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Formula for coefficients of interpolation polynomial

Who knows this formula for polynomial interpolation?For n=3 Lagrange interpolation why is it equal to 1?Coefficients of Newton interpolation polynomialInterpolation in 2-D co-ordinate systemEquivalent condition for interpolation polynomialNumerical mathematics, Lagrange interpolationNumerical mathematics, Lagrange interpolation..Who knows this formula for polynomial interpolation?find interpolation polynomial for $f(x) = x^4+3x^2$ of degree $le3$ , such that $max_xin[-1,1]|f(x)-p(x)|$ is minimalHow to derive the Newton interpolation polynomialWhat is the geometric meaning of null-determinant?

0

$begingroup$

(Question) What is the conventional formula for coefficients of interpolation polynomial?

Consider the interpolation problem: find the polynomial through a given set of points $(x_0,y_0),...,(x_n,y_n)$. Suppose we want the polynomial in canonical form (monomial basis): coefficients times powers of $x$.

Is there a standard solution to this problem? What would be the formula for the coefficients and how is it obtained?

Note: I found my own solutions for this problem, but I would like to know the ‘standard’ solution, if there is one. Maybe it involves some Linear Algebra that I am not aware of yet.

interpolation interpolation-theory

share|cite|improve this question

edited 2 days ago

Max

asked Feb 11 at 16:18

MaxMax

9211319

$endgroup$

add a comment | 

0

$begingroup$

(Question) What is the conventional formula for coefficients of interpolation polynomial?

Consider the interpolation problem: find the polynomial through a given set of points $(x_0,y_0),...,(x_n,y_n)$. Suppose we want the polynomial in canonical form (monomial basis): coefficients times powers of $x$.

Is there a standard solution to this problem? What would be the formula for the coefficients and how is it obtained?

Note: I found my own solutions for this problem, but I would like to know the ‘standard’ solution, if there is one. Maybe it involves some Linear Algebra that I am not aware of yet.

interpolation interpolation-theory

share|cite|improve this question

edited 2 days ago

Max

asked Feb 11 at 16:18

MaxMax

9211319

$endgroup$

add a comment | 

$begingroup$

(Question) What is the conventional formula for coefficients of interpolation polynomial?

Consider the interpolation problem: find the polynomial through a given set of points $(x_0,y_0),...,(x_n,y_n)$. Suppose we want the polynomial in canonical form (monomial basis): coefficients times powers of $x$.

Is there a standard solution to this problem? What would be the formula for the coefficients and how is it obtained?

Note: I found my own solutions for this problem, but I would like to know the ‘standard’ solution, if there is one. Maybe it involves some Linear Algebra that I am not aware of yet.

interpolation interpolation-theory

share|cite|improve this question

edited 2 days ago

Max

asked Feb 11 at 16:18

MaxMax

9211319

$endgroup$

share|cite|improve this question

edited 2 days ago

Max

asked Feb 11 at 16:18

MaxMax

9211319

share|cite|improve this question

edited 2 days ago

Max

asked Feb 11 at 16:18

MaxMax

9211319

asked Feb 11 at 16:18

MaxMax

9211319

asked Feb 11 at 16:18

MaxMax

9211319

1 Answer
1

active

oldest

votes

2

$begingroup$

Here is a general way to obtain it. Your unique polynomial will have to be of degree at most $n$, so assume we have
$$p(x) = a_0 + a_1x + +ldots + a_nx^n = sum_k=1^n a_k x^k.$$

Then, plugging in your points, you get the equations
$$
y_i = sum_k=1^n a_k x_i^k, quad forall i in [n].
$$
This is a linear system of $n+1$ equations in $n+1$ unknowns, guaranteed to be invertible due to the nature of the system. Hence, there is a unique solution, which you can find by Gaussian Elimination, for example.

UPDATE - EXAMPLE

Perhaps an example will help. Consider finding the unique linear polynomial through the points $(0,0)$ and $(1,1)$. We assume the form $p(x) = a_0 + a_1x$ which leads to the equations
$$
beginsplit
0 = a_0 + a_1 cdot 0\
1 = a_0 + a_1 cdot 1
endsplit
$$
and the first equation simplifies to $a_0=0$, plugging into the second yields $a_1=1$, so we have $$p(x) = 0 + 1 cdot x = x,$$
which is, of course, the unique line through $(0,0)$ and $(1,1)$, as expected.

share|cite|improve this answer

edited Feb 11 at 16:33

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you define $a_k$ specifically?
  $endgroup$
  – Max
  Feb 11 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max you are not defining $a_k$, they are variables for which you are solving
  $endgroup$
  – gt6989b
  Feb 11 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, instead of define, I meant how do you get $a_k$ directly (so a formula for $a_k$).
  $endgroup$
  – Max
  Feb 11 at 16:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max see update for an example. One can use something like Cramer's Rule if you insist on calculating coefficients directly via a formula instead of a general algorithmic procedure.
  $endgroup$
  – gt6989b
  Feb 11 at 16:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or both... a elegant closed form with cubic computation :)
  $endgroup$
  – Max
  Feb 12 at 16:08
  

  
  
  
  

 | 
show 7 more comments

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2

$begingroup$

Here is a general way to obtain it. Your unique polynomial will have to be of degree at most $n$, so assume we have
$$p(x) = a_0 + a_1x + +ldots + a_nx^n = sum_k=1^n a_k x^k.$$

Then, plugging in your points, you get the equations
$$
y_i = sum_k=1^n a_k x_i^k, quad forall i in [n].
$$
This is a linear system of $n+1$ equations in $n+1$ unknowns, guaranteed to be invertible due to the nature of the system. Hence, there is a unique solution, which you can find by Gaussian Elimination, for example.

