How to calculate interpolation error with repeated nodes
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Given $f(x)=cos(pi x)$
a) Calculate the interpolation polynomial such that $p(-1)=f(-1)$, $p'(-1)=f'(-1)$, $p(0)=f(0)$, $p(1)=f(1)$ and $p'(1)=f'(1)$.
b) Show that $|f(x)-p(x)|leqfrac{pi^5}{120}$ for every $xin[-1,1]$
So I managed to do item a) and obtain $p(x)=2x^4-4x^2+1$. However, when I try to do item b) and bound my error, I have repeated nodes, $x_0=-1$ and $x_2=1$ and therefore im not sure if I can use the bound I would like to use, say:
$$||W_{n+1}(x)||_{infty}leqleft(frac{b-a}{2}right)^{n+1}$$
Since my nodes are equidistant. Is there an aditional condition I should add because of the repetition of my nodes?
numerical-methods lagrange-interpolation
asked Nov 22 at 19:16
Arete the Mathematician
62
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up vote
1
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Given $f(x)=cos(pi x)$
a) Calculate the interpolation polynomial such that $p(-1)=f(-1)$, $p'(-1)=f'(-1)$, $p(0)=f(0)$, $p(1)=f(1)$ and $p'(1)=f'(1)$.
b) Show that $|f(x)-p(x)|leqfrac{pi^5}{120}$ for every $xin[-1,1]$
So I managed to do item a) and obtain $p(x)=2x^4-4x^2+1$. However, when I try to do item b) and bound my error, I have repeated nodes, $x_0=-1$ and $x_2=1$ and therefore im not sure if I can use the bound I would like to use, say:
$$||W_{n+1}(x)||_{infty}leqleft(frac{b-a}{2}right)^{n+1}$$
Since my nodes are equidistant. Is there an aditional condition I should add because of the repetition of my nodes?
numerical-methods lagrange-interpolation
asked Nov 22 at 19:16
Arete the Mathematician
62
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $f(x)=cos(pi x)$
a) Calculate the interpolation polynomial such that $p(-1)=f(-1)$, $p'(-1)=f'(-1)$, $p(0)=f(0)$, $p(1)=f(1)$ and $p'(1)=f'(1)$.
b) Show that $|f(x)-p(x)|leqfrac{pi^5}{120}$ for every $xin[-1,1]$
So I managed to do item a) and obtain $p(x)=2x^4-4x^2+1$. However, when I try to do item b) and bound my error, I have repeated nodes, $x_0=-1$ and $x_2=1$ and therefore im not sure if I can use the bound I would like to use, say:
$$||W_{n+1}(x)||_{infty}leqleft(frac{b-a}{2}right)^{n+1}$$
Since my nodes are equidistant. Is there an aditional condition I should add because of the repetition of my nodes?
numerical-methods lagrange-interpolation
asked Nov 22 at 19:16
Arete the Mathematician
62
Given $f(x)=cos(pi x)$
a) Calculate the interpolation polynomial such that $p(-1)=f(-1)$, $p'(-1)=f'(-1)$, $p(0)=f(0)$, $p(1)=f(1)$ and $p'(1)=f'(1)$.
b) Show that $|f(x)-p(x)|leqfrac{pi^5}{120}$ for every $xin[-1,1]$
So I managed to do item a) and obtain $p(x)=2x^4-4x^2+1$. However, when I try to do item b) and bound my error, I have repeated nodes, $x_0=-1$ and $x_2=1$ and therefore im not sure if I can use the bound I would like to use, say:
$$||W_{n+1}(x)||_{infty}leqleft(frac{b-a}{2}right)^{n+1}$$
Since my nodes are equidistant. Is there an aditional condition I should add because of the repetition of my nodes?
numerical-methods lagrange-interpolation
numerical-methods lagrange-interpolation
asked Nov 22 at 19:16
Arete the Mathematician
62
asked Nov 22 at 19:16
Arete the Mathematician
62
asked Nov 22 at 19:16
Arete the Mathematician
62
asked Nov 22 at 19:16
Arete the Mathematician
62
asked Nov 22 at 19:16
Arete the Mathematician
62
62
add a comment |
add a comment |
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