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Fri Jul 28, 2017 9:01 pm
Forum: Calculus
Topic: A series involving Harmonic numbers
Replies: 2
Views: 1007

### Re: A series involving Harmonic numbers

This is closely related to problem 11993 from American Mathematical Monthly Journal . Now the problem presented in that integral form can be dealt with rather easily and one avoids having to calculate the last Euler Sum I left off .. :) There's an old blog post of mine with spoilers for this problem...
Mon May 29, 2017 7:28 am
Forum: Calculus
Topic: Root integral
Replies: 1
Views: 775

Making the change of variable $x \mapsto x^2$, the integral changes to: $\displaystyle \mathcal{J}(k) = \frac{1}{2}\int_1^4 \frac{x^k}{\sqrt{(x-1)(4-x)}}\,dx$. Let, $\displaystyle f(z) = \frac{z^k}{\sqrt{(z-1)(4-z)}}$ and consider the integral of $f(z)$ about a positively oriented dumbell contour ($... Sat May 27, 2017 5:48 pm Forum: Functional Analysis Topic: Lemma Replies: 2 Views: 938 ### Re: Lemma Since,$U : H \to H$satisfies$\lVert U \rVert \le 1$, then $$\left<(I-U)h,h\right> = \lVert h \rVert^2 - \left< Uh,h \right> \ge \lVert h \rVert^2 (1 - \lVert U \rVert) \ge 0 \text{ for all } h \in H$$ We claim that,$N(I-U) = R(I-U)^{\perp}$If,$h \in N(I-U)$then,$\left<(I-U)(h - th'), h - th'...
Thu May 25, 2017 8:10 am
Forum: Calculus
Topic: An infinite product
Replies: 1
Views: 699

### Re: An infinite product

Using the infinite product for cosines: $\displaystyle \cos (\pi x) = \prod\limits_{n=1}^{\infty} \left(1-\frac{x^2}{\left(n-\frac{1}{2}\right)^2}\right)$, \begin{align*}\prod\limits_{n=1}^{\infty} \left(1 + \frac{x^2}{n^2+n-1}\right) &= \prod\limits_{n=2}^{\infty} \frac{\left(n-\frac{1}{2}\right)^2...
Wed May 24, 2017 8:11 pm
Forum: Measure and Integration Theory
Topic: A problem from Rudin's Real and Complex Analysis
Replies: 0
Views: 537

Problem : Let, $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}^k$, prove that there exists Borel functions $g$ and $h$ such that $\mu_k(\{x \in \mathbb{R}^k: g(x) \neq h(x)\}) = 0$ and $g(x) \le f(x) \le h(x)$ a.e. $[\mu_k]$. (where, $\mu_k$ is the Lebesgue measure on $\mathbb{R}^k... Sun May 21, 2017 8:48 am Forum: Linear Algebra Topic: Linear isometry Replies: 3 Views: 1063 ### Re: Linear isometry An alternative approach: It suffices to show that$f$is surjective and preserves mid-points, i.e., if$x,y \in \mathbb{R}^2$then, $$f\left(\frac{x+y}{2}\right) = \frac{f(x) + f(y)}{2}$$ then, by simple induction one can extend the above identity to$f(rx+(1-r)y) = rf(x) + (1-r)f(y)$where,$r$is ... Sat May 20, 2017 11:24 am Forum: Meta Topic: New sub-forum for Measure and Integration theory Replies: 1 Views: 752 ### New sub-forum for Measure and Integration theory Would it be a possible to have a sub-forum for 'Measure and Integration Theory' that is covered in most graduate classes? Sat May 20, 2017 11:07 am Forum: PDE Topic: An integral identity for Harmonic Functions Replies: 0 Views: 780 ### An integral identity for Harmonic Functions If$u$be a Harmonic Function in a open connected set$\Omega \subset \mathbb{R}^n$and$\overline{B(x_0,R)} \subset \Omega$(the closed ball of radius$R$centered at$x_0 \in \Omega$). (i) Show that:$$\int_{\partial B(0,1)} u(x_0 + ry)u(x_0 + Ry)\,d\sigma(y) = \int_{\partial B(0,1)} u^2(x_0 + cy)... Thu May 18, 2017 11:24 pm Forum: Functional Analysis Topic: Closed linear subspace Replies: 1 Views: 723 ### Re: Closed linear subspace If,$p = 1$, then$E_1$is clearly a closed subspace of$L^1([0,\infty))$as: If$f_n \overset{L^1}{\longrightarrow} f$for$f_n \in E_1$, then$\displaystyle \left|\int_{0}^{\infty} f\,dx\right| = \left|\int_{0}^{\infty} (f-f_n)\,dx\right| \le \int_{0}^{\infty} \left|f - f_n\right|\,dx \underset{n ...
Thu May 18, 2017 7:58 pm
Forum: Functional Analysis
Topic: An application of Banach-Steinhaus theorem
Replies: 0
Views: 532

### An application of Banach-Steinhaus theorem

Suppose $1 < p < \infty$ and $p,q$ are conjugate indices, i.e., $\displaystyle \frac{1}{p} + \frac{1}{q} = 1$. If $(\mathbb{R},\mu)$ be the Lebesgue measure. If the following properties hold: (i) $\displaystyle g \in L^{q}_{\text{loc}}(\mu)$. (ii) \$\displaystyle \int_{\mathbb{R}} |fg|\,d\mu < \infty...