This seems like it should be super simple but I can't figure it out. Prove that if $f(t)$ satisfies $f(t)=f'(t)$, then $f(t+s)=f(t)f(s)$ for all $t,s$ belonging to R without using the fact that $f(t)=c\cdot e^t$. The only hint I'm given is that I can use the fact that two functions satisfying the same ODE $x'=x$ with the same initial condition are equal. Any ideas?
Asked
Active
Viewed 194 times
4
-
http://math.stackexchange.com/a/1597885/272831 – Simply Beautiful Art Aug 31 '16 at 19:46
-
Oh, welcome to the community! Please refer here for a tutorial on how to format your math: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Simply Beautiful Art Aug 31 '16 at 19:53
-
4You also need to assume $f(0)=1$ (or $f(0)=0$), otherwise $f(s+t)=f(s)f(t)$ doesn't hold (just take $s=t=0$, for instance). – Arthur Aug 31 '16 at 19:53
2 Answers
6
It is false, unless you also assume that $f(0) = 1$ or that $f(0) = 0$.
So, let's additionally assume that $f(0) = 1$ and let's avoid the fact that this implies that $f(t) = e^t$.
Take $s \in {\mathbb R}$ and define functions $g$ and $h$ by $$g(t) := f(s + t)$$ and $$h(t) := f(s) f(t).$$ Then $$g'(t) = f'(s+t) = f(s+t) = g(t)$$ and $$h'(t) = f(s) f'(t) = f(s) f(t) = h(t),$$ so $g$ and $h$ both satisfy the ode $y' = y$.
Additionally, $g(0) = f(s)$ and $h(0) = f(s) f(0) = f(s)$, so they have the same initial condition.
Therefore $g$ and $h$ are the same function, so $f(s + t) = f(s) f(t)$.
Magdiragdag
- 15,049
4
It is false for all $c$ except $1$ or $0$. For instance, $f(t)=2e^t$ satisfies $f'=f$ but $$f(t+s)=\frac12 f(t)f(s)$$