### Archive

Archive for the ‘Math Proofs’ Category

## Is Periodicity Closed Under Arithmetic Operation?

No.

Consider + and – first. Let f(x)=sin(x), g(x)=1 iff x is rational and g(x)=0 iff x is irrational, then both f(x) and g(x) are periodic. Let F(x)=f(x)-g(x). We claim that F(x) is not periodical. Otherwise, Let T>0 be a period of F(x). Then we have F(0)=0=F(T)=sin(T)-g(T), So sin(T)=g(T). If sin(T)=1, then T=2k*pi+pi/2 is irrational. Contradiction. If sin(T)=0, then T=2k*pi. However, F(1)=sin(1)-g(1)=sin(1)-1, F(1+2k*pi)=sin(1+2k*pi)-g(1+2k*pi)=sin(1). Contradiction. So F(x) is not periodical.

Now let’s consider * and /. Let f(x)=x-[x], g(x)=sin(x), F(x)=f(x)*g(x). We claim that F(x) is not periodical. Otherwise let T>0 be a period of F(x). We have F(0)=0=F(T)=(T-[T])*sin(T). If T-[T]=0, then T is integral. So F(2*pi)=2*pi-[2*pi]=F(2*pi+T)=(2*pi-[2*pi])*sin(T), so sin(T)=1. Impossible. If sin(T)=0, then T=2k*pi. So F(1)=0=F(1+2k*pi)=(2k*pi-[2k*pi])*sin(1). Impossible. So F(x) is not periodical.

Categories: Math Proofs

## 欧拉公式的三种证明

$e^{i\pi}+1=0$

$e^{ix}=\cos x + i \sin x$

$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\frac{(ix)^6}{6!}+\frac{(ix)^7}{7!}+\frac{(ix)^8}{8!}+\ldots$

$=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\frac{x^8}{8!}+\ldots$

$=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots\right)$

$=\cos x + i \sin x$

$\frac{d}{dx}e^{ix}=\frac{d}{dx}\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}$

$=i\sum_{n=1}^{\infty}i^{n-1}\frac{x^{n-1}}{(n-1)!}$

$=i e^{ix}$

$f(x)=(\cos x - i \sin x )\cdot e^{ix}$

$1=(\cos x - i \sin x )\cdot e^{ix}$

$\frac{d}{dx}f(x)=-\sin x + i \cos x$

$=i(i \sin x + \cos x)=i f(x)$

$e^{ix}=\cos x + i \sin x$