[Math] Find $f(x)$ if $f(x+y)=f(x)+f(y) – 2xy + (e^x -1)(e^y -1) $

functional-equationsfunctions

Let $f$ be a differential function satisfying the relation $f(x+y)=f(x)+f(y) – 2xy + (e^x -1)(e^y -1)$$ \ \forall x , y \in\mathbb R $ and $f'(0)=1$

My work

Putting $y=0$
$$f(x)=f(x) + f(0)$$

$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to 0} \frac{f(x)+f(h) – 2xh + (e^x -1)(e^h -1)-f(x)}{h}$$
$$f'(x)=\lim_{h\to 0} \frac{f(h) – 2xh + (e^x -1)(e^h -1)}{h}$$
How to predict things after that?

Best Answer

As $f$ is differentiable, $$ \lim_{h \to 0} \frac{f(h)}{h}=f'(0)=1 $$ Also, $$\lim_{h \to 0} \frac{-2xh+(e^x-1)(e^h-1)}{h}=-2x+(e^x-1)\lim_{h \to 0} \frac{e^h-1}{h}=-2x+e^x-1$$

Putting this together, $$ f'(x)=1-2x+e^x-1=-2x+e^x $$ and you can integrate to find $f$.

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