# Prove if infinite product of $f(x)$ is $0$ then so is infinite product of $f(xvarphi)$

Mathematics Asked by Smorx on January 1, 2022

Prove or disprove that if $$prodlimits_{x=2}^{infty} f(x)=0$$ and $$f(x)neq0$$ for any $$xgeq0$$ then $$prodlimits_{x=2}^{infty} f(xvarphi)=0$$ for any constant $$varphigeq2$$

This seems true but I’m not quite sure how to prove it since the constant is inside a function f.

Let is say we want to disprove it. Then we need a $$varphi$$ for which this product is not zero. One way to obtain that is for $$f$$ to be $$1$$ at all integer multiples of $$varphi$$. So let's decide that $$varphi = 2$$ and $$f$$ is $$1$$ for all even integers. Then we need the values of $$f$$ on odd integers to give $$0$$ in the first product. One way is to have $$f(text{odd}) = 1/text{odd}$$. Sharpening that up, we obtain...

This is false as written. Let $$f(x) = begin{cases}1 ,& text{x is an even integer } \ 1/x ,& text{otherwise} end{cases}$$

Then $$prod_{x=2}^infty f(x) = 0$$, but $$prod_{x=2}^infty f(2x) = 1$$. In fact, for any $$varphi geq 2$$ an even integer, $$prod_{x=2}^infty f(varphi x) = 1$$.

Answered by Eric Towers on January 1, 2022

Not really: you can consider your favourite function $$f$$ such that $$f(n)=1-frac1n$$ for all odd $$nin Bbb N$$ and $$f(n)=1-2^{-n-1}$$ for all even $$ninBbb N$$. Then clearly $$prod_n f(n)$$ diverges to $$0$$, whereas $$prod_n f(2n)$$ converges to a positive number.

Answered by user239203 on January 1, 2022

This is not true. Consider any continuous function defined over $$mathbb{N}$$ by $$f(n)=cases{frac12&n odd\1&n even}$$ An explicit example is given by $$f(x)=frac{3+cos{(pi x)}}4$$ Then we have $$prod_{x=2}^infty f(x)=0$$ But $$prod_{x=2}^infty f(2x)=1$$

Answered by Peter Foreman on January 1, 2022

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