- #1

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## Homework Statement

Consider the Lagrange Polynomial approximation [tex]p(x) =\sum_{k=0}^n f(x_k)L_k(x)[/tex] where [tex]L_k(x)=\prod_{i=0,i\neq k}^n \frac{x-x_i}{x_k-x_i}[/tex]

Let [tex]\psi(x)=\prod_{i=0}^n x-x_i[/tex]. Show that [tex]p(x)=\psi(x) \sum_{k=0}^n\frac{f(x_k)}{(x-x_k)\psi^\prime(x)}[/tex]

## Homework Equations

None. Just plug in and see if it pops out.

## The Attempt at a Solution

I just evaluated what [itex]p(x) = \sum_{k=0}^n f(x_k)L_k(x)[/itex] and [itex]\psi^\prime(x)[/itex]. Writing out some terms of p(x):[tex]p(x)=f(x_0)L_0(x_0)+f(x_1)L_1(x_1)+\cdots+f(x_n)L_n(x_n)[/tex]

[tex]=f(x_0)\frac{\prod_{i=0,i\neq k}^n x-x_i}{\prod_{i=0,i\neq k}^n x_o-x_i}+\cdots+f(x_n)\frac{\prod_{i=0,i\neq k}^n x-x_i}{\prod_{i=0,i\neq k}^n x_n-x_i}[/tex]

I find that this is just

[tex]p(x)=\prod_{i=0,i\neq k}^nx-x_i \sum_{k=0}^n \frac{f(x_k)}{\prod_{i=0,i\neq k}^n x_k-x_i}[/tex]

The product outside of the sum is just [itex]\psi(x)[/itex]. I then evaluate [itex]\psi^\prime(x)[/itex]:

[tex]\psi^\prime(x)=\prod_{i=0,i\neq k=0}^nx-x_i +\prod_{i=0,i\neq k=1}^nx-x_i +\cdots +\prod_{i=0,i\neq k=n}^nx-x_i [/tex]

Evaluating this at [itex]x_k[/itex] we get exactly the denominator from above. I cannot figure out where the extra [itex]x-x_k[/itex] term comes from. I suspect my error is in taking the derivative but I have looked this over for many hours and cannot find my mistake.