if I want to prove the following: [imath]||A||_{\infty} \leq \sqrt{n}||A||_{2}[/imath] and the ones for Frobenius: [imath]||A||_{F} \leq \sqrt{n}||A||_{1}[/imath] [imath]||A||_{F} \leq \sqrt{n} ||A||_{2}[/imath] Any help is appreciated. Thank you for your time
1166288
[imath]\lim(x^2+y^2)/(x-y)[/imath] when [imath](x,y)\to(0,0)[/imath] How can I prove (without polar coordinates) that the [imath]\lim \frac{x^2+y^2}{x-y}[/imath] when [imath](x,y)\rightarrow (0,0)[/imath] does not exist? Moreover, can I use [imath](x^2+y^2)=(x+y)(x-y)[/imath] in this case? If I do it seems strange because I would get [imath]\frac{x^2+y^2}{x-y}=x+y[/imath] and the limit would be zero. Whereas if I try finding the limit along [imath]y=mx[/imath], the function becomes [imath]\frac{x(1+m^2)}{1-m}[/imath] which suggest that if m approaches 1 the function goes to [imath]\pm \infty[/imath]. If I try polar coordinates I get [imath]\frac{r}{\sin{a}-\cos{a}}[/imath] which suggests that the limit does not exist (and if a approaches [imath]\pi/4[/imath] the expression is [imath]\pm \infty[/imath]. Than you!
1053633
Two paths that show that [imath]\frac{x-y}{x^2 + y^2}[/imath] has no limit when [imath](x,y) \rightarrow (0,0)[/imath] I'm having a difficult time trying to find two different paths that give me different limits for the following: [imath]\lim_{(x,y) \rightarrow (0,0)} \frac{x-y}{x^2 + y^2}[/imath]
1163524
Proving a sequence formula using induction Suppose for [imath]T_n[/imath]: [imath]T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}[/imath] [imath]T_0=2,\quad T_1=3,\quad T_2=6[/imath] For integer, [imath]n \ge 3[/imath] I conjectured that: [imath]T_n = 2^n + n![/imath] The above is actually TRUE. Using induction I have to prove that. How do I go about proving: [imath]T_{n+1} = 2^{n+1} + (n+1)![/imath] Of course I will use: [imath]T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}[/imath] But can I for example change: [imath]T_{n} = 2^{n} + n![/imath] And let [imath]n \to n-1[/imath] to get: [imath]T_{n-1} = 2^{n-1} + (n-1)![/imath] I would say no? Because I havent yet proved [imath]T_n[/imath], so how can I change [imath]n \to n-1[/imath]? Thanks!
1163381
Putnam 1990 A1 Induction Help A1. [imath](150,9,1,0,0,0,0,0,1,1,6,33)[/imath] Let [imath]T_0=2,\quad T_1=3,\quad T_2=6,[/imath] and for [imath]n\ge3[/imath], [imath]T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}.[/imath] The first few terms are [imath]2,3,6,14,40,152,784,5168,40576,363392.[/imath] Find, with proof, a formula for [imath]T_n[/imath] of the form [imath]T_n=A_n+B_n[/imath], where [imath](A_n)[/imath] and [imath](B_n)[/imath] are well-known sequences. I found that: [imath]T_n - n! = 2^n[/imath] and verified that [imath]T_0 = 2^0 + 1 = 2[/imath] Hence, [imath]T_n = 2^n + n![/imath] I need to prove that: [imath]T_{n+1} = 2^{n+1} + (n+1)![/imath] Which I can't do. I went back and stated: [imath]T_{n+1} = nT_{n} + T_n + 4T_n - 4nT_{n-1} - 4T_{n-1} + 4T_{n-2} + T_{n-2} - 8T_{n-2}[/imath] What to do next?
1166134
Product of Sums: Show that the following is a Polynomial by converting it into standard form. [imath]\prod_{k=0}^n (1+x^{2^k})[/imath] The given expression simplifies to [imath](1+x)(1 + x^2)...(1 + x^{2^n})[/imath] I am not able to proceed further. How do I express this in Summation form?
1035360
Converting expressions to polynomial form My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook. Page 57. Exercise 12. Show that the following are polynomials by converting them to the form [imath]\sum_{k=0}^{m}a_kx^k[/imath] for a suitable [imath]m[/imath]. In each case [imath]n[/imath] is a positive integer. [imath]a)[/imath] [imath](1+x)^{2n}.[/imath] [imath]b)[/imath] [imath]\frac{1-x^{n+1}}{1-x}, x\not=1.[/imath] [imath]c)[/imath] [imath]\prod_{k=0}^{n}(1+x^{2^k}).[/imath] The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be [imath](1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.[/imath] Answer to part b) would be the following: [imath]\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.[/imath]thanks to @DiegoMath's hint. As for part c), we have [imath]\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})[/imath] and I have trouble "converting" this to sum which would be of a form of polynomial.
1166710
Show that the integral is purely imaginary Let f be a holomorphic function in a domain containing the closed unit disc Show that[imath] \int \overline f \frac{∂f}{∂z} dz[/imath] over the unit circle is purely imaginary I tried solving and got the answer to be purely real instead. What am I doing wrong?
1164196
integral of complex conjugate times the differential is purely imaginary I am trying to figure out why [imath]\int_C\overline{f(z)}f'(z)dz[/imath] is purely imaginary (C is a closed curve). I was told to use the Cauchy-Riemann equations to show that the real part of the integrand is an exact differential. I wrote down that [imath]\int_C\overline{f(z)}f'(z)dz = \int_C(u(x,y)-v(x,y))(\frac{du}{dx}+\frac{dv}{dx}) [/imath] The real part of this is [imath]u(x,y)\frac{du}{dx}+v(x,y)\frac{dv}{dx}[/imath]. I can apply the C-R equations, but I don't see how that would make this part dissapear. EDIT: For clarification, [imath]f(z)[/imath] is analytic in the region in which C lies. EDIT2: So integrals of exact differentials over closed curves always vanish. This means that I need to prove that [imath]u(x,y)\frac{du}{dx}+v(x,y)\frac{dv}{dx}[/imath] is a exact differential.
1166933
Multiple Integral Puzzle [imath]I=\int_1^2\int_1^2\int_1^2\int_1^2 {{x_1+x_2+x_3-x_4}\over{x_1+x_2+x_3+x_4}} \,dx_1\,dx_2\,dx_3\,dx_4[/imath] [imath]I=1/2[/imath] or [imath]1/3[/imath] or [imath]1/4[/imath] or [imath]1[/imath] ? [from ISI-Kolkatta Sample papers] I know that there must be a really clever trick to get this one. Just not able to see it. Any hint will be really useful for me.
1131769
[imath] \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 [/imath] Evaluate [imath]I= \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4[/imath] Answer Options: [imath]1[/imath] [imath]\frac{1}{2}[/imath] [imath]\frac{1}{3}[/imath] [imath]\frac{1}{4}[/imath] I need some suggestion here. I tried to evaluate one by one but it becomes messy. I think there some trick involved here.
1167242
There exists a sequence of Riemann integrable functions on [0, 1] whose pointwise limit is not Riemann integrable There exists a sequence of Riemann integrable functions on [imath][0, 1][/imath] whose pointwise limit is not Riemann integrable. I think I need to construct some sequences but I don't know where to start.
612098
An example of a sequence of Riemann integrable functions [imath](f_n)[/imath] that converges pointwise to a function [imath]f[/imath] that is not Riemann integrable. I have been given the example [imath]$$ f_n(x)= \begin{cases} 1, & \text{if}\; x\in\{(x_k)_{k=1}^{n}\}, \\ 0, & \text{otherwise.} \end{cases} $$[/imath] Here the sequence [imath](x_k)_{k=1}^{\infty}[/imath] is the sequence that "enumerates" elements of [imath]\mathbb{Q}[/imath]. The sequence of functions converges pointwise to [imath]$$ f(x)= \begin{cases} 1, & x\in\mathbb{Q}, \\ 0, & x\notin\mathbb{Q}, \end{cases} $$[/imath] which isn't Riemann integrable. However, I am not quite convinced that [imath]$f_n$[/imath] is integrable. I would like some help in seeing how [imath]$f_n$[/imath] is Riemann integrable. Also, I would like more examples of sequences of Riemann integrable functions [imath]$f_n$[/imath] that converge pointwise to a function [imath]f[/imath] that is not Riemann integrable.
1167334
Simply connected domain (Homotopy) Can anyone help me showing that : For any simply connected plain domain [imath]D_1, D_2 \subseteq \mathbb{C}[/imath] where [imath] D_1 \cap D_2 \neq \phi[/imath] is connected, show that [imath]D_1 \cup D_2[/imath] and [imath]D_1 \cap D_2[/imath] are simply connected. The restriction is I have to use homotopy notion. That is, a domain [imath]D \subseteq \mathbb{C}[/imath] is simply connected if every continuous closed curve is homotopic to a point [imath]z \in D[/imath]. I have no idea about this since homotopy is very new to me.
