# Thread: Polynomial Approximation Problem

1. ## Polynomial Approximation Problem

Suppose that $f:\mathbb{R}\longrightarrow\mathbb{R}$ is of class $C^2$. Given $b>0$ and a positive number $\epsilon$, show that there exists a polynomial $p$ such that:

$|p(x)-f(x)|<\epsilon$
$|p'(x)-f'(x)|<\epsilon\\$
$|p''(x)-f''(x)|<\epsilon$

For all $x\in[0,b]$.

The hint given was to choose a polynomial $q$ that approximates $f''$, and suppose that $|q-f''|<\eta$ throughout the interval $[0,b]$. Now say that $p''=q, p(0)=f(0), p'(0)=f'(0)$, and to bound $|p'-f'|$ and $|p-f|$ in terms of both $\eta$ and $b$.

So I'm thinking to choose that $q$ and try to find an antiderivative, and then find an antiderivative of that such that the other two conditions hold, but I'm having trouble seeing how those distances all relate to one another.

2. Originally Posted by mathematicalbagpiper
Suppose that $f:\mathbb{R}\longrightarrow\mathbb{R}$ is of class $C^2$. Given $b>0$ and a positive number $\epsilon$, show that there exists a polynomial $p$ such that:

$|p(x)-f(x)|<\epsilon$
$|p'(x)-f'(x)|<\epsilon\\$
$|p''(x)-f''(x)|<\epsilon$

For all $x\in[0,b]$.

The hint given was to choose a polynomial $q$ that approximates $f''$, and suppose that $|q-f''|<\eta$ throughout the interval $[0,b]$. Now say that $p''=q, p(0)=f(0), p'(0)=f'(0)$, and to bound $|p'-f'|$ and $|p-f|$ in terms of both $\eta$ and $b$.

So I'm thinking to choose that $q$ and try to find an antiderivative, and then find an antiderivative of that such that the other two conditions hold, but I'm having trouble seeing how those distances all relate to one another.
Just to start you in the right direction, take $q$ as in the hint, and define $r(x) = f'(0) + \int_0^xq(t)\,dt.$ Now apply the mean value theorem to the function $r - f'$ in the interval from 0 to x. Remembering that $r(0) - f'(0) = 0$ and that $r'(t) = q(t)$, you should be able to conclude that $|r(x) - f'(x)| = x|q(t) - f''(t)|$ for some t between 0 and x. But $x|q(t) - f''(t)|\leqslant b\eta$. So if $\eta$ has been chosen so that $b\eta<\varepsilon$, it will follow that $|r(x) - f'(x)| < \varepsilon$ (for all $x\in[0,b]$), as required.

Now you have to repeat a similar argument, by integrating and using the mean value theorem again, to get the corresponding result at the level of f.