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2881333 | Problem regarding Fourier Coefficients.
Question. Let [imath]~f \in C^1[-\pi,\pi][/imath] such that [imath]f(-\pi)=f(\pi)[/imath] . Define, for [imath]n\in \mathbb{N}[/imath], [imath]b_n=\int_{-\pi}^{\pi}f(t) \sin nt~ dt[/imath] Which of the following statements are true? a. [imath]b_n \to 0[/imath] as [imath]n\to \infty[/imath]. b. [imath]nb_n \to 0[/imath], as [imath]n\to \infty[/imath]. c. The series [imath]\sum n^3 b_n^3[/imath] is absolutely convergent. My Solution. a. True. Since by Riemann-Lebesgue Lemma we know for any [imath]f \in L(I)[/imath] and any [imath]\beta \in \mathbb{R}[/imath] we have:[imath] \lim_{\alpha \to +\infty}\int_I f(t) \sin (\alpha t+ \beta)=0[/imath] b. True. Using integration by parts we get [imath]~nb_n=\int_{-\pi}^{\pi}f(t) \cos nt~ dt[/imath]. Again using Riemann-Lebesgue Lemma we have the results. c. I can't figure out how to conclude (c). Please help. Thank you. | 2613382 | Convergence of the [imath]\sum_{n=1}^{\infty}n^3b_n^3[/imath] where [imath]b_n=\int_{-\pi}^{\pi}f(t)\sin (nt)\ dt[/imath].
Let [imath]f\in C^1[\pi,\pi][/imath]. Define, for [imath]n\in \mathbb{N}[/imath], [imath]b_n=\int_{-\pi}^{\pi}f(t)\sin (nt)\,\mathrm{d}t[/imath]. Which of the following statements are true?? a) [imath]b_n\to 0, [/imath]as [imath]n\to \infty[/imath] b) [imath]nb_n\to 0,[/imath] as [imath]n\to \infty[/imath] c) The series [imath]\sum_{n=1}^{\infty}n^3{b_n}^3[/imath] is absolutely convergent. Option a) is true by Riemann-Lebesgue Theorem but what about c) ???I think b) is false ... |
2881510 | Minimally dense subsets of [imath][0,1][/imath]?
I'm curious about the notion of a minimally dense subset of a given space (I've been using [imath][0,1][/imath], but if others are interesting, I'm interested in that too!). Here are two questions. Does there exist a set [imath]A[/imath] that is dense in [imath][0,1][/imath], but for all infinite [imath]B\subset A[/imath], [imath]A\setminus B[/imath] is not dense in [imath][0,1][/imath]? I suspect that the answer to this question is no, but I haven't been able to prove it. I'm also curious about what modifications to this statement can create a notion of "minimally dense" subsets. For all [imath]\alpha\in \mathbb{R}\setminus\mathbb{Q}[/imath], there exist infinitely many rational numbers [imath]\frac{m}{n}[/imath] with [imath]|\alpha-\frac{m}{n}|<\frac{1}{n^2}[/imath]. In his analysis book, Browder remarks that this theorem is a way that one can quantify density. Is there a way to use statements like this to come up with a "minimally dense" subset of [0,1]? Let me know if there are other ways of talking about this notion, or whether you think there is something wrong with the idea of a minimally dense set in general. Thanks! | 340585 | Minimal dense subset
Let [imath]X[/imath] a space; and [imath]Y[/imath] is a dense subset of [imath]X[/imath]. However there may be some subset [imath]Z[/imath] of [imath]Y[/imath] such that [imath]Z[/imath] is also dense in [imath]X[/imath]. Now I want to delete some elements of [imath]Y[/imath] which seem needless, so that I can make the new dense subset [imath]Z[/imath] of [imath]Y[/imath] be the smallest. How can I get it? My idea is this: If there exists [imath]y \in \overline{Y\setminus \{y\}}[/imath], then delete [imath]y[/imath], because [imath]Y\setminus \{y\}[/imath] is also dense in [imath]X[/imath]. However what should I do next step? Thanks ahead:) |
2881488 | Prove: [imath]F=\{x_n\}\cup \{x\}[/imath] Is closed
Let [imath]M[/imath] be a metric space, [imath]x_n\in M[/imath] a sequence which converges to [imath]x\in M[/imath] Prove: [imath]F=\{x_n\}\cup \{x\}[/imath] is a closed set So we have [imath]x_n\to x[/imath] such that [imath]x_n\in F[/imath] and [imath]x\in F[/imath] and we know that a set is closed if it contains all of its accumulation points, so [imath]F[/imath] is closed Or must I look at [imath]F^c[/imath] and prove that it is open? | 2676809 | Showing a Sequence and its Limit are Closed
Let [imath](X, d)[/imath] be a metric space, and [imath](x_n)_{n \in \mathbb{N}}[/imath] a sequence in [imath]X[/imath], with [imath]x_n \rightarrow x[/imath] w.r.t. [imath]d[/imath] for some [imath]x \in X[/imath]. Show that the set [imath]\{x_n : n \in \mathbb{N}\}\cup\{x\}[/imath] is closed in [imath](X, d)[/imath]. I figured I should use [imath]\overline{F} = F \cup F'[/imath] to show the above set is closed. Let [imath](x_n)_{n \in \mathbb{N}} = F[/imath]. Then, all elements of the sequence are in [imath]F[/imath], which is equal to [imath]\{x_n : n \in \mathbb{N}\}[/imath]. Clearly, the sequence (or [imath]F[/imath]) has only one accumulation point, [imath]x[/imath]. So, [imath]F' = \{x\}[/imath]. This shows [imath]\{x_n : n \in \mathbb{N}\}\cup\{x\} = \overline{F}[/imath], and thus is closed. Is this sufficient in proving the problem? Or would I have to show that [imath]x[/imath] is the only accumulation point as well (figured I don't have to because we know limits are unique)? Any advice would be very helpful. Thank you. |
2808213 | Proof that [imath](1^2 + 2^2 + \cdots + n^2)^n > n^n (n!)^2[/imath]
I need to prove that [imath]a_{n} = (1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2}=b_{n}[/imath] for any natural number n. Proof: Note that by the arithmetic-geometric mean inequality,[imath]\frac{1^{2} +2^{2} + ...+n^{2}}{n} > (1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2}[/imath]. Then [imath]1^{2} +2^{2} + ...+n^{2} > n(1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2} = n(n!)^{n}[/imath]. This implies that [imath]a_{n} =(1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2n} \geq n^{n}(n!)^{n}= b_{n}[/imath]. Then we conclude that [imath]a_{n} > b_{n}[/imath] for all natural numbers [imath]n[/imath]. Is there anything wrong with my proof? | 1266685 | Problem using the AM-GM inequality
I am getting crazy with this one. Suppose [imath]a_n=(1^2+2^2+3^2+\ldots+n^2)^n[/imath] and [imath]b_n=n^n(n!)^2[/imath]. Show that [imath]a_n>b_n[/imath] for all [imath]n[/imath]. They suggest to use the AM-GM inequality. |
2881607 | Why cannot [imath]\frac{0}{0}[/imath] (indeterminate element) be added to a Field extension to [imath]\mathbb{C}[/imath]?
Why cannot [imath]\frac{0}{0}[/imath] be added to a field extension to complex field [imath]\Bbb C[/imath]? I know the complex field [imath]\Bbb C[/imath] is closed. But can we define additional elements such as [imath]1*\frac{0}{0}[/imath], [imath]2*\frac{0}{0}[/imath], [imath]i*\frac{0}{0}[/imath] and extend [imath]\Bbb C[/imath] and study its properties? | 1370214 | What is [imath]0\div0\cdot0[/imath]?
We all know that multiplication is the inverse of division, and therefore [imath]x\div{x}\cdot{x}=x[/imath] But what if [imath]x=0[/imath]? [imath]0\div0[/imath] is undefined so [imath]0\div0\cdot0[/imath] should be too, but whatever happens when we divide that first [imath]0[/imath] by [imath]0[/imath] should be reversed when we multiply it by [imath]0[/imath] again, so what is the right answer? 0, undefined, or something else entirely? |
2881673 | Factorising fourth power polynomial with 5 terms
I've searched all over the internet and cannot seem to factorise this polynomial. [imath]x^4 - 2x^3 + 8x^2 - 14x + 7[/imath] The result should be [imath](x − 1)(x^3 − x^2 + 7x − 7)[/imath] What are the steps to get to that result? I've tried grouping but doesn't seem to work... | 684036 | Factoring a hard polynomial
This might seem like a basic question but I want a systematic way to factor the following polynomial: [imath]n^4+6n^3+11n^2+6n+1.[/imath] I know the answer but I am having a difficult time factoring this polynomial properly. (It should be [imath](n^2 + 3n + 1)^2[/imath]). Thank you and have a great day! |
2582650 | Does this integral converge when [imath]p<0[/imath]?
Does this integral converge when [imath]p<0[/imath]? [imath]\int\limits_0^\infty \sin(x^p)\ dx[/imath] | 1089043 | Convergence of [imath]\int_0^\infty [/imath]sin[imath] (x^p) dx[/imath]
Consider the [imath]\displaystyle \int_0^\infty [/imath]sin[imath] (x^p) dx[/imath]. Does it converge when [imath]p<0[/imath]? Does it converge when [imath]p>1[/imath]? My Work: Let [imath]x^p=y[/imath], then [imath]\displaystyle \int_0^\infty [/imath]sin[imath] \displaystyle (x^p) dx=\frac{1}{p}\sum_{n=1}^\infty \int_{(n-1)\pi}^{n\pi} \frac{\text{sin} \; y}{y^r} dy[/imath] , where [imath]r=\frac{p-1}{p}>1 [/imath]when [imath]p<0[/imath] and [imath]0<r=\frac{p-1}{p}\leq 1 [/imath]when [imath]p>1[/imath]. I know that [imath]\displaystyle \int_0^\infty [/imath]sin[imath] (\frac{1}{x}) dx[/imath] diverges which is a special case of first case. But failed to show it generally. Can anyone please give me a hint to preceed? |
2881651 | Prove [imath] \forall c \in \mathbb{R}^{+}, \forall B \in \mathbb{N}, \exists n \in \mathbb{N}, (n \geq B) \wedge (\sqrt{2n} > c \ln(n+2)) [/imath]
So far I've proved the following: [imath]2n \geq \ln(2n) > \ln(n+2)[/imath] for [imath]n \geq 3 [/imath] The proof must be without the use of calculus but I have no idea on how to proceed. | 2880820 | Prove [imath] \forall c \in \mathbb{R^{+}}, \forall B \in \mathbb{N}, \exists n \in \mathbb{N}, (n \geq B) \wedge (\sqrt{2n} > c \ln(n+2)) [/imath]?
So far I've proved the following: [imath]2n \geq \ln(2n) > \ln(n+2)[/imath] for [imath]n \geq 3 [/imath] The proof must be without the use of calculus but I have no idea on how to proceed. |
2872327 | Integral of 1-form [imath]\omega=\dfrac{-y \,dx + x \,dy}{x^2 +y^2}[/imath] over a triangle.
I'm trying to evaluate the integral of the [imath]1[/imath]-form [imath]\omega=\dfrac{-y \,dx +x\,dy}{x^2 +y^2}[/imath] through the corners of a triangle with the vertices [imath]A= (-5,-2)[/imath], [imath]B=(5,-2)[/imath], [imath]C=(0,3)[/imath]. I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral? | 1307738 | Application of Green's Theorem when undefined at origin
Problem: Let [imath]P={-y \over x^2+y^2}[/imath] and [imath]Q={x \over x^2+y^2}[/imath] for [imath](x,y)\ne(0,0)[/imath]. Show that [imath]\oint_{\partial \Omega}(Pdx + Qdy)=2\pi[/imath] if [imath]\Omega[/imath] is any open set containing [imath](0,0)[/imath] and with a sufficiently regular boundary. Working: Clearly, we cannot immediately apply Green's Theorem, because [imath]P[/imath] and [imath]Q[/imath] are not continuous at [imath](0,0)[/imath]. So, we can create a new region [imath]\Omega_\epsilon[/imath] which is [imath]\Omega[/imath] with a disc of radius [imath]\epsilon[/imath] centered at the origin excised from it. We then note [imath]{\partial Q \over \partial x} - {\partial P \over \partial y} = 0[/imath] and apply Green's Theorem over [imath]\Omega_\epsilon[/imath]. Furthermore, [imath]\oint_C(Pdx + Qdy)=2\pi[/imath] if [imath]C[/imath] is any positively oriented circle centered at the origin. I get the general scheme of how to approach this problem, however I am unsure of how to argue it in a rigorous manner. |
2882804 | Prove that [imath]\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\ldots[/imath]
Prove that [imath]\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots[/imath] [imath]x\geq{0}[/imath] How should I start? Any hints would be helpful. Also how would the expression change for [imath]x\lt{0}.[/imath] EDIT: Thanks to @Somos in the comments. I was able to come up with my own solution. (Adding here if someone ever stumbles upon here looking for a solution.) Subtract [imath]\lfloor{x}\rfloor[/imath] from both sides. Then we need to prove that, [imath]-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots=0[/imath] We know that, [imath]\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor[/imath] Therefore, [imath]-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor=-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor[/imath] Subsituting this, [imath]-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots[/imath] [imath]-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2}+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots[/imath] Similarly, I reach, [imath]\lim_{n\to\infty}-\bigg\lfloor{\frac{x}{2^n}}\bigg\rfloor[/imath] i.e. [imath]=0[/imath] Hence Proved. | 1821536 | Prove that [imath]\lfloor\frac{n+1}{2}\rfloor+\lfloor\frac{n+2}{4}\rfloor+\lfloor\frac{n+4}{8}\rfloor+\lfloor\frac{n+8}{16}\rfloor+ \dots=n[/imath]
Prove [imath]\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots=n[/imath] where [imath][x]=\lfloor x\rfloor[/imath] [imath]$$ It was suggested that somehow I use the identity $[x]=\left[\dfrac x2\right]+\left[\dfrac{x+1}{2}\right][/imath][imath]$After struggling for a while, I realised I wasn't getting anywhere using his hint, probably because I couldn't really understand how I was to use it. Instead I tried to use the Squeeze Theorem by rewriting the $nth$ term of the series (referred to later as S) as[/imath][imath] [/imath]t_n=\left[\dfrac{n+2^k}{2^{k+1}}\right] \text{ where } 0\le k<\infty[imath] [/imath]\Rightarrow \dfrac{n+2^k}{2^{k+1}}-1<\left[\dfrac{n+2^k}{2^{k+1}}\right]\le \dfrac{n+2^k}{2^{k+1}}[imath] [/imath][imath] [/imath] \lim_{k\to \infty}(k+1)\left(\dfrac{n+2^k}{2^{k+1}}-1\right)<S\le \lim_{k\to \infty}(k+1)\left( \dfrac{n+2^k}{2^{k+1}}\right)[imath][/imath] However these bounds are too loose as the limits diverge to $-\infty$ and $\infty$ respectively. [imath]$ Could somebody please show me how to prove the series is equal to [/imath]n$, either through the given hint, or through the selection of tighter bounds for the Squeeze Theorem? Many thanks! |
2879899 | Show that a proper locally invertible map is surjective
Let [imath]f:R^2\to R^2[/imath] be a continuously differentiable function such that [imath]Df(x)[/imath] is invertible for all [imath]x\in R^2[/imath] and [imath]f^{-1}(K)[/imath] is compact for every compact set [imath]K[/imath]. Show that [imath]f[/imath] is surjective. The first condition gives by the inverse function theorem that [imath]f[/imath] is invertible in a neighborhood of any point. But how exactly to use properness? | 1089321 | Application of inverse function theorem for several variable functions
Let [imath]f:\mathbb{R}^2\rightarrow \mathbb{R}^2[/imath] be continuously differentiable, and assume [imath]Df(x)[/imath] is invertible for all [imath]x\in \mathbb{R}^2[/imath]. Also for any compact [imath]K[/imath] in [imath]\mathbb{R}^2[/imath], [imath]f^{-1}(K)[/imath] is compact. prove [imath]f[/imath] is onto. My Work: This is an application of inverse function theorem because it satisfies the hypothesis of this thoerem. So [imath]A=f(\mathbb{R}^2)[/imath] is open in [imath]\mathbb{R}^2[/imath]. So, [imath]A^c[/imath] is closed. Afterwards I was stuck. please give me a hint. |
1086256 | Solution of the equation [imath]x^r=a[/imath]
Let [imath]F_{p^n}[/imath] be the field with [imath]p^n[/imath] elements. Suppose [imath]p^n-1=q_1^{a_1}...q_k^{a_k}[/imath] where [imath]q_i[/imath] are distinct primes. Find the no. of integers [imath]r\in\{0,1,...,p^n-2\}[/imath] for which the equation [imath]x^r=a[/imath] has a solution for every [imath]a\in F_{p^n}[/imath]. My Work: [imath]a^{p^n}=a[/imath] for all [imath]a\in F_{p^n}[/imath]. When [imath]r=0[/imath], [imath]x^0=1=a[/imath]. So the equation has no solution unless [imath]a=1[/imath]. When [imath]r=1[/imath], [imath]x=a[/imath]. So the equation has a solution for all [imath]a[/imath]. But I was stuck in finding solutions for other [imath]r[/imath] values. Can anyone please give me a hint? | 1054594 | Find the number of integers [imath]r[/imath] such that the polynomial [imath]x^{r}-a[/imath] has a linear factor over [imath]\mathbb{F}_{p^{n}}[/imath]
If we have a finite field [imath]\mathbb{F}_{p^{n}}[/imath], how does one determine the number of integers [imath]r[/imath] in [imath]\{0,1, \ldots, p^{n}-2 \}[/imath] for which the equation: [imath]x^{r}=a[/imath] has a solution for every [imath]a \in \mathbb{F}_{p^{n}}[/imath]. The problem also mentions that [imath]p^{n}-1=q_{1}^{a_{1}}\ldots{q_{n}^{a_{n}}}[/imath] for distince primes [imath]q_{i}[/imath]. My attempt to understand the problem: I tried to understand the problem by considering the finite field [imath]\mathbb{F}_{9}[/imath] as the splitting field of [imath]x^{2}+1[/imath], whose root in [imath]\mathbb{F}_{9}[/imath] is regarded as [imath]\alpha[/imath]. In this case we have [imath]r \in \{1, \ldots, 7 \}[/imath]. I have been listing possibles [imath]r[/imath] and [imath]a[/imath]'s trying to find a solution in [imath]\mathbb{F}_{p^{n}}[/imath], for example when I set [imath]r=2[/imath] and [imath]a=\alpha[/imath] then [imath]x^{2}=\alpha[/imath] when [imath]x=\alpha+2[/imath]. I've doing this, but this process seems laborious and I haven't gotten much insight out of it yet. Is there a obvious way to do the above mentioned problem? |
2877912 | Show that every nonzero prime ideal in [imath]\mathbb{C}[x, y]/(x^2 - y^2 - 1)[/imath] is maximal
Show that the ring [imath]A := \mathbb{C}[x, y]/(x^2 - y^2 - 1)[/imath] is an integral domain. Further show that every nonzero prime ideal in A is maximal. I proved that [imath]A[/imath] is an integral domain by showing that [imath]x^2 - y^2 - 1[/imath] is an irreducible in the UFD [imath]\mathbb{C}[x, y][/imath] and hence a prime. To prove that every nonzero prime ideal in [imath]A[/imath] is maximal I need to prove either that: i) A is a PID, so in a sense I need to be able to express [imath]x[/imath] in terms of [imath]y[/imath] (or vice versa) but I only know that [imath]x^2 = y^2 + 1[/imath] in [imath]A[/imath]. ii)If [imath]I[/imath] is a prime ideal in [imath]A[/imath] then every non zero element in [imath]A/I[/imath] is invertible. | 2394930 | Prove that [imath]\frac{\mathbb{C}[x,y]}{\langle x^2-y^2-1\rangle}[/imath] is an integral domain such that all the nonzero prime ideals are maximal.
