Editing Newton polynomial (section)

=== Polynomial interpolation ===
{{Main|Polynomial interpolation}}
For a polynomial <math>p_n</math> of degree less than or equal to n, that interpolates <math>f</math> at the nodes <math>x_i</math> where <math>i = 0,1,2,3,\cdots,n</math>. Let <math>p_{n+1}</math> be the polynomial of degree less than or equal to n+1 that interpolates <math>f</math> at the nodes <math>x_i</math> where <math>i = 0,1,2,3,\cdots,n, n+1</math>. Then <math>p_{n+1}</math> is given by:

<math display="block">p_{n+1}(x) = p_n(x) +a_{n+1}w_n(x) </math>

where <math display="inline">w_n(x) := \prod_{i=0}^n (x-x_i)  </math> and <math display="inline">a_{n+1} :={f(x_{n+1})-p_n(x_{n+1}) \over w_n(x_{n+1})}    </math>.

'''Proof:'''

This can be shown for the case where <math>i = 0,1,2,3,\cdots,n</math>:<math display="block">p_{n+1}(x_i) = p_n(x_i) +a_{n+1}\prod_{j=0}^n (x_i-x_j) = p_n(x_i) </math>and when <math>i = n+1</math>:

<math display="block">p_{n+1}(x_{n+1}) = p_n(x_{n+1}) +{f(x_{n+1})-p_n(x_{n+1}) \over w_n(x_{n+1})}   w_n(x_{n+1}) = f(x_{n+1})  </math>

By the uniqueness of interpolated polynomials of degree less than <math>n+1</math>, <math display="inline">p_{n+1}(x) = p_n(x) +a_{n+1}w_n(x) </math> is the required polynomial interpolation. The function can thus be expressed as:

<math display="inline">p_{n}(x) = a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)+\cdots + a_n(x-x_0)\cdots(x-x_{n-1}) </math>

where the factors <math>a_i</math> are [[divided differences]]. Thus, Newton polynomials are used to provide polynomial interpolation formula of n points.<ref>{{Cite book |last=Epperson |first=James F. |title=An introduction to numerical methods and analysis |date=2013 |publisher=Wiley |isbn=978-1-118-36759-9 |edition=2nd |location=Hoboken, NJ}}</ref>


Taking <math>y_i=f(x_i)</math> for some unknown function in Newton divided difference formulas, if the representation of x in the previous sections was instead taken to be <math>x=x_j+sh</math>, in terms of [[Divided differences#Forward and backward differences|forward differences]], the '''Newton forward interpolation formula''' is expressed as:<math display="block">f(x) \approx N(x)=N(x_j+sh) = \sum_{i=0}^{k}{s \choose i}\Delta^{(i)} f(x_j)
</math>whereas for the same in terms of [[Divided differences#Forward and backward differences|backward differences]], the '''Newton backward interpolation formula''' is expressed as:<math display="block">f(x) \approx N(x) =N(x_j+sh)=\sum_{i=0}^{k}{(-1)}^{i}{-s \choose i}\nabla^{(i)} f(x_j). 
</math>This follows since relationship between divided differences and forward differences is given as:<ref>{{cite book |last1=Burden |first1=Richard L. |url=https://archive.org/details/numericalanalysi00rlbu |title=Numerical Analysis |last2=Faires |first2=J. Douglas |date=2011 |isbn=9780538733519 |edition=9th |page=[https://archive.org/details/numericalanalysi00rlbu/page/n146 129] |publisher=Cengage Learning |url-access=limited}}</ref><math display="block">[y_j, y_{j+1}, \ldots , y_{j+n}] = \frac{1}{n!h^n}\Delta^{(n)}y_j,</math>whereas for backward differences, it is given as:{{Citation needed|date=December 2023}}<math display="block">[{y}_{j}, y_{j-1},\ldots,{y}_{j-n}] = \frac{1}{n!h^n}\nabla^{(n)}y_j.   </math>