4
$\begingroup$

Let $f$ be a differentiable function satisfying

$$f(x + y) = e^xf(y) + e^yf(x)$$

for all $x, y \in \mathbb{R}$. Find $f'(0)$.

I tried to use the definition of $f'(0)$ to do this:

$$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0}\frac{f(h)}{h}$$

The problem here is that I would have to apply L'Hopital's rule on the RHS , which would give me

$$f'(0) = \frac{\lim_{h\to0}f'(h)}{\lim_{h\to0}1} = f'(0)$$

This doesn't really help much. Any ideas?

$\endgroup$
6
$\begingroup$

It is a trick question. A function satisfies that if $f(x) = cxe^x$ for any constant $c$. So $f'(0)=c$, and this can be any real number.

You can get the above functional form by: $\frac{f(x+h)-f(x)}{h} = \frac{e^xf(h)}{h} + \frac{f(x)(e^h-1)}{h}$
and so, taking limits and defining $\lim_{h\rightarrow 0} f(h)/h=c$ gives:
$f'(x) = ce^x + f(x)$.
So then we just solve the differential equation.

$\endgroup$
1
$\begingroup$

Given $f(x+y) = e^{y}f(x)+e^{x}f(y)$

Divide both side by $e^{x+y}\;,$ we get

$\displaystyle e^{-(x+y)}f(x+y) = e^{-x}f(x)+e^{-y}f(y)$

Now Let $\displaystyle e^{-x}f(x) = g(x)$, Then equation Convert into $g(x+y)=g(x)+g(y)$

Which is a Cauchy Functional equation whose solution is $g(x)=cx$.

So $\displaystyle g(x)=e^{-x}f(x)=cx\Rightarrow f(x)=cxe^{x}$

So $f{'}(x)=c\left\{xe^{x}+e^{x}\right\}$, So $f^{'}(0)=c$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.