Prove that $\sin(x^2+x)$ is not periodic using f(x+T) = f(x)

Question

How can I prove that $\sin(x^2+x)$ is not periodic using $f(x+T) = f(x)$?

Answer 1

For $f(x)=\sin(x^2+x)$, suppose there exists some number $T$ such that for all $x\in\mathbb{R}$, $f(x)=f(x+T)$. We will show that any such $T$ must be equal to zero, and hence that $f$ is not periodic.

Observe $f(0)=\sin(0^2+0)=0$, so by the periodicity of $f$ we have $$f(0+kT)=0=\sin\big((kT)^2+kT\big)$$ for all $k\in\mathbb{Z}$. It is a property of the $\sin$ function that $\sin(y)=0$ iff $y=m\pi$ for some $m\in\mathbb{Z}$. Thus for every $k\in\mathbb{Z}$, there exists some $m_k\in\mathbb{Z}$ such that $$m_k\pi=(kT)^2+kT$$ which actually tells us that $$ m_k=\frac{k^2T^2+kT}{\pi}$$ Remember that $k\in\mathbb{Z}$ is ours to choose.

First choose $k=1$, so $$m_1=\frac{T^2+T}{\pi}\in\mathbb{Z}$$ Next choose $k=2$, so $$m_2=\frac{4T^2+2T}{\pi}\in\mathbb{Z}$$ The set of rational numbers is closed under addition and multiplication, hence $$\frac{1}{2}(4m_1-m_2)=\frac{T}{\pi}\in\mathbb{Q}$$ and $$\frac{1}{2}(m_2-2m_1)=\frac{T^2}{\pi}\in\mathbb{Q}$$ You can show by contradiction that the product of a non-zero rational number and an irrational number is irrational, and further that the quotient of a non-zero rational number and an irrational number is irrational. Observe that by closure, $(\frac{T}{\pi})(\frac{T}{\pi})=\frac{T^2}{\pi^2}\in\mathbb{Q}$. But then we have $\frac{1}{\pi}(\frac{T^2}{\pi})\in\mathbb{Q}$, the product of the irrational number $\frac{1}{\pi}$ with the rational number $\frac{T^2}{\pi}$ yielding a rational number. Therefore the only possibility is that $\frac{T^2}{\pi}=0$, which forces $T=0$.