Let [imath]R=\frac{\mathbb{C}[x,y]}{\langle x^2-y^2-1 \rangle}[/imath]. Prove that (1) [imath]R[/imath] is an integral domain. (2) Any nonzero prime ideal of [imath]R[/imath] is maximal. My idea: (1) The first part is easy. Since [imath]x^2-y^2-1\in \mathbb{C}[y][x][/imath] is Eisenstein with respect to the prime [imath]y+i[/imath], it is irreducible. Hence the ideal generated by it is a prime ideal. (2) I am struggling with this part. I have one approach (not sure if correct). Approach : Let [imath]I=\langle x^2-y^2-1 \rangle[/imath]. Then [imath]R=\mathbb{C}[x,y]/I[/imath]. Any ideal of [imath]R[/imath] is of the form [imath]J/I[/imath] where [imath]J\subseteq \mathbb{C}[x,y][/imath] is an ideal containing [imath]I[/imath]. If we can show that any prime ideal [imath]J[/imath] containing [imath]I[/imath] is maximal, then we are done. How to prove this? Is there any other way to prove the second part? Edit: I am trying to avoid Hilbert Nullstellensatz and the notion of dimension of a ring, as these concepts are not there in the syllabus of the exam from which I found this question. |
2882643 | Evaluate [imath]\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}[/imath]
Evaluate [imath]I=\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}[/imath] My book gave the substitution for [imath]\int \frac{dx}{P\sqrt{Q}}[/imath] as [imath]\frac{Q}{P}=t^2[/imath] when [imath]P[/imath] and [imath]Q[/imath] are quadratic expressions So accordingly i used [imath]\frac{x^2+x+1}{x^2-x+1}=t^2 \tag{1}[/imath] we get [imath]\frac{(1-x^2) \, dx}{(x^2-x+1)^2}=t \,dt[/imath] Then [imath]I=\int \frac{\sqrt{x^2-x+1}\:dt}{1-x^2}[/imath] By Componendo Dividendo in [imath](1)[/imath] we get [imath]\frac{x^2+1}{x}=\frac{t^2+1}{t^2-1}[/imath] But how to express integrand purely in terms of [imath]t[/imath]? | 482360 | Compute the integral [imath]\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}[/imath]
Compute the indefinite integral [imath] \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}} [/imath] My Attempt: [imath] \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx [/imath] Now define [imath]t[/imath] such that [imath]t^2=\dfrac{x^2+x+1}{x^2-x+1}[/imath] to get [imath] \begin{align} 2t\,dt &= \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\,dx\\ 2tdt &= \frac{-4x^2+2x+2}{(x^2-x+1)^2}dx \end{align} [/imath] I don't know how to proceed from here. |
2328336 | Problem relating to circumscribing the escribed sircles
The triangle DEF circumscribes the three escribed circles of the triangle ABC. I have to prove that [imath]\dfrac{EF}{a \cos A} = \dfrac{FD}{b \cos B}= \dfrac{DE}{c \cos C}.[/imath] Also, point D is opposite to A. E is opposite to B. F is opposite to C I am not able to find the angles of the triangle DEF. Help me | 995262 | prove this properties of triangles trigonometric question
The triangle [imath]DEF[/imath] circumscribes the three escribed circles of triangle [imath]ABC[/imath]. Prove that [imath]\frac{EF}{a\,\cos A} = \frac{FD}{b\,\cos B} = \frac{DE}{c\,\cos C}[/imath] |
2883023 | Finding the number of zeros of a polynomial in the closed disk
Find the number of zeros of [imath]f(z)=z^6-5z^4+3z^2-1[/imath] in [imath]|z|\leq1[/imath]. My attempts have not gotten far. I know we can examine the related equation [imath]f(w)=w^3-5w^2+3w-1[/imath] in [imath]|w|\leq1[/imath], letting [imath]w=z^2[/imath]. It is clear that [imath]f(w)=0[/imath] for [imath]|w|=1[/imath] if and only if [imath]w=-1[/imath]. My main problem is that this seems obviously like an application of Rouché's theorem, where we let [imath]g(w)=5w^2[/imath], whereupon we have the relation [imath]|f(w)|< |g(w)|[/imath] for all [imath]|w|=1[/imath], with the sole exception that equality holds when [imath]w=-1[/imath]. I would like to use Rouché to equate the number of zeros of [imath]f[/imath] and [imath]g[/imath], but my understanding of Rouché is that the inequality should be strict with no exceptions. | 2245699 | Find the zeros of [imath]h(z)=z^6-5z^4+3z^2-1[/imath] within the unit disc - Verification
Let [imath]h(z)=z^6-5z^4+3z^2-1[/imath]. Using Rouche's theorem, with [imath]f(z)=-5z^4[/imath] and [imath]g(z)=z^6+3z^2-1[/imath]. On the unit disc [imath]\lvert f(z) \rvert =5 > \lvert 1+3-1 \rvert=\lvert g(z)\rvert [/imath] And the number of zeros is 4 for [imath]f(z)[/imath] in the unit disc, so it is also five for [imath]h(z)=f(z)+g(z)[/imath]. Is this correct? Thanks. Found an old answer of mine, I chose [imath]g(z)=z^6-1[/imath] and [imath]f(z)=-5z^4+3z^2[/imath], and got the answer [imath]4[/imath] roots again. |
2393824 | Dimension of of the subspace formed by all matrices which commute with a given matrix
Let [imath]A\in M_n(\mathbb{C})[/imath] be a matirx. Consider the subspace [imath]C(A)=\{X\in M_n(\mathbb{C})\text{ }|\text{ }AX=XA\}[/imath] How to prove that the dimension of [imath]C(A)[/imath] is at least [imath]n[/imath]? The only idea I could get was that the dimension is at least [imath]d[/imath] where [imath]d[/imath] is the degree of the minimal polynomial of [imath]A[/imath]. | 1379878 | [imath]C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}[/imath] is a subspace of dimension at least [imath]n[/imath].
Let [imath]M_n(\mathbb{C})[/imath] denote the vector space over [imath]\mathbb{C}[/imath] of all [imath]n\times n[/imath] complex matrices. Prove that if [imath]M[/imath] is a complex [imath]n\times n[/imath] matrix then [imath]C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}[/imath] is a subspace of dimension at least [imath]n[/imath]. My Try: I proved that [imath]C(M)[/imath] is a subspace. But how can I show that it is of dimension at least [imath]n[/imath]. No idea how to do it. I found similar questions posted in MSE but could not find a clear answer. So, please do not mark this as duplicate. Can somebody please help me how to find this? EDIT: Non of the given answers were clear to me. I would appreciate if somebody check my try below: If [imath]J[/imath] is a Jordan Canonical form of [imath]A[/imath], then they are similar. Similar matrices have same rank. [imath]J[/imath] has dimension at least [imath]n[/imath]. So does [imath]A[/imath]. Am I correct? |
2879563 | Prove that [imath]\det Df\equiv 0[/imath] if [imath]g\circ f \equiv 0[/imath]
Let [imath]f:R^n\to R^n[/imath] be a smooth function and [imath]g:R^n\to R[/imath] [imath]g(x_1,\dots,g_n)=x_1^5+\dots+x_n^5[/imath] Assume that [imath]g\circ f\equiv 0[/imath]. Prove that [imath]\det Df\equiv 0[/imath]. The only idea I have is to apply the chain rule [imath]d(g\circ f)_p=dg_{f(p)}\cdot df_p[/imath], where each term is the corresp. matrix of partial derivatives. Then since [imath]g\circ f=0[/imath], [imath]d(g\circ f)_p=0[/imath] for any [imath]p[/imath]. But this doesn't imply that either of the two matrices on the RHS is zero. Is it still true that the determinant of one of them is zero? Why do we need the explicit form of [imath]g[/imath]? | 1019339 | Show that [imath]g\circ f\equiv 0[/imath] implies [imath]\det Df\equiv 0[/imath], where [imath]g(x_1,...,x_n)=x_1^5+...+x_n^5[/imath]
Let [imath]f:\mathbb{R}^n \rightarrow \mathbb{R}^n[/imath] be a smooth function and let [imath]g:\mathbb{R}^n \rightarrow \mathbb{R}[/imath] be defined by [imath]g(x_1,...,x_n)=x_1^5+...+x_n^5[/imath]. Suppose [imath]g\circ f\equiv 0[/imath]. Show that [imath]\det Df\equiv 0[/imath]. I was going to start the solution from [imath]n=2[/imath] and then use induction on [imath]n[/imath]. So, for [imath]n=2[/imath], [imath]f:\mathbb{R}^2 \rightarrow \mathbb{R}^2[/imath] is a smooth function and [imath]g:\mathbb{R}^2 \rightarrow \mathbb{R}[/imath] where [imath]g(x_1,x_2)=x_1^5+x_2^5[/imath]. Then [imath]Df=\begin{pmatrix} f_{x_1x_1} & f_{x_1x_2} \\ f_{x_2x_1} & f_{x_2x_2} \end{pmatrix}[/imath] But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue? Ok so since [imath]g\circ f[/imath] is [imath]0[/imath], the image of [imath]f[/imath] is the line [imath]x_1+x_2=0[/imath] which has no min, max or saddle. If det[imath]Df(x_0)>0[/imath] and [imath]f_{x_1x_1}(x_0)>0[/imath] OR det[imath]Df(x_0)>0[/imath] and [imath]f_{x_1x_1}(x_0)<0[/imath], we have contradiction. But if det[imath]Df(x_0)>0[/imath] and [imath]f_{x_1x_1}(x_0)=0[/imath], then det[imath]Df(x_0)=0[/imath] or, det[imath]Df(x_0)<0[/imath](Since [imath]f_{x_1x_2}=f_{x_2x_1}[/imath]). So this is impossible. If det[imath]Df(x_0)<0[/imath] we have saddle and it is impossible. So finally we conclude that det[imath]Df(x)=0[/imath] for all [imath]x[/imath]. Am I correct? |
2883472 | [imath]S^n[/imath] is not homeomorphic to [imath]S^{n-1}[/imath]
My geometric intuition is clear about the fact that [imath]S^n \ncong S^{n-1}[/imath] [imath]\forall n \geq 2[/imath] . It's very easy to do it for lower dimensions using simple Analysis arguments. ([imath]S^n[/imath] is the n-sphere in [imath]\Bbb R^{n+1}[/imath]) But I really want to learn an Elementary proof for the general case (i.e. [imath]\forall n \geq 2[/imath]) without using tools from Algebraic Topology, using basic General Topology arguments. How to come up with an Elemenatry proof for the fact? The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question! | 226878 | How do I prove that [imath]S^n[/imath] is homeomorphic to [imath]S^m \Rightarrow m=n[/imath]?
This is what I have so far: Assume [imath]S^n[/imath] is homeomorphic to [imath]S^m[/imath]. Also, assume [imath]m≠n[/imath]. So, let [imath]m>n[/imath]. From here I am not sure what is implied. Of course in this problem [imath]S^k[/imath] is defined as: [imath]S^k=\lbrace (x_0,x_1,⋯,x_{k+1}):x_0^2+x_1^2+⋯+x_{k+1}^2=1 \rbrace[/imath] with subspace topology. |
2883490 | Prove that maximal number of monomials ...
I stumbled across this problem and I realize that it's propably easy but somehow I can't imagine the problem properly. Prove that the maximal number of monomials (that are not similar) of polynomial [imath]n[/imath] variables of a degree [imath]d[/imath] is equal to [imath]\binom{n+d}{n}[/imath] | 168807 | Maximal number of monomials of multivariate polynomial
If we consider a multivariate polynomial of [imath]n[/imath] variables with degree [imath]N[/imath], how do we show that the maximal number of monomials is [imath]\binom{N+n}{n} = \frac{(N+n)!}{n!\, N!}\quad ?[/imath] The hint our teacher gave was that the number of monomials is the number of independent coefficients. So how many independent coefficients does a multivariate polynomial have (at most)? Thank you. |
2883505 | Solve for θ: [imath]y = \arctan\Bigl(\frac{x· \sin(θ)- \sin(θ·x)}{x·\cos(θ)-\cos(θ·x)}\Bigr)[/imath]
I'm stuck here: [imath]\tan(y)=\frac{x·\sin(θ)-\sin(θ·x)}{x·\cos(θ)-\cos(θ·x)}[/imath] | 2881078 | Find θ: [imath]β=\arctan\frac{(R+r)·\sin(θ)-r·\sin(θ(1+\frac{R}{r}))}{(R+r)·\cos(θ)-r·\cos(θ(1+\frac{R}{r}))}[/imath]
The parametric formula for an epicycloid is: [imath]x=(R+r)·\cos(θ)-r·\cos\Biggl(θ\biggl(1+\frac{R}{r}\biggr)\Biggr)\\ y=(R+r)·\sin(θ)-r·\sin\Biggl(θ\biggl(1+\frac{R}{r}\biggr)\Biggr)[/imath] The angle θ is the angle between the positive x-axis and a line running from the origin through the center of the generating circle (with radius r). If instead of the angle [imath]\theta[/imath], you are given the angle between the positive x-axis and the end of the curve (we will call this angle [imath]\beta[/imath]), how do I find [imath]\theta[/imath]? [imath]β = \arctan\frac{(R+r)·\sin(θ)-r·\sin\Biggl(θ\biggl(1+\frac{R}{r}\biggr)\Biggr)}{(R+r)·\cos(θ)-r·\cos\Biggl(θ\biggl(1+\frac{R}{r}\biggr)\Biggr)}[/imath] |
625218 | For which [imath]{n\in{\Bbb Z}}[/imath] does there exist a matrix [imath]P\in{\Bbb C}^{4\times 4}[/imath] such that [imath]P^n=M[/imath]?
Consider the matrix [imath] M=\left(\begin{matrix} 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{matrix}\right). [/imath] For which [imath]{n\in{\Bbb Z}}[/imath] does there exist a matrix [imath]P\in{\Bbb C}^{4\times 4}[/imath] such that [imath]P^n=M[/imath]? Since [imath]P[/imath] must be not invertible, one must have [imath]n\geq 1[/imath]. For [imath]n=1[/imath], it is trivial. When [imath]n=3[/imath], we have (thanks to answers to this question) [imath]P^3=M[/imath] where [imath]P[/imath] is the [imath]4\times 4[/imath] Jordan black with [imath]\lambda=0[/imath]: [imath] P=\left(\begin{matrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{matrix}\right). [/imath] How can I deal with the general cases? | 388970 | Does there exist a matrix [imath]P[/imath] such that [imath]P^n=M[/imath] for a special matrix [imath]M[/imath]?
Consider the matrix [imath] M=\left(\begin{matrix} 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{matrix}\right). [/imath] Is there a matrix [imath]P\in{\Bbb C}^{4\times 4}[/imath] such that [imath]P^n=M[/imath] for some [imath]n>1[/imath]? One obvious fact is that if such [imath]P[/imath] exists, then [imath]P[/imath] must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if [imath]M[/imath] is an arbitrary nilpotent matrix with index [imath]k[/imath]? |
2236693 | Solving [imath]ax=e^{bx}[/imath]
I've been asked by a friend to help him solve this equation, but since we couldn't find the right answer, I thought about posting it here. Firstly, I thought about derivating both sides and get: [imath]a=be^{bx}[/imath] and from here we could find the answer pretty quickly, but then I noticed that what I did was incorrect because I can't derivate both sides - the functions aren't equal. | 1584123 | How can one find the zeroes of [imath]f(x)=ae^{bx}+cx+d[/imath]?
A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form [imath]f(x)=ae^{bx}+cx+d = 0[/imath] where [imath]a, b, c,[/imath] and [imath]d[/imath] are constant, real numbers that come from the physics problem and [imath]x[/imath] will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where [imath]c=0[/imath], the solution is simply [imath]x = \frac{\ln(\frac{-d}{a})}{b}[/imath] but for other values of [imath]c[/imath], I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots? |
2883869 | If [imath]a, b, c[/imath] are positive real numbers such that [imath]abc=1[/imath] prove that [imath]\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3[/imath]
If [imath]a, b, c[/imath] are distinct positive real numbers such that [imath]abc=1[/imath], prove that [imath]\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3.[/imath] I tried to do this problem by assuming that [imath]a<b<c[/imath]. By using this, the first and the third term of the inequality are positive and the second is negative. Thus, we can obtain the minimum value of the expression by minimizing the sum of the first and the third and maximizing the second. I'm stuck at this part and would appreciate if someone could help me. | 942594 | How find this value [imath]\frac{a^3}{(a-p)(a-q)}+\frac{b^3}{(b-p)(b-q)}+\frac{c^3}{(c-p)(c-q)}[/imath]
let [imath]p,q[/imath] is a [imath](x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0[/imath] roots,find this value [imath]\dfrac{a^3}{(a-p)(a-q)}+\dfrac{b^3}{(b-p)(b-q)}+\dfrac{c^3}{(c-p)(c-q)}[/imath] where [imath]a,b,c[/imath] is give numbers. I know this Lagrange interpolation [imath]\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-a)(b-c)}+\dfrac{c^3}{(c-a)(c-b)}=a+b+c[/imath] But my problem is different.so how find it? |
1082455 | The no. of subgroups which are not normal
Let [imath]G[/imath] be a group. Let [imath]t[/imath] be the no. of subgroups of [imath]G[/imath] which are not normal. Prove that [imath]t\neq 1[/imath]. My Work: Case 1: [imath]G[/imath] is abelian. Then all subgroups are normal. Hence, [imath]t=0[/imath] Case 2: [imath]G[/imath] is non abelian. Assume [imath]t=1[/imath]. Then there is a subgroup [imath]H[/imath] which is not normal in [imath]G[/imath]. Then if there is another subgroup [imath]K[/imath] such that [imath]K\neq H,G,\{0\}[/imath] then [imath]K\triangleleft G[/imath]. Then I was going to get a contradiction using 2nd isomorphism theorem. But how can I prove the existence of such [imath]K[/imath]? I am stuck now. Can anyone please help me? | 624594 | The number of non normal subgroups of a group cannot be [imath]1[/imath]?
Let [imath]G[/imath] be a group. Let [imath]n[/imath] be the number of subgroups of [imath]G[/imath] that are not normal. Prove that [imath]n\not=1[/imath]. When [imath]G[/imath] is Abelian, then this is trivial since [imath]n=0[/imath]. For the non-Abelian group [imath]G[/imath], I have to show that [imath]n\geq 2[/imath]. But I don't see how to go on. Any hint? |
2848049 | Prove that [imath]\sum_{k\ge 1}\frac{kg(x^2)}{1+k^3[g(x^2)]^2}[/imath] does not converge uniformly on the real line
Suppose [imath]g: \mathbb R_{\ge 0}\to \mathbb R_{\ge 0}[/imath] is a continuous bijective function. Prove that the series [imath]\sum_{k\ge 1}\frac{kg(x^2)}{1+k^3[g(x^2)]^2}[/imath] is not uniformly convergent on [imath]\mathbb R[/imath]. If I knew the sum of the series, I could use the negation of the definition of the uniform convergence of partial sums, but since I don't know the limit of partial sums (and how to find one), I cannot do this. How can I proceed? | 1889840 | Uniform convergence of [imath]\sum\limits_{k=1}^{\infty}\dfrac{kf(x^2)}{1+k^3f(x^2)^2}[/imath]
Let [imath]f:[0,\infty) \rightarrow [0,\infty][/imath] be a continuous bijection and the series [imath]\sum\limits_{k=1}^{\infty}\dfrac{kf(x^2)}{1+k^3f(x^2)^2}[/imath]. Prove that the series converges uniformly on [imath][\epsilon, \infty)[/imath] when [imath]\epsilon>0[/imath] and not converge uniformly on [imath]\mathbb{R}[/imath]. I tried using M-test for the first part so I suppose to prove [imath]f[/imath] is a increasing function. It seems to be very difficult. Please give me a hand. Thank you in advance for your help. |
1798985 | Which exponents r>0 is the limit finite
I am trying to find values of [imath]r>0[/imath] such that [imath]\lim\limits_{n\rightarrow \infty} \sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}[/imath] is finite. I have tried to use integral methods for this limit such as [imath]\sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}=\sum\limits_{k=1}^{n}\frac{n^{r-1}}{n^r+k^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac1n\sum\limits_{k=1}^{n}\frac{1}{1+(\frac kn)^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r}[/imath]. So the first term, I could use the integral method to estimate. However, the second one is hard to evaluate. Could you please help me with this? Thanks | 624179 | When do we have [imath] \lim_{n\to\infty}\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}<+\infty[/imath] for [imath]r>0[/imath]?
