Search found 177 matches
- Sun Jul 03, 2016 6:45 pm
- Forum: Real Analysis
- Topic: A limit with arctan
- Replies: 1
- Views: 1784
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 {\...
- Sun Jul 03, 2016 6:41 pm
- Forum: Real Analysis
- Topic: Example of sequence of functions (2)
- Replies: 1
- Views: 1718
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...
- Sun Jul 03, 2016 6:31 pm
- Forum: Number theory
- Topic: Dirichlet series
- Replies: 1
- Views: 3511
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: $$F(z)G(z) = \sum_{m=1}^{\infty} \fra...
- Wed May 25, 2016 1:44 pm
- Forum: General Mathematics
- Topic: Period of sum of functions
- Replies: 0
- Views: 1892
Period of sum of functions
Let $f:\mathbb{Z} \rightarrow \mathbb{R}$ be a function with period $T>0$, that is $f\left(x+T\right)=f(x)$ forall $x \in \mathbb{Z}$. The least period of a function divides every other period of it. Let ${\rm a, b}$ be two natural coprime numbers and let $f, g$ be two functions that are onto $\math...
- Wed May 18, 2016 9:02 pm
- Forum: General Mathematics
- Topic: A determinant involving a triangle
- Replies: 1
- Views: 3311
A determinant involving a triangle
Let a triangle have angles $A, B,C$. Evaluate the determinant:
$$\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}$$
$$\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}$$
- Wed May 18, 2016 8:09 pm
- Forum: Calculus
- Topic: An integral resulting in harmonic number
- Replies: 1
- Views: 1771
Re: An integral resulting in harmonic number
Let $m \in \mathbb{N}$. Prove that: $$\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...
- Wed May 18, 2016 7:51 pm
- Forum: General Mathematics
- Topic: Completeness
- Replies: 2
- Views: 3259
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.
- Tue May 03, 2016 12:25 pm
- Forum: Real Analysis
- Topic: Fourier series and a known identity
- Replies: 1
- Views: 2180
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}...
- Wed Apr 20, 2016 4:37 pm
- Forum: Calculus
- Topic: Three similar improper integrals
- Replies: 7
- Views: 5511
Re: Three similar improper integrals
A relative post can be found here.
- Tue Apr 12, 2016 9:47 pm
- Forum: Number theory
- Topic: On Euler's $\phi$ function
- Replies: 1
- Views: 3600
Re: On Euler's $\phi$ function
If $\gcd(2,n)=1$ (that is $n$ is odd) then: $$\phi(2n)=\phi(2)\phi(n)=\phi(n)$$ holds. Now, suppose that $n$ is even, that is $n=2^r m$ and $\gcd(2,m)=1$. Then: $$\phi(n)=\phi\left(2^r \right) \phi(m)=2^r \left (1 - \frac{1}{2} \right) \phi(m)=2^{r-1} \phi(m)$$ while $$\phi(2n)=\phi \left (2^{r+1} m...