# Consider a family of functions $f_b(x)$ such that $f_b(0)=b$ and $f_b(x)=2^af_b(x-a)$. Find an expression for $f_b(2x)$ in terms of $f_b(x)$

Consider a family of functions $f_b(x)$ such that $f_b(0)=b$ and $f_b(x)=2^af_b(x-a)$. Find an expression for $f_b(2x)$ in terms of $f_b(x)$.

I haven't clear in my mind how to approach the problem.

From the problem I have that $f_b(a)=2^a \cdot b$ , $f_b(2x)=2^a f_b(2x-a)$ and $f_b(2x+a)=2^a f_b(2x)$, so I have that $$f_b(2x) (1+2^a)=2^a f_b(2x-a) + f_b(2x+a)$$ $$f_b(2x)=\cfrac{2^a f_b(2x-a) + f_b(2x+a)}{1+2^a}$$

I am kind of clueless now,it seems to me as I haven't even touched the problem but I am just scribbling around...

Can you guys give me a hint ?

• Your question hits the spot: What is $c$? I suspect it might be a typo. - Then again, you might find that $f_c(2x)/f_b(x)^2$ is a simple expression in $b$ and $c$ (provided $b\ne 0$) Dec 23 '15 at 10:10
• If they meant $f_b(2x)$ then I think I've found the solution ,but if $c$ is another family of functions I don't see how to solve the problem.One question:Where does your last expression follow from ?Thanks for your comment. Dec 23 '15 at 10:15
• $f_c$ would merely be another member of the family Dec 23 '15 at 17:02

Take $a = -x$. See what happens.
If we take $a = -x$, then $f_b(x) = 2^a f_b(x-a) = 2^{-x} f_b(2x)$. Multiplying both sides by $2^x$ we get $2^x f_b(x) = f_b(2x)$. This is an expression of $f_b(2x)$ in terms of $f_b(x)$.
• Neat +1. However why can we let $a=-x$ ?$a$ is some fixed number while $-x$ isn't.Also given some $x$ we are supposing that $a=-x$.I don't see why this must be the case,or why we can make this case. Dec 23 '15 at 10:33