Let [imath]r>0[/imath] and [imath] a_n(r)=\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} [/imath] When do we have [imath] \lim_{n\to\infty}a_n(r)<+\infty? [/imath] One can write [imath] a_n(r)=\frac1n\cdot\sum_{k=1}^{n^2}\frac{1}{1+\left(\frac{k}{n}\right)^r} [/imath] which looks simpler but I don't see how it would help. Also I tried to bounded [imath]\{a_n(r)\}[/imath] but I don't see how. [ADDED:](Thanks to Daniel) It does look like a Riemann sum. I think this is not an integral on a finite integral. Instead it looks like [imath]\int_0^\infty\frac{1}{1+x^r}dx.[/imath] But I'm not sure if this the corresponding integral or not. |
2821013 | Existence of elements of order [imath]2[/imath] in [imath]S_n[/imath] whose Product has order [imath]n[/imath]
Show that all for [imath]n>2[/imath], [imath]S_n[/imath] contains two elements [imath]x,y[/imath], both of order [imath]2[/imath], such that their product is of order [imath]n[/imath]. I can find such elements for [imath]n[/imath] small, but I don't know how to make up an algorithm to produce these [imath]x,y[/imath] for large [imath]n[/imath]. Any hints are appreciated. | 2395015 | For [imath]n\geq 3[/imath] there exist [imath]x,y\in S_n[/imath] such that [imath]x[/imath] and [imath]y[/imath] have order [imath]2[/imath] and [imath]xy[/imath] has order [imath]n[/imath].
Assume that [imath]n\geq 3[/imath]. Let [imath]S_n[/imath] be the symmetric group with [imath]n![/imath] elements. Prove that there exists [imath]x[/imath] and [imath]y[/imath] in [imath]S_n[/imath] such that [imath]x[/imath] and [imath]y[/imath] have order [imath]2[/imath] and [imath]xy[/imath] has order [imath]n[/imath]. For [imath]n=3[/imath], it is easy to see that [imath]x=(1, 2)[/imath] and [imath]y=(1, 3)[/imath] works. How to prove it for the general case? |
2879766 | Sharing a pie evenly among an unknown number of people.
This is a question inspired by the question "Nine gangsters and a gold bar" on the Puzzling Stack Exchange. Suppose you're throwing a party, and you know that either 7, 8, or 9 people will arrive. Before anyone arrives, you want to cut the pie into the fewest number of pieces such that you can give everyone an equal amount. In this case, you can cut the pie into 18 pieces of the following sizes (this is conjectured to be the minimal number of pieces). [imath]\frac{1}{168}, \frac{1}{72}, \frac{1}{56}, \frac{1}{36}, \frac{2}{63}, \frac{1}{28}, \frac{23}{504}, \frac{1}{21}, \frac{25}{504}, \frac{5}{84}, \frac{31}{504}, \frac{4}{63}, \frac{19}{252}, \frac{5}{63}, \frac{1}{12}, \frac{47}{504}, \frac{7}{72}, \frac{1}{9}[/imath] Then you can divide this into nine parts: [imath] \left(\frac{1}{168}, \frac{23}{504}, \frac{5}{84}\right), \left(\frac{1}{72}, \frac{7}{72}\right), \left(\frac{1}{56}, \frac{47}{504}\right), \left(\frac{1}{36}, \frac{1}{12}\right), \left(\frac{2}{63}, \frac{5}{63}\right), \left(\frac{1}{28}, \frac{19}{252}\right), \left(\frac{1}{21}, \frac{4}{63}\right), \left(\frac{25}{504}, \frac{31}{504}\right), \text{ and } \left(\frac{1}{9}\right). [/imath] Into eight parts: [imath] \left(\frac{1}{168}, \frac{1}{28}, \frac{1}{12}\right), \left(\frac{1}{72}, \frac{1}{9}\right), \left(\frac{1}{56}, \frac{1}{21}, \frac{5}{84}\right), \left(\frac{1}{36}, \frac{7}{72}\right), \left(\frac{2}{63}, \frac{47}{504}\right), \left(\frac{23}{504}, \frac{5}{63}\right), \left(\frac{25}{504}, \frac{19}{252}\right), \text{ and } \left(\frac{31}{504}, \frac{4}{63}\right) [/imath] And into seven parts: [imath] \left(\frac{1}{168}, \frac{1}{72}, \frac{1}{21}, \frac{19}{252}\right), \left(\frac{1}{56}, \frac{1}{36}, \frac{1}{28}, \frac{31}{504}\right), \left(\frac{2}{63}, \frac{1}{9}\right), \left(\frac{23}{504}, \frac{7}{72}\right), \left(\frac{25}{504}, \frac{47}{504}\right), \left(\frac{5}{84}, \frac{1}{12}\right), \text{ and } \left(\frac{4}{63}, \frac{5}{63}\right) [/imath] The question I'm interested in an algorithm that can produce such a minimal pie cut. In this case, it would take in the set of the possible numbers of guests, [imath]\{7,8,9\}[/imath], and output the list of fractions above—or a different list of minimal length. I don't particularly care about the computational complexity of the program. I am interested in proving that the program outputs the minimal number of "slices". More examples [imath] \begin{align*} f(\{2\}) = f(\{1,2\}) &= \left(\frac{1}{2},\frac{1}{2}\right) \\ f(\{n\}) = f(\{1,n\}) &= \underbrace{\left(\frac{1}{n},\frac{1}{n},\cdots,\frac{1}{n}\right)}_n \\ f(\{2,3\}) &= \left(\frac{1}{3},\frac{1}{3},\frac{1}{6},\frac{1}{6}\right) \end{align*} [/imath] | 1383406 | Minimum Cake Cutting for a Party
You are organizing a party. However, the number of guests to attend your party can be anything from [imath]a_1[/imath], [imath]a_2[/imath], [imath]\ldots[/imath], [imath]a_n[/imath], where the [imath]a_i[/imath]'s are positive integers. You want to be prepared, so you want to cut a cake into smaller pieces. The pieces are not necessarily of equal size. The requirement is that, no matter how many guests come (which you will have known before distributing the cake), you will be able to give each of them some pieces of the cake without having to cut the cake any further so that everybody will get the same amount of cake. What is the minimum number of pieces of your cake you will have to cut it into? Trivially, if [imath]n=1[/imath], then the answer is [imath]a_1[/imath]. The answer is also known for [imath]n=2[/imath], which is [imath]a_1+a_2-\gcd\left(a_1,a_2\right)[/imath] (if you wonder why graph theory is in the tag, it is because the only proof known to me for the [imath]n=2[/imath] case is done via a graph-theorical argument). I conjecture that the answer, in general, is [imath]m:=\sum_{j=1}^n\,(-1)^{j-1}\,g_j\,,[/imath] where [imath]g_j:=\sum_{1\leq k_1<k_2<\ldots<k_j\leq n}\,\gcd\left(a_{k_1},a_{k_2},\ldots,a_{k_j}\right)[/imath] for [imath]j=1,2,\ldots,n[/imath] (here, [imath]g_1[/imath] is simply [imath]a_1+a_2+\ldots+a_n[/imath]). It is easy to cut the cake into [imath]m[/imath] pieces to satisfy the required condition. Apparently, the conjecture is false for [imath]n>2[/imath] (see Dividing the whole into a minimal amount of parts to equally distribute it between different groups.). However, I expect that my guess is not far away from the correct answer. EDIT: The [imath]n=2[/imath] case with [imath]\gcd\left(a_1,a_2\right)=1[/imath] appeared as a problem for the Spring Contest, A Level, of the Tournament of the Towns of 1990. See https://keoserey.files.wordpress.com/2012/12/imtot-book-3.pdf (page 35 of the PDF). Related Topics: Dividing the whole into a minimal amount of parts to equally distribute it between different groups. https://puzzling.stackexchange.com/questions/19870/nine-gangsters-and-a-gold-bar/19881#19881 https://mathoverflow.net/questions/214477/minimal-possible-cardinality-of-a-a-1-a-k-distributable-multiset |
2831775 | Integrating [imath]F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}[/imath] along a circle
Let [imath]F:\mathbb R^3 - \{0\}\to \mathbb R[/imath] be given by [imath]F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}.[/imath] Compute [imath]\int_C F\cdot ds[/imath] where [imath]C[/imath] is the unit cirlce on the plane [imath]x+y+z=3[/imath] with the orientation from the point [imath]\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)[/imath] to [imath]\left(1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)[/imath] to [imath]\left(1 + \frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)[/imath] and back to [imath]\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)[/imath] What I've done: By Green's theorem, the integral is [imath]\iint_D \operatorname{curl}(F)\cdot k\ dA[/imath] I got that the curl is [imath]\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}}[/imath] so the integral is [imath]\int\int_D \frac{z}{(x^2+y^2+z^2)^{3/2}} \ dA[/imath] But I'm having difficulties with determining how to define the region [imath]D[/imath] to compute the last integral. | 615803 | Alternative approach for a line integral?
Let [imath]F[/imath] be the vector field in [imath]{\Bbb R}^3\setminus\{0\}[/imath] defined by [imath] F(x,y,z):=\left(0,\frac{xz}{(y^2+z^2)\sqrt{x^2+y^2+z^2}},\frac{-xy}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}\right). [/imath] Compute the line integral [imath]\int_CF\cdot ds[/imath], where [imath]C[/imath] is the unit circle centered at the point [imath](1,1,1)[/imath] that lies on the plane [imath]x+y+z=3[/imath] and has the orientation from the point [imath]\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}}\right)[/imath] to [imath]\left(1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}},1-\frac{1}{\sqrt{6}}\right)[/imath] to [imath]\left(1+\frac{2}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}\right)[/imath] and back to [imath]\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}}\right)[/imath]. A key step is to parameterize the curve: [imath] \begin{cases} (x-1)^2+(y-1)^2+(z-1)^2=1,\\ x+y+z=3. \end{cases} [/imath] Let [imath] \begin{cases} x=1+\sin\theta\cos\phi\\ y=1+\sin\theta\sin\phi\\ z=1+\cos\theta \end{cases} [/imath] Substituting it into [imath]x+y+z=3[/imath], we have [imath]\theta=g(\phi)[/imath] for some function [imath]g[/imath], which can be written down explicitly. Then we have [imath] (x,y,z)=(x(\phi),y(\phi),z(\phi)). [/imath] Here is my question: Is there an alternative approach which might be more efficient than using the parameterization above to calculate the line integral? [Added] I'm not sure if Stokes' theorem would help. Some calculations render that [imath] \nabla\times F(x,y,z)=\left(\frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}}, \frac{z}{(x^2+y^2+z^2)^{3/2}}\right). [/imath] It seems that this would not simplify the calculation when we do integration on the "upper" sphere (since the center of the circle is not at the origin), of which the unit circle in the problem is the boundary. |
632697 | Convergence of [imath] \sum_{n=1}^\infty\left(\exp(\frac{(-1)^n}{n})-1\right)[/imath]?
Consider the series [imath] \sum_{n=1}^\infty a_n [/imath] where [imath] a_n:=\left(\exp\left(\frac{(-1)^n}{n}\right)-1\right). [/imath] Show that it is convergent but not absolutely convergent. I tried the alternating test but I'm not sure if [imath]\{|a_n|\}[/imath] is decreasing or not. And I don't see how other tests might help. Any ideas? [Disclaimer: this question, asked on Jan 9 '14 at 16:01, is NOT a duplicate of the linked one, which was asked much later on Jul 19 '15 at 20:32] | 1366873 | Show that the series [imath]\sum(\exp(\frac{(-1)^n}{n})-1)[/imath]converges, but not absolutely.
Show that the series converges, but not absolutely. [imath]\sum_{n=1}^{\infty}( [/imath]exp[imath](\frac{(-1)^n}{n})-1)[/imath]. My Try: Let [imath]a_n=[/imath]exp[imath](\frac{(-1)^n}{n})-1[/imath]. I was going to use alternating series test because the sequence [imath]\{a_n\}[/imath] is alternating. But [imath]\{|a_n|\}[/imath] is not decreasing to [imath]0[/imath]. So, I am stuck now. Can anybody please give me a hint? |
1093821 | Find all possible Jordan canonical forms of [imath]A[/imath], where [imath](A-2I)^3(A+2I)^2=0[/imath]
A [imath]5\times 5[/imath] matrix [imath]A[/imath] satisfies the equation [imath](A-2I)^3(A+2I)^2=0[/imath]. Assuming there are at least [imath]2[/imath] linearly independent eigenvectors for [imath]2[/imath], write all possible Jordan canonical forms. My Question: The eigenvalues of [imath]A[/imath] are [imath]2[/imath] and [imath]-2[/imath]. If we know the number of linearly independent eigenvectors for [imath]2[/imath], can we determine the number of linearly independent eigenvectors for [imath]-2[/imath]? | 594108 | If [imath](A-2I)^3(A+2I)^2=0[/imath], then what are the possible Jordan canonical forms of [imath]A[/imath]?
Here is the exercise: Let [imath]A[/imath] be a [imath]5\times5[/imath] complex matrix such that [imath](A-2)^3(A+2)^2=0[/imath], where we define [imath]A-\mu:=A-\mu I[/imath] for scalar [imath]\mu[/imath]. Assume that [imath]\lambda=2[/imath] is an eigenvalue of [imath]A[/imath] and its geometric multiplicity is at least [imath]2[/imath]. What are the possibilities for the Jordan canonical form (JCF)? What I know so far from the assumption is that the minimal polynomial of [imath]A[/imath] is of the form [imath]f(x)=(x-2)^i(x+2)^j[/imath] where [imath]1\leq i\leq 3[/imath] and [imath]0\leq j\leq 2[/imath]. The number of blocks in the Jordan segment [imath]J(2)[/imath] is at least [imath]2[/imath]. One can write the possible minimal polynomial one by one, which gives the information of the size of the largest block in each Jordan segment, and use the possible geometric multiplicity of [imath]\lambda=2[/imath] to find JCF. Here are my questions: Is there an alternative approach? Can we use the characteristic polynomial of [imath]A[/imath] here? |
2884670 | Integration for special sine-function
I am having trouble integrating the following: [imath]\int_{0}^{\infty} \sin(t)\cdot t^{x-1} ~dt[/imath] for [imath]0 < x < 1[/imath]. Does anybody know how that can be done. Ive tried to set it up with powerseries for [imath]\sin(t)[/imath], and by a substitution followed with integration by parts, non of that gave anything useful. Does anybody know how to integrate this, using the gamma function? Thanks | 437582 | I'm looking for several ways to prove that [imath]\int_{0}^{\infty }\sin(x)x^mdx=\cos(\frac{\pi m}{2})\Gamma (m+1)[/imath]
I'm looking for several ways to prove that [imath]\int_{0}^{\infty }\sin(x)x^mdx=\cos\left(\frac{\pi m}{2}\right)\Gamma (m+1)[/imath] for [imath]-2< Re(m)< 0[/imath] |
2885088 | Are finite extensions of global function fields always simple?
A global function field is a finite extension of [imath]\mathbb{F}_p(T)[/imath] for some prime number [imath]p[/imath]. A finite field extension [imath]L/K[/imath] is called simple if it has a primitive element, i.e. there exists an [imath]x \in L[/imath] such [imath]L[/imath] is generated (as a [imath]K[/imath]-algebra) by [imath]x[/imath]. Finite extensions of global functions fields definitely do not need to be separable, but are they always simple? | 101406 | Is a field perfect iff the primitive element theorem holds for all extensions, and what about function fields
Let [imath]L/K[/imath] be a finite separable extension of fields. Then we have the primitive element theorem, i.e., there exists an [imath]x[/imath] in [imath]L[/imath] such that [imath]L=K(x)[/imath]. In particular, the primitive element theorem holds for all finite extensions of a perfect field. Question 1. Is a field [imath]K[/imath] perfect if and only if the primitive element theorem holds for all finite extensions of [imath]K[/imath]? Question 2. Suppose that [imath]K[/imath] is a field extension of [imath]\mathbf{F}_p[/imath] of transcendence degree 1, i.e., a function field over a finite field. Does the primitive element theorem hold for any finite extension of [imath]K[/imath]? In Question 2, I am actually only interested in the case [imath]K=\mathbf{F}_q(t)[/imath]. |
2884967 | Indefinite integral involving normal cdf
In this answer, the following integral identity is used, where [imath]\Phi(t)[/imath] is the cdf of the standard normal variable. [imath]\int_{-\infty}^{\infty}t\frac d{dt}\Phi(t)^2dt=1/\sqrt{\pi}.[/imath] I tried to simplify to understand how. \begin{align*} \int_{-\infty}^{\infty}t\frac d{dt}\Phi(t)^2dt=\int_{-\infty}^{\infty}2t\Phi(t)\phi(t)dt=\int_{-\infty}^{\infty}2t\Phi(t)d\Phi(t)=\int_0^12x\Phi^{-1}(x)dx \end{align*} How do I proceed to get the answer? Is there a simpler way? | 2855794 | Showing [imath]2\int_{-\infty}^{\infty} x f(x) F(x) \ dx = \frac{1}{\sqrt{\pi}}[/imath] for standard normal pdf and cfd
i am trying to prove the identity in the title. I strongly think I need to use the error function [imath]\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp\{-t^2\}\ dt[/imath] in some way. Best I have so far is replacing [imath]F(x) = \frac{1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)}{2}[/imath] to end up with [imath]2\int_{-\infty}^{\infty} x \, f(x) \, F(x) \ dx = \int_{-\infty}^{\infty} \text{erf}\left(\frac{x}{\sqrt{2}}\right) \, x\, \frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{1}{2}x^2\right\}\ dx.[/imath] My attempts on partial integration have failed, any other ideas or does anyone succeed? I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof. Very thankful for any help. |
2884444 | Strictly decreasing and concave down takes on negative values
Visually it seems that a strictly decreasing twice-differentiable map [imath]f: [0, \infty) \to \mathbb{R}[/imath] with [imath]f''(x) < 0[/imath] for all [imath]x[/imath] should be negative at some point, but I can't come up with a formal proof. Any help? | 2389312 | Does there exist some strictly positive function such that both the derivative and second derivative are strictly negative?
If [imath]f(x) > 0[/imath] over the reals is it possible to have [imath]f'(x) < 0[/imath] and [imath]f''(x)< 0 [/imath] over the reals? Assuming [imath]f[/imath] can be differentiated twice. |
2884719 | Sequence of polynomials converging to [imath]\frac{1}{z}[/imath]
Is there a sequence of polynomials converging uniformly to [imath]\frac{1}{z}[/imath] in [imath]K:=\{z\in\mathbb{C}\mid 1<|z|<2\}[/imath]? My first attempt was to use the theorem of Runge which would apply if [imath]K[/imath] would be compact and [imath]\mathbb{C}\setminus K[/imath] connected. As [imath]K[/imath] is not closed, it is not compact. But if I consider the closure [imath]\bar{K}[/imath], then [imath]\mathbb{C}\setminus\bar{K}[/imath] is not connected? So the Theorem can not be applied here? Any other hints? | 2125350 | Sequence of polynomials which converge uniformaly to 1/z
Does not there a sequence of polynomials which converges uniformly to [imath]1/z[/imath] on [imath]\{z\in\Bbb C\;:\;|z|=1\}[/imath]? Explain. Here is where I'm having problems with the question. I understand what a sequence is, however I don't know what a sequence of polynomials is. Does this mean that a you come up with a generic sequence that represents a polynomial and all these polynomials reduce down to [imath]1/z[/imath]? There is also uniform convergence in real analysis. Is this the same type of uniform converge? My class mate that this the answer comes from some type of approximation of polynomials theorem. |
850442 | An interesting identity involving powers of [imath]\pi[/imath] and values of [imath]\eta(s)[/imath]
It happens that, for any [imath]m\geq 1[/imath], [imath]\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{1}[/imath] where [imath]E_{2m}[/imath] is an integer number. My proof works through the following lines: the LHS is: [imath]\frac{1}{(2m)!}\int_{0}^{1}\frac{(\log x)^{2m}}{1+x^2}dx = \frac{1}{2\cdot(2m)!}\int_{0}^{+\infty}\frac{(\log x)^{2m}}{1+x^2}dx,[/imath] so we just need to compute: [imath]\left.\frac{d^{2m}}{dk^{2m}}\int_{0}^{+\infty}\frac{x^k}{1+x^2}\right|_{k=0},\tag{2}[/imath] but: [imath] \int_{0}^{+\infty}\frac{x^{1/r}}{1+x^2}\,dx = r\int_{0}^{+\infty}\frac{y^r}{1+y^{2r}}\,dy = \frac{\pi/2}{\cos(\pi/(2r))}[/imath] by the residue theorem, so [imath]E_{2m}[/imath] is just the absolute value of an Euler number, that belongs to [imath]\mathbb{N}[/imath]. Can someone provide a real-analytic proof of [imath](1)[/imath]? | 762813 | Infinite Series [imath]\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}[/imath]
I'm looking for a way to prove [imath]\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}[/imath] I know that [imath]\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)[/imath] so maybe I could simplify the above more? |
2885205 | Is [imath]f\left(x\right)=nx^{n}\left(1-x\right)[/imath] uniformly convergent on [imath]x\in\left[0,\:1\right][/imath]
Is [imath]f_{n}\left(x\right)=nx^{n}\left(1-x\right)[/imath] uniformly convergent on [imath]x\in\left[0,\:1\right][/imath] We can see [imath]f_{n}\left(x\right)\rightarrow f\left(x\right)=0[/imath] point-wise, but [imath]f_{n}\left(x\right)\geq f_{n+1}\left(x\right)[/imath] does not seem hold ? | 808835 | Uniform convergence of [imath]f_n(x)=nx^n(1-x)[/imath] for [imath]x \in [0,1][/imath]?
