Search found 179 matches
- Wed Apr 20, 2016 4:37 pm
- Forum: Calculus
- Topic: Three similar improper integrals
- Replies: 7
- Views: 6128
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: 4931
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...
- Tue Apr 12, 2016 8:31 pm
- Forum: Algebraic Structures
- Topic: Cyclic group
- Replies: 2
- Views: 3030
Re: Cyclic group
Let $G$ a group of order $n$. Prove that if ${\rm{gcd}}(\varphi(n), n)=1$, where \(\varphi\) is the Euler totient function, then $G$ must be a cyclic group. $\newcommand{ord}{{\rm ord}}$ In fact both directions hold. That is: Theorem: Let $n$ be a positive integer. Then every finite group of order ...
- Tue Apr 12, 2016 8:43 am
- Forum: Real Analysis
- Topic: Series sum
- Replies: 4
- Views: 3262
Re: Series sum
A relative question can be found here.jacks wrote:I am getting $\displaystyle \sum^{\infty}_{n=1}\frac{\sin n}{n} = \frac{\pi-1}{2}.$
- Thu Apr 07, 2016 7:46 am
- Forum: General Mathematics
- Topic: Hermite - Hadamard's inequality
- Replies: 2
- Views: 3831
Re: Hermite - Hadamard's inequality
To finish off the exercise it remains to prove the first inequality. Well, \begin{align*} \frac{1}{b-a} \int_{a}^{b}f(x)\,dx &=\frac{1}{b-a} \left [ \int_{a}^{(a+b)/2} f(x)\,dx + \int_{(a+b)/2}^{b} f(x)\,dx \right ] \\ &=\frac{1}{2}\int_{0}^{1}\left [ f\left ( \frac{a+b- t (b-a)}{2} \right )...
- Wed Apr 06, 2016 7:58 pm
- Forum: Real Analysis
- Topic: Inequality and integrals
- Replies: 1
- Views: 2025
Re: Inequality and integrals
Let $\mathcal{P}_n = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \cdots, \frac{n}{n}=1 \right\}$ be a partition of the given interval. Since $f$ is convex it follows from Jensen's inequality that: $$f\left ( \sum_{i=1}^{n}\frac{1}{n}g\left ( \xi_i \right ) \right )\leq \sum_{i=1}^{n}\frac{1}{n}f\left ( g(\...
- Wed Apr 06, 2016 7:43 pm
- Forum: General Mathematics
- Topic: Binomial sum
- Replies: 1
- Views: 2747
Re: Binomial sum
For the second sum let $\omega=e^{2\pi i /3}$ be a third root of unity. The binomial expansion tells us $$\left ( 1+x \right )^n = \sum_{k=0}^{n}\binom{n}{k}x^n$$ Subbing $x$ with $1, \omega, \omega^2$ respectively at the binomial expansion we get: $$\begin{matrix} \displaystyle \sum_{k=0}^{n}\binom...
- Wed Apr 06, 2016 7:41 pm
- Forum: General Mathematics
- Topic: Sum with complex numbers
- Replies: 1
- Views: 3108
Re: Sum with complex numbers
The sum is written as: $$\sum_{n=1}^{2017}\frac{1}{1+z^n}= \frac{1}{1+z}+ \frac{1}{1+z^2}+\cdots + \frac{1}{1+z^{2017}}$$ Now we are "changing" the first $1008$ terms. We note that $$\frac{1}{1+z}= \frac{z^{2017}}{z+z^{2017}}= \frac{z^{2016}}{1+z^{2016}}$$ and similarly $\displaystyle \fra...
- Wed Apr 06, 2016 7:37 pm
- Forum: General Mathematics
- Topic: Hadamard's inequality
- Replies: 1
- Views: 3033
Re: Hadamard's inequality
By the Gramm - Schmidt process we can establish the existence of an orthonormal basis $\mathbf{b_1, b_2, \dots, b_N}$ for $\mathbb{R}^N$ such that \begin{equation} {\rm span}_{\mathbb{R}} \left \{ \mathbf{a_1, a_2, \dots, a_N} \right \}={\rm span}_{\mathbb{R}}\left \{ \mathbf{a_1, a_2, \dots, b_N} \...