Search found 77 matches
- Fri Jan 15, 2016 10:39 pm
- Forum: Number theory
- Topic: Divisibility
- Replies: 1
- Views: 2707
Re: Divisibility
Suppose that \((a^n-b^n)|(a^n+b^n)\). We may assume that \(a > b\). Then there is an integer \(k > 1\) such that \(a^n+b^n = k(a^n-b^n)\) which gives \((k+1)b^n = (k-1)a^n\). Suppose \(p>2\) is a prime such that \(p|(k+1)\). Since \((k+1,k-1) = (k+1,2) \leqslant 2\) then \(p \nmid (k-1)\). Looking a...
- Sat Jan 09, 2016 11:21 am
- Forum: Calculus
- Topic: Series with binomial coefficient
- Replies: 4
- Views: 3629
Re: Series with binomial coefficient
We have that \[ \binom{3n}{n} = \frac{1}{2\pi i}\oint_C \frac{(1+z)^{3n}}{z^{n+1}} \, dz\] where $C$ is any simple closed curve enclosing the origin. [This follows by the residue theorem as the coefficient of $z^n$ in $(1+z)^{3n}$ is $\binom{3n}{n}$.] In what follows, we will take $C$ to be the unit...
- Fri Jan 08, 2016 9:13 pm
- Forum: General Mathematics
- Topic: Arithmetic - Harmonic progression
- Replies: 1
- Views: 2403
Re: Arithmetic - Harmonic progression
Just observe that $\displaystyle{\frac{1}{n!}, \frac{2}{n!}, \ldots, \frac{n}{n!}}$ are all distinct terms of the sequence.
- Fri Jan 01, 2016 2:53 pm
- Forum: General Mathematics
- Topic: Functions preserving order
- Replies: 1
- Views: 2290
Re: Functions preserving order
Hi Apostole. The right word is "order" not "provision".
I know the answer but it is too nice to spoil this problem by giving it. Let's hope that somebody who does not know the answer will work it out.
I know the answer but it is too nice to spoil this problem by giving it. Let's hope that somebody who does not know the answer will work it out.
- Fri Jan 01, 2016 2:50 pm
- Forum: General Mathematics
- Topic: $1-1$ functions
- Replies: 1
- Views: 2202
Re: $1-1$ functions
Let \(q_1,q_2,\ldots\) be an enumeration of the rationals. Suppose we have already defined \(g,h\) on \(q_1,\ldots,q_k\) such that their sum is equal to \(f\) on those points and they are 1 to 1 so far. When defining \(g\) and \(h\) on \(q_{k+1}\) it is enough to pick any \(q \in \mathbb{Q}\) such t...
- Sat Dec 26, 2015 12:08 am
- Forum: General Mathematics
- Topic: Example of a function
- Replies: 3
- Views: 3373
Re: Example of a function
Why try something complicated while something simple would do? Maybe you wanted to demand a strict inequality? In that case my simple example does not work any more but the non-continuous solutions of Cauchy's equation do work. However there are still simpler functions which also work. For example \...
- Thu Dec 24, 2015 3:08 pm
- Forum: General Mathematics
- Topic: Example of a function
- Replies: 3
- Views: 3373
Re: Example of a function
The function which is equal to $1$ everywhere except at $0$ on which it is equal to $0$ is such a function.
- Thu Dec 24, 2015 3:04 pm
- Forum: Competitions
- Topic: Α perfect square
- Replies: 1
- Views: 3328
Re: Α perfect square
We know that the sum of all terms is \[ \frac{n(n+1)(2n+1)}{6}\] and thus their average is \[ \frac{(n+1)(2n+1)}{6}\] Since $2$ must divide $(n+1)(2n+1)$ we must have $n = 2m-1$ for some natural number $m$. So we need $\displaystyle{ \frac{m(4m-1)}{3}}$ to be a perfect square. Since $m$ and $4m-1$ a...
- Tue Dec 15, 2015 11:06 am
- Forum: Number theory
- Topic: Congruency of binomial coefficient
- Replies: 1
- Views: 2970
Re: Congruency of binomial coefficient
Well, \[\binom{2n}{n} = \frac{(2n)!}{n!n!}.\] We have that $p|(2n!)$ as $p < 2n$, but $p \nmid (n!)^2$ as $p$ is prime and $p > n$. So $\binom{2n}{n} \equiv 0 \bmod p$.
- Sun Dec 06, 2015 7:09 pm
- Forum: Real Analysis
- Topic: An infinite series
- Replies: 2
- Views: 2514
Re: An infinite series
Let us write \(a_n\) for the \(n\)-th term of the series and \(s_n\) for the partial sum of the first \(n\) terms. Observe that \[ s_{6n} = \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{8n-1} \right) - \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{4n} \right) = H_{8n} - \frac{1}{2}H_...