Mixing
Calculation of a interpolating polynomial
Let $x_i=i+1$ for all $i=0,1,...,20$ and let $p(x)$ be a polynomial of degree at most 20 satisfying the following property:
$$p(x_i)=(x_i)^{21}$$ for all $i=0,1,...,20$
I need to compute $p(0)$
I realised that $x^{21} - p(x)$ gives me the remainder term $-R_n(x)=(x-x_0)(x-x_1)...(x-x_{20})f[x,x_0,x_1,...,x_{20}]$
So $R_n(0)=21!f[x,x_0,x_1,...,x_{20}]$ where $f[x,x_0,x_1,...,x_{20}]$ represents the divided difference.
I am not really sure how to go on with finding the divided difference, any help is really appreciated!
polynomials interpolation
asked Nov 20 '18 at 12:55
Derp
228214
add a comment |
Let $x_i=i+1$ for all $i=0,1,...,20$ and let $p(x)$ be a polynomial of degree at most 20 satisfying the following property:
$$p(x_i)=(x_i)^{21}$$ for all $i=0,1,...,20$
I need to compute $p(0)$
I realised that $x^{21} - p(x)$ gives me the remainder term $-R_n(x)=(x-x_0)(x-x_1)...(x-x_{20})f[x,x_0,x_1,...,x_{20}]$
So $R_n(0)=21!f[x,x_0,x_1,...,x_{20}]$ where $f[x,x_0,x_1,...,x_{20}]$ represents the divided difference.
I am not really sure how to go on with finding the divided difference, any help is really appreciated!
polynomials interpolation
asked Nov 20 '18 at 12:55
Derp
228214
add a comment |
Let $x_i=i+1$ for all $i=0,1,...,20$ and let $p(x)$ be a polynomial of degree at most 20 satisfying the following property:
$$p(x_i)=(x_i)^{21}$$ for all $i=0,1,...,20$
I need to compute $p(0)$
I realised that $x^{21} - p(x)$ gives me the remainder term $-R_n(x)=(x-x_0)(x-x_1)...(x-x_{20})f[x,x_0,x_1,...,x_{20}]$
So $R_n(0)=21!f[x,x_0,x_1,...,x_{20}]$ where $f[x,x_0,x_1,...,x_{20}]$ represents the divided difference.
I am not really sure how to go on with finding the divided difference, any help is really appreciated!
polynomials interpolation
asked Nov 20 '18 at 12:55
Derp
228214
Let $x_i=i+1$ for all $i=0,1,...,20$ and let $p(x)$ be a polynomial of degree at most 20 satisfying the following property:
$$p(x_i)=(x_i)^{21}$$ for all $i=0,1,...,20$
I need to compute $p(0)$
I realised that $x^{21} - p(x)$ gives me the remainder term $-R_n(x)=(x-x_0)(x-x_1)...(x-x_{20})f[x,x_0,x_1,...,x_{20}]$
So $R_n(0)=21!f[x,x_0,x_1,...,x_{20}]$ where $f[x,x_0,x_1,...,x_{20}]$ represents the divided difference.
I am not really sure how to go on with finding the divided difference, any help is really appreciated!
polynomials interpolation
polynomials interpolation
asked Nov 20 '18 at 12:55
Derp
228214
asked Nov 20 '18 at 12:55
Derp
228214
asked Nov 20 '18 at 12:55
Derp
228214
asked Nov 20 '18 at 12:55
Derp
228214
asked Nov 20 '18 at 12:55
Derp
228214
228214
add a comment |
add a comment |
1 Answer
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Hint: $x^{21} - p(x)$ is a monic polynomial of degree $21$ and roots $x_0,dots,x_{20}$.
answered Nov 20 '18 at 13:34
lhf
163k10167386
add a comment |
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1 Answer
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1 Answer
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Hint: $x^{21} - p(x)$ is a monic polynomial of degree $21$ and roots $x_0,dots,x_{20}$.
answered Nov 20 '18 at 13:34
lhf
163k10167386
add a comment |
Hint: $x^{21} - p(x)$ is a monic polynomial of degree $21$ and roots $x_0,dots,x_{20}$.
answered Nov 20 '18 at 13:34
lhf
163k10167386
add a comment |
Hint: $x^{21} - p(x)$ is a monic polynomial of degree $21$ and roots $x_0,dots,x_{20}$.
answered Nov 20 '18 at 13:34
lhf
163k10167386
Hint: $x^{21} - p(x)$ is a monic polynomial of degree $21$ and roots $x_0,dots,x_{20}$.
answered Nov 20 '18 at 13:34
lhf
163k10167386
answered Nov 20 '18 at 13:34
lhf
163k10167386
answered Nov 20 '18 at 13:34
lhf
163k10167386
answered Nov 20 '18 at 13:34
lhf
163k10167386
163k10167386
add a comment |
add a comment |
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