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Estimate approximation error of a function
The Next CEO of Stack OverflowWhat is the motivation for this theorem on Polynomial Interpolation Error?Why does the interpolation error go to zero if we increase the number of sampling points?Cubic polynomial interpolationError formula when using a polynomial interpolationError of Lagrange Interpolating polynomial is zero for polynomials of low degree$g$ ($h$) interpolates $f$ at $x_0,ldots x_{n-1}$ ($x_{1},ldots,x_n$), then $g(x)+frac{x_0-x}{x_n-x_0}[g(x)-h(x)]$ interpolates $f$Upper bounding polynomial interpolation errorHow does one find a polynomial approximation of a non-analytic function?Minimize interpolation error for sin$(x)$ on $[0,pi]$Error estimate in the approximation of Incomplete Beta Function
$begingroup$
It is supposed to approximate a function $f$ on the interval $[a, b]$ by a function $p$ that fits piecewise a polynom of degree $n$.
I've noted the following steps:
- Decompose $[a, b]$ into $N$ subintervals $[t_j, t_{j+1}], j = 0,.. , N - 1$, the length $h = frac{b-a}{N}$ with $t_j = a + jh$.
In each subinterval $[t_j, t_{j + 1}]$ select the interpolation points $x_{i, j}: = t_j + frac{1}{n}ih, i = 0,... , n$ and approximate $f$ to $[t_j, t_{j + 1}]$ by using the polynom that interpolates in the points $x_{i, j}, i = 0,..., n$.
That's how we get one Function $p$, which is a polynomial of degree $n$ on every subinterval.
Now I have to show that $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}h^{n+1}||f^{(n+1)}||_{infty, [a,b]}$
I found a formula that says: $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}* ||w_{n+1}||_{infty, [a,b]}*||f^{(n+1)}||_{infty, [a,b]}$ where $w_{n+1}$ is defined as $(x-x_0)...(x-x_n)$
So how can I prove that $h^{n+1} = (frac{b-a}{N})^{n+1} $ is bigger or equals $w_{n+1} = (x-x_0)...(x-x_n)$. Because in this case I can just use this formula to prove the given relation?
functional-analysis polynomials approximation error-propagation
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
$endgroup$
add a comment |
$begingroup$
It is supposed to approximate a function $f$ on the interval $[a, b]$ by a function $p$ that fits piecewise a polynom of degree $n$.
I've noted the following steps:
- Decompose $[a, b]$ into $N$ subintervals $[t_j, t_{j+1}], j = 0,.. , N - 1$, the length $h = frac{b-a}{N}$ with $t_j = a + jh$.
In each subinterval $[t_j, t_{j + 1}]$ select the interpolation points $x_{i, j}: = t_j + frac{1}{n}ih, i = 0,... , n$ and approximate $f$ to $[t_j, t_{j + 1}]$ by using the polynom that interpolates in the points $x_{i, j}, i = 0,..., n$.
That's how we get one Function $p$, which is a polynomial of degree $n$ on every subinterval.
Now I have to show that $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}h^{n+1}||f^{(n+1)}||_{infty, [a,b]}$
I found a formula that says: $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}* ||w_{n+1}||_{infty, [a,b]}*||f^{(n+1)}||_{infty, [a,b]}$ where $w_{n+1}$ is defined as $(x-x_0)...(x-x_n)$
So how can I prove that $h^{n+1} = (frac{b-a}{N})^{n+1} $ is bigger or equals $w_{n+1} = (x-x_0)...(x-x_n)$. Because in this case I can just use this formula to prove the given relation?
functional-analysis polynomials approximation error-propagation
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
$endgroup$
add a comment |
$begingroup$
It is supposed to approximate a function $f$ on the interval $[a, b]$ by a function $p$ that fits piecewise a polynom of degree $n$.
I've noted the following steps:
- Decompose $[a, b]$ into $N$ subintervals $[t_j, t_{j+1}], j = 0,.. , N - 1$, the length $h = frac{b-a}{N}$ with $t_j = a + jh$.
In each subinterval $[t_j, t_{j + 1}]$ select the interpolation points $x_{i, j}: = t_j + frac{1}{n}ih, i = 0,... , n$ and approximate $f$ to $[t_j, t_{j + 1}]$ by using the polynom that interpolates in the points $x_{i, j}, i = 0,..., n$.
That's how we get one Function $p$, which is a polynomial of degree $n$ on every subinterval.
Now I have to show that $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}h^{n+1}||f^{(n+1)}||_{infty, [a,b]}$
I found a formula that says: $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}* ||w_{n+1}||_{infty, [a,b]}*||f^{(n+1)}||_{infty, [a,b]}$ where $w_{n+1}$ is defined as $(x-x_0)...(x-x_n)$
So how can I prove that $h^{n+1} = (frac{b-a}{N})^{n+1} $ is bigger or equals $w_{n+1} = (x-x_0)...(x-x_n)$. Because in this case I can just use this formula to prove the given relation?
functional-analysis polynomials approximation error-propagation
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
$endgroup$
It is supposed to approximate a function $f$ on the interval $[a, b]$ by a function $p$ that fits piecewise a polynom of degree $n$.
I've noted the following steps:
- Decompose $[a, b]$ into $N$ subintervals $[t_j, t_{j+1}], j = 0,.. , N - 1$, the length $h = frac{b-a}{N}$ with $t_j = a + jh$.
In each subinterval $[t_j, t_{j + 1}]$ select the interpolation points $x_{i, j}: = t_j + frac{1}{n}ih, i = 0,... , n$ and approximate $f$ to $[t_j, t_{j + 1}]$ by using the polynom that interpolates in the points $x_{i, j}, i = 0,..., n$.
That's how we get one Function $p$, which is a polynomial of degree $n$ on every subinterval.
Now I have to show that $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}h^{n+1}||f^{(n+1)}||_{infty, [a,b]}$
I found a formula that says: $||f-p||_{infty, [a,b]} leq frac{1}{(n+1)!}* ||w_{n+1}||_{infty, [a,b]}*||f^{(n+1)}||_{infty, [a,b]}$ where $w_{n+1}$ is defined as $(x-x_0)...(x-x_n)$
So how can I prove that $h^{n+1} = (frac{b-a}{N})^{n+1} $ is bigger or equals $w_{n+1} = (x-x_0)...(x-x_n)$. Because in this case I can just use this formula to prove the given relation?
functional-analysis polynomials approximation error-propagation
functional-analysis polynomials approximation error-propagation
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
edited Mar 18 at 20:43
mrs fourier
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
asked Mar 18 at 1:07
mrs fouriermrs fourier
9011
9011
add a comment |
add a comment |
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