1161630
Intersection and union of simply connected domain I have some basics concept that, in [imath]\mathbb{C}[/imath], a "simply connected domain [imath]D[/imath]" is a region in [imath]\mathbb{C}[/imath] with no holes. I am not sure whether it has various formal definition , but the one used in my complex analysis class looks very complicated. I barely understand what it means : [imath]\textbf{Definition}[/imath] Let [imath]\gamma_0, \gamma_1[/imath] be two continuous closed path in a domain [imath]D \subseteq \mathbb{C}[/imath] parameterized by [imath]I = [0,1][/imath];that is, [imath]\gamma_i(0) = \gamma_i(1)[/imath] for all [imath]i = 0, 1.[/imath] We say [imath]\gamma_0, \gamma_1[/imath] are [imath]\textbf{homotopic as closed paths on D}[/imath] if there exists function [imath]\delta : I \times I \rightarrow D[/imath] such that [imath]1) \ \delta(t,0) = \gamma_0(t) \ \forall t \in I \ \ [/imath] [imath]2) \ \delta(t,1) = \gamma_1(t) \ \forall t \in I \ \ [/imath] [imath]3) \ \delta(0,u) = \delta(1,u) \ \forall u \in I.[/imath] A continuous closed path is [imath]\textbf{homotopic to a point}[/imath] if it is homotopic to a constant path(as a closed path). [imath]\textbf{Definition}[/imath] A region [imath]D \subseteq \mathbb{C}[/imath] is called [imath]\textbf{simply conected}[/imath] if every continuous closed path in [imath]D[/imath] is homotopic to a point in [imath]D[/imath]. [imath]\textbf{Problem :}[/imath] 1) Let [imath]D_1, D_2[/imath] be simply connected plane domains whose intersection is nonempty and connected. Prove that their intersection and union are both simply connected. 2) Let [imath]P, Q[/imath] be smooth functions on a domain [imath]D \subseteq \mathbb{C}[/imath], Find necessary and sufficient condition for the form [imath]P dz + Q d\bar{z}[/imath] to be closed.
1167345
Is an empty set equal to another empty set? I have a definition that claims that two sets are equal A = B, if and only if: [imath]\forall x ( x \in A \leftrightarrow x \in B)[/imath] An empty set contains no elements. If I define the sets: A = [imath]\emptyset[/imath] B = [imath]\emptyset[/imath] then can I conclude that A [imath]\neq[/imath] B? Since they contain no elements? Or would I have to show that an elements exists in one but not the other? To me it seems that these are not equal, but I'm struggling to wrap my head around this. Thanks, Paul
987545
Why there is a unique empty set? I Have a question. Given only the definition of equality of two sets, how can we prove that there is one and only one empty set. I mean by equality of two sets the following: [imath]A=B \iff \forall x (x \in A \iff x \in B)[/imath]
1167993
How can I obtain a solution for the equation [imath]a^2 + b^2 = c^2 + 1[/imath]? For the equation [imath]a^2 + b^2 = c^2[/imath], the solution is: [imath]a = m^2 - n^2, b= 2mn, c = m^2 + n^2[/imath] [imath]m,n\in\mathbb{Z}[/imath] and [imath]m > n[/imath], free to choose How is a similar solution obtained for the equation [imath]a^2 + b^2 = c^2 + 1[/imath]? I'm familiar with it, but don't know how to obtain it.
351491
Integral solutions of hyperboloid [imath]x^2+y^2-z^2=1[/imath] Are there integral solutions to the equation [imath]x^2+y^2-z^2=1[/imath]?
948138
By an example, show that.. Let [imath]X_1[/imath] ,[imath]X_2[/imath] be two r.v.'s with joint and marginal ch.f.'s [imath]\phi_{X_1,X_2}[/imath], [imath]\phi_{X_1}[/imath] and [imath]\phi_{X_2}[/imath]. By an example, show that [imath]\phi_{X_1,X_2} (t,t) = \phi_{X_1} (t) \phi_{X_2} (t)[/imath] [imath]\quad \forall t \in \Bbb R[/imath] does not imply independence of [imath]X_1[/imath],[imath]X_2[/imath]
376511
A criterion for independence based on Characteristic function Let [imath]X[/imath] and [imath]Y[/imath] be real-valued random variables defined on the same space. Let's use [imath]\phi_X[/imath] to denote the characteristic function of [imath]X[/imath]. If [imath]\phi_{X+Y}=\phi_X\phi_Y[/imath] then must [imath]X[/imath] and [imath]Y[/imath] be independent?
1168249
Can you check my proof that [imath]H[/imath] is characteristic? Suppose [imath]|G|=pm[/imath],where [imath]p\nmid m[/imath]. Then if [imath]H\mathrel{\unlhd}G[/imath] and [imath]|H|=p[/imath], than [imath]H[/imath] has to be characteristic. We call a subgroup characteristic when [imath]\varphi(T)\subset T[/imath], [imath]\forall \varphi \in Aut(G)[/imath]. My proof: Suppose [imath]g\in \varphi(H)[/imath] and [imath]g \notin H[/imath]. Since [imath]\varphi[/imath] is bijective [imath]|\varphi(H)|=|H|=p[/imath]. So [imath]\langle g\rangle|p[/imath] i.e. [imath]\langle g\rangle =p[/imath]. Since [imath]H[/imath] is a normal group and [imath]\langle g\rangle[/imath] is another subgroup, [imath]\langle g\rangle H[/imath] is also subgroup of [imath]G[/imath]. So [imath]p^2||G|=pm[/imath] which contradicts to the fact that [imath]p\nmid m[/imath]. Is the idea right? Which parts would you advice to fix?
112116
Normal subgroup of prime order is characteristic I'm stuck on this exercise from Herstein's topics in algebra: Suppose that [imath]G[/imath] is a group and [imath]|G| = pm[/imath], where [imath]p \not| ~ m[/imath] and [imath]p[/imath] is a prime. If [imath]H[/imath] is a normal subgroup of order [imath]p[/imath] in [imath]G[/imath], prove that [imath]H[/imath] is characteristic. How to prove that? What we know: [imath]H[/imath] and [imath]\varphi(H)[/imath] have prime order [imath]p[/imath], so they must be cyclic and abelian They are both normal subgroups of [imath]G[/imath] how to show that, in fact, we have [imath]H = \varphi(H)[/imath] for every automorphism [imath]\varphi[/imath] ?
1168362
Method of ascent to prove that [imath]x^2 − 3y^2 = 1[/imath] has infinitely many solutions Use the method of ascent to prove there are infinitely many solutions to the Diophantine equation: [imath]x^2 − 3y^2 = 1[/imath] We can do this by showing how, given one solution [imath](u, v)[/imath], we can compute another solution [imath](w, z)[/imath] that is larger is some suitable sense. Then my proof will involve finding a pair of formulas, something like: [imath]w = x + y[/imath] and [imath]z = x − y[/imath]. However I tried these formulas and they don't work. So I asked my teacher and she said that there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3.
648781
Proof that the equation [imath]x^2 - 3y^2 = 1[/imath] has infinite solutions for [imath]x[/imath] and [imath]y[/imath] being integers I have seen the Pell's equation wiki page but I need to prove this from scratch without mentioning any formula. I have also seen multiple answers on this site but the answers tend to skip over and assume formulas. This is what people do -> [imath]x^2 - 3y^2 = 1[/imath] has solution 1, 0. Next they say we "observe" that [imath](x',y') = (2x + 3y, x + 2y)[/imath] is also a solution. What I'm asking is how they get this value. Can someone help?
1168595
Prove that [imath]\gcd(2^{2^m}+1,2^{2^n}+1)=1[/imath] if [imath]m,n[/imath] are positive integers. Prove that [imath]\gcd(2^{2^m}+1,2^{2^n}+1)=1[/imath] if [imath]m,n[/imath] are positive integers. Let [imath]d=\gcd(2^{2^m}+1,2^{2^n}+1)[/imath], then [imath]d\mid 2^{2^m}+1[/imath] and [imath]d\mid2^{2^n}+1[/imath] and then [imath]d\mid2^{2^m}+1-2^{2^n}-1[/imath], i.e. [imath]d\mid2^{2^m}-2^{2^n}[/imath] where we have taken [imath]m>n[/imath]. Thus [imath]d\mid2^{2^n}(2^{2^{m-n}}-1)[/imath], but [imath]d \nmid 2^{2^n}[/imath] thus [imath]d\mid2^{2^{m-n}}-1[/imath]. Then how will I proceed??
2092080
[imath]gcd\left(2^{2^n}+1, 2^{2^m}+1\right)[/imath] Given two Fermat's number [imath]a=2^{2^n}+1[/imath] and [imath]b=2^{2^m}+1[/imath] with [imath]n,m\in\mathbb{Z}, ~n,m\ge0~\wedge~n\ne m[/imath]. Prove [imath]\gcd(a,b)=1[/imath].
