Interpolation- Lagrange polynomial
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Let $x_0,x_1,...,x_n$ will be different real numbers.
Show, that: $f[x_0,x_1,...,x_n]=sum_{i=0}^mfrac{f(x_i)}{Phi '(x)}$ where $Phi (x)=(x-x_0)(x-x_1)...(x-x_m)$
So, I have some problems.How to start?
lagrange-interpolation interpolation-theory
asked Dec 1 at 20:24
PabloZ392
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up vote
0
down vote
favorite
Let $x_0,x_1,...,x_n$ will be different real numbers.
Show, that: $f[x_0,x_1,...,x_n]=sum_{i=0}^mfrac{f(x_i)}{Phi '(x)}$ where $Phi (x)=(x-x_0)(x-x_1)...(x-x_m)$
So, I have some problems.How to start?
lagrange-interpolation interpolation-theory
asked Dec 1 at 20:24
PabloZ392
1386
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $x_0,x_1,...,x_n$ will be different real numbers.
Show, that: $f[x_0,x_1,...,x_n]=sum_{i=0}^mfrac{f(x_i)}{Phi '(x)}$ where $Phi (x)=(x-x_0)(x-x_1)...(x-x_m)$
So, I have some problems.How to start?
lagrange-interpolation interpolation-theory
asked Dec 1 at 20:24
PabloZ392
1386
Let $x_0,x_1,...,x_n$ will be different real numbers.
Show, that: $f[x_0,x_1,...,x_n]=sum_{i=0}^mfrac{f(x_i)}{Phi '(x)}$ where $Phi (x)=(x-x_0)(x-x_1)...(x-x_m)$
So, I have some problems.How to start?
lagrange-interpolation interpolation-theory
lagrange-interpolation interpolation-theory
asked Dec 1 at 20:24
PabloZ392
1386
asked Dec 1 at 20:24
PabloZ392
1386
asked Dec 1 at 20:24
PabloZ392
1386
asked Dec 1 at 20:24
PabloZ392
1386
asked Dec 1 at 20:24
PabloZ392
1386
1386
add a comment |
add a comment |
1 Answer
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I guess $f$ is defined as the minimum polynomial with $f(x_i)=y_i$. You get two polynomials of same order $n$, equal on $n+1$ points. They are equal.
answered Dec 1 at 21:43
Damien
2844
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Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I guess $f$ is defined as the minimum polynomial with $f(x_i)=y_i$. You get two polynomials of same order $n$, equal on $n+1$ points. They are equal.
answered Dec 1 at 21:43
Damien
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
I guess $f$ is defined as the minimum polynomial with $f(x_i)=y_i$. You get two polynomials of same order $n$, equal on $n+1$ points. They are equal.
answered Dec 1 at 21:43
Damien
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
I guess $f$ is defined as the minimum polynomial with $f(x_i)=y_i$. You get two polynomials of same order $n$, equal on $n+1$ points. They are equal.
answered Dec 1 at 21:43
Damien
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I guess $f$ is defined as the minimum polynomial with $f(x_i)=y_i$. You get two polynomials of same order $n$, equal on $n+1$ points. They are equal.
answered Dec 1 at 21:43
Damien
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 1 at 21:43
Damien
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 1 at 21:43
Damien
2844
answered Dec 1 at 21:43
Damien
2844
2844
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Damien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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