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.

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    $\begingroup$ 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

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 $f$.

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  • $\begingroup$ 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

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 $\,f(x+1) = 2f(x).$

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  • $\begingroup$ 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

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