1168726
Prove that a sequence is increasing A city's population in the [imath]n^{th}[/imath] year is denoted by [imath]x_n[/imath] (in millions). If, [imath]\forall n \in \mathbb N^+[/imath], we have: [imath]x_1 = \frac34[/imath], [imath]x_{n+1} = 2x_n - x_n^2[/imath], show that as [imath]n \to \infty[/imath], the population will tend to a finite limit. What is this limit? For this question, I am supposed to use the Monotone Convergence Theorem, so I decided to start off by showing that that sequence [imath](x_n)[/imath] is bounded. I wrote: For each [imath]n \in \mathbb N^+[/imath], [imath]x_n \ge 0[/imath] (population cannot be negative). Also, [imath]x_{n+1} = x_n(2 - x_n)[/imath]. If [imath]x_n > 2[/imath] then [imath]x_{n+1} < 0[/imath] which is an impossibility. Therefore [imath]0 \le x_n \le 2[/imath] and so [imath](x_n)[/imath] is bounded. For the "monotone" part of the proof, I computed a couple of terms of the sequence and realised that it was increasing towards the value [imath]1[/imath]. I tried induction but got stuck. Here's what I wrote: Let [imath]P(n)[/imath] be the proposition that [imath]x_n \le x_{n+1}[/imath]. [imath]P(1)[/imath] is true (not gonna show the working here). Assume that [imath]P(k)[/imath] is true for some [imath]k \in \mathbb N^+[/imath], i.e. [imath]x_k \le x_{k+1}[/imath]. Then I need to show [imath]x_{k+1} \le x_{k+2} = 2x_{k+1} - x_{k+1}^2[/imath]. We have [imath]x_{k+1} = 2x_k - x_k^2 \le 2x_{k+1} - x_k^2[/imath]. But from the inductive hypothesis we get [imath]x_k^2 \le x_{k+1}^2[/imath], which means [imath]2x_{k+1} - x_k^2 \ge 2x_{k+1} - x_{k+1}^2[/imath] and so we can't conclude [imath]x_{k+1} \le x_{k+2}[/imath]? Any help in this question is appreciated.
676414
Got stuck in proving monotonic increasing of a recurrence sequence I'm stuck with the prove of the following recurrence sequence which was part of and old exam. [imath]a_1:=\frac{1}{2}, a_{n+1}:=a_n(2-a_n)[/imath] for [imath]n \in \mathbb{N}[/imath] I have to show that [imath]0 <a_n < 1[/imath] and as a second part (b), that its monotonic increasing (That's where I got stuck) and as a thid part (c) the convergence and limit of the sequence a) My approach was to use induction to show that the sequence is upper and lower bounded: Base case: [imath] n=1, a_1 < 1 = \frac{1}{2} <1 [/imath] ok Inductive step: [imath]n\rightarrow n+1[/imath] [imath]a_{n+2} = a_{n+1}(2-a_{n+1}) < 1[/imath] Then my idea was to rewrite it as using the given definition from [imath]a_{n+1}[/imath]: [imath]a_{n+2}=a_n(2-a_n)(2-(a_n(2-a_n))[/imath] and I know that [imath]a_n<1[/imath] Then I used the base case that [imath]a_n=\frac{1}{2}<1[/imath] so that I've got [imath] 0 < \frac{3}{4}*(2-\frac{3}{4}) = 0.9375 <1[/imath] Question: Is this right? b) For the second part of the task, to show montonic increasing I know that [imath]a_{n+1} \leq a_n[/imath] or [imath]a_{n+1}-a_n\leq 0[/imath] [imath]a_{n+1}=a_n(2-a_n) \geq 0 \Leftrightarrow a_n-a_n^2 \geq 0 \Rightarrow a_n^2-a_n \geq 0[/imath] Here I got stuck: Question: How do I make the final conclusion? c) Limes: [imath]a_n \rightarrow a[/imath] [imath]a=a(2-a) \Leftrightarrow a=2a-a^2 \Rightarrow a^2-a=0[/imath] Solving gives [imath]a_1=1[/imath] and [imath]a_2=0[/imath] Because the sequence is monotonic increasing it converges and it must be [imath]>0[/imath] so the [imath]\lim(a_n) =1[/imath] Question: Is that conclusion "exact", I mean can I use this conlcusion or is there something missing? Best Regards, Christoph
1168644
[imath](x + y)^s \le x^s + y^s[/imath] for [imath]0 < s \le 1[/imath] I'm reading a book on valuations on a field(Hasse's number theory). He says [imath](x + y)^s \le x^s + y^s[/imath] for [imath]0 < s \le 1[/imath] and [imath]x, y \ge 0[/imath]. I tried to prove it but failed. Another question: Is [imath](x + y)^s \ge x^s + y^s[/imath]? where, [imath]s \gt 1[/imath] and [imath]x, y \ge 0[/imath]. OK, the first questions is duplicate, but IS THE SECOND QUESTION DUPLICATE, TOO?
1155225
How to prove for [imath]s<1,|a+b|^s\le|a|^s+|b|^s[/imath] How to prove for $s<1[imath]$|a+b|^s\le|a|^s+|b|^s[/imath] I tried to prove $|a+b|\le (|a|^s+|b|^s)^{\frac{1}{s}}[imath], but [/imath]\frac{1}{s}$ may not be integer, so I do not know how to expand it if it is not integer. Is it the right way to prove it by trying to expand it using binomial expansion? How to prove it? Thanks!
1167930
Interchange summation and differentiation I asked this question already on math.stackexchange, but did not receive any answers see here Let [imath]f = \sum_{n=0}^{\infty} a_n e_n [/imath] where [imath]e_n[/imath] are an ONB of [imath]L^2[0,1].[/imath] Now assume we have that [imath]\frac{d}{dx}e_n = \lambda_n e_n.[/imath] Assume [imath]f \in H^1[0,1],[/imath] so i.e. [imath]||f'||_{L^2} < \infty[/imath] I want to say now that [imath]\frac{d}{dx} f = \sum_{n} a_n \lambda_n e_n[/imath] but I am not sure how to justify the interchange of summation and differentiation. How exactly does this follow?
1166721
Interchange summation and differentiation for ONB Let [imath]f = \sum_{n=0}^{\infty} a_n e_n [/imath] where [imath]e_n[/imath] are an ONB of [imath]L^2[0,1].[/imath] Now assume we have that [imath]\frac{d}{dx}e_n = \lambda_n e_n.[/imath] Assume [imath]f \in H^1[0,1],[/imath] so i.e. [imath]||f'||_{L^2} < \infty[/imath] I want to say now that [imath]\frac{d}{dx} f = \sum_{n} a_n \lambda_n e_n[/imath] but I am not sure how to justify the interchange of summation and differentiation. So I am trying to make rigorous here what physicists always do.
1168760
Irrationality proof trick with Mod You will see here: Bill Dubuque's Slick [imath]\sqrt{3}[/imath] irrationality proof What is the trick with modulus for proving irrationality? What about [imath]\sqrt{2}[/imath] Can you prove this is irrational by that trick?
598295
Is there a proof of the irrationality of [imath]\sqrt{2}[/imath] that involves modular arithmetic? I was reading Ian Stewart's Concepts of Modern Mathematics. Using congruences, It's possible to explain why all perfect squares end in [imath]0,1,4,5,6,9[/imath] but not in [imath]2,3,7,8[/imath]. With this I had the idea of exploring the congruences for both sides of [imath]n^2=2m^2[/imath] in Mathematica: Table[Mod[n^2, 9], {n, 0, 20}] Table[Mod[2 m^2, 9], {n, 0, 20}] And had the results: {0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4} {0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8} But I'm still not sure if the outputs really show what I'm looking for, I have also tried [imath]mod \;10[/imath]. The idea is still pretty loose in my mind, I'm stuck on deciding if this proves something or what directions I could take in this enterprise.