I want to decide whether or not [imath]f_n(x)=nx^n(1-x)[/imath] is uniformly convergent or not. I have shown that [imath]\lim_{n\to\infty} f_n(x)=0[/imath] for [imath]x \in [0,1][/imath]. Now [imath]f_n(0)=f_n(1)=0[/imath]. And in [imath](0,1)[/imath], we have [imath]f'_n(x)=n^2x^{n-1}-n(n+1)x^n[/imath]. Putting this to zero gives [imath]x=\frac{n}{n+1}[/imath]. Also [imath]f_n(\frac{n}{n+1})=(\frac{n}{n+1})^{n+1}[/imath]. But that tends to [imath]\frac{1}{e}[/imath] as [imath]n[/imath] tends to [imath]\infty[/imath] which is not [imath]0[/imath] so we don't have uniform convergence, right? Nonetheless we have [imath]\lim_{n\to\infty} \int_0^1 f_n(x)dx=\lim_{n\to\infty} (\frac{n}{n+2}-\frac{n}{n+1})=0=\int_0^1 \lim_{n\to\infty} f_n(x)dx[/imath]. Is all this right? I have some doubts, I mostly doubt my intuition in analysis... Thank you! |
2879473 | Geometric interpretation of the tensor product of two projectif spaces [imath]\mathbb{R} P^n \otimes \mathbb{R} P^n[/imath]
We define a Projective space of a vector space as follow : http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces Given a vector space [imath]V[/imath] over a field [imath]\mathbb{K}[/imath], its associated projective space [imath]\mathbb{P}(V)[/imath] is by definition : [imath]\mathbb{P}(V) = V - \{0\}/\sim[/imath] where [imath]\sim[/imath] is the equivalence relation [imath]u \sim v[/imath] iff [imath]u = \lambda v[/imath] for [imath]u, v\in V - \{0\}[/imath] and [imath]\lambda \in \mathbb{K}[/imath]. In particular, The real projective space [imath]\mathbb{R} P^n[/imath] is the quotient space of [imath]\mathbb{R}^{n+1}-\{0\}[/imath] by the equivalence relation defined on [imath]\mathbb{R}^{n+1}-\{0\}[/imath] by [imath]x\sim y \iff y=tx[/imath] for some nonzero real number [imath]t[/imath], where [imath]x,y\in\mathbb{R}^{n+1}-\{0\}[/imath]. Geometrically, two nonzero points in [imath]\mathbb{R}^{n+1}[/imath] are equivalent if and only if they lie on the same line through the origin, so [imath]\mathbb{R} P^n[/imath] can be interpreted as the set of all lines through the origin in [imath]\mathbb{R}^{n+1}[/imath]. In other hand, For two real vector spaces [imath]V[/imath] and [imath]W[/imath], the Tensor-Product [imath]V \otimes W[/imath] is the Quotient space [imath]K^{V\times W} / R(V,W)[/imath], where [imath]R(V,W)[/imath] is generated by the vectors \begin{align*} (v,\lambda w) - \lambda (v,w) & \\ (v,w+w')-(v,w)-(v,w') & \\ (\lambda v,w) - \lambda(v,w) & \\ (v+v',w)-(v,w)-(v',w) & \\ \text{and} \quad K^{V\times W} := \{f: V\times W \rightarrow\ K\}. & \end{align*} for any basis [imath](v_i)_{i\in I}[/imath] of V and [imath](w_j)_{j\in J}[/imath] of W, [imath](v_i\otimes w_j)_{i\in I,j\in J}[/imath] is a basis of [imath]V\otimes W[/imath], so any [imath]a\in V\otimes W[/imath] is a linear combination of some vectors [imath]v_i\otimes w_j[/imath]. And a tensor of type [imath](n, m)[/imath] is an element of [imath]V^{\ast \otimes m} \otimes V^{\otimes n}[/imath]. For example, - A tensor of type [imath](0, 0)[/imath] is a scalar. - A tensor of type [imath](1, 0)[/imath] is a vector. How can Intemperate geometrically the tensor product of two projectifs space ? What is interpretation of the tensor product of two Grassmannians ? Thanks in advance. | 486025 | Geometric intuition for the tensor product of vector spaces
First of all, I am very comfortable with the tensor product of vector spaces. I am also very familiar with the well-known generalizations, in particular the theory of monoidal categories. I have gained quite some intuition for tensor products and can work with them. Therefore, my question is not about the definition of tensor products, nor is it about its properties. It is rather about the mental images. My intuition for tensor products was never really geometric. Well, except for the tensor product of commutative algebras, which corresponds to the fiber product of the corresponding affine schemes. But let's just stick to real vector spaces here, for which I have some geometric intuition, for example from classical analytic geometry. The direct product of two (or more) vector spaces is quite easy to imagine: There are two (or more) "directions" or "dimensions" in which we "insert" the vectors of the individual vector spaces. For example, the direct product of a line with a plane is a three-dimensional space. The exterior algebra of a vector space consists of "blades", as is nicely explained in the Wikipedia article. Now what about the tensor product of two finite-dimensional real vector spaces [imath]V,W[/imath]? Of course [imath]V \otimes W[/imath] is a direct product of [imath]\dim(V)[/imath] copies of [imath]W[/imath], but this description is not intrinsic, and also it doesn't really incorporate the symmetry [imath]V \otimes W \cong W \otimes V[/imath]. How can we describe [imath]V \otimes W[/imath] geometrically in terms of [imath]V[/imath] and [imath]W[/imath]? This description should be intrinsic and symmetric. Note that SE/115630 basically asked the same, but received no actual answer. The answer given at SE/309838 discusses where tensor products are used in differential geometry for more abstract notions such as tensor fields and tensor bundles, but this doesn't answer the question either. (Even if my question gets closed as a duplicate, then I hope that the other questions receive more attention and answers.) More generally, I would like to ask for a geometric picture of the tensor product of two vector bundles on nice topological spaces. For example, tensoring with a line bundle is some kind of twisting. But this is still some kind of vague. For example, consider the Möbius strip on the circle [imath]S^1[/imath], and pull it back to the torus [imath]S^1 \times S^1[/imath] along the first projection. Do the same with the second projection, and then tensor both. We get a line bundle on the torus, okay, but how does it look like geometrically? Perhaps the following related question is easier to answer: Assume we have a geometric understanding of two linear maps [imath]f : \mathbb{R}^n \to \mathbb{R}^m[/imath], [imath]g : \mathbb{R}^{n'} \to \mathbb{R}^{m'}[/imath]. Then, how can we imagine their tensor product [imath]f \otimes g : \mathbb{R}^n \otimes \mathbb{R}^{n'} \to \mathbb{R}^m \otimes \mathbb{R}^{m'}[/imath] or the corresponding linear map [imath]\mathbb{R}^{n n'} \to \mathbb{R}^{m m'}[/imath] geometrically? This is connected to the question about vector bundles via their cocycle description. |
2884581 | How is this equation obtained? [imath]\int_0^\infty \frac{x^{s-1} dx}{e^x-1}=\Pi(s-1)\zeta(s) [/imath]
This equation: [imath]\int_0^\infty \frac{x^{s-1}}{e^x-1} dx=\Pi(s-1)\zeta(s) [/imath] was used by Riemann in his famous paper from 1859. Seemingly it follows from: [imath] \int_0^\infty e^{-nx}x^{s-1}dx= \frac{\Pi(s-1)}{n^s}[/imath] a result achieved by repeated integration by parts. How does the first equation follows from the second? | 1103139 | Integral Representation of the Zeta Function: [imath]\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx[/imath]
How does one get from this [imath]\zeta(s)=\sum_{k=1}^{\infty}\frac1{k^s}[/imath] to the integral representation [imath]\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx[/imath] of the Riemann Zeta function? I can see that it can be rewritten as [imath]\Gamma(s)\zeta(s)=\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx[/imath] and the Gamma function as an integral yields [imath]\zeta(s)\int_{0}^\infty \frac{x^{s-1}}{e^x}dx=\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx[/imath] But this approach does not work as the right integral does not converge. So how does one go from the summation to the integral representation? |
2885646 | Proving that [imath]\{0,1\}^\mathbb N[/imath] and [imath]\mathbb{N^N}[/imath] have the same cardinality
Problem Let [imath]D[/imath] be the set of all the functions [imath]f\colon\mathbb N\to \mathbb N[/imath] , where [imath]\mathbb N[/imath] is the set of natural numbers. Let [imath]E[/imath] be the set of all functions [imath]f\colon \mathbb N\to\{\,0,1\,\}[/imath]. Prove both the sets have equal cardinality. Attempt (failed) [imath]E[/imath] is the set of all infinite binary sequences. By Cantor's second diagonal argument [imath]E[/imath] is uncountable. Since [imath]E\subset D[/imath], [imath]D[/imath] is also uncountable. If i prove there exists injections from [imath]D[/imath] to [imath]E[/imath] and also from [imath]E[/imath] to [imath]D[/imath], then Schroeder-Bernstein theorem can be used. Attempt 2 Consider two functions [imath]\alpha\colon D \to E[/imath] and [imath]\beta\colon E \to D [/imath] Consider a function [imath]f \in E[/imath] is a function from [imath]\mathbb N[/imath] to [imath]\{\,0,1\,\}[/imath]. [imath]\beta(f) [/imath] is a function that assigns to [imath]n[/imath] the [imath]f(n)+1[/imath] . This is a injective function . I have difficulty finding [imath]\alpha[/imath]. Any hint or help will be appreciated. | 750367 | Proving that [imath]\mathrm{card}(2^{\mathbb{N}})=\mathrm{card}(\mathbb{N}^\mathbb{N})[/imath]
I'd like to prove that [imath]\mathrm{card}(2^{\mathbb{N}})=\mathrm{card}(\mathbb{N}^\mathbb{N})[/imath], I have the following 'sketch' but I'm not sure if this works. [imath]|2^{\mathbb{N}}|\leq|\mathbb{N}^{\mathbb{N}}|\leq|2^\mathbb{N^{\mathbb{N}}}|=|2^{\mathbb{N}\times\mathbb{N}}|=|2^\mathbb{N}|[/imath], then [imath]|2^{\mathbb{N}}|=|\mathbb{N}^\mathbb{N}|[/imath] I'm taking for granted the first inequality, (i.e: [imath]|2^{\mathbb N}|\leq|\mathbb{N}^{\mathbb{N}}|[/imath]), could be done a further proof about this. Would it be enough to point out that the functions in [imath]2^{\mathbb{N}}[/imath] are in [imath]\mathbb{N}^{\mathbb{N}}[/imath] but there are functions in the last one that are not in the first one? Should I try to give a more formal proof? |
2884885 | Graph Theory Problem S-C
Let [imath]G[/imath] be a self-complementary graph of order [imath]n[/imath], where [imath]n ≡ 1 \pmod{4}[/imath]. Prove that [imath]G[/imath] contains an odd number of vertices of degree [imath]\frac{(n − 1)}{2}[/imath]. My approach Let [imath]G[/imath] be a self-complementary graph of order n, where [imath]n=4k+1[/imath] i.e. [imath]n[/imath] is odd. Now size of [imath]G[/imath] is [imath]\frac{n(n-1)}{4}[/imath]. Now if [imath]V(G)=\{v_i:1\leq i \leq n\}[/imath]. Now [imath]\sum d(v_i)=\frac{n(n-1)}{2}[/imath]=even. Then what can i do? | 1708075 | Show that a self-complementary graph contains an odd number of vertices with degree [imath](n-1)/2[/imath]
Does anyone know how to prove that a self-complementary graph contains an odd number of vertices with degree [imath](n-1)/2[/imath]? Thanks for the help :D |
2885826 | The height of the Mandelbrot Set
A simple question, but one I am unable to find the answer to; What is [imath]\sup\{y:x+y\cdot i \in M\}[/imath] Where [imath]M[/imath] is the Mandelbrot set. How is this constant calculated? Is it a known constant? Etc... | 936462 | Supremum of all y-coordinates of the Mandelbrot set
Let [imath]M\subset \mathbb R^2[/imath] be the Mandelbrot set. What is [imath]\sup\{ y : (x,y) \in M \}[/imath]? Is this known? To be more descriptive: What is the supremum of all y-coordinates of all black points in the following picture: Picture File:Mandel zoom 00 mandelbrot set.jpg by Wolfgang Beyer licensed under CC-BY-SA 3.0 |
2885935 | Proving minimal sufficiency using Lehmann-Scheffe approach when joint pdf is only an indicator function
Let [imath]X_1,...,X_n[/imath] be iid and uniformly distributed on the interval [imath](-\theta, \theta)[/imath] where [imath]\theta>0[/imath]. I am asked to derive the set of minimal sufficient statistics. Here's my work: For each [imath]X_i[/imath], the pdf is given by [imath]f(x)=\frac{1}{2\theta}I(-\theta<x_i<\theta).[/imath] Hence, the joint pdf for [imath]X[/imath] can be given by [imath]g(x)=(\frac{1}{2\theta})^nI(-\theta<x_i<\theta)[/imath] for all [imath]i=1, 2,...,n[/imath]. Now let [imath]h(x,y)=\frac{g(x)}{g(y)}[/imath] for a particular combination of [imath]X[/imath] and [imath]Y[/imath] belonging to the sample space. Then [imath]h(x,y)[/imath] can be rewritten as follows:[imath]h(x,y)=\frac{(\frac{1}{2\theta})^nI(-\theta<x_i<\theta)}{(\frac{1}{2\theta})^nI(-\theta<y_i<\theta)}=\frac{I(-\theta<x_i<\theta)}{I(-\theta<y_i<\theta)}=\frac{I(max(|x_i|)<\theta)}{I(max(|y_i|)<\theta)} [/imath] I know that a statistic [imath]T[/imath] is minimally sufficient for [imath]\theta[/imath] if the following holds: [imath]h(x,y)[/imath] is independent of [imath]\theta[/imath] for all possible values of [imath]\theta[/imath] if and only if [imath]T(x)=T(y)[/imath]. My question is as follows: if [imath]max(|x_i|)=max(|y_i|)[/imath], would my final expression above for [imath]h(x,y)[/imath] be free of [imath]\theta[/imath] for all possible values of [imath]\theta[/imath]? I want to say that [imath]h(x,y)[/imath] would equal 1 for all [imath]\theta[/imath], but couldn't this expression also be undefined? Thanks! | 2116770 | Minimal sufficient statistic of [imath]\operatorname{Uniform}(-\theta,\theta)[/imath]
I am seeking clarification on why both the vector [imath](X_{(1)},X_{(n)})^T[/imath] and [imath]\max\{-X_{(1)},X_{(n)}\}[/imath] are sufficient for [imath]\operatorname{Unif}(-\theta,\theta)[/imath], but only [imath]\max\{-X_{(1)},X_{(n)}\}[/imath] is minimal sufficient, as stated here. Can this be explained by the fact that while both can describe the data adequately, [imath]\max\{-X_{(1)},X_{(n)}\}[/imath] is of lesser dimension and is thus minimal sufficient? Using the definition of minimal sufficiency ([imath]T(X)[/imath] is minimal sufficient if [imath]T(x)=T(y) \iff \frac{\mathcal{L}(x;\theta)}{\mathcal{L}(y;\theta)}[/imath] does not depend on [imath]\theta[/imath]), I run into issues as with either choice of statistic, I need to analyze [imath]\frac{\mathbb{1}_{[\max\{-X_{(1)},X_{(n)}\}<\theta]}}{\mathbb{1}_{[\max\{-Y_{(1)},Y_{(n)}\}<\theta]}}[/imath] or [imath]\frac{\mathbb{1}_{[-\theta<X_{(1)}]}\mathbb{1}_{[X_{(n)}<\theta]}}{\mathbb{1}_{[-\theta<Y_{(1)}]}\mathbb{1}_{[Y_{(n)}<\theta]}}[/imath] which may be [imath]\frac{0}{0}[/imath]. Even when not [imath]\frac{0}{0}[/imath], I'm having trouble seeing why the former is not dependent on [imath]\theta[/imath], but the latter is dependent on [imath]\theta[/imath] if [imath]T(X)=T(Y)[/imath]. |
2886037 | Why a matrix cannot be divided?
We know that we can operate addition[imath](+)[/imath] , subtraction[imath](-)[/imath] , multiplication[imath](×)[/imath] on a matrix. But why we can't operate division [imath](÷)[/imath] on a matrix?... Sorry for my bad english | 228229 | Is division of matrices possible?
Is it possible to divide a matrix by another? If yes, What will be the result of [imath]\dfrac AB[/imath] if [imath] A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}, B = \begin{pmatrix} w & x \\ y & z \\ \end{pmatrix}? [/imath] |
2885667 | Help evaluating [imath] \int \frac {\sqrt{x^2-1}}x \ dx[/imath]
Before anything, I'd like to clarify that I have no background in calculus (I'm still in school). I only try to learn calculus as a hobby. Please be gentle. I'm trying to evaluate the aforementioned integral but the answer I got doesn't seem to match with 3 sources different sources that I found online. My process was to substitute in [imath]x = \sec u, \ dx = \tan u\sec u\, du[/imath]. This led me to the answer [imath]\tan (\mathrm{arcsec}\ x) - \mathrm{arcsec}\ x[/imath]. However after trying to check my answer I saw that WolframAlpha says it's [imath]\arctan \left(\frac 1{\sqrt{x^2-1}}\right) + \sqrt{x^2-1}[/imath], Symbolab says it's [imath]-\arctan (\sqrt{x^2-1}) + \sqrt{x^2-1}[/imath], and this video says it's [imath]-\mathrm{arcsec} \,x + \sqrt{x^2-1}[/imath]. I plotted all four results on a graph. Symbolab and WolframAlpha only differ by a constant so I assume that both of them have the correct result whereas the video and I have the wrong ones. Also it can be seen that all the results are the same on the positive side of the x axis but things break down on the negative side. My final questions are: Where did the video creator and I go wrong? What is happening on the negative x axis? EDIT: It seems this post might be a duplicate of this one. I hope this post doesn't get removed because I believe what I asked is a bit different from what was asked there. EDIT 2: Upon further inspection, I guess it's reasonable for this post to be removed. | 757059 | Evaluate [imath]\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x[/imath]
My try, using [imath]x = \sec(u)[/imath] substitution: [imath] \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &=& \int \tan^2(u) \mathrm{d}u \\ &=& \tan(u) - u + C \\ &=& \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} [/imath] However, according to Wolfram Alpha, the answer should be: [imath] \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C [/imath] When I derive this last answer I don't get back the integrand, but rather: [imath] \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} [/imath] I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of [imath]arcsec(x)[/imath]. Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand? |
2886670 | Linear dependence over the null field
The set {0} with the + and × operations defined as 0+0=0 and 0×0=0 sarisfies the properties to be a field, so a vector space can be constructed over it, with the property [imath]0x = x [/imath] where x is a vector. In my book it states that the vector [imath]x [/imath] by itself forms a linearly dependent system iff [imath]x=0[/imath]. This seems to lead to a contradiction: if we assume that the vector space is over {0} with the properties said before, then [imath]x =0[/imath] is also linearly independent since the only way to get the null vector from it is multiply it by 0... So does this mean that it is linearly dependent and independent at same time? Which would mean that lin. dependence is not the negation of lin. independence... Or is the book wrong? Or am I making a mistake? | 427078 | Is [imath]\{0\}[/imath] a field?
