Search found 179 matches

by Riemann
Tue Jul 05, 2016 6:25 am
Forum: Calculus
Topic: A series with cosine
Replies: 1
Views: 1894

Re: A series with cosine

Evaluate the series $$\sum_{n=1}^{\infty} \frac{\cos \frac{n \pi}{3}}{9-4n^2}$$ We have seen here that the cosine series of $\sin 3x , \; x \in (0, \pi)$ is $$\sin 3x = \frac{2}{3\pi} + \frac{12}{\pi} \sum_{n=1}^{\infty} \frac{\cos 2nx}{9-4n^2}$$ Hence for $x=\frac{\pi}{6}$ we have that: \begin{ali...
by Riemann
Sun Jul 03, 2016 6:59 pm
Forum: Linear Algebra
Topic: A value of a determinant
Replies: 1
Views: 3101

Re: A value of a determinant

$A$ satisfies the polynomial equation $x^3+3x-4=(x^2+x+4)(x-1)=0.$ The minimal polynomial is a factor of that, and the characteristic polynomial contains no factors that aren't also present in the minimal polynomial. And a real matrix has a real characteristic polynomial. So the characteristic polyn...
by Riemann
Sun Jul 03, 2016 6:45 pm
Forum: Real Analysis
Topic: A limit with arctan
Replies: 1
Views: 1788

Re: A limit with arctan

Succesively we have that: \begin{align*} \lim_{n\to + \infty} \sum_{k=1}^n \frac{\arctan k}{n+k} &=\lim_{n\to +\infty} \sum_{k=1}^n \frac{\frac{\pi}{2}-\arctan {\frac{1}{k}}}{n+k} \\ &=\frac{\pi}{2}\lim_{n\to+\infty} \sum_{k=1}^n \frac{1}{n+k}-\lim_{n\to+\infty} \sum_{k=1}^n \frac{\arctan {\...
by Riemann
Sun Jul 03, 2016 6:41 pm
Forum: Real Analysis
Topic: Example of sequence of functions (2)
Replies: 1
Views: 1722

Re: Example of sequence of functions (2)

Here is an example. Take for instance $f_n:[0, 2\pi] \rightarrow \mathbb{R} $ such that $f_n(x)=\frac{\sin nx}{\sqrt{n}}$. Then we note that $f_n \xrightarrow{n \rightarrow +\infty} 0$. Now note that the derivative of $f$ is $f'_n(x)=\sqrt{n} \cos nx $ which does not converge uniformly to any functi...
by Riemann
Sun Jul 03, 2016 6:31 pm
Forum: Number theory
Topic: Dirichlet series
Replies: 1
Views: 3948

Re: Dirichlet series

Let $\displaystyle F(z)=\sum_{m=1}^{\infty} \frac{f(m)}{m^z}$ and $\displaystyle G(z)=\sum_{n=1}^{\infty} \frac{g(n)}{n^z}$ be two complex series that converge absolutely somewhere in the complex plane then we define the convolution Dirichlet product as follows: $$F(z)G(z) = \sum_{m=1}^{\infty} \fra...
by Riemann
Wed May 25, 2016 1:44 pm
Forum: General Mathematics
Topic: Period of sum of functions
Replies: 0
Views: 1983

Period of sum of functions

Let $f:\mathbb{Z} \rightarrow \mathbb{R}$ be a function with period $T>0$, that is $f\left(x+T\right)=f(x)$ forall $x \in \mathbb{Z}$. The least period of a function divides every other period of it. Let ${\rm a, b}$ be two natural coprime numbers and let $f, g$ be two functions that are onto $\math...
by Riemann
Wed May 18, 2016 9:02 pm
Forum: General Mathematics
Topic: A determinant involving a triangle
Replies: 1
Views: 3477

A determinant involving a triangle

Let a triangle have angles $A, B,C$. Evaluate the determinant:

$$\mathscr{D}=\begin{vmatrix}
1 &\sin A & \cot \frac{A}{2} \\\\
1& \sin B & \cot \frac{B}{2}\\\\
1& \sin C & \cot \frac{C}{2}
\end{vmatrix}$$
by Riemann
Wed May 18, 2016 8:09 pm
Forum: Calculus
Topic: An integral resulting in harmonic number
Replies: 1
Views: 1777

Re: An integral resulting in harmonic number

Let $m \in \mathbb{N}$. Prove that: $$\int_0^1 x^m \log(1-x)\, {\rm d}x=-\frac{\mathcal{H}_{m+1}}{m+1}$$ Here is a solution. Recalling the fact that: \begin{equation} \mathcal{H}_m = \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ] \end{equation} Proposition : It holds that: $$\math...
by Riemann
Wed May 18, 2016 7:51 pm
Forum: General Mathematics
Topic: Completeness
Replies: 2
Views: 3341

Re: Completeness

Papapetros Vaggelis wrote:Does the ordered field of the rational functions satisfy the completeness theorem : " All non-empty

sets have a supremum" .

No, it does not, because the ordered field of the rational functions does not satisfy the Archimedean property. This is quite known.
by Riemann
Tue May 03, 2016 12:25 pm
Forum: Real Analysis
Topic: Fourier series and a known identity
Replies: 1
Views: 2186

Re: Fourier series and a known identity

The function is expanded (fair and square) in a Fourier series in the given interval. Let us evaluate the coefficients: $\displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \, {\rm d}x = \frac{2\sinh \pi a}{ \pi a}$ $\begin{aligned} a_n&= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos nx \, {\rm d}...