Home > Math Proofs > Is Periodicity Closed Under Arithmetic Operation?

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
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