An interesting twist on Cauchy's functional equation I was looking at Cauchy's functional equation when I happened upon an interesting result.
Consider a differentiable function $f:\mathbb R\to\mathbb R$ with $f(1)=1$ and $f(x+y)=f(x)+f(y)$.
With $x=y=0$, it follows that $f(0)=0$.
...and then I do something not so orthodox:
$$f(x+h)=f(x)+f(h)$$
$$f(x+h)-f(x)=f(h)=f(0+h)-f(0)$$
$$\frac{f(x+h)-f(x)}h=\frac{f(0+h)-f(0)}h$$
as $h\to0$, we end up with, nicely,
$$f'(x)=f'(0)$$
Thus, $f(x)$ is linear, and interpolating with $f(0)=0$ and $f(1)=1$, we end up with
$$f(x)=x$$
just as suspected.
Now, I don't usually see this done with functional equations, using derivatives and all, so I was wondering 2 questions:


*

*Is what I did ok, given the conditions on $f$?

*Could someone produce an interesting scenario of when a hard to solve functional equation reduces nicely with derivatives?  Thanks :-)
 A: Here is an interesting example of a functional equation that can be solved more easily using derivatives:
$$f(x+1)-f(1-x)+f(2x)=x^2+x+1$$
By taking the derivative of both sides, we have
$$f'(x+1)+f'(1-x)+2f'(2x)=2x+1$$
and by differentiating again,
$$f''(x+1)-f''(1-x)+4f''(2x)=2$$
At this point it seems safe to assume that $f$ is constant, or that $f(x)=a$ for some constant $a$. Then we have
$$a-a+4a=2$$
$$a=\frac{1}{2}$$
and so
$$f''(x)=\frac{1}{2}$$
$$f'(x)=\frac{1}{2}x+b$$
for some other constant $c$. Then we can discover from the equation
$$f'(x+1)+f'(1-x)+2f'(2x)=2x+1$$
that
$$\frac{1}{2}+b+\frac{1}{2}+b+2b=1$$
$$4b+1=1$$
$$b=0$$
and so we have
$$f'(x)=\frac{1}{2}x$$
and
$$f(x)=\frac{1}{4}x^2+c$$
for some constant $c$. Finally, from the original functional equation
$$f(x+1)-f(1-x)+f(2x)=x^2+x+1$$
we have
$$\frac{1}{4}+c-\frac{1}{4}-c+c=1$$
$$c=1$$
And, at long last, we have
$$\color{red}{f(x)=\frac{1}{4}x^2+1}$$
Most polynomial functional equations can be solved easily this way. A solution can be obtained to the original equation by assuming that $f$ was quadratic and solving for $a$, $b$, and $c$, but that would involve a lot of messy algebra, and a neatly partitioned answer involving derivatives is highly preferable.
More examples like this can be found here.