1166708
Permutation(?) mapping Problem Statement: Let [imath]G[/imath] be a finite group, say a group with [imath]n[/imath] elements, and let [imath]S[/imath] be a nonempty subset of [imath]G[/imath]. Suppose [imath]e \in S[/imath], and that [imath]S[/imath] is closed with respect to multiplication. Prove that [imath]S[/imath] is a subgroup of [imath]G[/imath]. (HINT: It remains to prove that [imath]G[/imath] is closed with respect to inverses. Let [imath]G = \{a_1, \ldots ,a_n\}[/imath]; one of these elements is [imath]e[/imath]. If [imath]a_i \in G[/imath], consider the distinct elements [imath]\{a_ia_1, a_ia_2, \ldots a_ia_n\}[/imath]. Incomplete Solution: Let [imath]G = \{a_1, \ldots, a_n\}.[/imath] Let [imath]a_i \in G.[/imath] Then any number of [imath]a_ia_1, a_ia_2, \ldots, a_ia_n \in S,[/imath] since [imath]S[/imath] is closed under multiplication. One of those [imath]a_ia_j[/imath] is [imath]e[/imath] since [imath]e \in S.[/imath] Then either [imath](a_i)^{-1}[/imath] or [imath](a_j)^{-1} \in S.[/imath] To complete the proof I was told: You need to argue why one of them must be [imath]e[/imath]. Show [imath]f(x)=a_i \cdot x[/imath] is an injection, then since it's a map from a finite set to itself that implies it's also a surjection. Let [imath]a_i \cdot x = a_i \cdot y[/imath]. Then [imath]x = y[/imath]. So, [imath]f[/imath] is injective. Not too sure if that's correct, but I have a follow up question: For [imath]x = 1, 2, 3, \ldots, n[/imath]; [imath]f(x) \in \{a_i, a_i \cdot 2, a_i \cdot 3, \ldots a_i \cdot n\}[/imath]. So, [imath]f: \{1, 2, 3, ..., n\} \to \{a_i, a_i \cdot 2, a_i \cdot 3, ..., a_i \cdot n\}[/imath]. Is that correct? If so, how do we know [imath]f[/imath] maps from a finite set to itself?
1166317
Showing a set is closed under inverses Let [imath]G[/imath] be a finite group, say a group with [imath]n[/imath] elements, and let [imath]\emptyset \neq S \subseteq G[/imath]. Suppose [imath]e \in S[/imath], and that [imath]S[/imath] is closed with respect to multiplication. Prove that [imath]S[/imath] is a subgroup of [imath]G[/imath]. Let [imath]G = \{a_1, \ldots, a_n\}.[/imath] Let [imath]a_i \in G.[/imath] Then any number of [imath]a_ia_1, a_ia_2, \ldots, a_ia_n \in S,[/imath] since [imath]S[/imath] is closed under multiplication. One of those [imath]a_ia_j[/imath] is [imath]e[/imath] since [imath]e \in S.[/imath] Then either [imath]a_i[/imath] or [imath]a_j[/imath] is [imath]a^{-1}.[/imath] Would this argument work?
753042
Finitely many prime ideals lying over [imath]\mathfrak{p}[/imath] Let [imath]A[/imath] be a commutative ring with identity and [imath]B[/imath] a finitely generated [imath]A[/imath]-algebra that is integral over [imath]A[/imath]. If [imath]\mathfrak{p}[/imath] is a prime ideal of [imath]A[/imath], there are only finitely many prime ideals [imath]P[/imath] of [imath]B[/imath] such that [imath]P\cap A=\mathfrak{p}[/imath]. Let me say that I am aware of this answer, but I can't follow through the hint. Also, I don't know how to extend to work for algebras rather than extension rings.
2055952
Finiteness of the fibers of the prime spectrum Let [imath]A\rightarrow B[/imath] be a ring homomorphism such that [imath]B[/imath] is a finitely generated [imath]A[/imath]-module. How one shows that the (set-theoretic) fibers of the map [imath]\operatorname{Spec}B\rightarrow\operatorname{Spec}A[/imath], where the spectra are considered as topological spaces, are always finite?
1169251
prove that [imath]f(x,y)= \frac{x^3y}{x^4+y^2}[/imath] for [imath](x,y) \neq (0,0)[/imath] and [imath]f(0,0) = 0[/imath] is continous at [imath](0,0)[/imath] How to prove that [imath]f: \mathbb R^2 \to \mathbb R[/imath] [imath]f(x,y)= \frac{x^3y}{x^4+y^2}[/imath] for [imath](x,y) \neq (0,0)[/imath] and [imath]f(0,0) = 0[/imath] is continous at [imath](0,0)[/imath]? By definition: Let [imath]\epsilon>0[/imath] I need to prove that [imath]\exists \delta>0[/imath] so that [imath]\forall \vec x\in B_\delta(0,0)[/imath] then [imath]|f(x,y)-0|<\epsilon[/imath] [imath]|f(x,y)|=|{x^3y\over x^4+y^2}|[/imath]=[imath]{|x^3y|\over x^4+y^2}[/imath] I know that [imath]|x|,|y|\le ||\vec x||[/imath] so [imath]|x^3y|\le ||\vec x||^4[/imath] then [imath]{|x^3y|\over x^4+y^2}\le {||\vec x||^4\over x^4+y^2}[/imath] but I don´t know how to proceed from here Any ideas?
674095
Prove [imath]\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0[/imath] How would you prove the following limit? [imath]\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0[/imath] I think the best way is using the squeeze theorem but I can't find left expression. [imath]0 \le \frac{x^3y}{x^4 + y^2} \le \frac{x^3y}{x^4} \le \frac{x^3y}{x^3} \le y = 0.[/imath] But I'm not sure I'm right (especially at [imath]\frac{x^3y}{x^4} \le \frac{x^3y}{x^3}).[/imath] If I'm right - I'd glad if you can accept it. If I'm wrong - can you please correct me? Thanks in advance!
1169333
What is the value of this Infinite Product of prime numbers expression? What is the value of: [imath]\prod_1^\infty \frac{p_i^2}{p_i^2 -1 }[/imath] Where [imath]p_i[/imath] are the prime numbers: 2, 3, ...
621918
Solving infinite sums with primes. Let [imath]p_n[/imath] denote the [imath]n[/imath]'th prime number. How would one go about proving that infinite products like: [imath]\prod_{k=1}^\infty1 - \frac{1}{(p_k)^2} = \frac{6}{\pi^2}[/imath] or [imath]\prod_{k=1}^\infty\frac{{p_k}^2}{{p_k}^2 - 1} = \frac{\pi^2}{6}[/imath] are correct? Is there any way to prove it except by exhaustion?
1169666
Why can't [imath]A_4[/imath] has a subgroup of order [imath]6[/imath]? Can anyone provide with an explanation of why the group [imath]A_4[/imath], which is the group formed by the set of even permutations of [imath]S_4[/imath] under the operation of composition of functions, can not have an order of [imath]6[/imath]? I know Lagrange's Theorem tells us that the orders of possible subgroups of [imath]A_4[/imath] are [imath]1,2,3,4,6,12[/imath], and I can find a subgroup a subgroup of all of the orders listed except [imath]6[/imath], and I'm pretty sure there is not one, but cannot come up with a solid explanation as to why this cannot happen? Thanks for the help!
544610
Use a lemma to prove that [imath]A_4[/imath] has no subgroup of order [imath]6[/imath]. Lemma: If [imath]H\le G[/imath] has index [imath]2[/imath], i.e. [imath][G:H]=2[/imath], then for any [imath]a\in G[/imath] we have [imath]a^2\in H[/imath]. The [imath]12[/imath] elements of [imath]A_4[/imath] are [imath](1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143)[/imath], [imath](234)[/imath], and [imath](243)[/imath].
1169565
evaluating Polar Integrals. Cartesian to Polar? I can't for the life of me figure out this problem. There's not example in my textbook. I'm suppose to convert this into a polar integral and evaluate it [imath]\int_0^6 \int_0^y x \;dx \;dy[/imath] I have my graph for this, but the upper bound of the y-axis I have no clue on how to convert that into polar
269868
Transform to polar the following integral [imath]\int_0^6\int_0^y x \, dx \, dy[/imath] I need to transform this integral [imath]\int_0^6\int_0^y x \, dx \, dy[/imath] to polar and then find its value. I'm stuck finding the r-limits of integration.
318136
The range of [imath]T^*[/imath] is the orthogonal complement of [imath]Ker(T)[/imath] How can I prove that, if [imath]V[/imath] is a finite-dimensional vector space with inner product and [imath]T[/imath] a linear operator in [imath]V[/imath], then the range of [imath]T^*[/imath] is the orthogonal complement of the null space of [imath]T[/imath]? I know what I must do (for a [imath]v[/imath] in the range of [imath]T^*[/imath], I have to show that [imath]v\perp w[/imath] for every [imath]w[/imath] in [imath]\ker(T)[/imath] and then do the opposite), but I don't know how to show that this inner product is zero.
642418
range of adjoint operator Let [imath]V[/imath] be an inner product space and [imath]T:V \to V[/imath] be a linear operator. Show that [imath]N(T)^\perp=R(T^*)[/imath]. Trial: [imath]N(T)=R(T^*)^\perp[/imath] since [imath]y \in N(T) \iff Ty=0 \iff \langle x,Ty \rangle=0 \quad \forall x \iff \langle T^*x,y \rangle=0 \quad \forall x \iff y \in R(T^*)^\perp[/imath]. Taking [imath]\perp[/imath] both sides, [imath]N(T)^\perp=(R(T^*)^\perp)^\perp[/imath]. I know that for finite dimensional [imath]W[/imath], [imath](W^\perp)^\perp=W[/imath]. But in this case it may not be finite dimensional. Does it need another approach?
