Here's a question I found while studying functional equations.
$f(x+y)=f (x)f (y)$ (for all $x,y$ belonging to the set of real numbers) and $f'(0)=2.$ We have to find $f (x).$
So the method we've been taught in class goes like this
-
Partially differentiate both sides of functional equation w.r.t. $x$ or $y$
$f'(x+y)=f'(x)f (y)$ (Partial differentiating w.r.t. $x$) -
Put $x=0$ to get $f'(y)=f'(0)f (y).$
Now since $f '(0)$ is known, we have a differential equation which we can easily solve to give the answer which comes out to be $f (x)=e^{2x}.$
But I am not getting it why this method works?
Secondly, in the first step while partial differentiating we assumed that $y$ is a constant. But in the second step we are integrating w.r.t. $y.$ This doesn't make sense. How can $y$ be treated as a constant and a variable in a single problem?
Best Answer
We have $$f(x+y) = f(x)f(y)$$ Partial differentiate respect to x $$f'(x+y) = f'(x)f(y) + f(x)f'(y) \frac{dy}{dx}$$ Since there is no restriction on x or y. y is independent of x. $\frac{dy}{dx}=0$ $$f'(x+y) = f'(x)f(y)$$ The definition of function is for all x,y it applies the above $\forall x,y$. We can choose any x we want. Since we know $f'(0)$, we choose $x=0$ $$ f'(y) = f'(0)f(y)=2f(y)$$ So, $f'(y)=2f(y)$ solve using seperation of variables $$\frac{df(y)}{dx}=2f(y)$$ $$\frac{df(y)}{f(y)}=2dx$$
$$ln(f(y))=2x+C$$
$$f(y)=e^{2x+C}=Ke^{2x}$$