Is there a general way to find a $f(x)$ such that $f(x+1)/f(x)=2,~f(0)=A$?

Question

I know a function that satifies $\frac{f(x+1)}{f(x)}=2,~f(0)=A$ is $f(x)=A\cdot 2^x$. However, that is because this question is arise in a chapter of exponentiation in a junior-high school textbook, so I can successfully guess it. However, is there a general way to find the unknown function given by such functional equations? Is this related to differential equations?

PS: The question is: If the amount of a kind of bacteria will double every month, and if the original amount is $A$, then write the amount function with respect to month.

Answer 1

I'm interpreting your request for $x \in \mathbb{N}$, rather than $x \in \mathbb{R}$; Michael Burr's answer covers the real case.

---

Functional equations are in general not easy to solve, but there are various techniques. For your particular example, just list out the first few values of $f$, spot that they look like powers of $2$, and prove by induction that they are powers of $2$.

A more systematic approach to your question might involve generating functions. Consider $$F(y) = \sum_{n=1}^{\infty} f(n) y^n$$ (treating this sum purely formally, without considering questions of convergence; as long as we never attempt to evaluate $F$ at any value of $y$, our resulting manipulations are safe whether or not the sums converge).

Then $$F(y) = \sum_{n=1}^{\infty} 2 f(n-1) y^n = 2y \sum_{n=1}^{\infty} f(n-1) y^{n-1} = 2y (f(0) + F(y))$$ where for the first step we used that $f(n) = 2 f(n-1)$, and for the last step we used that $$f(0) + F(y) = f(0) \times y^0 + \sum_{n=1}^{\infty} f(n)y^n = \sum_{n=0}^{\infty} f(n) y^n = \sum_{n=1}^{\infty} f(n-1) y^{n-1}$$

Therefore $$F(y)(1-2y) = 2y f(0) = 2yA$$ so $$F(y) = \frac{2y}{1-2y} A$$

Finally, $f(n)$ is the coefficient of $y^n$ in that expression, which is easy by the binomial expansion of $(1-2y)^{-1}$: it's $A \times 2^n$.

Answer 2

Let $g:[0,1)\rightarrow\mathbb{R}$ be any function such that $g(0)=A$. Then, we can define $f$ as follows: $$ f(x)=2^{\lfloor x\rfloor}g(x-\lfloor x\rfloor). $$ In this case, $$ \frac{f(x+1)}{f(x)}=\frac{2^{\lfloor x+1\rfloor}g((x+1)-\lfloor x+1\rfloor)}{2^{\lfloor x\rfloor}g(x-\lfloor x\rfloor)}=\frac{2^{\lfloor x\rfloor+1}g((x+1)-(\lfloor x\rfloor+1))}{2^{\lfloor x\rfloor}g(x-\lfloor x\rfloor)} =\frac{2^{\lfloor x\rfloor+1}g(x-\lfloor x\rfloor)}{2^{\lfloor x\rfloor}g(x-\lfloor x\rfloor)}=2. $$ So, there are lots of possible functions. If you want $f$ to be continuous, then $g$ must be continuous. As you add more conditions, like differentiability, there are additional restrictions on $g$.

Answer 3

Discretization of continuous dynamical system $$\dot{x}=\ln2\cdot x$$ is $$x\left(t\right)=e^{\ln{2} \cdot t}x_{0}$$ or $$x_{t+1}=2^{t+1}x_{0}=2x_{t}$$ So, your equation $f\left(x+1\right)=2f\left(x\right)$ is just a discretization of the differential equation $$\frac{df}{dx}=\ln2\cdot f\left(x\right)$$