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answered Inequality $\left(\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}\right)\left(\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}\right)\leq1$ for all $x\in\mathbb R$
2d
comment Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$
Have you tried the sine series?
Aug
30
comment Evaluate the summation $\sum_{k=1}^{n}{\frac{1}{2k-1}}$
Not directly related to the problem, but I want to point out that the formula you gave is only asymptotically true. Indeed, the harmonic number $H_n = 1 + (1/2) + \cdots + (1/n)$ has the following asymptotic expansion $$ H_n \sim \log(n) + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \frac{1}{120n^4} + \cdots. $$
Aug
28
comment Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$?
I think you are correct since $f$ is the liminf of Borel measurable functions. And also an upvote for this splendid example!
Aug
28
comment Check whether the sum of the series $\sum^{\infty}_{n=1}\frac{\sin(nx)}{nx}\cos\frac{x}{n}$ is continous on $(0,\pi)$
This does not converge uniformly on all of $(0, \pi)$ due to the Gibbs phenomenon. Rather, you may show that it converges uniformly on any compact subset of $(0, \pi)$, which is indeed enough for establishing continuity of the sem. When doing so, it is convenient to split the series int two parts using $$\cos(x/n) = 1 - 2\sin^2(x/2n)$$. Then you may apply Dirichlet test and Weierstrass $M$-test seprartely.
Aug
28
awarded  Good Answer
Aug
28
awarded  Nice Answer
Aug
28
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
I was able to reach the same conclusion, although a slightly different sum of the form $$\sum_{k=1}^{(m-1)/2} \operatorname{Li}_2(-\tan^2(\pi k/m)).$$ And I was also able to check Vladimir's result for $m = 5$ using pentagon identity. In our case, it reads $$ \mathrm{Li}_2\left(-1-\tfrac{2}{\sqrt{5}}\right) + \mathrm{Li}_2\left(-1+\tfrac{2}{\sqrt{5}}\right) = \mathrm{Li}_2\left(\tfrac{1}{5}\right) + \mathrm{Li}_2\left(-\tfrac{3+\sqrt{5}}{2}\right) + \mathrm{Li}_2\left(-\tfrac{2}{3+\sqrt{5}}\right) + \text{[some log term]}. $$
Aug
28
answered A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Aug
27
revised Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$
added 1 character in body
Aug
27
comment Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$
@tired, Oh, you are right. I will fix that!
Aug
27
comment Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$
@Lucian, Thank you, your comment made my answer meaningful now :)
Aug
27
answered Show that there exists holomorphic $f$ such that $f^2=\frac{\sin z}{z}$
Aug
26
revised Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$
added 107 characters in body
Aug
26
answered Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$
Aug
26
comment Evaluating $\int_0^{\pi /2}\left(\frac{1}{\sqrt{\tan(x)}}+\frac{1}{\sqrt{\arctan(x)}}\right) dx$
The tangent part is easy to calculate; you can use the beta function. The arctangent part is... well, I will be surprised if it has either an antiderivative or a closed form.
Aug
26
comment Limit of $\frac{1}{2^n}\sum_{n=0}^{2^n-1}f(e^\frac{2k\pi}{2^n}i)$ for a complex-analytic function
Also, it seems that you already know that $F(t) = \sum_n a_n e^{int}$ is a Fourier series on $\Bbb{R}/2\pi \Bbb{Z}$. Have you tried some Fourier analysis on it?
Aug
26
comment Limit of $\frac{1}{2^n}\sum_{n=0}^{2^n-1}f(e^\frac{2k\pi}{2^n}i)$ for a complex-analytic function
I doubt that $\sum_{k\neq 0} z^k k^2$ converges on $C$. (Actually, this converges to $-2\pi \delta''_1(z)$ in distribution sense, where $\delta_1(z)$ is the Dirac mass at $z = 1$.)
Aug
26
comment If $\Pr\{Y<t \mid X\} < f(t)$, where $f(t)$ doesn't depend on $X$, then unconditionally, $\Pr\{Y<t\} < f(t)$, right?
If $\Bbb{P}(Y < t \mid X) < f(t)$ a.s. in $\Bbb{P}$, then taking expectation to both sides gives $\Bbb{P}(Y < t) < f(t)$.
Aug
25
comment Holomorphic function with reals to reals
@m.deslauriers There are many ways to see that. You may look at the Cauchy-Riemann equation. Or, you can expand $f$ as power series and check that $g$ has the McLaurin series with coefficients conjugated.