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The question is about scaling the period of a periodic function. This question has been asked and answered. What bothers me is something else:

For $k \ne 0$

$f(x + p) = f(x) \Rightarrow f(kx + p) = f(kx)$ (1)

$f(kx) = f(k (x + p / k))$ (2)

$f(x) = f(kx / k) = f(x + p / k)$ (3)

I must have misunderstood something, because (3) is definitely not valid for every $k\ne0$. Counter example for (3): $cos(0) \ne cos(0 + 2\pi/100)$.

tmaric
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2 Answers2

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You're mixing up $f$ and $g$ from that answer. It's true that $g(0) = g(0 + 2\pi/100)$ does not hold for $g = \cos$, but it holds for $g(x) = \cos(100x)$.

Mees de Vries
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  • Thanks. I wanted to see what would happen if I wouldn't use $g(x)$ in the proof, hence this question. I still can't see the specific error in the statements and I know that 3 is not valid. If the statements that lead up to 3 are true, how come 3 is false? – tmaric Mar 14 '18 at 10:41
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    @tmaric, because (the second equation in) 3 does not follow from any of the previous statements. It seems that you are reaching that conclusion by applying an implication like $f(x) = f(y) \implies f(x/k) = f(y/k)$, which is certainly not true. In fact, you have given the counterexample yourself, with $f = \cos, x = 0, y = 2\pi, k = 100$. – Mees de Vries Mar 14 '18 at 10:58
  • Thanks a lot! That is what I was missing. – tmaric Mar 14 '18 at 11:03
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$(3)$ is not valid of course, because it claims that $f$ is $\frac{p}{k}$-periodic $\forall k \ne 0$, which is cleary an absurd condition.

On the contrary the new function $g(x) := f(kx)$ has period $\frac{p}{k}$ because $g$ is simply a dilation/contraction of $f$: $$g\left(x+\frac{p}{k}\right) = f\left(k\left( x+\frac{p}{k}\right)\right) = f\left(k x+p\right) = f\left(k x \right) = g(x)$$

Vexx23
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