Polynomial interpolation

**math8** · April 10th 2010, 01:05 PM

Let $x_{0}, x_{1}, \cdots , x_{n}$ be distinct points in the interval [a,b] and $f \in C^{1}[a,b]$ .

We show that for any given $\epsilon >0$ there exists a polynomial p such that

$\left\| f-p \right\|_{\infty} < \epsilon$ and $p(x_{i}) = f(x_{i})$ for all $i=1,2, \cdots , n$

I know $\left\| f\right\| _{\infty}= max_{x \in [a,b]}|f(x)|$ and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes $x_{0}, x_{1}, \cdots , x_{n}$ . If yes, I am not sure how to prove the problem.

**Deadstar** · April 10th 2010, 01:50 PM

Lol you might wanna check out your post again... Whatever you copied of physics forum has been saved as an image so won't show up I guess.

**maddas** · April 10th 2010, 02:09 PM

Its probably cached by their browser so they can't tell. I'll try to repair it...

Originally Posted by math8

Let $x_0,x_1,\cdots x_n$ be distinct points in the interval [a,b] and $f\in C^1([a,b])$ .

We show that for any given $\epsilon > 0$ there exists a polynomial p such that $||f-p||_\infty < \epsilon$ and $p(x_i)=f(x_i)$ for all $i=1,\cdots, n$ .

I know $||f||_\infty = \max_{x\in[a,b]}|f(x)|$ and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes $x_1,\cdots,x_n$ . If yes, I am not sure how to prove the problem.

**maddas** · April 10th 2010, 08:01 PM

I'll give it a shot I guess...

Assume WLOG that both a and b are in the $x_i$ , and replace f with f-g, where g is the polynomial which interpolates the nodes $x_i$ . Clearly if f-g can be approximated by a polynomial, f can be approximated by that polynomial plus g. So we have to find a polynomial which is within $\epsilon$ of f and vanishes at $x_i$ .

Chuse as g a polynomial which is everywhere within $\epsilon\over 2n$ of f by the Stone–Weierstrass theorem. Let $d_i = f(x_i) - g(x_i)$ which is no more than $\epsilon\over 2n$ in magnitude. We will chuse polynomials $h_i$ to add to g to correct these errors so that the final approximating polynomial is $g+\sum h_i$ . Then $||f-g-\sum h_i||_\infty \le ||f-g||_\infty + \sum ||h_i|| \le {\epsilon\over 2n} + n{\epsilon \over 2n} \le \epsilon$ if all the $||h_i||_\infty \le {\epsilon\over 2n}$ .

Therefore we have only to prove the following claim: for every $\epsilon>0$ and $x,y\in[a,b]$ ( $y<\epsilon$ ), there is a polynomial h which is everywhere less than $\epsilon$ in magnitude on [a,b] and h(x)=y.

To do this, chuse a cosine wave centered at x and scale it down to amplitude y. We can take its Taylor polynomial to enough terms that inside [a,b] it is closer to zero at every point than the cosine wave [perhaps someone should check this, as I only have a rough sketch], so that this Taylor polynomial is y at x and it is $\le \epsilon$ on [a,b], as desired.

...god, what a mess. Its probably wrong too. Be warned!

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