# Polynomial of Degree 2

• Jan 19th 2011, 12:35 PM
zebra2147
Polynomial of Degree 2
My professor has this proof in his notes and I was wondering if anyone could help get me rolling...

Let $\displaystyle f :\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Suppose
$\displaystyle f(x + y) = f(x) + f(y) + 2xy$ for all $\displaystyle x,y\in \mathbb{R}$.

Show by completing the two parts below that $\displaystyle f$ is a polynomial of degree 2.
(1) Prove $\displaystyle f'(x) = f'(0)+2x$, for all $\displaystyle x\in \mathbb{R}$.
(2) Prove $\displaystyle f(x) = x^2+f'(0)x+f(0)$,for all $\displaystyle x\in \mathbb{R}$.

I think my biggest problem is coming up with the best way to represent $\displaystyle f'(x)$ which is obviously a big roadblock.
• Jan 19th 2011, 12:44 PM
TheEmptySet
Quote:

Originally Posted by zebra2147
My professor has this proof in his notes and I was wondering if anyone could help get me rolling...

Let $\displaystyle f :\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Suppose
$\displaystyle f(x + y) = f(x) + f(y) + 2xy$ for all $\displaystyle x,y\in \mathbb{R}$.

Show by completing the two parts below that $\displaystyle f$ is a polynomial of degree 2.
(1) Prove $\displaystyle f'(x) = f'(0)+2x$, for all $\displaystyle x\in \mathbb{R}$.
(2) Prove $\displaystyle f(x) = x^2+f'(0)x+f(0)$,for all $\displaystyle x\in \mathbb{R}$.

I think my biggest problem is coming up with the best way to represent $\displaystyle f'(x)$ which is obviously a big roadblock.

If you take the derivative with respect to x you get

$\displaystyle \displaystyle f'(x+y)=f'(x)+2y \implies f'(x)=f'(x+y)-2y$

This must hold for all $\displaystyle y \in \mathbb{R}$ so what choice of $\displaystyle y$ will give the result that you want?
• Jan 19th 2011, 01:01 PM
zebra2147
$\displaystyle y=-x$??

So that we get $\displaystyle f'(x) = f'(0)+2x$?
• Jan 19th 2011, 01:23 PM
TheEmptySet
Quote:

Originally Posted by zebra2147
$\displaystyle y=-x$??

So that we get $\displaystyle f'(x) = f'(0)+2x$?

Yes that is the one:)
• Jan 19th 2011, 01:53 PM
zebra2147
So is the information that you have helped me come up sufficient for the proof because it only has to work for all $\displaystyle x$ and not all $\displaystyle y$?

And do you have any hints for part 2?

Thanks!
• Jan 19th 2011, 02:20 PM
Drexel28
Quote:

Originally Posted by zebra2147
So is the information that you have helped me come up sufficient for the proof because it only has to work for all $\displaystyle x$ and not all $\displaystyle y$?

And do you have any hints for part 2?

Thanks!

Let $\displaystyle g(x)=f(x)-x^2-f'(0)x$. Note that $\displaystyle g'(x)=f'(x)-2x-f'(0)=0$ and since $\displaystyle g'(x)=0$ on an interval (in this case all of $\displaystyle \mathbb{R}$) we may conclude that $\displaystyle f(x)-x^2-f'(0)x=g(x)=c$ for some $\displaystyle c$. Noting that $\displaystyle g(0)=f(0)$ gives the desired results.