Interpolation- Lagrange polynomial
$begingroup$
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 '18 at 20:24
PabloZ392PabloZ392
686
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:24
PabloZ392PabloZ392
686
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:24
PabloZ392PabloZ392
686
$endgroup$
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 '18 at 20:24
PabloZ392PabloZ392
686
asked Dec 1 '18 at 20:24
PabloZ392PabloZ392
686
asked Dec 1 '18 at 20:24
PabloZ392PabloZ392
686
asked Dec 1 '18 at 20:24
PabloZ392PabloZ392
686
asked Dec 1 '18 at 20:24
PabloZ392PabloZ392
686
686
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 '18 at 21:43
DamienDamien
59714
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 '18 at 21:43
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
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 '18 at 21:43
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
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 '18 at 21:43
DamienDamien
59714
$endgroup$
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 '18 at 21:43
DamienDamien
59714
answered Dec 1 '18 at 21:43
DamienDamien
59714
answered Dec 1 '18 at 21:43
DamienDamien
59714
answered Dec 1 '18 at 21:43
DamienDamien
59714
59714
add a comment |
add a comment |
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