Consider the set [imath]F[/imath] consisting of the single element [imath]I[/imath]. Define addition and multiplication such that [imath]I+I=I[/imath] and [imath]I \times I=I[/imath] . This ring satisfies the field axioms: Closure under addition. If [imath]x, y \in F[/imath], then [imath]x = y = I[/imath], so [imath]x + y = I + I = I \in F[/imath]. Closure under multiplication. [imath]x \times y = I \times I = I \in F[/imath] Existence of additive identity. [imath]\forall x \in F[/imath] (i.e., for [imath]x=I[/imath]), [imath]x + I = x[/imath], so [imath]I[/imath] is the additive identity. Existence of mulitiplicative identity. [imath]\forall x \in F, x \times I = x[/imath], so [imath]I[/imath] is the multiplicative identity. Additive inverse. [imath]\forall x \in F, \exists y = I \in F: x + y = I[/imath] Multiplicative inverse. [imath]\forall x \in F, \exists y = I \in F: x \times y = I[/imath]. However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. Commutativity of addition. [imath]\forall x, y \in F, x + y = I = y + x[/imath] Commutativity of multiplication. [imath]\forall x, y \in F, x \times y = I = y \times x[/imath] Associativity of addition. [imath]\forall x, y, z \in F[/imath], [imath](x + y) + z = I + I = I[/imath] and [imath]x + (y + z) = I + I = I[/imath], so [imath](x + y) + z = x + (y + z)[/imath] Associativity of multiplication. [imath]\forall x, y, z \in F[/imath], [imath](x \times y) \times z = I \times I = I[/imath] and [imath]x \times (y \times z) = I \times I = I[/imath], so [imath](x \times y) \times z = x \times (y \times z)[/imath] Distributivity of multiplication over addition. [imath]I \times (I + I) = I[/imath] and [imath]I \times I + I \times I = I[/imath], so [imath]\forall x,y,z \in F, x(y+z) = xy+xz[/imath] Based on the above, [imath]\{I\}[/imath] seems to qualify as a field. If [imath]I[/imath] is assumed to be a real number, then the unique solution of [imath]I + I = I[/imath] and [imath]I \times I = I[/imath] is, of course, [imath]I = 0[/imath]. So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it? Specifically, is it required for the multiplicative identity to be distinct from the additive identity? |
2886753 | Limit of the Product [imath]\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)[/imath]
I know I can transform the above product as follows \begin{eqnarray*}\lim_{n\rightarrow\infty}\left(\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}\right)&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)\\&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right) \end{eqnarray*} My thoughts are this is going to go to [imath]e^{-x/2}[/imath], but I can't figure out how to go from [imath]\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right)[/imath] to [imath]\lim_{n\rightarrow\infty}\left(1+\frac{-1/2}{n}\right)^n[/imath] | 2906986 | How can I find the limit of the following sequence [imath]\frac{1}{2} . \frac {3}{4} ....... \frac {2n - 1}{2n} [/imath]?
How can I find the limit of the following sequence: [imath]\frac{1}{2} . \frac {3}{4} ....... \frac {2n - 1}{2n} [/imath] The problem for me is that they are multiplied and not added, also I can see that the numerator is the set of odd numbers and the denominator is the set of even numbers but then what? |
2886522 | Separable Commutative [imath]C*[/imath] algebra
Consider the algebra [imath]C(X)[/imath] of continuous complex functions over a compact space [imath]X[/imath]. On what conditions this algebra is separable? What if [imath]X[/imath] is a compact subset of [imath]\mathbb{R}^n[/imath]? | 1331321 | [imath]C(X)[/imath] is separable when [imath]X[/imath] is compact?
[imath]X[/imath] is a compact metric space, [imath]C(X)[/imath] is separable when [imath]X[/imath] is compact where [imath]C(X)[/imath] denotes the space of continuous functions on [imath]X[/imath]. How to prove it? And if [imath]X[/imath] is just a compact Hausdorff space, then [imath]C(X)[/imath] is still separable? Or if [imath]X[/imath] is just a compact (not necessarily Hausdorff) space, then [imath]C(X)[/imath] is still separable? Please help me. Thanks in advance. |
2886825 | How many ways are there to arrange [imath]n[/imath] married couples in a line so that a husband and his wife are not together?
There are [imath]2n[/imath] people. There are [imath]2n![/imath] ways to arrange them. The number of ways to arrange them such that couples are always together is [imath]n! \cdot 2^n[/imath] How do you calculate the number of ways to arrange them such that no couples are together? | 1033191 | Arranging couples (husband and wife) on a bench so no wife would sit next to her husband?
Find the number of ways of arranging n couples [imath]\{H_i,W_i\}[/imath], [imath]i=1,2,...,n,[/imath] in a row such that [imath]\{H_i\}[/imath] is not adjacent to [imath]\{W_i\}[/imath] for each [imath]i=1,2,...,n[/imath]. I have included part of the solution to the question. An arrangement in [imath]S[/imath] satisfies the property [imath]P_i[/imath], where [imath]i=1,2,...,n,[/imath] if and only if [imath]H_i[/imath] and [imath]W_i[/imath] are adjacent. [imath]\omega(1) = \sum_{i=1}^{n} \omega(P_i) = \binom{n}{1}(2)(2n-1)![/imath] My question is about the above statement. I don't understand why finding [imath]\omega(1)[/imath], which is the number of arrangements where only one couple is adjacent, is given by [imath]\binom{n}{1}(2)(2n-1)![/imath]. I do understand that there are (2n−1)! ways to arrange the remaining people after choosing a couple [imath]\binom{n}{1}[/imath] and that they can be permutated in [imath]2![/imath] ways. But I don't see why the (2n−1)! would ensure that the rest of the couples won't be adjacent. |
376907 | Inclusion of orthogonal complements: if [imath]U_1\subset\ U_2[/imath], then [imath]U_2^\bot\subset\ U_1^\bot[/imath]
Show if [imath]U_1\subset\ U_2[/imath], then [imath]U_2^\bot\subset\ U_1^\bot[/imath] I'm thinking using contradiction. [imath]\exists v\in U_2^\bot[/imath] s.t. [imath]v\notin U_1^\bot[/imath] [imath]\Rightarrow[/imath] [imath]\exists u\in U_1[/imath] s.t. [imath]u\notin U_2[/imath] Let [imath]v\in U_2^\bot[/imath] and [imath]v\notin U_1^\bot[/imath], then [imath]\exists u_2\in U_2[/imath] s.t. [imath]\lt v,u_2\gt =0[/imath] and [imath]\forall u_1 \in U_1, \lt v,u_1\gt \neq 0[/imath] This implies, [imath]\forall u_1 \in U_1, u_1\neq u_2[/imath]. Hence, [imath]U_1\not\subset U_2[/imath] I feel like this should be really easy with direct proof, but I keep getting [imath]U_1^\bot = U_2^\bot[/imath]. What am I doing wrong? Any hint will be helpful. Thank you! | 1911645 | Proving that if [imath]V[/imath] is an inner product space and [imath]U_1, U_2[/imath] are subsets, then [imath]U_1 \subset U_2[/imath] if and only if [imath]U^{\perp}_2 \subset U^{\perp}_1[/imath]
Suppose [imath]V[/imath] is an inner product space and [imath]U_1, U_2[/imath] are subsets, then I'd like to show that [imath]U_1 \subset U_2[/imath] if and only if [imath]U^{\perp}_2 \subset U^{\perp}_1[/imath]. I can show the first forward direction, by first taking [imath]w \in U^{\perp}_2[/imath]. Then, [imath]w[/imath] must be a vector where [imath]<w, u_2> = 0[/imath] for all [imath]u_2 \in U_2[/imath]. But, since [imath]U_1 \subset U_2[/imath], then [imath]<w, u_1> = 0[/imath] for all [imath]u_1 \in U_1[/imath]. Hence, [imath]w \in U^{\perp}_1[/imath]. I am not sure how to show the opposite direction. My proof is as follows: Take [imath]u_1 \in U_1[/imath]. Now, let [imath]w \in U^{\perp}_2[/imath]. Then, [imath]<w,u_2> = 0[/imath] for all [imath]u_2 \in U_2[/imath]. But, [imath]U^{\perp}_2 \subset U^{\perp}_1[/imath], and so [imath]w \in U^{\perp}_1[/imath] as well. Hence, [imath]<w,u_1> = 0[/imath] for all [imath]u_1 \in U_1[/imath]. After this, I am not sure how to complete the proof. Am I in the right direction? Thanks! |
2886466 | How could I prove this?
[imath]\newcommand{\intd}{\,\mathrm{d}}[/imath]On an AOPS forum a user solved that [imath]\int\frac{1}{x(x+1)(x+2)\ldots(x+n)}\intd{x}=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\ln|x+k|+C[/imath] This to me is absolutely amazing and their solution interesting enough, I am curious as to how I would go about proving this. To me this shouts induction, it is obvious for [imath]n=0[/imath] ([imath]\int\frac1x\intd{x}[/imath]) however on the inductive step I could not figure out what to do. Integration by parts did not help and beyond that I am at a loss. I look forward to your responses | 2783915 | How to prove the closed form of the integral [imath]\int \frac {dx}{\prod_{r=0}^n (x+r)}[/imath]
I want to derive a closed formula for the integral [imath]I_n= \int \frac {dx}{\prod_{r=0}^n (x+r)}[/imath] On writing out first few terms we get For [imath]n=0[/imath], [imath]I_0=\ln \vert x\vert+C[/imath] For [imath]n=1[/imath], [imath]I_1=\ln \vert x\vert-\ln \vert x+1\vert+C[/imath] For [imath]n=2[/imath] [imath]I_2=\frac {1}{2!}\left(\sum_{r=0}^2 (-1)^r\binom {2}{r} \ln \vert x+r\vert\right)+C[/imath] For [imath]n=3[/imath] [imath]I_3=\frac {1}{3!}\left(\sum_{r=0}^3 (-1)^r\binom {3}{r} \ln \vert x+r\vert\right)+C[/imath] Hence for generalized [imath]n[/imath] we have [imath]I_n=\frac {1}{n!}\left(\sum_{r=0}^n (-1)^r\binom {n}{r} (\ln \vert x+r\vert)\right)+C[/imath] Now this is just an observation but I want to prove that it is correct. I have tried lot of methods but not useful. Partial fractions would have been most useful but would go out on tedious task which is nearly impossible. Also integration by parts won't help nor any trig substitution. So any ideas are welcome. And ya, this is not a homework question, it's a question which I just saw in a integral challenge paper. |
2887257 | Maximum prime divisor of [imath]13^4+16^5-172^2[/imath]
My approach: I tried it by remainder theorem as there 13,16 which are very close and raised to the power some small powers. So if it becomes like : [imath]13^4+(13+3)^5-(13*13+3)²[/imath] Now how should I proceed, [imath]13[/imath] will obviously not be the answer. | 2886489 | Find the largest of the three prime divisors of the number [imath]13^4 + 16^5 - 172^2[/imath]
I was able to factor out only the prime 13,thus [imath]13^4 + 16^5 - 172^2=13\cdot 80581[/imath] What should be done to solve it? (Maybe some clever factorization, modulo, or anything else?) |
2887231 | How to get a closed form for [imath]\sum_{i=1}^n \frac{3^i}{2^i}[/imath]
How can I get a closed form for: [imath]\sum_{i=1}^n \frac{3^i}{2^i}[/imath] I have just started studying closed forms for summations and I am still lost on this matter so I would appreciate if you guys could explain that step by step or suggest me some references for me to study. Thanks in advance. | 2645248 | Partial Sums of Geometric Series
This may be a simple question, but I was slightly confused. I was looking at the second line [imath]S_n(x)=1-x^{n+1}/(1-x)[/imath]. I was confused how they derived this. I know the infinite sum of a geometric series is [imath]1/(1-x)[/imath]. I just can't figure out how the partial sums, [imath]S_n(x)[/imath], have [imath]1-x^{n+1}[/imath] on the numerator. How was this derived? Thank you. Example 5.20. The geometric series [imath] \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + \dotsb [/imath] has partial sums [imath] S_n(x) = \sum_{k=0}^n x^k = \frac{1 - x^{n+1}}{1 - x} \cdotp [/imath] Thus, [imath]S_n(x) \to 1/(1-x)[/imath] as [imath]n \to \infty[/imath] if [imath]|x| < 1[/imath] and diverges if [imath]|x| \geq 1[/imath], meaning that [imath] \sum_{n=0}^\infty x^n = \frac{1}{1-x} \qquad \text{pointwise on $(-1,1)$}. [/imath] (Original image here.) |
2887239 | Intuition behind higher derivatives
While I can easily imagine the second derivative conveying the concavity, and the first derivative conveying the slope of any function in a graph. How do I visually understand the meaning of higher derivatives apart from the fact that they represent the rate of [imath](n-1)^{th}[/imath] derivatives. | 274789 | Meaning of different Orders of Derivative
I have been trying to analyse the meaning of higher order derivatives and their geometrical significance. Given a function [imath]f(x)[/imath] what are the unique geometric interpretation of its higher orders? [imath]f'(x)[/imath] - slope (rate of change of x w.r.t y) [imath]f''(x)[/imath] - convexity / concavity of [imath]f(x)[/imath] based on [imath]sign(f''(x))[/imath] [imath]f'''(x)[/imath] - increasing / decreasing slope of [imath]f'(x)[/imath] [imath]f''''(x)[/imath] - ?? ... [imath]f^n(x)[/imath] - ?? I apologize if this question is rudimentary. |
2887619 | Does the series for [imath]\cos(x)/x[/imath] converges?
The sequence of [imath] a_x ={\cos (x)\over x} [/imath] does converge to zero. As a result, intuitively [imath] \sum_{x=1}^\infty {\cos (x)\over x} [/imath] should also converge right? But I've been told that the series diverges. This shouldn't be true... right? | 2289319 | For what values of [imath]\theta[/imath] does [imath]\sum_{1}^{\infty} \frac{e^{i n\theta}}{n}[/imath] Converge?
I just came across the problem of what happens in the boundary cases of What is convergence interval of this series [imath]\sum\limits_{n =1}^{\infty}\frac{(z- 1 - i)^{n}}{n\cdot2^{n}}[/imath]. More concisely, for what [imath]\theta[/imath] will the following sum converge? [imath]\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}[/imath] It is well known that this sum will diverge for [imath]\theta=0[/imath] and will converge for [imath]\theta=\pi[/imath]. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge. Another idea I had was to split the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge [imath]\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}[/imath] Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with [imath]\theta=1[/imath], but I don't see how to generalize this. |
2887223 | a decreasing sequence of sets with [imath]\mu(A_1)=\infty[/imath]
Let [imath](A_n)[/imath] be a decreasing sequence of sets. I am looking for an example that if [imath]\mu^*(A_1) = \infty[/imath], [imath]\mu^*(\cap_{n \in N} A_n)\not= \lim_{n\to \infty}\mu^*(A_n).[/imath] I saw from another post that if we define sets to be [imath]A_1 = [0, \infty)[/imath] and [imath]A_n =\emptyset[/imath] for [imath]n \ge 2[/imath], this sequence of sets can be an example which satisfies the above conditions. But, I do not understand why. I think that [imath]\lim_{n\to \infty}\mu^*(A_n)= 0[/imath] as [imath]A_n[/imath] is empty set for [imath]n \ge 2[/imath]. But, I also think that [imath]\cap_{n \in N} A_n = \emptyset[/imath]. That is, [imath]\mu^*(\cap_{n \in N} A_n)=0[/imath] as well. Could you elaborate on this? | 1492847 | Example of decreasing sequence of sets with first set having infinite measure
I was wondering if someone could please give me an example of a sequence of decreasing sets where the first set has infinite Lebesgue measure; i.e., [imath]\{B_{n}\}_{n=1}^{\infty}[/imath] such that [imath]m(B_{1}) = \infty[/imath] but [imath]m(\cap_{n=1}^{\infty} B_{n}) \neq \lim_{n \to \infty}m(B_{n})[/imath]? thank you. |
2886608 | Power Series Solution
Consider the initial value problem [imath]xu''+\sin(x)u=0 \ \ \ \ u(0)=0, u'(0)=2[/imath] Derive the first [imath]4[/imath] non-zero terms of a power series solution to this problem about the point [imath]x=0[/imath]. I know the solution will have the from [imath]u(x)=\sum_{k=0}^{\infty}A_kx^k[/imath] But upon differentiating, I get that [imath]\sum_{k=0}^{\infty}k(k-1)A_kx^{k-2}+\sin(x)\sum_{k=0}^{\infty}A_kx^{k-1}=0[/imath] and do not know how to proceed. Would an initial substitution help? | 2887278 | Convergence of the Power Series for [imath]xu''+\sin(x)u=0[/imath]
Consider the initial value problem [imath]xu''+\sin(x)u=0 \ \ \ \ u(0)=0, u'(0)=2[/imath] What can be said about the radius of convergence of this series? I have determined that the first four nonzero terms (about the point [imath]x=0[/imath]) of the power series are [imath]2x,-x^2,\frac{1}{6}x^3,-\frac{1}{72}x^4[/imath] How can I test for convergence? Intuitively it looks not to converge, but how can I show this? edit The solution will have the from [imath]u(x)=\sum_{k=0}^{\infty}A_kx^k.[/imath] Differentiating, \begin{align} \sum_{k=2}^{\infty} k(k-1)A_kx^{k-1}+\sin(x)\sum_{k=0}^{\infty}A_kx^k&=0 \\ \sum_{k=2}^{\infty} k(k-1)A_kx^{k-1}+\left(\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}\right)\sum_{k=0}^{\infty}A_kx^k&=0 \\ (2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+\left(2x^2+A_2x^3+\left(A_3-\frac{1}{3}\right)x^4+\left(A_4-\frac{A_2}{6!}\right)x^5+\left(A_5-\frac{A_3}{3!}+\frac{2}{5!}\right)x^6+..\right)=0 \end{align} Equating coefficients, [imath]x^{2k}=0[/imath] for [imath]k\in\mathbb{Z^+}[/imath]. [imath]6A_3+2=0\implies A_3=-\frac{1}{3}[/imath] [imath]20A_5+A_3-\frac{1}{3}=0\implies A_5=\frac{1}{30}[/imath] [imath]42A_7+A_5-\frac{A_3}{3!}+\frac{2}{5!}=0\implies A_7=-\frac{19}{7560}[/imath] Hence [imath]u=2x-\frac{1}{3}x^3+\frac{1}{30}x^5-\frac{19}{7560}x^7+..[/imath] |
2887902 | Proximal Operator of an $ {L}_{\infty} $ Norm with a Constraint
Dear Convex Optimization Experts, My question is related to this post: The Proximal Operator of the [imath] {L}_{\infty} [/imath] (Infinity Norm), but not really same, I think, as I have a constraint. Apologies if it is obvious to extend the answer. So, I am seeking a proximal operator [imath]\textrm{prox}_{\lambda f}(x)[/imath] of this function [imath]f(x) = I\left\{x \in C \right\} \ ,[/imath] [imath]C = \left\{x : \lVert x \rVert_{\infty} \leq \gamma \right\} \ ,[/imath] where [imath]x \in \mathbb{R}^{n \times 1}[/imath], [imath]I\left\{\cdot \right\}[/imath] is an indicator/characteristic function, [imath]\gamma \in \mathbb{R}^{n \times 1}[/imath], and the constraint [imath]\lVert x \rVert_{\infty} \leq \gamma[/imath] is element-wise. EDIT: The desired constraint is: [imath]C = \left\{x : |x_i| \leq \gamma_i \right\} \ ,[/imath] | 1825747 | Orthogonal Projection onto the [imath] {L}_{\infty} [/imath] Unit Ball
What is the Orthogonal Projection onto the [imath] {\ell}_{\infty} [/imath] Unit Ball? Namely, given [imath] x \in {\mathbb{R}}^{n} [/imath] what would be: [imath] {\mathcal{P}}_{ { \left\| \cdot \right\| }_{\infty} \leq 1 } \left( x \right) = \arg \min_{{ \left\| y \right\| }_{\infty} \leq 1} \left\{ {\left\| y - x \right\|}_{2}^{2} \right\} [/imath] I managed to get an answer using the Moreau Decomposition. Yet I would be happy to see if someone can derive the answer directly. Thank You. |
2379223 | Convergence of [imath]\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p}[/imath]
Consider the series [imath]\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p}[/imath] where [imath]p>0[/imath]. For what [imath]p[/imath], the series converges? This question appears in the Tier 1 Analysis exam (2014/08) of Indiana University Bloomington. And I modify it a little to make it look more tight... | 1019514 | For which [imath]p[/imath] does the series [imath]\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p}[/imath] converge?
