Finding Expression for terms in Lagrange Polynomial Interpolation
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So for a paper I am writing I am using Lagrange polynomial interpolation:
$$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$
And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..
Any help is greatly appreciated, Thank you!
analysis functions polynomials interpolation lagrange-interpolation
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add a comment |
$begingroup$
So for a paper I am writing I am using Lagrange polynomial interpolation:
$$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$
And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..
Any help is greatly appreciated, Thank you!
analysis functions polynomials interpolation lagrange-interpolation
$endgroup$
add a comment |
$begingroup$
So for a paper I am writing I am using Lagrange polynomial interpolation:
$$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$
And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..
Any help is greatly appreciated, Thank you!
analysis functions polynomials interpolation lagrange-interpolation
$endgroup$
So for a paper I am writing I am using Lagrange polynomial interpolation:
$$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$
And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..
Any help is greatly appreciated, Thank you!
analysis functions polynomials interpolation lagrange-interpolation
analysis functions polynomials interpolation lagrange-interpolation
edited Feb 5 at 8:20
mrtaurho
6,08771641
6,08771641
asked Jan 1 at 7:55
ToadfutureToadfuture
505
505
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2 Answers
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$begingroup$
To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
$$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
$$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
$$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.
Now taking into account the product that we initially ignored and the sum out front we find
$$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$
I hope this helps!
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The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
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– Max
Feb 5 at 8:20
add a comment |
$begingroup$
To find the coefficients in monomial basis.
$C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$
Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$
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2 Answers
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2 Answers
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active
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active
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votes
$begingroup$
To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
$$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
$$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
$$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.
Now taking into account the product that we initially ignored and the sum out front we find
$$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$
I hope this helps!
$endgroup$
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
add a comment |
$begingroup$
To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
$$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
$$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
$$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.
Now taking into account the product that we initially ignored and the sum out front we find
$$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$
I hope this helps!
$endgroup$
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
add a comment |
$begingroup$
To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
$$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
$$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
$$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.
Now taking into account the product that we initially ignored and the sum out front we find
$$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$
I hope this helps!
$endgroup$
To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
$$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
$$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
$$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.
Now taking into account the product that we initially ignored and the sum out front we find
$$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$
I hope this helps!
answered Jan 1 at 9:02
Cameron KuchtaCameron Kuchta
12
12
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
add a comment |
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
$begingroup$
The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
$endgroup$
– Max
Feb 5 at 8:20
add a comment |
$begingroup$
To find the coefficients in monomial basis.
$C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$
Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$
$endgroup$
add a comment |
$begingroup$
To find the coefficients in monomial basis.
$C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$
Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$
$endgroup$
add a comment |
$begingroup$
To find the coefficients in monomial basis.
$C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$
Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$
$endgroup$
To find the coefficients in monomial basis.
$C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$
Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$
edited Feb 6 at 14:44
answered Feb 4 at 15:50
MaxMax
9211319
9211319
add a comment |
add a comment |
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