1169989
Find the number of trailing zeroes. Find the number of trailing zeroes. [imath]k=1^1\times 2^2\times 3^3\times \cdots \times100^{100}[/imath] It usually involves calculating number of [imath]5[/imath]'s in [imath]5^5\times 10^{10}\times 15^{15}\times \cdots\times 100^{100}[/imath] calulating 5's one by one is pretty boring and time consuming are their any other methods.
1128755
How many 0's are in the end of this expansion? How many [imath]0's[/imath] are in the end of: [imath]1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4.... 99^{99}[/imath] The answer is supposed to be [imath]1100[/imath] but I have absolutely NO clue how to get there. Any advice?
1168234
Making values of functionals simultaneously positive We work with real Banach spaces. Let [imath]X[/imath] be an infinite-dimensional Banach space and let [imath](f_n)[/imath] be a sequence of norm-one functionals on [imath]X[/imath] that are linearly independent. Can we find an element [imath]x\in X[/imath] such that [imath]f_n(x)>0[/imath] for all [imath]n[/imath]? I guess not. What if [imath]X[/imath] is reflexive?
313425
A certain element which makes functionals positive Suppose [imath]X[/imath] is a possibly non-separable Banach space and let [imath]X^*[/imath] be its dual. Also, let [imath](f_n)_{n=1}^\infty[/imath] be a sequence of [EDIT: linearly independent] norm-one functionals in [imath]X^*[/imath]. Does there exist an element [imath]x\in X[/imath] such that [imath]f_n(x) \in (0,\infty)[/imath] for all [imath]n[/imath]?
1170531
Probability of the maximum of a reflecting Brownian motion Let [imath]\{W_t\}_{t\geq 0}[/imath] be a standard Brownian motion (starting at [imath]0[/imath]). For [imath]T[/imath] large enough, I would like to prove that [imath]P(\max_{t\in[0,T]} |W_t| \leq c T^{1/3})[/imath] is bigger than a negative power of [imath]T[/imath] (like [imath]a T^{-1/6}[/imath] for example, for an appropriate constant [imath]a > 0[/imath]) or at least bigger than [imath]e^{-b T^{1/3}}[/imath] if [imath]b[/imath] can be made arbitrarily close to [imath]0[/imath]. We know that without the absolute value, i.e. [imath]P(\max_{t\in[0,T]} W_t \leq c T^{1/3})[/imath], this property is true because [imath]\max_{t\in [0,T]} W_t[/imath] has the same law as [imath]|W_T|[/imath]. So the result should be true intuitively because keeping [imath]|W_t|[/imath] in a cylinder should not be much less probable than restricting a standard Brownian motion to be lower than a bound of the same order. I've tried to prove this result without success. Thanks for any help.
1169233
Lower bound on the probability of the maximum of a reflecting Brownian motion Let [imath]\{W_t\}_{t\geq 0}[/imath] be a standard Brownian motion (starting at [imath]0[/imath]). For [imath]T[/imath] large enough, I would like to prove that [imath]P(\max_{t\in[0,T]} |W_t| \leq c T^{1/3})[/imath] is bigger than a negative power of [imath]T[/imath] (like [imath]a T^{-1/6}[/imath] for example, for an appropriate constant [imath]a > 0[/imath]) or at least bigger than [imath]e^{-b T^{1/3}}[/imath] if [imath]b[/imath] can be made arbitrarily close to [imath]0[/imath]. We know that without the absolute value, i.e. [imath]P(\max_{t\in[0,T]} W_t \leq c T^{1/3})[/imath], this property is true because [imath]\max_{t\in [0,T]} W_t[/imath] has the same law as [imath]|W_T|[/imath]. So the result should be true intuitively because keeping [imath]|W_t|[/imath] in a cylinder should not be much less probable than restricting a standard Brownian motion to be lower than a bound of the same order. I've tried to prove this result without success. Thanks for any help.
1171084
Evaluate [imath]\sum_{n=1}^\infty (-1)^n/(n^4)[/imath] Given that [imath]\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}[/imath] find [imath]\sum_{n=1}^\infty \frac{(-1)^n}{n^4}[/imath] The answer should be [imath]\frac{-7\pi^4}{720}[/imath].
1169861
Use [imath]\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}[/imath] to compute [imath]\sum_{n=1}^\infty \frac{(-1)^n}{n^4}[/imath] Is it possible to use the fact that [imath]\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}[/imath] to compute [imath]\sum_{n=1}^\infty \frac{(-1)^n}{n^4}[/imath]?
1171106
If a sequence converges, so does its norm Let [imath]\{u_k\}[/imath] be a sequence in [imath]\mathbb{R}^n[/imath]. Prove that if [imath]u_k[/imath] converges, then [imath]\{||u_n||\}[/imath] converges.
279762
Convergence of a sequence implies convergence of the norm of the sequence If [imath]\langle X_n \rangle[/imath] is a sequence in [imath]\mathbb R^n[/imath] then [imath]\lim\limits_{n\to \infty}X_n=A[/imath]. Show that [imath]\lim\limits_{n\to \infty} \| X_n \| = \| A \|[/imath] . My thought process, [imath]\| X_i \|= (\sum_{i=1}^n x_i^2)^{(1/2)}[/imath] therefore be the multiplicative property of limits [imath]x_i^2 = a_i^2[/imath]. Therefore the limit of norm of the components is equivalent to the limit of the corresponding limit of each component of the sequence.
1084967
Functional equation [imath]f(ax)=bf(x)[/imath] What are all the solutions to the functional equation [imath]f(ax)=bf(x)[/imath], where [imath]a,b>0[/imath], and [imath]f[/imath] is continuous, strictly monotone and increasing, and [imath]x[/imath] ranges over the reals? references? proof? Thanks Additional details following the first response: It is easy to see that the function [imath]f(x)=c\cdot x^\alpha[/imath], with [imath]\alpha =\log b/\log a[/imath] and any constant [imath]c[/imath] is a solution for the functional equation. Also, the [imath]c[/imath] can be different for the positives and the negatives. So, if [imath]x^\alpha[/imath] is not monotone itself, we can create a monotone solution by gluing together a positive [imath]c[/imath] for the positive side with a negative [imath]c[/imath] for the negative side. Is this correct? My question is if these are all the solutions, and if this is appears in the literature. Thanks.
1033898
Solve the functional equation [imath]2f(x)=f(ax)[/imath] for some [imath]a[/imath]. I am trying to solve the following functional equation, and could use some help.[imath] 2f(x)=f(ax)[/imath] For some [imath]a\in\mathbb{R}[/imath]. By repeated adding [imath]2f(x)[/imath] together we notice that [imath]2nf(x)=f(a^nx).[/imath] Also [imath]2mf(a^{-m}x)=f(x)\Rightarrow \frac{1}{2m}f(x)=f(a^{-m}x).[/imath] Putting it all together I have for all [imath]m,n\in\mathbb{N}[/imath], [imath]\frac{n}{m}f(x)=\frac{2n}{2m}f(x)=f(a^{n-m}x)[/imath] I am unsure of how to proceed from here.
1170674
About solving a second order difference equation Let [imath]r>4[/imath] be a positive integer. I want to solve this difference equation: [imath]u_{n+1}-r²(1+r²ⁿ⁺¹)u_{n}+r²r²ⁿ⁺¹u_{n-1}-2r²r²ⁿ⁺¹=0[/imath] but I have no a good idea to start.
1138033
Solving a difference equation with several parameters Let [imath]r>4[/imath] be a positive integer. Let us consider this difference equation: [imath]u_{q+1}=(r^{2q+1}+(c/a))u_{q}-(c/a)r^{2q-1}u_{q-1} +2c+d-(bc/a)[/imath] where [imath]a,b,c,d[/imath] are integers. I want to find a closed form, but I am not able to find the good idea. Or at least how we can prove that a solution exists.
957206
difference between independent binomial variables It is well known that if [imath]X \sim B(m, p)[/imath] and [imath]Y \sim B(n, p)[/imath] are independent then [imath]X+Y \sim B(m+n, p)[/imath] but what is the distribution of [imath]X-Y[/imath]? Here is what I have tried. [imath]\Pr[X-Y = c] = \sum_{i=0}^n \Pr[X = c+i] \Pr[Y = i][/imath] which can be simplified to [imath]\left(\frac{p}{1-p}\right)^c (1-p)^{m+n} \sum_{i=0}^n \binom{m}{c+i} \binom{n}{i} \left(\frac{p}{1-p}\right)^{2i}[/imath]. Letting [imath]q = \left(\frac{p}{1-p}\right)^2[/imath], simplifying this expression reduces to simplifying [imath]\sum_{i=0}^n \binom{m}{c+i} \binom{n}{i} q^i[/imath]. But I am stuck here and perhaps there is a better approach.