Suppose [imath]\displaystyle a_n=\frac{1}{(\text{log}2)^p+(\text{log3})^p+...+(\text{log}n)^p}[/imath] for [imath]n\geq 2[/imath] and [imath]p>0[/imath]. For which values of [imath]p[/imath] does [imath]\sum_{n=2}^\infty a_n[/imath] converge? So, when [imath]p\leq 1[/imath], [imath]\displaystyle a_n\geq \frac{1}{(n-1)(\text{log}n)^p}[/imath] for [imath]n\geq 2[/imath]. Let [imath]\displaystyle b_n=\frac{1}{(n-1)(\text{log}n)^p}[/imath]. Then [imath]\sum b_n[/imath] diverges since [imath]p\leq 1[/imath]. Therefore by comparison test [imath]\sum a_n[/imath] diverges when [imath]p\leq 1[/imath]. When [imath]p>1[/imath], [imath]\displaystyle a_n\leq \frac{1}{(\text{log}n)^p}[/imath]. Let [imath]\displaystyle c_n= \frac{1}{(\text{log}n)^p}[/imath]. If [imath]\sum c_n[/imath] converges then I will win. But now I am stuck. Is [imath]\sum c_n[/imath] converges? Then how can I prove it? If not how can I conclude the state of convergence of [imath]\sum a_n[/imath] for [imath]p>1[/imath]? Can anybody please help? |
2028811 | Easy way to fit a power function [imath]y=a+b*x^c[/imath]
I am writing a program in C++/CLI in which I fit certain functions to some data input (certain number of [imath]x[/imath] and [imath]y[/imath] pairs). I have [imath]7[/imath] models from which one is [imath]y = a*x^b[/imath] and I have the solution for calculating both [imath]a[/imath] and [imath]b[/imath] factors (I have the sum of [imath]ln(x)[/imath], sum of [imath]ln(y)[/imath], sum of [imath]ln(x) * ln(y)[/imath], sum of [imath]ln(x) * ln(x)[/imath] and then [imath]a[/imath] and [imath]b[/imath] are just results of 2 equations). Then I have the second model which is [imath]y = a + b*x^c[/imath] and I don't know how to get it. Could anyone help? I also have troubles with models [imath]y = a + b*sinh^{-1}(x/c)[/imath] and [imath]y = a*x +b*sinh^{-1}(x/c)[/imath]. | 2025491 | Best fitting in a curve of the form [imath]Ax^B+C[/imath]
I am trying to fit data of the form [imath](x_i,y_i)[/imath], [imath]i=1,\ldots,n[/imath], in a curve of the form [imath]y=Ax^B+C[/imath], where [imath]B\in (0,1)[/imath]. All three constants [imath]A,B,C[/imath] are to be determined optimally (no particular norm for the moment). Also, the [imath]x_i[/imath]'s are positive integers, if that helps. Is there any standard method to attack such a problem? |
2885190 | Let [imath]a_n \neq 0 \;\; \forall n \in \mathbb{N}[/imath] and [imath]\exists L = \lim |\frac{a_{n+1}}{a_n}|[/imath]. Show that, if [imath]L<1[/imath], then [imath]\lim a_n = 0[/imath]
Let [imath]a_n \neq 0 \;\; \forall n \in \mathbb{N}[/imath] and [imath]\exists L = \lim |\frac{a_{n+1}}{a_n}|[/imath]. Show that, if [imath]L<1[/imath], then [imath]\lim a_n = 0[/imath] Please verify if my attempt makes sense and how could I fix it. I also would appreciate to see another ways of solving this exercise. Choose [imath]c[/imath] such that [imath]L<c<1[/imath]. Then [imath]\exists n_1 \in \mathbb{N}[/imath] such that [imath]n \geq n_1 \implies |\frac{a_{n+1}}{a_n}|<c[/imath] (already verified this) [imath]\implies|a_{n+1}| < c|a_n| \implies\frac{|a_{n+1}|}{c^{n+1}} < \frac{|a_n|}{c^n}[/imath] Then [imath]\frac{|a_n|}{c^n}[/imath] is monotonically decreasing for [imath]n \geq n_0[/imath]. Since [imath]\frac{|a_n|}{c^n} \geq 0[/imath], this sequence is limited, then there is [imath]n_1\geq n_0[/imath] such that [imath]\exists M>0[/imath] such that [imath]0 \leq \frac{|a_n|}{c^n} \leq M \implies 0 \leq |a_n| < Mc^n[/imath]. By squeeze theorem, [imath]\lim |a_n| = 0[/imath]. But I'm failing to conclude [imath]\lim a_n = 0[/imath], all I can see is that [imath]\lim a_n \leq 0[/imath]. | 1513393 | [imath]\lim a_n = 0[/imath] if [imath]\left|\frac{a_{n +1}}{a_n}\right|\to L < 1[/imath]
Suppose that [imath]a_n \neq 0[/imath], for every n and that [imath]L = \lim |\frac{a_{n +1}}{a_n}|[/imath] exists. Show that if L < 1, then [imath]\lim a_n = 0[/imath]. What I did so far: If L < 1 and [imath]L = \lim |\frac{a_{n +1}}{a_n}|[/imath], there exists [imath]n_0[/imath] such that for [imath]n \geq n_0[/imath], [imath]0 < |a_{n+1}| < |a_n|[/imath]. That means that the sequence [imath](|a_n|)_{n \geq n_0}[/imath] is decreasing. Consider the set [imath]S = \{ |a_0|, ..., |a_{n_0} \}[/imath]. S is finite. Let [imath]\beta = \max_{0 \leq i \leq n_0} |a_ i|[/imath]. Thus, [imath](|a_n|)[/imath] is limited (because, for every n, [imath]0 \leq |a_n| \leq \beta).[/imath] Now I have that [imath](|a_n|)[/imath] is limited and, throwing away a finite number of terms (the [imath]n_0[/imath] firsts) I can assume that it is decreasing. So I know that [imath](|a_n|) converges. [/imath] How can I prove that it converges to 0[imath]? [/imath] I also know that if I prove that \lim |a_n| = 0[imath], then I have that [/imath] \lim a_n = 0$. |
2887597 | Integral of [imath]\int\ e^{-x/2}\sqrt(sin-1)/(cosx+1)[/imath]
[imath]\int\ e^{-x/2}\sqrt(sin-1)/(cosx+1)[/imath] The result that I'm getting here contains a cosecx term but the answer has secx term! Please help. | 390080 | Definite Integral of square root of polynomial
I need to learn how to find the definite integral of the square root of a polynomial such as: [imath]\sqrt{36x + 1}[/imath] or [imath]\sqrt{2x^2 + 3x + 7} [/imath] EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares. This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function. |
2888158 | Is there a bijection [imath]\big\{x:\mathbb{N}\to\{0,1\}:\{i\in\mathbb{N}:x(i)=1\}\text{ is finite}\big\}\to\big\{x:\mathbb{N}\to\{0,1\}\big\}[/imath]?
Can we use Cantor's theorem to figure this out? Intuitively the domain seems to be a subset of the codomain, but where do powersets come in? Thanks in advance! | 612036 | Is the collection of finite subsets of [imath]\mathbb{Z}[/imath] countable?
The collection of all subsets of [imath]\mathbb{Z}[/imath] is uncountable, due to Cantor's theorem But how can I prove that the collection of all finite subsets of [imath]\mathbb{Z}[/imath] is countable? |
2888711 | Evaluate [imath]\int x^3 \sqrt{4-x^2} dx[/imath]
Evaluate [imath]\int x^3 \sqrt{4-x^2} dx[/imath] Is my solution correct? In the book, some other (final)solution is given. | 1729189 | Solve: [imath]\int x^3\sqrt{4-x^2}dx[/imath]
[imath]\int x^3\sqrt{4-x^2}dx[/imath] I tried changing the expression like this: [imath]\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx[/imath] And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression. |
2888880 | Is [imath]\phi(n) + \sigma(n) \geq 2n[/imath] always true?
Suppose [imath]\phi[/imath] is Euler totient function and [imath]\sigma[/imath] is divisor sum. Is [imath]\phi(n) + \sigma(n) \geq 2n[/imath] true for every natural [imath]n[/imath]? I manually checked the inequality for all numbers between [imath]1[/imath] and [imath]20[/imath] - and it holds on them. I do not know, however, how to prove this fact in general. Also, it is not hard to see, that for prime [imath]p[/imath], [imath]\phi(n) + \sigma(n) = 2n[/imath]. That means, that if a counterexample exists, it has to be composite. Any help will be appreciated. | 2875495 | Can a composite number [imath]n[/imath] be the arithmetic mean of [imath]\sigma(n)[/imath] and [imath]\varphi(n)[/imath]?
Let [imath]\varphi(n)[/imath] be the totient-function [imath]\sigma(n)[/imath] be the divisor-sum-function It is clear that every prime number [imath]n[/imath] is the arithmetic mean of [imath]\varphi(n)[/imath] and [imath]\sigma(n)[/imath], in other words , the equality [imath]\varphi(n)+\sigma(n)=2n[/imath] holds. Can the given equality hold for a composite number [imath]n[/imath] ? Upto [imath]10^8[/imath], I did not find an example. I conjecture that the equality can only hold for primes or [imath]n=1[/imath]. How can I prove it ? |
2888867 | Find the cardinality of the set [imath]A_p[/imath] defined as the following :
For any prime number [imath]p[/imath], [imath]A_p[/imath]=the set of integers [imath]d\in \{1,2,3,\dots, n\}[/imath] such that the power of [imath]p[/imath] in the prime factorization of [imath]d[/imath] is odd. Then \begin{align*} A_p= & \lfloor\dfrac{n}{p}\rfloor-\lfloor\dfrac{n}{p^2}\rfloor+\lfloor\dfrac{n}{p^3}\rfloor-\lfloor\dfrac{n}{p^4}\rfloor+\dots \end{align*} Any one can give me any idea how can I show this ? Update: I have gone through [imath]1p,2p,3p,\dots, kp\leq n<(k+1)p\implies \lfloor \dfrac{n}{p}\rfloor=k[/imath], but I can not understand after this step. | 2885105 | For any prime number [imath]p[/imath], let [imath]A_p[/imath] be the set of integers [imath]d\in \{1,2,\dots, 999\}[/imath] s. Then what is the cardinality of [imath]A_p[/imath]?
For any prime number [imath]p[/imath], let [imath]A_p[/imath] be the set of integers [imath]d\in \{1,2,\dots, 999\}[/imath] such that the power of [imath]p[/imath] in the prime factorization of [imath]d[/imath] is odd. Then what is the cardinality of [imath]A_p[/imath]? I have used to the following result, but failed to show the required result. The largest exponent [imath]e[/imath] of a prime [imath]p[/imath] such that [imath]p^e[/imath] is a divisor of [imath]n![/imath] is given by [imath] e=\left\lfloor \dfrac{n}{p} \right\rfloor + \left\lfloor \dfrac{n}{p^2} \right\rfloor + \left\lfloor \dfrac{n}{p^3} \right\rfloor [/imath] How can I do next? |
2888793 | Proof explanation: Prove that [imath]\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2[/imath]
Two days ago, I asked a question Prove that [imath]\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2[/imath] but was answered just once. However, I am finding it hard to understand the proof provided. Please, I'll need thorough explanation or another proof. As to the proof, I don't understand why triple sum was used and not double. If we differentiate [imath]g(x)=f(b)-f(x)-f'(x)(b-x)[/imath], what do we get? I am asking because I find it hard to comprehend the proof. Please, can anyone explain these to me? Alternative proofs are welcome. Thanks. | 2885937 | Prove that [imath]\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2[/imath]
Let [imath]\Omega[/imath] be an open set in [imath]\Bbb{R}^n[/imath] and [imath]f:\Omega\to \Bbb{R}^m [/imath] be of class [imath]\Bbb{C}^2.[/imath] Let [imath]a,b[/imath] be in [imath]\Omega[/imath] such that [imath][a,b]\subset \Omega.[/imath] Prove that \begin{align}\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2.\end{align} where [imath]f'(a)(b-a)=\big<\nabla f(x),(b-a)\big>=\sum^{n}_{k=1}\frac{\partial f_i}{\partial x_k}(a)(b_k-a_k).[/imath] I think the idea of Mean Value Theorem for [imath]C'[/imath] functions in [imath]\Bbb{R}^n[/imath] (Reference for Mean Value Theorem in several variables) will help me in showing it but I don't know how to! Please, can anyone show me a proof or reference? |
422159 | If a prime [imath]p[/imath] is divided by 30, remainder is either prime or 1
Show that if a prime number [imath]p[/imath] is divided by 30, then the remainder is either a prime or 1. I did the sum sum but cannot complete it. I took [imath]p=6k+1[/imath] and [imath]p=6k-1[/imath] form. now for any [imath]k=5m[/imath] we get [imath]6k=30m[/imath] form so we can say that the remainder comes 1 but then I can't give proper proof of the rest. | 308355 | What can primes, except 2, 3, and 5, be congruent to [imath]\pmod {30}[/imath]?
After some trials, I found out that a prime [imath]p \gt 5[/imath] is congruent to [imath]q\pmod{30}[/imath], where [imath]q[/imath] is also a prime, and [imath]1 \le q \lt 30 \;[/imath] (i.e. [imath]p \equiv q\pmod{30}.[/imath] Is there a way to write a formal proof to show this? Is that an acceptable answer. |
2889438 | Why doesn't (12)(34) commute with any odd ordered element in [imath]A_5[/imath]
In Dummit and foote, for the proof for the simplicity of [imath]A_5[/imath], it claims that it is "easy" to see that [imath](12)(34)[/imath] doesn't commute with any odd elements in [imath]A_5[/imath] and it commutes with [imath](13)(24)[/imath], I have absolutely not idea why this is considered "easy", what method did the author use to quickly see this? It further claimed that "it follows that [imath]|C_{A_5}((12)(34)| = 4[/imath], which again, I don't see how it follows from the above statement without also checking other two cycles do not commute with [imath](12)(34)[/imath]. | 504866 | [imath](1 2)(3 4)[/imath] does not commute with any nonidentity element of odd order in [imath]A_5[/imath].
On Dummit's Abstract Algebra on p. 128, it says: "It is easy to see that [imath](1 2)(3 4)[/imath] ... does not commute with any non-identity element of odd order in [imath]A_5[/imath]." But I don't find it easy. Any help, please? |
2227635 | Evaluate [imath]\sum_{k=0}^{n} {3n \choose 3k}[/imath]
Evaluate [imath]\sum_{k=0}^{n} {3n \choose 3k}[/imath]. It's not hard to evaluate by putting roots of unity. Therefore, I would like to see some solutions using elementary mathematical stuffs but not roots of unity. And the answer is [imath]\frac{1}{3}(2^{3n}+2(-1)^n)[/imath] | 213583 | Show that [imath]\sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3}[/imath]
The other day a friend of mine showed me this sum: [imath]\sum_{k=0}^n\binom{3n}{3k}[/imath]. To find the explicit formula I plugged it into mathematica and got [imath]\frac{8^n+2(-1)^n}{3}[/imath]. I am curious as to how one would arrive at this answer. My progress so far has been limited. I have mostly been trying to see if I can somehow relate the sum to [imath]\sum_{k=0}^{3n}\binom{3n}{k}=8^n[/imath] but I'm not getting very far. I have also tried to write it out in factorial form, but that hasn't helped me much either. How would I arrive at the explicit formula? |
2889571 | Physical meaning of transformation with symmetric matrices
[imath]A[/imath] is a symmetric matrix and [imath]v[/imath] is a vector. I wonder if there is a physical/geometric meaning to the transformation [imath]Av[/imath] due to the symmetry in [imath]A[/imath]. | 2727082 | What does a symmetric matrix transformation do, geometrically?
I need some geometric intuition behind what exactly a symmetric matrix transformation does. In a [imath]2 \times 2[/imath] and [imath]3 \times 3[/imath] vector space, what are they generally? |
2889051 | [imath]X>0[/imath], [imath]Y>0[/imath] and [imath]X^2+Y ^3\geq X^3+Y^4[/imath]. Prove that [imath]X^3+Y ^3\leq 2[/imath]
[imath]X>0[/imath], [imath]Y>0[/imath] and [imath]X^2+Y ^3\geq X^3+Y^4[/imath]. Prove that [imath]X^3+Y ^3\leq 2[/imath] First I tried: [imath]0<X\leq1[/imath] and [imath]0<Y\leq1[/imath], but this in not the only case for [imath]X^2+Y ^3\geq X^3+Y^4[/imath] and get nowhere. Can somobody give me an idea? | 2178764 | If [imath]x,y>0[/imath] and [imath]x^2+y^3\ge x^3+y^4[/imath], prove that [imath]x^3+y^3 \le 2[/imath].
As in the title. If [imath]x,y>0[/imath] and [imath]x^2+y^3\ge x^3+y^4[/imath], prove that [imath]x^3+y^3 \le 2.[/imath] This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here. I'd greatly appreciate any hints. |
2889466 | Show that [imath]f[/imath] is locally constant function.
[imath]f: D \to C[/imath] is analytic, Denote [imath]f = u+iv[/imath] and if [imath]\exists[/imath] an [imath]\mathbb{R}[/imath] differentiable function [imath]g: R \to R[/imath] where [imath]u = g\circ v[/imath]. Show that [imath]f[/imath] is locally constant function. My idea is to partial differentiate [imath]u[/imath] wrt [imath]x[/imath] and [imath]y[/imath] respectively and get [imath]u_{xx} =g_{xx}(v) \cdot v_{x}^2+g_x(v) \cdot v_{xx}[/imath] [imath]u_{yy} = g_{yy}(v) \cdot v_{y}^2+g_{y}(v) \cdot v_{yy}[/imath] And since [imath]f[/imath] is analytic, we know that [imath]u,v[/imath] are harmonic. I tried adding up both equations and got stuck, any ideas on how to continue? | 2748301 | Show that the holomorphic function [imath]u + iv[/imath] must be constant if [imath]u = h \circ v[/imath]. Problem from Remmert.