UPDATE - EXAMPLE

Perhaps an example will help. Consider finding the unique linear polynomial through the points $(0,0)$ and $(1,1)$. We assume the form $p(x) = a_0 + a_1x$ which leads to the equations
$$
beginsplit
0 = a_0 + a_1 cdot 0\
1 = a_0 + a_1 cdot 1
endsplit
$$
and the first equation simplifies to $a_0=0$, plugging into the second yields $a_1=1$, so we have $$p(x) = 0 + 1 cdot x = x,$$
which is, of course, the unique line through $(0,0)$ and $(1,1)$, as expected.

share|cite|improve this answer

edited Feb 11 at 16:33

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you define $a_k$ specifically?
  $endgroup$
  – Max
  Feb 11 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max you are not defining $a_k$, they are variables for which you are solving
  $endgroup$
  – gt6989b
  Feb 11 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, instead of define, I meant how do you get $a_k$ directly (so a formula for $a_k$).
  $endgroup$
  – Max
  Feb 11 at 16:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max see update for an example. One can use something like Cramer's Rule if you insist on calculating coefficients directly via a formula instead of a general algorithmic procedure.
  $endgroup$
  – gt6989b
  Feb 11 at 16:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or both... a elegant closed form with cubic computation :)
  $endgroup$
  – Max
  Feb 12 at 16:08
  

  
  
  
  

 | 
show 7 more comments

2

$begingroup$

Here is a general way to obtain it. Your unique polynomial will have to be of degree at most $n$, so assume we have
$$p(x) = a_0 + a_1x + +ldots + a_nx^n = sum_k=1^n a_k x^k.$$

Then, plugging in your points, you get the equations
$$
y_i = sum_k=1^n a_k x_i^k, quad forall i in [n].
$$
This is a linear system of $n+1$ equations in $n+1$ unknowns, guaranteed to be invertible due to the nature of the system. Hence, there is a unique solution, which you can find by Gaussian Elimination, for example.

UPDATE - EXAMPLE

Perhaps an example will help. Consider finding the unique linear polynomial through the points $(0,0)$ and $(1,1)$. We assume the form $p(x) = a_0 + a_1x$ which leads to the equations
$$
beginsplit
0 = a_0 + a_1 cdot 0\
1 = a_0 + a_1 cdot 1
endsplit
$$
and the first equation simplifies to $a_0=0$, plugging into the second yields $a_1=1$, so we have $$p(x) = 0 + 1 cdot x = x,$$
which is, of course, the unique line through $(0,0)$ and $(1,1)$, as expected.

share|cite|improve this answer

edited Feb 11 at 16:33

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you define $a_k$ specifically?
  $endgroup$
  – Max
  Feb 11 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max you are not defining $a_k$, they are variables for which you are solving
  $endgroup$
  – gt6989b
  Feb 11 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, instead of define, I meant how do you get $a_k$ directly (so a formula for $a_k$).
  $endgroup$
  – Max
  Feb 11 at 16:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max see update for an example. One can use something like Cramer's Rule if you insist on calculating coefficients directly via a formula instead of a general algorithmic procedure.
  $endgroup$
  – gt6989b
  Feb 11 at 16:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or both... a elegant closed form with cubic computation :)
  $endgroup$
  – Max
  Feb 12 at 16:08
  

  
  
  
  

 | 
show 7 more comments

$begingroup$

Here is a general way to obtain it. Your unique polynomial will have to be of degree at most $n$, so assume we have
$$p(x) = a_0 + a_1x + +ldots + a_nx^n = sum_k=1^n a_k x^k.$$

Then, plugging in your points, you get the equations
$$
y_i = sum_k=1^n a_k x_i^k, quad forall i in [n].
$$
This is a linear system of $n+1$ equations in $n+1$ unknowns, guaranteed to be invertible due to the nature of the system. Hence, there is a unique solution, which you can find by Gaussian Elimination, for example.

UPDATE - EXAMPLE

Perhaps an example will help. Consider finding the unique linear polynomial through the points $(0,0)$ and $(1,1)$. We assume the form $p(x) = a_0 + a_1x$ which leads to the equations
$$
beginsplit
0 = a_0 + a_1 cdot 0\
1 = a_0 + a_1 cdot 1
endsplit
$$
and the first equation simplifies to $a_0=0$, plugging into the second yields $a_1=1$, so we have $$p(x) = 0 + 1 cdot x = x,$$
which is, of course, the unique line through $(0,0)$ and $(1,1)$, as expected.

share|cite|improve this answer

edited Feb 11 at 16:33

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

$endgroup$

share|cite|improve this answer

edited Feb 11 at 16:33

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557

answered Feb 11 at 16:24

gt6989bgt6989b

35.1k22557