# How can I determinine all functions or a function f?

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.

• 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. – Paul Aug 1 at 15:03
• Hint: $f(n) = 2f(n-1) = 4f(n-2) = \ldots = 2^k f(n-k)$. – mwt Aug 1 at 15:08
• 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? – Billy Claxtone Aug 2 at 2:53

## 2 Answers

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 $$f$$.

• Yes. Let say I have a restriction: for x ∈ (0, 1], f(x) = x(x − 1). – Billy Claxtone Aug 2 at 2:47
• 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. – 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 $$\,f(x+1) = 2f(x).$$

• How did you come up with that definition? – Billy Claxtone Aug 2 at 2:48
• @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. – Somos Aug 3 at 18:45