The problem as stated in R. Remmert's Theory of Complex Functions p. 62., is this: Let [imath]f = u + iv[/imath] be holomorphic in the region [imath]G \subseteq \mathbb{C}[/imath] and satisfy [imath]u = h \circ v[/imath] for some differentiable function [imath]h : \mathbb{R} \to \mathbb{R}[/imath]. Show that [imath]f[/imath] is constant. I've attacked this computationally in each way I can think of. I've tried to show that [imath]\frac{\partial f}{\partial z}[/imath] is zero; played with the Laplace equation, knowing that [imath]h \circ v[/imath] and [imath]v[/imath] must be hamonically conjugate, etc. I even considered spinning an argument based on the maximum principle. It seems obvious, but I have not succeeded in a proof yet. Perhaps I am missing a concept? |
2889855 | Calculate, [imath]f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)[/imath]
If [imath]f(x)=\frac{4^x}{4^x+2}[/imath] Calculate, [imath]f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)[/imath] My Attempt: I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated, [imath]f\bigg(\frac{1}{2}\bigg)=\frac{1}{2}[/imath] [imath]f\bigg(\frac{1}{3}\bigg)+f\bigg(\frac{2}{3}\bigg)=1[/imath] [imath]f\bigg(\frac{1}{4}\bigg)+f\bigg(\frac{2}{4}\bigg)+f\bigg(\frac{3}{4}\bigg)=\frac{3}{2}[/imath] I could see that, [imath]f\bigg(\frac{1}{n}\bigg)+f\bigg(\frac{2}{n}\bigg)+f\bigg(\frac{3}{n}\bigg)\ldots f\bigg(\frac{n-1}{n}\bigg)=\frac{n-1}{2}[/imath] So, [imath]f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)=998[/imath] which is indeed the right answer. But I am not satisfied with my method. How else can I solve it? | 2401137 | find [imath]f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})[/imath] of [imath]f(x)=\frac{2}{2+4^x}[/imath]
[imath]f(x)=\frac{2}{2+4^x}[/imath] find [imath]f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})[/imath] Please guide me through it, the only step I know is probably to eliminate the denominator ps. Not a homework |
2886109 | Prove that [imath]| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1[/imath] where [imath]|x|\leq1[/imath] and [imath]|y|\leq1[/imath]
Prove that [imath]| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1[/imath] where [imath]|x|\leq1[/imath] and [imath]|y|\leq1[/imath] I tried: [imath]x=\sin\alpha[/imath] and [imath]y=\cos\beta[/imath] [imath]\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}[/imath] but if I write [imath]\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta[/imath], it's not true because [imath]\cos\alpha \sin\beta[/imath] can be negative. Can someone give me an idea? | 375260 | Prove this inequality [imath]a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 [/imath]
Prove that for [imath]a,b\in [-1,1][/imath]: [imath]a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1[/imath] |
2890106 | Show that any ball (open or closed) in [imath]\mathbb{R}^n[/imath] is a convex set.
I tried to show that the set [imath]X = \{(x, y)\in \mathbb{R}^2 : x^2 + y^2<1\}[/imath] is convex. | 137738 | Proving that closed (and open) balls are convex
Let [imath]X[/imath] be a normed linear space, [imath]x\in X[/imath] and [imath]r>0[/imath]. Define the open and closed ball centered at [imath]x[/imath] as [imath] B(x, r) = \{y \in X : \Vert x − y\Vert < r\} [/imath] [imath] \overline{B}(x, r) = \{y \in X : \Vert x − y\Vert \leq r\}. [/imath] Then [imath]B(x, r)[/imath] and [imath]\overline{B}(x, r)[/imath] are convex. I tried to prove this, but either my calculation is incorrect, or I am on the wrong path: I aim to show for the closed ball [imath]\overline{B}(x,r)[/imath] (for open ball I assume the proof is similar). Suppose [imath]y,z \in \overline{B}(x, r)[/imath]. Then [imath]\Vert x − y\Vert \leq r[/imath] and [imath]\Vert x − z\Vert \leq r[/imath]. We must show that [imath]t \in [0,1][/imath] implies [imath]ty + (1-t)z \in \overline{B}(x,r)[/imath]. But [imath]t \in [0,1][/imath] implies [imath] \Vert ty + (1-t)z - x\Vert = \Vert t(y-z) + z - x\Vert \leq |t| \Vert y-z\Vert + \Vert z-x\Vert \leq |t|(\Vert y-x\Vert + \Vert x-z\Vert) + \Vert z-x\Vert < |t|(2r) + r = r(2|t| + 1), [/imath] which is not necessarily [imath]\leq r[/imath]. We probably wanted to end up with [imath]< |t|r[/imath] or [imath]\leq |t|r[/imath] as our final inequality. Thanks in advance. |
2890094 | Determine the limit distribution of [imath]\frac{\max\{X_1, X_2,...,X_n \}}{n}[/imath] as [imath]n \to\infty[/imath]
Let [imath]X_1, X_2, . . .[/imath] be independent, [imath]C(0, 1)[/imath]-distributed random variables (Cauchy). Determine the limit distribution of [imath]Y_n=\frac{\max\{X_1, X_2,...,X_n \}}{n}\qquad\text{ as }\quad n\to\infty[/imath] This is how far I have come: [imath]F_{Y_n}(y)=P(\max\{X_1, X_2,...,X_n \}<ny)=F_{\max(X_i)}(ny)=(F_x(ny))^n[/imath] Since the cumulative distribution function of [imath]X[/imath] is [imath]F_X(x)=\frac{1}{\pi}\arctan(x)+\frac{1}{2}, x \in \mathbb{R}[/imath], we get : [imath]F_{Y_n}(y)=\left(\frac{1}{\pi}\arctan(ny)+\frac{1}{2}\right)^n[/imath] Correct so far? Now, we seek [imath]\lim_{n\to\infty}F_{Y_n}(y)[/imath] and its here im stuck. I have tried to use the identity [imath]\arctan(x)+\arctan(1/x)=\frac{\pi}{2}[/imath] but I just get complicated sums inside sums leading me nowhere. Any help is appreciated! | 1595202 | Limit distribution of [imath]\frac1n\cdot \max(X_1,X_2,...,X_n)[/imath] for [imath](X_n)[/imath] i.i.d. standard Cauchy
Let [imath]X_1, X_2,\ldots [/imath] be independent, Cauchy [imath]C(0,1)-[/imath]distributed random variables. Determine the limit distribution of [imath]Y_n=\frac 1 {n} \cdot \max{(X_1,X_2,\dots,X_n)}[/imath] as [imath]n \rightarrow \infty[/imath]. Remark. It may be helpful to know that [imath]\arctan x + \arctan \frac 1 {x} = \frac \pi 2[/imath] and that [imath]\arctan y = y - \frac {y^3} 3 + \frac {y^5} 5 - \frac {y^7} 7 + \dots[/imath] I not sure how to attack this question, nor how I should handle the [imath]\max{(X_1,X_2,\ldots,X_n)}[/imath] in order to find the limit distribution. Any advice appreciated. |
2889733 | A smooth function separating two disjoint closed subsets
I'm trying this problem (2.14) in John M Lee's book on smooth manifolds: For any two disjoint closed sets [imath]A[/imath] and [imath]B[/imath] we can find a smooth function such that [imath]f^{-1}(1)=A[/imath] and [imath]f^{-1}(0)=B[/imath]. Can someone give a hint on how to do this? I tried to use normality and partition of unity and the fact that for any closed set [imath]K[/imath] there is a smooth function [imath]g[/imath] such that [imath]g^{-1}(0)=K[/imath]. But none of these seem to work. | 963209 | Finding bump function on a smooth manifold using partitions of unity.
Let [imath]M[/imath] be a smooth manifold. Let [imath]A[/imath] and [imath]B[/imath] be disjoint closed sets of [imath]M[/imath]. Show there exists a smooth function [imath]f[/imath] such that [imath]f^{-1}(0)=A[/imath] and [imath]f^{-1}(1)=B[/imath]. This is my idea so far, Since [imath]A[/imath] and [imath]B[/imath] are disjoint closed subsets [imath]\{M-A,M-B\}[/imath] is an open cover for [imath]M[/imath]. Therefore there exists a partition of unity [imath]\{\psi_{1},\psi_{2}\}[/imath] with [imath]\psi_{1}[/imath] supported in [imath]M-A[/imath] and [imath]\psi_{2}[/imath] supported in [imath]M-B[/imath]. Furthermore [imath]\psi_{1}+\psi_{2}=1[/imath]. Then [imath]\psi_{1}(1-\psi_{2})[/imath] is zero on [imath]A[/imath] and [imath]1[/imath] on [imath]B[/imath]. I get the inclusions [imath]A\subset f^{-1}(0)[/imath] and [imath]B\subset f^{-1}(1)[/imath]. I've tried adding a few more open sets to the cover like [imath]M-(A\cup B)[/imath] and [imath](M-A)-B[/imath] and [imath](M-B)-A[/imath] but I still can't ensure that the function only vanishes at [imath]A[/imath]. Then I moved on to attempt to find any bump function that is 1 only on [imath]A[/imath] and ran into the same problem. Am I missing something here or do I just need to be clever with how I define a function. Any hints would be much appreciated. Thanks. |
2880796 | Any subset [imath]X[/imath] of [imath]\mathbb{R}^n[/imath] satisfying property C is compact.
Define a subset [imath]X[/imath] of [imath]\mathbb{R}^n[/imath] to have property C if every sequence with exactly one accumulation point in [imath]X[/imath] converges in [imath]X[/imath]. (Recall that [imath]x[/imath] is an accumulation point of a sequence [imath](x_n)[/imath] if every neighborhood of [imath]x[/imath] contains infinitely many [imath]x_n[/imath].) Show that any subset [imath]X[/imath] of [imath]\mathbb{R}^n[/imath] satisfying property C is compact. Since [imath]\mathbb{R}^n[/imath] is Euclidean, compactness [imath]\iff[/imath] closed and bounded. To show [imath]X[/imath] is closed, we can take any limit point [imath]x[/imath] of [imath]X[/imath]. By definition of a limit point, we can find a sequence in [imath]X[/imath] converging to [imath]x[/imath]. Then [imath]x[/imath] is the unique accumulation point of the sequence and hence by hypothesis [imath]x\in X[/imath] How can I show that [imath]X[/imath] is bounded? Alternatively, I could show that [imath]X[/imath] is sequentially compact, but I don't know how to proceed. | 2306842 | Proving that a set in which any sequence having exactly one accumulation point converges is compact
I found the following question about compactness: A set [imath]S\subseteq \mathbb{R}^n[/imath] has the following property: if a sequence of [imath]S[/imath] has exactly one accumulation point in [imath]S[/imath], then it converges in [imath]S[/imath]. Prove that [imath]S[/imath] is compact. We know that in [imath]\mathbb{R}^n[/imath] compact sets correspond to closed and bounded sets. (1) I am trying to prove that [imath]S[/imath] is closed. The condition in the question says if a sequence of [imath]S[/imath] has an accumulation point in [imath]S[/imath] then it converges in [imath]S[/imath]. I think instead the condition should be if a sequence of [imath]S[/imath] has an accumulation point( in [imath]\mathbb{R}^n[/imath], not necessarily in [imath]S[/imath] )then it converges in [imath]S[/imath]. Am I correct? (2) How to prove that [imath]S[/imath] is bounded? P.S. I have made an edit (there is exactly one accumulation point). Source of the question: Question 1 of this problem set. |
2890450 | [imath]\sum_{1\leq l \lt m\lt n} \dfrac{1}{5^l3^m2^n}[/imath]
I need a help to deal with the serie [imath]\sum_{1\leq l \lt m\lt n} \dfrac{1}{5^l3^m2^n}[/imath] (OBM level university, 2018). Some ideia? Thanks very much. | 2806034 | The value of [imath]\sum_{1\leq l< m [/imath]
How to solve this summation ? Also, I'm not sure what does [imath]1\leq l< m <n[/imath] supposed to imply in the development of summation form. [imath]\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}[/imath] This one is from the Galois-Noether Contest in 2018: Galois-Contest |
2885645 | How to use Rolle's Theorem to show that [imath]f[/imath] has at most one fixed point? NOT DUPLICATE
Let [imath]a,b \in \mathbb{R}[/imath] be such that [imath]a\lt b [/imath]. Suppose that [imath]f:[a,b]→\mathbb{R}[/imath] is a continuous function on [imath][a,b][/imath] and is differentiable on [imath](a,b)[/imath] and that [imath]f'(x) \gt 1[/imath] , for all [imath]x\in(a,b)[/imath]. Use Rolle's Theorem to show that [imath]f[/imath] has at most one fixed point in (a,b), that is a point [imath]x_o\in(a,b)[/imath] such that [imath]f(x_0)=x_0[/imath]. Hint: You may find it helpful to argue using proof by contradiction. My solution so far: Rolle's Theorem states that if [imath]f[/imath] is a continuous function on [imath][a,b][/imath] and is differentiable on [imath](a,b)[/imath] and that [imath]f(a)=f(b)[/imath], then there is a point [imath]c\in(a,b)[/imath] such that [imath]f'(c)=0[/imath]. Assume for a contradiction, there are two fixed points in [imath](a,b)[/imath], that is, [imath]x_0,x_1\in(a,b)[/imath] such that [imath]f(x_0)=x_0[/imath] and [imath]f(x_1)=x_1[/imath]. Now, I don't know how to proceed from this because [imath]f(x_0)\ne f(x_1)[/imath], so surely you can't apply Rolle's Theorem here. Note: This is not a duplicate question as this question explicitly asks to use Rolle's Theorem. | 217026 | Show that [imath]f[/imath] has at most one fixed point
Let [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] be a differentiable function. [imath]x\in\mathbb{R}[/imath] is a fixed point of [imath]f[/imath] if [imath]f(x)=x[/imath]. Show that if [imath]f'(t)\neq 1\;\forall\;t\in\mathbb{R}[/imath], then [imath]f[/imath] has at most one fixed point. My biggest problem with this is that it doesn't seem to be true. For example, consider [imath]f(x)=x^2[/imath]. Then certainly [imath]f(0)=0[/imath] and [imath]f(1)=1 \Rightarrow 0[/imath] and [imath]1[/imath] are fixed points. But [imath]f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}[/imath]. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition. |
2889157 | Prove formally that multivariate function has global extrema
Consider the following function: [imath]f(x_1, \dots, x_n) = p_1 \log x_1 + \dots + p_n\log x_n[/imath] subject to the constraint that [imath]\sum_i p_i = \sum_i x_i = 1[/imath]. It is also known that [imath]p_i \in [0,1][/imath] and [imath]x_i \in [0,1][/imath]. I need to prove formally that the function has global maximum when [imath]\frac{p_1}{x_1} = \dots = \frac{p_n}{x_n}[/imath], i.e, [imath]x_i = p_i[/imath] gives the global maximum. I can prove this formally for two variables by using the standard first and second derivative test (eliminate one of the variables). I can also verify empirically that [imath]x_i = p_i[/imath] is the global maximum of this function when [imath]n > 2[/imath] but want to prove this formally. Is there a suitable technique for multivariate functions that I can use? | 611012 | [imath]\sum a_i \ln(b_i) \leq \sum a_i \ln(a_i)[/imath] with [imath]\sum a_i = \sum b_i = 1[/imath]
Okay, this is my another try on the question [imath]\sup \sum a_i \ln(b_i)[/imath] with [imath]\sum a_i = \sum b_i =1[/imath] which I unfortunately mis-stated and actually asked for a problem different than the one I have to solve. I wish to show that given the constraints [imath]0 < a_i, b_j < 1[/imath] and [imath]\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} b_i = 1[/imath] we have the following inequality: [imath]\sum a_i \ln(b_i) \leq \sum a_i \ln(a_i)[/imath] The problem arose when I was trying to compute some topological pressures and [imath]a_i, b_j[/imath] are actually measures of some sets, but this inequality (if, hopefully, true) is purely algebraic. Can anyone provide some suggestions? |
2890374 | How can I show that these two polynomials are coprime?
Let [imath]n , m \in \mathbb{N}[/imath] and let [imath]k = \gcd(n , m)[/imath]. Then there exist [imath]p , q \in \mathbb{N}[/imath] coprime such that [imath]n = p k[/imath] and [imath]m = q k[/imath]. And it is easy to see that [imath] \frac{X^n - 1}{X^k - 1} = \sum_{i = 0}^{p - 1} X^{k i} = f(X) \qquad \mbox{ and } \qquad \frac{X^m - 1}{X^k - 1} = \sum_{j = 0}^{q - 1} X^{k j} = g(X)\mbox{,} [/imath] being [imath]X[/imath] an indeterminate. How can I show that [imath]\gcd(f(X) , g(X)) = 1[/imath]? | 2888718 | Prove that [imath]\gcd(f(x),g(x)) = 1[/imath].
Let [imath]F[/imath] be a field, let [imath]p , q[/imath] be two coprime natural numbers and we consider the two polynomials [imath] f(X) = \sum_{i = 0}^{p - 1} X^{k i} \qquad \mbox{ and } \qquad g(X) = \sum_{j = 0}^{q - 1} X^{k j} [/imath] in [imath]F[X][/imath], where [imath]k \in \mathbb{N}[/imath]. If [imath]x[/imath] is an arbitrary element in [imath]F[/imath], how can I show that [imath]\gcd(f(x) , g(x)) = 1[/imath]? |
327023 | Prove that every self-adjoint unitary linear operator can be expressed in the form [imath]U\alpha = \beta - \gamma[/imath]
This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me? Suppose [imath]U[/imath] is a self-adjoint unitary linear operator on a finite-dimensional inner product space [imath]V[/imath]. Prove that we can find a subspace [imath]W[/imath] such that, if we express every [imath]\alpha[/imath] in [imath]V\,[/imath] in the form [imath]\alpha = \beta + \gamma[/imath], here [imath]\beta[/imath] in [imath]W[/imath] and [imath]\gamma[/imath] in [imath]W^{\bot}[/imath], then we have [imath]U(\alpha) = \beta - \gamma[/imath] If we denote [imath]W = range(U+I)[/imath], [imath]W' = range(U - I)[/imath],then must have [imath]W \subset null(U-I)[/imath] and [imath]W \subset W^{'\bot}[/imath]. We can finish the proof if we can prove that [imath]W' = null(U+I)[/imath] or [imath]W' = W^{\bot}[/imath]. But I can't prove this... Thanks | 2054132 | A decomposition associated with a self-adjoint isometry
Let [imath]V[/imath] be a finite dimensional real inner product space and [imath]L: V \rightarrow V[/imath] be a self-adjoint isometry. Show that there exists a subspace [imath]U[/imath] of [imath]V[/imath] such that [imath]L(x +y) = x -y[/imath] whenever [imath]x \in U[/imath] and [imath]y \in U^{\perp}[/imath]. Here is my solution (it is a problem from a qual I took an hour ago since the posting time and I will appreciate any critique or an alternatively better argument): The primary decomposition theorem says that if [imath]p(t) = p_{1}(t) ... p_{k}(t)[/imath] is a polynomial so that [imath]\text{gcd}(p_{1}, ..., p_{k}) = 1[/imath] and [imath]p(L) = 0[/imath], where [imath]L[/imath] is a linear map on a finite dimensional vector space [imath]V[/imath], then [imath]V = \ker p_{1}(L) \oplus \ker p_{2}(L) \oplus ... \oplus \ker p_{k}(L)[/imath]. Note that [imath]L: V \rightarrow V[/imath] is self-adjoint [imath]\implies L^{*} = L[/imath] and [imath]L[/imath] being isometry [imath]\implies L[/imath] is unitary so that [imath] LL^{*} = I[/imath]. It follows that [imath]I = LL^{*} = L^{2}[/imath] so that [imath]L[/imath] satisfies the polynomial [imath]p(t) = (t-1)(t+1)[/imath]. By the primary decomposition theorem, [imath]V = \ker (L -I) \oplus \ker (L +I)[/imath]. Let [imath]U = \ker (L -I)[/imath]. I did assume without showing that [imath]U^{\perp} = \ker (L + I)[/imath]. Let [imath]v = x +y[/imath] be written uniquely, then [imath]L(v) = L(x) + L(y) = x - y[/imath]. |
2891378 | A ring is commutative if the solution of [imath]ab=ca[/imath] is always unique
Gallian 12.8: Show that a ring is commutative if [imath]ab=ca \implies b=c[/imath] when [imath]a\neq 0[/imath]. It's an odd statement. I tried to approach the problem by proving it directly or by contradiction, but I honestly got nowhere. Looking forward to some hint. | 1214268 | Show that a ring is commutative if it has the property that ab = ca implies b = c when [imath]a\neq 0[/imath]
Show that a ring is commutative if it has the property that ab = ca implies b = c when [imath]a\neq 0[/imath]. This is my proof to show that a ring is commutative if it has the property that ab = ca implies b = c. We need to show that if x, y ∈ R then xy = yx. Let a = x, b = yx and c = xy. Then ab = x(yx) = (xy)x = ca. Thus, by the hypothesis, b = c, or xy = yx. Thus, for x, y ∈ R we have xy = yx. How would I show when [imath]a\neq 0[/imath]? |
2891466 | What's the 2017th derivative of [imath]f(x)=\frac{1}{1-x^2}[/imath]?