1065487
Difference between two independent binomial random variables with equal success probability Let [imath]X[/imath] ~ [imath]Bin(n,p)[/imath] and [imath]Y[/imath] ~ [imath]Bin(m,p)[/imath] be two independent random variables. Find the distribution of [imath]Z=X-Y[/imath]. see also Difference of two binomial random variables I figured this out: [imath] P(Z=z)=\cases{\sum_{i=o}^{min(m,n)} Bin(k+i,n,p)*Bin(i,m,p), &if $z\ge0$;\cr \sum_{i=0}^{min(m,n)} Bin(i,n,p) * Bin(i-z, m, p),&otherwise. \cr}[/imath] I also validated it by Monte Carlo simulation. For [imath]n=30[/imath], [imath]m=20[/imath] and [imath]p=0.5[/imath], I get the following distribution, where the circles are the analytical probabilities and the line connects the MC estimates. Because that looked to me pretty much like a binomial distribution, I gave it a try and figured out that its actually a binomial, just shifted by m to the left. This can be simply written as [imath]P(Y=y) = Bin(y+m, m+n, p)[/imath]. Hence, given equal success probabilities, the sum of two independent binomially distributed random variables is binomial, but also their difference, just shifted to the left. This question here difference between independent binomial variables is actually the same as mine, but received no answer and only the comment that there would be no simple formula. But the above formula looks pretty simple to me. Is it correct that for the case of equal success probabilities, the above equations actually describes the distribution of [imath]Z=X-Y[/imath]? I read in a book that [imath]Z[/imath] could not be binomial distributed because it had negative support. Is it right to call it a shifted binomial?
1171853
Intuition behind the substitution method of integration I want to know what is the rationale behind substitution method of integration.I'm very familiar with the following sort of integration but I don't understand why we substitute i.e use one variable in terms of another variable. For the integral, [imath]\int 2x\sqrt{x^2+3}\,dx[/imath] we can use the substitution method in the following manner. If [imath]z=x^2+3,[/imath] then [imath]dz=2xdx[/imath]. We have [imath]\int\sqrt{z}dz[/imath]. My question is that why [imath]\int\sqrt{z}dz[/imath]=[imath]\int 2x\sqrt{x^2+3}\,dx[/imath]? Is there rationale behind it?
77306
Why does substitution work in antiderivatives? I'm not entirely sure what I want to ask here, so please bear with me! I think the explanation they give us in school for how finding the antiderivative by substitution works is: [imath]\int f(g(t))g'(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du[/imath] But I never really understood the equality [imath]F(u)=\int f(u)du[/imath]. Why can we behave as if the identity [imath]u=g(t)[/imath] doesn't exist, compute the integral [imath]\int f(u)du[/imath], and then 'substitute' [imath]u[/imath] in the result yielding the correct answer? In a sense, it feels like the variable [imath]u[/imath] switches from being 'meaningful' in [imath]F(u)[/imath] (where it stands for [imath]g(t)[/imath]) and being insignificant in [imath]\int f(u)du[/imath], since as I perceive it the "variable name" here is meaningless and we could say [imath] \int f(u)du = \int f(k)dk = \int f(x)dx = ... [/imath] all the same. Another way to ask the same thing: why [imath](Fg)(t)=\int f(u)du[/imath] and not, say, [imath](Fg)(t)=\int f(x)dx[/imath]? Where am I getting confused? Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!
1147103
Homology of [imath]\mathbb{R}\setminus A_+[/imath]. Let [imath]A[/imath] be the unit circle in the [imath]xy[/imath] plane in [imath]3[/imath]-dimensional real space and let [imath]A_+[/imath] be a semicircle. I have to compute the homology of [imath]\mathbb{R}^3\setminus A_+[/imath]. I was thinking that [imath]\mathbb{R}^3\setminus A_+[/imath] is homotopically equivalent to [imath]\mathbb{R}^3\setminus \{(1,0,0)\}[/imath] which can than be proved to be homeomorphic to [imath]S^2[/imath]. However, I have trouble finding a retraction [imath]g:\mathbb{R}^3\setminus \{(1,0,0)\}\rightarrow \mathbb{R}^3\setminus A_+[/imath] to construct the equivalence, since homotopy equivalence is not stable under subtractions, i.e. if [imath]A,B\subset X[/imath], [imath]A[/imath] homotopically equivalent to [imath]B[/imath] it does not follow that [imath]X\setminus A[/imath] homotopically equivalent to [imath]X\setminus B[/imath].
1171261
Homology groups of [imath]\mathbb{R}^3\setminus \ S^1[/imath] and [imath]\mathbb{R}^3\setminus E^1_+[/imath] Here is a problem and my attempt at the solution. If my conclusion or the proof is incorrect I would appreciate a pointer in the right direction. Thanks in advance. Let [imath]S^1[/imath] be the unit circle in the xy plane in [imath]\mathbb{R^3}[/imath] and let [imath]E^1_+[/imath] and [imath]E^1_-[/imath] be two of its semicircles. Find the homology groups (with integer coefficients) of a) [imath]\mathbb{R^3}\setminus E^1_+[/imath] and b) [imath]\mathbb{R^3}\setminus \ S^1[/imath]. Attempt of solution: a) [imath]H_n(\mathbb{R^3},S^1)\cong H_n(\mathbb{R^3}\setminus E^1_+, E^1_-)[/imath], by excision. Then [imath]H_n(\mathbb{R^3}\setminus E^1_+, E^1_-) \cong H_n(\mathbb{R^3}\setminus E^1_+, {pt})[/imath], since [imath]E^1_-[/imath] is contractible. Then we know [imath]H_n(\mathbb{R^3}\setminus E^1_+, {pt}) \cong \tilde{H_n}(\mathbb{R^3}\setminus E^1_+)\cong H_n(\mathbb{R^3}\setminus E^1_+)[/imath] for [imath]n>0[/imath]. So we get [imath]H_n(\mathbb{R^3}\setminus E^1_+) \cong H_n(\mathbb{R^3},S^1)[/imath]. From the long exact sequence of relative homology we know that [imath]H_n(\mathbb{R^3},S^1) \cong H_{n-1}(S^1)[/imath], so [imath]H_n(\mathbb{R^3}\setminus E^1_+)\cong \mathbb{Z}[/imath] for [imath]n=1,2[/imath]. For [imath]n=0[/imath] we also get [imath]\mathbb{Z}[/imath] since [imath]\mathbb{R^3}\setminus E^1_+[/imath] is arcwise connected. b) the same reasoning but starting with [imath]H_n(\mathbb{R}^3,E^2_+)[/imath], where [imath]E^2_+[/imath] is the upper hemisphere of [imath]S^2[/imath], gives [imath]H_n(\mathbb{R}^3 \setminus S^1) \cong H_{n-1}(E^2_+)[/imath], but [imath]E^2_+[/imath] is contractible so [imath]H_n(\mathbb{R}^3 \setminus S^1)\cong \mathbb{Z}[/imath] for [imath]n=1[/imath] and for [imath]n=0[/imath] it is also [imath]\mathbb{Z}[/imath] since [imath]\mathbb{R}^3 \setminus S^1[/imath] is arcwise connected. For [imath]n>1[/imath], [imath]H_n(\mathbb{R}^3 \setminus S^1)=0[/imath]. EDIT @msteve pointed out that b) is wrong. Excision can't be used like in a) since [imath]S^1[/imath] isn't in the interior of [imath]E^2_+[/imath]. So a different approach is needed here.
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Does [imath](1093)^2|2^{1092}-1[/imath]? I have this problem: Show that [imath](1093)^2|2^{1092}-1[/imath]. By this | I mean divide. I would like to write here my progress but I have absolutely no clue how to do show that it divides. Any ideas? Thanks
97557
Is [imath]2^{1093}-2[/imath] a multiple of [imath]1093^2[/imath]? I can't solve this problem; it may be easy though. Is the number [imath]2^{1093} -2[/imath] a multiple of [imath]1093^2[/imath]? I do appreciate any kind of help.
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Analytic solution of the non-linear ODE: [imath]{u}''+\frac{1}{x}{u}'=-\delta e^{u}[/imath] I am trying to find the analytic solution of [imath]{u}''+\frac{1}{x}{u}'=-\delta e^{u}[/imath] given the homogeneous mixed boundary conditions [imath]{u'(0)}=0[/imath] [imath]u(1)=0[/imath] How would one attack such a problem? I have been given that the analytic solution is [imath]u=ln\left ( \frac{8a/\delta }{(ax^2+1)^2} \right )[/imath] where [imath]a[/imath] solves [imath]8a=\delta (a+1)^2[/imath]. I am guessing the only way to solve it analytically is to introduce [imath]x=e^{-y}[/imath], but I don't see how would this workout all the way. The equation cannot be reduced to a 1st-order by a change of variable, nor is separable. Any ideas?
