Search found 597 matches
- Mon Jun 19, 2017 8:59 pm
- Forum: Real Analysis
- Topic: Multiple integral
- Replies: 1
- Views: 3586
Multiple integral
Let $\langle \cdot, \cdot \rangle$ denote the usual inner product of $\mathbb{R}^m$. Evaluate the integral $$\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left( - ( \langle x, \mathcal{S} ^{-1} x \rangle )^a \right) \, {\rm d}x$$ where $\mathcal{S}$ is a positive symmetric $m \times m$ matrix and ...
- Wed Jun 14, 2017 8:33 am
- Forum: Calculus
- Topic: A generating function involving harmonic number of even index
- Replies: 1
- Views: 3712
A generating function involving harmonic number of even index
Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Prove that forall $|x|<1$ it holds that
\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1}}{2n+1} = \frac{\arctan x \log (1+x^2)}{2}\]
\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1}}{2n+1} = \frac{\arctan x \log (1+x^2)}{2}\]
- Tue Jun 13, 2017 12:49 pm
- Forum: Calculus
- Topic: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
- Replies: 3
- Views: 4060
Re: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
Well,Grigorios Kostakos wrote:\(\psi\big(n+\frac{1}{2}\big)-\psi(n)\),
$$\psi^{(0)}\left( n + \frac{1}{2} \right) = -\gamma +2 \ln 2 +2 \sum_{k=1}^{n} \frac{1}{2k-1}$$
and
$$\psi^{(0)}(n) = - \gamma +\sum_{k=1}^{n-1} \frac{1}{k}$$
- Tue Jun 13, 2017 12:44 pm
- Forum: Calculus
- Topic: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
- Replies: 3
- Views: 4060
Re: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
I suppose you don't want Gauss digamma theorem to be used, e? , but rather a more elegant one.
- Mon May 29, 2017 11:46 am
- Forum: Calculus
- Topic: Double Euler sum
- Replies: 0
- Views: 2378
Double Euler sum
Here is something that came up today while evaluating something else.
Prove that
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )} = \frac{13 \pi^8}{113400}$$
Prove that
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )} = \frac{13 \pi^8}{113400}$$
- Sat May 20, 2017 6:18 pm
- Forum: Calculus
- Topic: On a series with cosine
- Replies: 2
- Views: 3298
On a series with cosine
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \cos \left( \frac{n \pi}{12} \right)}{n 3^n}$$
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \cos \left( \frac{n \pi}{12} \right)}{n 3^n}$$
- Fri May 12, 2017 10:09 pm
- Forum: Archives
- Topic: A collection of problems in Analysis
- Replies: 4
- Views: 8790
Re: A collection of problems in Analysis
The file is updated! Check first post for a direct download.
- Sun Apr 30, 2017 9:51 am
- Forum: Real Analysis
- Topic: Fourier series of sec
- Replies: 1
- Views: 2470
Fourier series of sec
We have seen (somewhere in the past) the Fourier series of $\cos x$ as well as $\sin ax$. How about the Fourier series of $\sec x$?
Let $x \in \left( - \frac{\pi}{4} , \frac{\pi}{4} \right)$. Expand $\sec x$ in a Fourier series.
Let $x \in \left( - \frac{\pi}{4} , \frac{\pi}{4} \right)$. Expand $\sec x$ in a Fourier series.
- Thu Apr 27, 2017 8:31 pm
- Forum: Competitions
- Topic: 16 th Cuban Mathematical Competition of Universities [Problem 5]
- Replies: 1
- Views: 5559
16 th Cuban Mathematical Competition of Universities [Problem 5]
Let $\alpha \in \mathbb{R} \setminus \mathbb{Z}$ and let us denote with $\lfloor \cdot \rfloor$ the floor function. Prove that the series
$$\mathcal{S}= \sum_{n=1}^{\infty} \left(\alpha-\frac{\lfloor n\alpha \rfloor}{n}\right)$$
diverges.
$$\mathcal{S}= \sum_{n=1}^{\infty} \left(\alpha-\frac{\lfloor n\alpha \rfloor}{n}\right)$$
diverges.
- Tue Apr 25, 2017 7:16 am
- Forum: Real Analysis
- Topic: Convergence of alternating series
- Replies: 0
- Views: 2106
Convergence of alternating series
The following series is an interesting one because of its slow convergence which you are asked to show! It was a question at École Polytechnique.
Prove that the series
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} |\sin n|}{n}$$
converges but not absolutely.
Prove that the series
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} |\sin n|}{n}$$
converges but not absolutely.