I started solving the first few derivatives to see if I can find a pattern but I can't find it. I tried to do something with general Leibniz rule but I don't really know if it's correct. [imath]f'(x)=\frac{2x}{(1-x^2)^2}=2xf(x)^2[/imath] [imath]f^{(n)}(x)=[\text{Leibniz}]=2x(f(x)^2)^{(n-1)} + 2(n-1)(f(x)^2)^{(n-2)}[/imath] Is this correct? What can I do next or is there another way of solving it? | 1973093 | What is the [imath]k-[/imath]th derivative of [imath] \frac{1}{1-x^2}[/imath]?
Let [imath]f(x) = \frac{1}{1-x^2}[/imath]. How to find the [imath]k-[/imath]th derivative of [imath]f(x)[/imath]? I thought it's an easy application of Taylor series. [imath]f(x) = \frac{1}{1-x^2}=\sum x^{2k}[/imath]. On the other hand, [imath]f(x) = \sum \frac{f^{(n)}}{n!}x^n[/imath] So I thought I just need to compare each terms. For example, [imath]\frac{f^{(2)}}{2}x^2=x^2\Longrightarrow f^{(2)}=2[/imath] So I concluded that [imath]f^{(2n)}=(2k)!,f^{(2n+1)}=0[/imath] But the solution gives me [imath]f^{(n)}=\frac{n!}{2}\left( \frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+n)^{n+1}}\right)[/imath] Where did I do wrong? |
2889977 | Evaluate [imath]\lim _{x \to 0} [2x^{-3}(\sin^{-1}x - \tan^{-1}x )]^{2x^{-2}}[/imath]=?
[imath]\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]^{\frac{2}{x^2}}[/imath] How to find this limit? My Try: I tried to evaluate this [imath]\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right][/imath] to understand the nature of the problem. I used L'Hopital. But it became too tedious to calculate. Can anyone please give me suggestion to solve it? Edit: I used the hint given By lab bhattacharjee. I expand the inverse functions and I got [imath]\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right] =1[/imath]. Now I think it remains to find the value of [imath]\lim _{x \to 0}e^{f(x).g(x)}[/imath] where [imath]f(x) = \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right][/imath] and [imath]g(x) = 2/x^2[/imath] | 2618002 | Evaluate the limit containing [imath]\arctan{x}[/imath] and [imath]\arcsin{x}[/imath]
Evaluate: [imath]\lim_{x\to{0}}\bigg(\frac{2}{x^3}.(\arcsin{x}-\arctan{x})\bigg)^{2/x^2}[/imath] I can just expand [imath]\arcsin{x}[/imath] and [imath]\arctan{x}[/imath] using their taylor expansions, but is there any other method? |
2891966 | A faithful ideal in a Noetherian ring is a regular ideal
Recall that an ideal [imath]I[/imath] of a commutative ring with identity [imath]R[/imath] is called regular if it contains a regular element. An element of [imath]R\setminus Z(R)[/imath] is said to be regular, where [imath]Z(R)[/imath] denotes the set of zero divisors of [imath]R[/imath]. Also, an ideal [imath]I[/imath] is faithful if [imath](0):I=(0)[/imath]. There is a well-known fact that: In a Noetherian ring, every faithful ideal is regular. But I can't prove it. | 1269660 | Annihilators of elements of a finitely generated faithful module over a noetherian reduced ring
Lately I've been thinking to annihilator of modules and I've conjectured a proposition I can't prove, so I'll expose my claim. Let [imath]A[/imath] be a noetherian reduced (commutative) ring and let [imath]M[/imath] be a faithful finitely generated module on [imath]A[/imath]. Is it true that there exists an element [imath]m \in M[/imath] such that [imath]\operatorname{Ann}(m)=0[/imath]? |
2892056 | Example of closed strict subspace of [imath]L^p[/imath] that is dense in [imath]L^r[/imath] for [imath]r[/imath]
This is a Stanford Analysis qual question. Any hints\ solutions\ references are highly appreciated. Let [imath]\mu[/imath] be the Lebesgue measure on [imath][0,1][/imath]. For [imath]1<p<\infty[/imath] construct a subspace of [imath]L^p([0,1],\mu)[/imath], which is not dense in [imath]L^p[/imath] but is dense in [imath]L^r[/imath] for all [imath]r<p[/imath]. a subspace of [imath]L^\infty([0,1],\mu)[/imath] which is dense in [imath]L^p[/imath] but not in [imath]L^s[/imath] for any [imath]s>p[/imath] I find is interesting that such a thing would hold because: For 1. by Holder's inequality, for [imath]1\leq q<p\leq \infty[/imath], we have [imath]L^p \subset L^q[/imath]. But of course, the norms are different so I guess what we need is a subspace [imath]M[/imath] closed in [imath]L^p[/imath] but dense in [imath]L^r[/imath] for [imath]r<p[/imath]. Any help will be appreciated. Thanks in advance | 480383 | Subspaces of [imath]L^p[/imath]
So studying Qualifying Exam problems in Analysis I cam across this one: For [imath]1\lt r \lt p \lt s \lt \infty[/imath] where [imath]\mu[/imath] denotes Lebesgue measure, a) Construct a subspace of [imath]L^p([0,1],\mu)[/imath] such that [imath] \forall r[/imath] the subspace is dense in [imath]L^r([0,1],\mu)[/imath] but not [imath]L^p[/imath]. b) Construct a subspace of [imath]L^\infty([0,1,\mu)[/imath] such that [imath] \forall s[/imath] the subspace is dense in [imath]L^p[/imath] but not [imath]L^s[/imath]. Now [imath]L^s\subset L^p \subset L^r[/imath] since [imath]\mu([0,1])\lt\infty[/imath], so the issue is recognizing the norms are not equivalent for this to be possible. I'm just not terribly familiar with subspaces of [imath]L^p[/imath] for [imath]p\neq2[/imath]. Is this just a matter of better knowing [imath]L^p[/imath] spaces or is there something bigger that I'm missing? |
2884340 | Proof of inequality and determining convergence/divergence
Prove that [imath]\frac{1}{\sqrt x}\ge\frac{2}{x+1}[/imath] for all [imath]x\gt0[/imath] I started by letting [imath]\sqrt x=n[/imath], giving me [imath]\frac{1}{n}\ge\frac{2}{n^2+1}[/imath], then taking the inverse and rearranging to get the quadratic [imath]n^2-2n+1\ge0[/imath], which when I solve it, gives me [imath]n\ge1[/imath] and therefore [imath]x\ge1[/imath]. This proves that the inequality holds for all [imath]x\ge1[/imath], but doesn't account for when [imath]0\lt x\lt1[/imath]? How can I account for that? Thanks very much in advance | 2883341 | Proving [imath]\frac{1}{\sqrt{x}}\ge \frac{2}{x+1}[/imath] for [imath]x> 0[/imath]
Prove: [imath]\frac{1}{\sqrt{x}}\ge \frac{2}{x+1}, \quad\forall x>0[/imath] Yeah, pretty much it. I've tried all manner of rearranging and just can't seem to get it. Thanks. |
2892049 | Calculus limit with sum of radicals
I am trying to calculate the following limit, but to be honest I've been failing miserably. [imath]\lim\limits_{n\rightarrow\infty}\frac{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+...+\sqrt{1+\frac{n}{n}}}{n}[/imath] I've been trying to use the fact below but got me no useful result: [imath]\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}[/imath] Also I've been trying to use the logarithm to obtain a product but still no results. Any hints, please? Update 1: I followed your advice and identified the function [imath]f(x)=\sqrt{1+x}, f:[0,1][/imath] So I guess now the result of that limit is [imath]\int_0^1{\sqrt{1+x}}[/imath] ? | 1582557 | Prove: [imath]\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)[/imath]
Prove: [imath]\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)[/imath] What method to use in order to find the closed form of summation [imath]\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}[/imath]? |
2892161 | About idempotent element
Problem: Let [imath]R[/imath] be a ring with [imath]1[/imath] and [imath]e\in R[/imath] such that [imath]e^2=e[/imath]. If [imath]R[/imath] is finite and does not contain any nonzero nilpotent element then show that [imath]Re=eRe[/imath]. I think I solved the problem yet I do not use the finiteness of [imath]R[/imath]. Thus, I doubt that my solution might be wrong. Solution: Claim 1: [imath]R=ker(f)\oplus im(f)[/imath] where [imath]f[/imath] is a homomorphism (with respect to [imath]+[/imath]) from [imath]R[/imath] to [imath]eR[/imath] by [imath]f(r)=er[/imath]. For any [imath]r\in R[/imath], we can write [imath]r=(r-er)+er[/imath] where [imath]r-er\in ker(f)[/imath] and [imath]er\in im(f)[/imath]. Thus, clearly we have [imath]R=ker(f)+ im(f)[/imath]. Now let [imath]k\in ker(f)\cap im(f)[/imath] then [imath]k=ex[/imath] for some [imath]x\in R[/imath] and [imath]ek=0[/imath]. [imath]0=ek=eex=ex=k.[/imath] Thus the sum is direct as claimed. Claim 2: Set [imath]K=\{r-er \mid r\in R\}[/imath]. Then [imath]K=ker(f)[/imath]. Note that [imath]K\leq ker(f)[/imath] as [imath]r-er\in ker(f)[/imath] for each [imath]r\in R[/imath]. Thus, the sum [imath]K+im(f)[/imath] is also direct. On the other hand, [imath]r=(r-er)+er[/imath] for all [imath]r\in R[/imath], and hence [imath]R=K\oplus im(f)[/imath]. Now pick [imath]x\in ker(f)[/imath]. Then [imath]x=k+i[/imath] where [imath]k\in K[/imath] and [imath]i\in im(f)[/imath]. Then [imath]i=x-k\in ker(f)[/imath], and hence [imath]i=0[/imath]. We obtain [imath]x=k[/imath] and [imath]ker(f)=K[/imath] as desired. Claim 3: [imath]eR=eRe[/imath]. We have [imath]R=ker(f)\oplus eR[/imath], and hence [imath]Re=ker(f)e\oplus eRe[/imath]. Pick [imath]t\in ker(f)e[/imath] then [imath]t=re-ere[/imath] for some [imath]r\in R[/imath]. [imath]t^2=((re)^2-re^2re-erere+ere^2re)=0[/imath] and hence [imath]t=0[/imath] by hypothesis. Then [imath]ker(f)e=0[/imath], and [imath]Re=eRe[/imath] as desired. Similarly, we would have [imath]eR=eRe[/imath]. | 2747791 | Proving lemma about centrality of idempotent elements in a Ring with no nilpotent elements.
Reading the class notes I stumbled upon an unproved lemma that I am having issues proving. The lemma is, if [imath]R[/imath] is a ring with no non-zero nilpotent elements and [imath]e[/imath] is idempotent then [imath]e[/imath] is central. In other words, if [imath]e[/imath] is idempotent in such a ring, then [imath]ex=xe[/imath] for all [imath]x\in R[/imath]. I am however not sure how to actually prove the statement. So far I have been playing around trying to find polynomials that can be factored into linear identical factors (e.g (x+e)^2) to see if I can make them be zero by converting an [imath]e^k[/imath] coefficient into [imath]e[/imath]. But something isn't fully clicking about this lemma. |
2892218 | How to show [imath]\frac{\sinθ +\tanθ}{2}>\theta[/imath] using geometry?
I could show in calculus where [imath]0<θ<\frac{\pi}{2}[/imath] like this [imath]\frac{d(θ)}{dθ}=1<\frac{1}{2}(\cosθ+\sec^2θ)[/imath] But I curioused about would it could prove by geometric way likewise trigonometric identities. I try to show with the area of circular sector but I failed... So this is question: How to show [imath]\frac{\sinθ +\tanθ}{2}>\theta[/imath] where [imath]0<\theta<\frac{\pi}{2}[/imath] in geometric way? | 2391978 | Is there a geometrical method to prove [imath]x<\frac{\sin x +\tan x}{2}[/imath]?
Suppose [imath]x \in (0,\frac{\pi}{2})[/imath] and we want to prove [imath]x<\frac{\sin x +\tan x}{2}[/imath]I tried to prove it by taking [imath]f(x)=\sin x+ \tan x -2x[/imath] and show [imath]f(x) >0 ,when\space x \in (0,\frac{\pi}{2})[/imath] take f'[imath]f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)[/imath] I get stuck here ,because the last line need to be proved [imath]\tan ^2 x>2sin^2(\frac x2) ,when\space x \in (0,\frac{\pi}{2})[/imath] [imath]\bf{Question}:[/imath] Is there a geometrical method to prove the first inequality ? (or other idea) Thanks in advance. [imath]\bf{Remark}: [/imath]I can see the function is increasing [imath]when\space x \in (0,\frac{\pi}{2})[/imath] like below :https://www.desmos.com/calculator/www2psnhmu |
2892965 | Prove or disprove that [imath][L:K] = \deg(p)[/imath]
Let [imath]L/K[/imath] be a finite extension and let [imath]\beta \in L[/imath]. If [imath]p[/imath] is the minimal polynomial of [imath]\beta[/imath] then is it true that [imath][L:K] = \deg(p)[/imath]? If not, give a counterexample. Can someone help me figure out if this is true or not? | 1208442 | How is the degree of the minimal polynomial related to the degree of a field extension?
I was reading through some field theory, and was wondering whether the minimal polynomial of a general element in a field extension L/K has degree less than or equal to the degree of the field extension? i.e. [imath]deg[/imath](m[imath]_{a}[/imath]) [imath]\leq[/imath] [imath][L:K][/imath] [imath]\forall[/imath] a [imath]\in[/imath] [imath]L[/imath] If it is true, then could anyone give me an idea as to how to prove this please? |
2452756 | If [imath]A[/imath] and [imath]a[/imath] are the radii of a triangle's circum- and in-circles, and [imath]b[/imath] the distance between the circles' centers, prove [imath]A^2-b^2=2 A a[/imath]
Given a triangle, denote by [imath]A[/imath] the radius of the circumscribed circle and denote by [imath]a[/imath] the radius of the inscribed circle. The distance between the centers of those circles is [imath]b[/imath]. Prove that [imath]A^2 - b^2 = 2Aa[/imath]. I found this in a prelim paper. I really dont know where to start. (Sadly, I can only use tangent properties, arc properties and trignometry) | 2893164 | A geometry problem from high school Olympiad selection round
Consider an isosceles triangle. Let r be the radius of the circumscribed circle and p be the radius of the inscribed circle. Prove that the distance d between the centers of these two circles is [imath]d = \sqrt{r(2p-r)}[/imath] |
2893435 | Show that if [imath]x^n=y^n[/imath] and [imath]n[/imath] is odd, then [imath]x=y[/imath]
This is an excercise from Spivak's Calculus. Show that if [imath]x^n=y^n[/imath] and [imath]n[/imath] is odd, then [imath]x=y[/imath]. Hint: first explain why it suffices to consider only the case x and y greater than 0, then show that x smaller than y or greater y are both impossible. I try to prove this by using induction. The base case [imath]n=1[/imath] is correct. Assume that [imath]n=2k+1[/imath] is true, prove that [imath]n=2k+3[/imath] is also true. But now I don't know how to proceed. | 903966 | If [imath]x^n=y^n[/imath] and [imath]n[/imath] is odd then [imath]x=y[/imath]
Here, we suppose that [imath]x,y\in\mathbb{R}[/imath] and that [imath]x^n=y^n[/imath], where [imath]n[/imath] is odd. I want to prove that [imath]x=y[/imath]. Maybe we can use that [imath]x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})[/imath] So, it suffices to show that if [imath]x^n=y^n[/imath] then [imath]x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0[/imath] Any hint? |
588976 | To prove that [imath]f_n(x) = \frac{nx}{1+n^2x^2}[/imath] does not uniformly converge to [imath]f(x) = 0[/imath] on [imath][0,1][/imath]
My Approach. To prove the given statement it is sufficient to show that [imath]\exists \epsilon > 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n > N [/imath] [imath] \Rightarrow | f_n(x) - f(x)| > \epsilon[/imath] Let [imath]\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1[/imath]. Hence [imath]| f_n(x) - f(x)| = \frac{1}{2} > \epsilon = \frac{1}{10}[/imath]. Hence, we have shown for every [imath]N \in \mathbb{N}[/imath], how to choose [imath]n[/imath] such that [imath]| f_n(x) - f(x)| > \epsilon[/imath]. Thus, [imath]f_n(x)[/imath] does not converge uniformly on [imath][0,1][/imath]. Is there any logical flaw or wrong steps in my proof. Any help would be appreciated. | 1669418 | Limits: [imath]\lim_{n\rightarrow\infty}\frac{nx}{1+n^2x^2}[/imath] on [imath]I=[0,1][/imath]
I'm doing an assignment for my analysis course on the uniform convergence. And I have to assess the uniform convergence of the sequence [imath]f_n(x)=\frac{nx}{1+n^2x^2}[/imath] on the intervals [imath]I=[1,2][/imath] and [imath]I=[0,1][/imath]. Before assessing whether there is uniform convergence, we need to provide the pointwise limit of the sequences. Now, for the first interval everything is okay, but I'm running into issues with finding the pointwise limit for the second interval. It's clear to me that for [imath]0<x\leqslant1[/imath] the limit of the sequence should be zero. But when [imath]x=0[/imath] I'm not sure if this is also the case. Clearly filling in [imath]x=0[/imath] for any finite [imath]n[/imath] would give zero for [imath]f_n(0)=0[/imath], but how can we be sure this is still the case when [imath]n\rightarrow\infty[/imath]? Also, for the uniform convergence part of the assignment, I've gotten as far as seeing that I should use the fact that [imath]f_n(\frac{1}{n})=\frac{1}{2}[/imath], but I'm not really sure yet how to use this. Any help is much appreciated! |
2893830 | Show [imath]\frac1{k-1}+\frac 6k-\frac{7}{k+1}=\frac{8k-6}{k(k^2-1)}[/imath], deduce [imath]\sum\limits_{k = 2}^n\frac{4k-3}{k(k^2-1)}[/imath]
Prove that if [imath]k > 1[/imath], then [imath] \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{7}{k + 1} = \dfrac{8k - 6}{k(k^2 - 1)}[/imath] Hence simplify [imath] \sum\limits_{k = 2}^n \dfrac{4k - 3}{k(k^2 -1)} [/imath] My Work [imath] \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{7}{k + 1} = \dfrac{8k - 6}{k(k^2 - 1)} = \dfrac{7k - 6}{k(k - 1)} - \dfrac{7}{k + 1}[/imath] [imath]= \dfrac{(7k - 6)(k + 1) - 7(k(k - 1))}{k(k - 1)(k + 1)}[/imath] [imath]= \dfrac{8k - 6}{k(k^2 - 1)} = 2 \left[ \dfrac{4k - 3}{k(k^2 -1)} \right][/imath] [imath]\therefore \sum_{k = 2}^n \dfrac{4k - 3}{k(k^2 - 1)} = \dfrac{1}{2} \sum_{k = 2}^n \left( \dfrac{1}{k -1} + \dfrac{6}{k} - \dfrac{7}{k + 1} \right)[/imath] [imath]= \dfrac{1}{2} \left( \sum_{k = 2}^n \dfrac{1}{k -1} + 6 \sum_{k = 2}^n \dfrac{1}{k} - 7 \sum_{k = 2}^n \dfrac{1}{k + 1} \right)[/imath] [imath]= \dfrac{1}{2} \left( \sum_{k = 1}^{n - 1} \dfrac{1}{k - 1} + 6 \sum_{k = 2}^n \dfrac{1}{k} - 7 \sum_{k = 3}^{n + 1} \dfrac{1}{k + 1} \right)[/imath] I would appreciate it if people could please take the time to review my work and explain how I should proceed. | 2892591 | If [imath]k > 1[/imath], then [imath]\frac1{(k - 1)^2}-\frac1{(k + 1)^2}=\frac{4k}{(k^2 - 1)^2}[/imath], hence simplify [imath]\sum\limits_{k = 2}^n\frac k{(k^2 - 1)^2}[/imath]
I have the following problem: Prove that if [imath]k > 1[/imath], then [imath]\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}[/imath] Hence simplify [imath]\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}[/imath] My Work: [imath]\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}[/imath] I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything. I would appreciate it if someone could please take the time to explain this problem. |
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