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Analytic solution of: [imath]{u}''+\frac{1}{x}{u}'=-\delta e^{u}[/imath] I am trying to find the analytic solution of [imath]{u}''+\frac{1}{x}{u}'=-\delta e^{u}[/imath] given the homogeneous mixed boundary conditions [imath]{u'(0)}=0[/imath] [imath]u(1)=0[/imath] How would one attack such a problem? I have been given that the analytic solution is [imath]u=ln\left ( \frac{8a/\delta }{(ax^2+1)^2} \right )[/imath] where [imath]a[/imath] solves [imath]8a=\delta (a+1)^2[/imath]. My approach (EDITED): I was given a hint: use the relation [imath]x=e^{-y}[/imath]. I proceeded with that and used the chain-rule as follows [imath](u\circ y)'(x)=\frac{d}{dx}u(y(x))=u'(y(x))y'(x)=u'x=-u'e^{y}[/imath] [imath](u\circ y)''(x)=\frac{d}{dx}[u'(y(x))y'(x)]=u''(y(x))[y'(x)]^2+u'(y(x))y''(x)=u''e^{2y}+u'e^{2y}[/imath] substituting these relations into the ODE yields [imath]e^{2y}u''=-\delta e^{u}[/imath] or in an alternative form [imath]u''=-\delta e^{-2y} e^{u}[/imath] which is separable (thanks for the help).
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Find a function [imath]f :\Bbb R^2\to \Bbb R[/imath] which is not differentiable at [imath](0,0)[/imath] even though all directional derivatives of [imath]f[/imath] exist at [imath](0,0).[/imath] This is a practice exam question and I have no idea how to start it. Find a function [imath]f :\Bbb R^2\to \Bbb R[/imath] which is not differentiable at [imath](0,0)[/imath] even though all directional derivatives of [imath]f[/imath] exist at [imath](0,0).[/imath]
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Example of a function [imath]f:\mathbb{R}^2\to\mathbb{R}[/imath] not differentiable at [imath](0,0)[/imath], but has a directional derivative at [imath](0,0)[/imath] in all directions Give an example of a function [imath]f:\mathbb{R}^2\to\mathbb{R}[/imath] such that [imath]f'_u(0,0)[/imath] exists in all directions [imath]\|u\| = 1[/imath], but [imath]f[/imath] is not differentiable at [imath](0,0)[/imath]. You have to show that your example satisfy the above requirement.
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How to prove: if [imath]\forall a,b,c[/imath], [imath]ab=ac\Rightarrow b=c[/imath] and [imath]ba=ca\Rightarrow b=c[/imath] then [imath](G,\circ)[/imath] is a group I'm reading basic algebra notes and met this exercise: Let [imath]\circ[/imath] be an operator defined on [imath]G[/imath], if [imath]\forall a,b,c[/imath], [imath]ab=ac\Rightarrow b=c[/imath] and [imath]ba=ca\Rightarrow b=c[/imath] then [imath](G,\circ)[/imath] is a group. I couldn't figure this out. Could you pls give me some hint?
510157
Show that G is a group, if G is finite, the operation is associative, and cancellation law hols Let [imath]G[/imath] be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. [imath]ab=ac[/imath] or [imath]ba=ca[/imath] implies [imath]b=c[/imath], for any choices of [imath]a,b,c[/imath] in [imath]G[/imath]. Assume that there is an identity element [imath]e[/imath] in [imath]G[/imath]. Show that [imath]G[/imath] is a group. Proof: To show [imath]G[/imath] is a group, conditions must hold. Suppose that [imath]ab=ac[/imath], then [imath]b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c[/imath]. So left cancellation holds in [imath]G[/imath]. 2) Suppose [imath]ba=ca[/imath], then [imath]b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c[/imath]. So right cancellation holds in [imath]G[/imath]. 3) If [imath]a[/imath] exists in [imath]G[/imath] then [imath]aa^{-1}=a^{-1}a=e[/imath], where [imath]a[/imath] is an inverse of [imath]a^{-1}[/imath]. Since inverses are unique, [imath](a^{-1})^{-1}=a[/imath]. 4) Let [imath]x[/imath] be the inverse of [imath]ab[/imath]. Then [imath](ab)x=e[/imath]. By associativity we have [imath]a(bx)=aa^{-1}[/imath]. Through left cancellation we have [imath]bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})[/imath] and [imath]x=b^{-1}a^{-1}[/imath]. Thus [imath](ab)^{-1}=b^{-1}a^{-1}[/imath]. So all conditions hold, [imath]G[/imath] is a group. Is this proof correct? I know to show [imath]G[/imath] is a group these conditions have to be met. This is all I have to show right?
1173032
How do I find the antiderivative of [imath]sin(u)e^u[/imath]? So, I have the following indefinite integral: [imath]\int sin(u)e^u \ du[/imath] I tried solving it using integration by parts, but it just kept repeating itself. What should I do?
438444
What is the integral of [imath]\int e^x\,\sin x\,\,dx[/imath]? I'm trying to solve the integral of [imath]\left(\int e^x\,\sin x\,\,dx\right)[/imath] (My solution): [imath]\int e^x\sin\left(x\right)\,\,dx=[/imath] [imath]\int \sin\left(x\right) \,e^x\,\,dx=[/imath] [imath]\left(\sin(x)\,\int e^x\right)-\left(\int\sin^{'}(x)\,\left(\int e^x\right)\right)[/imath] [imath]\left(\sin(x)\,e^x\right)-\left(\int\cos(x)\,e^x\right)[/imath] [imath]\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(\int-\sin\left(x\right)\,e^x\right)\right)[/imath] [imath]\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(-\sin\left(x\right)\,e^x-\int-\cos\left(x\right)\,e^x\right)\right)[/imath] I don't know how to complete because the solution gonna be very complicated.
1087752
Find the ring of homomorphisms [imath]\mathbb{Z} \to \mathbb{Z}[/imath] Show, which well-known ring is isomorphic to ring [imath]End(\mathbf{Z})(+,ͦ,-,0,id)[/imath] of homomorphisms [imath]\mathbf{Z}[/imath] -> [imath]\mathbf{Z}[/imath], where [imath]\mathbf{Z}[/imath] is commutative group [imath]\mathbf{Z}(+)[/imath] and 0 constant zero function. I struggle with solving this, including the problem of what is considered a "known" ring.
64525
Showing a Ring of endomorphisms is isomorphic to a Ring Im trying to show that [imath]\mathrm{End}(\langle \mathbb{Z},+\rangle)[/imath] is naturally isomorphic to [imath]\langle \mathbb{Z},+,\cdot\rangle[/imath], but I'm not quite sure which ring homomorphism to use. Thank you
1172967
Is the sum of two independent geometric random variables with the same success probability a geometric random variable? Is the sum of two independent geometric random variables with the same success probability parameter a geometric random variable? What is it's distribution? My approach is as follows: [imath]Z=X+Y[/imath] [imath]P(X+Y=z)=\sum\limits_{x} P(X=x)P(Y=z-x)[/imath] [imath]=\sum\limits_{x} p(1-p)^{(x-1)}p(1-p)^{(z-x-1)}[/imath] [imath]=\sum\limits_{x} p^{2}(1-p)^{z-2}[/imath] [imath]= p^{2}(1-p)^{z-2} \sum\limits_{x=0}^\infty 1[/imath] I am not sure how to turn this into a distribution. It looks like binomial with n = z and x = 2, but I don't know how to get the coefficient from this.
548525
How to compute the sum of random variables of geometric distribution Let [imath]X_{i}[/imath], [imath]i=1,2,\dots, n[/imath], be independent random variables of geometric distribution, that is, [imath]P(X_{i}=m)=p(1-p)^{m-1}[/imath]. How to compute the PDF of their sum [imath]\sum_{i=1}^{n}X_{i}[/imath]? I know intuitively it's a negative binomial distribution [imath]P\left(\sum_{i=1}^{n}X_{i}=m\right)=\binom{m-1}{n-1}p^{n}(1-p)^{m-n}[/imath] but how to do this deduction?
1172789
what function fulfills these conditions? So I know that if [imath]f(x) = x^{-1}[/imath], than [imath]f(f(x)) = x[/imath] but [imath]f(x)[/imath] is not necessarily [imath]x[/imath]. So now, is there [imath]g(x)[/imath] such that [imath]g(g(x)) \neq g(x) \neq x[/imath] but [imath]g(g(g(x))) = x[/imath]? If so what is it, else why not?
195852
Can the Identity Map be a repeated composition one other function? Consider the mapping [imath]f:x\to\frac{1}{x}, (x\ne0)[/imath]. It is trivial to see that [imath]f(f(x))=x[/imath]. My question is whether or not there exists a continuous map [imath]g[/imath] such that [imath]g(g(g(x)))\equiv g^{3}(x)=x[/imath]? Furthermore, is there a way to find out if there is such a function that [imath]g^{p}(x)=x[/imath] for a prime [imath]p[/imath]? Edit: I realise I was a little unclear - I meant to specify that it was apart from the identity map. The other 'condition' I wanted to impose isn't very precise; I was hoping for a function that didn't seem defined for the purpose. However, the ones that are work perfectly well and they certainly answer the question.
