3
$\begingroup$

Determine all functions $f$ in $f(x+1)=2f(x)$, for all $x$ in real number.

So I let $x$ be $x+1$. Then I have $f((x+1)+1)=2f(x+1)$. But since there is a new function $f(x+2)$, I couldn't determine the function $f$ using elimination. Any idea would be a great help.

$\endgroup$
  • 1
    $\begingroup$ Put $f(0) = a$. Can you find values for f(1), f(2) etc. Do you see a pattern? How about f(-1), f(-2)... Make a conjecture about f(x) when x is an integer. $\endgroup$ – Paul Aug 1 at 15:03
  • $\begingroup$ Hint: $f(n) = 2f(n-1) = 4f(n-2) = \ldots = 2^k f(n-k)$. $\endgroup$ – mwt Aug 1 at 15:08
  • $\begingroup$ Yes, I see that there's a pattern with the function, but the thing is, how can I able to graph it on a coordinate plane? $\endgroup$ – Billy Claxtone Aug 2 at 2:53
5
$\begingroup$

There are very many such functions if those constraints you have given are the only constraints.

For example, you can define an arbitrary function $F:[0,1)\to \mathbb R$. Then you can "extend" it to the whole real axis: on $[1,2)$ make it twice $F$, on $[2,3)$ make it $4$ times $F$, on $[-1,0)$ make it half $F$ etc. The way to write it in one formula:

$$f(x)=2^{\lfloor x\rfloor}F(x-\lfloor x\rfloor)$$

where $\lfloor x\rfloor$ denotes "the biggest integer not bigger than $x$".

One can prove that $f(x)=F(x)$ for $x\in[0,1)$, and any such function $f$ must be obtained this way, by extending its own restriction on $[0,1)$ - thus the above construction produces all such functions $f$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes. Let say I have a restriction: for x ∈ (0, 1], f(x) = x(x − 1). $\endgroup$ – Billy Claxtone Aug 2 at 2:47
  • $\begingroup$ Well, it will be $f(x)=2^{\lfloor x\rfloor}(x-\lfloor x\rfloor)(x-\lfloor x\rfloor-1)$, if you are looking for a formula. $\endgroup$ – Stinking Bishop Aug 2 at 7:54
1
$\begingroup$

Define $\,g(x) := f(x)2^{-x}.\,$ Then $\,g(x+1) = g(x)\,$ for all real $x$. Thus, $\,g(x)\,$ is a period $1$ real function. Conversely, any period 1 real function $\,g(x)\,$ determines uniquely the correspoinding $\,f(x)\,$ satisfying $\,f(x+1) = 2f(x).$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you come up with that definition? $\endgroup$ – Billy Claxtone Aug 2 at 2:48
  • $\begingroup$ @BillyClaxtone An obvious solution to the original equation is $2^x$, and if we divide $f(x)$ by that we get a solution to $g(x+1)=g(x)$ as I wrote in my answer. $\endgroup$ – Somos Aug 3 at 18:45

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.