0
$\begingroup$

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 ?

$\endgroup$
3
  • $\begingroup$ 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 $b\ne 0$) $\endgroup$ Dec 23 '15 at 10:10
  • $\begingroup$ 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. $\endgroup$
    – Mr. Y
    Dec 23 '15 at 10:15
  • $\begingroup$ $f_c$ would merely be another member of the family $\endgroup$ Dec 23 '15 at 17:02
2
$\begingroup$

Take $a = -x$. See what happens.

Spoilers:

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 $f_b(x)$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Mr. Y
    Dec 23 '15 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.