1173559
Prove it. How do I start? Given [imath]a,b >0[/imath] [imath] \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} = max\{a,b\}[/imath] I tried to start with taking a common but could reach the conclusion , please provide a solution.
807759
Show that [imath]\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} = \max\{c_1,c_2,\ldots,c_m\}[/imath] Let [imath]m\in \mathbb{N}[/imath] and [imath]c_1,c_2,\ldots,c_m \in \mathbb{R}_+[/imath]. Show that [imath]\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} = \max\{c_1,c_2,\ldots,c_m\}[/imath] My attempt: Since [imath]\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} \leq \lim_{n\rightarrow \infty}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}} = \lim_{n\rightarrow \infty} \sqrt[n]{n}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}}=\lim_{n \rightarrow \infty} \max\{\sqrt[n]{c_1^n},\sqrt[n]{c_2^n},\ldots,\sqrt[n]{c_m^n}\}=\lim_{n \rightarrow \infty}\max\{c_1,c_2,\ldots,c_m\}=\max\{c_1,c_2,\ldots,c_m\}[/imath] it follows that [imath]\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n}[/imath] is bounded, but I don't think it's monotonically decreasing, at least I can't prove this. Can anybody tell me whether the approach I have chosen is a good one, whether what I have done is correct and how to finish the proof?
1111089
How to show [imath]\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}[/imath]? Let [imath]a\geq 0[/imath] and [imath] b\geq 0[/imath]. Prove that [imath]\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}[/imath]. [Hint: Use the identity [imath](a^n -b^n)=(a-b)(\sum_{i=0}^{n-1}a^ib^{n-1-i})[/imath]] I need some help! I cannot do it even with the hint... :(
702498
Proof Verification: [imath]\lim (a^n + b^n)^{\frac 1 n} = b[/imath] for [imath] 0 \lt a \le b[/imath] Found the following exercise in Bartle's Elements of Real Analysis in the section on combinations of sequences. Am unsure about my solution and would really appreciate it if someone could verify it. If [imath] \; 0 \lt a \le b \; [/imath] and [imath]x_n = (a^n + b^n)^{\frac 1 n}, \; [/imath] then [imath]\lim (x_n) = b[/imath] My Attempt: [imath]x_n = (a^n + b^n)^{\frac 1 n} = b\left[\left({\frac a b}\right)^n + 1\right]^{\frac 1 n} [/imath] where [imath]\frac a b \le 1[/imath]. Now let us define two more sequences in [imath]\Bbb R, \;\; (y_n)[/imath] and [imath](z_n)[/imath] such that [imath]y_n = b[/imath] and [imath]\; z_n = \left[\left({\frac a b}\right)^n + 1\right]^{\frac 1 n}[/imath] for all [imath]n \in \Bbb N[/imath]. And furthermore, since [imath]\frac a b \le 1, [/imath] let us equate the fraction to [imath]\dfrac {1}{1 + t}[/imath] for some [imath]t \ge 0. [/imath] Then: [imath]|z_n - 1| = \left|{ \left[{\left({\dfrac {1}{1 + t}}\right)^n +1}\right]^{\frac 1 n} - 1}\right| = \left[{\left({\dfrac {1}{1 + t}}\right)^n +1}\right]^{\frac 1 n} - 1 [/imath] By Bernoulli's Inequality: [imath] |z_n - 1| \le \left[{\left({\dfrac {1}{1 + nt}}\right) +1}\right]^{\frac 1 n} - 1^{\frac 1 n} \le \dfrac {1}{(1 + nt)^{\frac 1 n}} \lt \dfrac{1}{(nt)^{\frac 1 n}}[/imath] Given any [imath]\epsilon \gt 0[/imath] there is a natural number [imath]m[/imath] such that [imath] m \gt \dfrac{1}{t\epsilon^m}[/imath]. Then; [imath]n \ge m \implies \epsilon \gt \dfrac{1}{(tm)^{\frac 1 m}} \ge \dfrac{1}{(tn)^{\frac 1 n}} \gt |z_n - 1| \implies \lim (z_n) = 1[/imath] We know that [imath]\lim (y_n) = b[/imath]. Therefore [imath](x_n) = (y_n.z_n)[/imath] converges to [imath]\lim(y_n).\lim(z_n) = b[/imath] Q.E.D. The following identities have also been used: [imath]\left|\sqrt[n]y-\sqrt[n]x\right|\le\sqrt[n]{|y-x|} \;\; \text{for every $x, y \gt 0$} [/imath] [imath]n^n \lt (n + 1 )^{n+ 1} \;\; \forall n \in \Bbb N[/imath]
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Putnam 2005 A1 Solution Show that every positive integer is a sum of one or more numbers of the form [imath]2^r3^s,[/imath] where [imath]r[/imath] and [imath]s[/imath] are nonnegative integers and no summand divides another. (For example, [imath]23=9+8+6.)[/imath] Suppose for [imath]k = \{1, 2, 3... ,n-1\}[/imath] this equality holds. Let [imath]n[/imath] be an even integer. Since [imath]\frac{n}{2} \le n -1[/imath] if [imath]n \ge 2[/imath] then, [imath]\frac{n}{2} = \sum 2^x 3^y \implies n = \sum 2^{x+1} 3^{y}[/imath] Proof hold for [imath]P(n)[/imath] from [imath]P(n-1)[/imath]. With odds, it gets trickier. Suppose [imath]n[/imath] is a odd integer, NOT PRIME. So: [imath]n = \{1, 9, 15, ...\}[/imath] Since, [imath]\frac{n}{3} \le n-1 [/imath] if [imath]n \ge 3[/imath] It follows, [imath]\frac{n}{3} = \sum 2^x e^y \implies n = \sum 2^x 3^{y+1}[/imath] Suppose [imath]n[/imath] is PRIME. [imath]n = \{3, 5, 7, 11, 13, 17, ... \}[/imath] There exists [imath]t[/imath] such that: [imath]n < 3^{t-1} \implies \log_{3}(n) < t-1 = \sum 2^x 3^{y}[/imath] By strong induction, [imath]\log_{3}(n) = \sum 2^x 3^y[/imath] Is there any thing I can do now ?I am stuck!
1172035
How to prove this form of [imath]n[/imath]? Show that every positive integer is a sum of one or more numbers of the form [imath]2^r3^s,[/imath] where [imath]r[/imath] and [imath]s[/imath] are nonnegative integers and no summand divides another. From: AOPS Putnam A1 Solution I see they mean that: [imath]n-1 = \sum_{k=1}^{x} 2^{r_k} 3^{s_k}[/imath] By induction hypothesis. Then: [imath]n = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} + 1[/imath] Then how to proceed by [imath]n/2[/imath]? How can you say that: [imath]\frac{n}{2} = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} [/imath] Where did the [imath]1[/imath] go? And how can you assume it works for [imath]\frac{n}{2}[/imath] (the hypothesis)?
1173597
Is this function a joint distribution function for two random variables? Is the following function [imath]F(x,y) = 1-e^{-xy} [/imath] for [imath]x,y \geq 0[/imath] [imath]F(x,y) = 0 [/imath] otherwise a joint distribution function for two random variables [imath]X[/imath] and [imath]Y[/imath]? Why?
1172854
Is the function [imath]F(x,y)=1−e^{−xy}[/imath] [imath]0 ≤ x[/imath], [imath]y < ∞[/imath], the joint cumulative distribution function of some pair of random variables? Is the function [imath]F(x,y)=1−e^{−xy}[/imath] [imath]0 ≤ x[/imath], [imath]y < ∞[/imath], the joint cumulative distribution function of some pair of random variables? How do I show this? I am confused about cumulative distribution functions. My thoughts so far: [imath]F(x,y)=1−e^{−xy}[/imath] [imath]F(x,y)=1−\frac{1}{e^{xy}}[/imath] so when x or y=0, F(x,y)=0, when x>0 and y>0, F(x,y)=1, and when y<0, [imath]F(x,y)=1−e^{xy}[/imath] so F(x)=0? I know the joint cdf is related to the joint pdf, maybe I have to do something with that?
1173583
Hensen inequality in trigonometry: [imath]\sin A + \sin B + \sin C \leq \frac{3}{2} \cdot \sqrt[2]{3} [/imath] Can anyone help me how to prove [imath]\sin A + \sin B + \sin C \leq \frac{3}{2} \cdot \sqrt[2]{3} [/imath] I have idea use Jensen but how to use it here?
990418
Maximum value of [imath]\sin A+\sin B+\sin C[/imath]? What is the maximum value of [imath]\sin A+\sin B+\sin C[/imath] in a triangle [imath]ABC[/imath]. My book says its [imath]3\sqrt3/2[/imath] but I have no idea how to prove it. I can see that if [imath]A=B=C=\frac\pi3[/imath] then I get [imath]\sin A+\sin B+\sin C=\frac{3\sqrt3}2[/imath]. And also according to WolframAlpha maximum is attained for [imath]a=b=c[/imath]. But this does not give me any idea for the proof. Can anyone help?