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

Question

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?

Answer 1

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 $f$.

Answer 2

Alternatively, write the functional equation as $$f(x+y)-e^{x+y}+(x+y)^2+1=f(x)-e^x+x^2+1 + f(y) - e^y +y^2+1$$

Substitute $g(x)=f(x)-e^x+x^2+1$. This gives

$$g(x+y)=g(x)+g(y)$$

which is the Cauchy functional equation. Because $f$ is differentiable at zero, all solutions are given by $g(x)=cx$ for some $c \in \mathbb R$.

So $f(x)=e^x+cx-x^2-1$. So $f'(x) = e^x+c-2x$, so $f'(0)=1+c=1$, hence $c=0$.

So $f(x)=e^x-x^2-1$.

Answer 3

$$f(x+y)= f(x)+f(y)-2xy+(e^x-1)(e^y-1)$$ Do partial differential with respect to $x$ keeping $y$ constant $$f'(x+y) = f'(x)-2y+(e^y-1)(e^x)$$ Put $x=0$ $$f'(y) = f'(0)-2y+(e^y-1)$$ Now integrate with respect to $y$ \begin{align} f(y) &= e^y -x^2 +c \\ f(0) &= 0 \\ c&= -1 \\ f(x) &= e^x -x^2 -1 \end{align}