# Solve $f(x)=c \times f(\frac{x}{2})$ for $c$

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

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

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 $$c = 2$$.

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

$$f(x) = c^{\ln(x)/\ln(2) - 1}$$

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

• Won't work from "given 2" C is const outside the fun.
– user119510
Apr 26 '19 at 12:54

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 $$|c|>1$$?)

$$f(x)=c \times f\left(\frac{x}{2}\right)$$ 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 $$x^n$$: $$\implies \frac{c^n}{2^nn!} = \frac{1}{n!}$$ $$\implies c = 2$$