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Given:

  1. Function $f(x)$ is infinitely differentiable

  2. equation (1) $f(x)=c \times f(\frac{x}{2})$

We have to find all $c$, for which the (1) has non-zero solutions

Any hints on theorems to apply here, I reckon it's somehow related to ODEs

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  • $\begingroup$ Just to clarify is the question asking you to find the possible values for the constant $c$ such that $f(x)$ is non-zero? $\endgroup$
    – 1123581321
    Apr 21 '19 at 5:08
  • $\begingroup$ @1123581321 such that eq 1 has solutions $\endgroup$
    – user119510
    Apr 21 '19 at 5:22
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If $ f(x)=c*f(x/2) $ then

$\begin{array}\\ f(x) &=cf(x/2)\\ &=c^2f(x/4)\\ &=c^3f(x/8)\\ &...\\ &=c^nf(x/2^n)\\ \end{array} $

If $|c| < 1$ then $f(x) \to 0$ so $f(x) = 0$ for all $x$.

If $f(0) \ne 0$, $\dfrac{f(x)}{c^n} \to f(0) $. If $|c| > 1$, $\dfrac{f(x)}{c^n} \to 0 $ which contradicts $f(0) \ne 0$.

If $f(0) = 0$, then, for small $x$, $f(x) = xf'(0)+O(x*2) $ so $f(x/2^n) =xf'(0)/2^n+O(x^2/4^n) $ so $f(x) =c^n(xf'(0)/2^n+O(x^2/4^n)) =xf'(0)(c/2)^n+O(x^2(c/4)^n)) $.

This only works if $c=2$; it goes to zero if $|c| < 2$ and to $\infty$ is $|c| > 2$.

Therefore we must have $c = 2$.

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    $\begingroup$ But $c=1$, has the solution $f(x)=1$. Even more, $c=2^n$ has the solution $f(x)=x^{n}$. $\endgroup$ Apr 21 '19 at 5:32
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    $\begingroup$ what about f=x^2? x^2=c*(x/2)^2 therefore for c=4 there are infinitely many solutions. $\endgroup$
    – user119510
    Apr 21 '19 at 5:34
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    $\begingroup$ Guess I should have considered $f(0) = f'(0) = ...=f^{(n)}(0) = 0$. I may do this tomorrow - too late now. $\endgroup$ Apr 21 '19 at 5:38
  • $\begingroup$ Any updates on the assumption? $\endgroup$
    – user119510
    Apr 23 '19 at 6:22
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$f(x) = c^{\ln(x)/\ln(2) - 1}$

By graphing $f(x)$ for different values of $c$, you can observe that $f(x)$ is non-zero for all $c > 0$.

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  • $\begingroup$ Won't work from "given 2" C is const outside the fun. $\endgroup$
    – user119510
    Apr 26 '19 at 12:54
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We have $f(x)=c^nf(\frac{x}{2^n})$. If $|c|<1$, then by taking $n\to\infty$, we get that $f(x)=\lim_{n\to\infty}c^nf(\frac{x}{2^n})=0f(0)=0.$ So, if $|c|<1$ the only solution is $f=0$.

Ideas:(I don't know if this helps in solving the problem) Note that since $f(x)=cf(\frac{x}{2})$, we have that $f$ is determined by the values of the function on $[1,2]$ and $[-2,-1]$. So, we may define $f$ to be a bump function on these intervals and extend it to the entire real line by using $f(x)=cf(\frac{x}{2})$ (does this work for $|c|>1$?)

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$$f(x)=c \times f\left(\frac{x}{2}\right)$$ Since $f(x)$ is infinitely differentiable, let us consider the Maclaurin series expansion: $$f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f^{(2)}(0) + \dots$$

Also, $$f^{(n)}(x) = \frac{c}{2^n}f^{(n)}\left(\frac{x}{2}\right)$$ $$\implies f(x) = f(0) + x\frac{c}{2}f'(0) + x^2\frac{c^2}{2^22!}f^{(2)}(0)+\dots$$

If the two series converge, we can equate the coefficients of $x^n$: $$\implies \frac{c^n}{2^nn!} = \frac{1}{n!}$$ $$\implies c = 2$$

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