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Newton's Divided Difference Method

Has anyone had ay experience in doing theese types?
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Avatar for joostan
joostan
13
Original post by Mihael_Keehl
Has anyone had ay experience in doing theese types?


Got a particular question?
Original post by joostan
Got a particular question?


I have done this type by doing interpolation took forever lmaoo

I have dabbled with this type but it seems largerly muddled:

For this question how do I formulate an equation.

Spoiler

Reply 3
Avatar for joostan
joostan
13
Original post by Mihael_Keehl
I have done this type by doing interpolation took forever lmaoo

I have dabbled with this type but it seems largerly muddled:

For this question how do I formulate an equation.

Spoiler



Label your interpolation values, (xi,fi)(x_i,f_i) for i=1,...,ni=1,. . .,n. Then the polynomial you seek is then a linear combination of the Newton basis polynomials, which are given by: pk+1(x)=∏i=0k(x−xi)p_{k+1}(x)=\displaystyle\prod_{i=0}^k(x-x_i) with p0(x)≡1p_0(x) \equiv 1.
The coefficients are then given by the Newton divided differences which you can calculate, building up from the given functional values fif_i.

By hand these calculations are, needless to say tedious, though not difficult, the benefit of this method is that the recursive algorithm is useful when writing computer programs to execute the interpolation.
(edited 8 years ago)
Original post by joostan
Label your interpolation values, (xi,fi)(x_i,f_i) for i=1,...,ni=1,. . .,n. Then the polynomial you seek is then a linear combination of the Newton basis polynomials, which are given by: ∏i=1n(x−xi)\displaystyle\prod_{i=1}^n(x-x_i).
The coefficients are then given by the Newton divided differences which you can calculate, building up from the given functional values fif_i.

By hand these calculations are, needless to say tedious, though not difficult, the benefit of this method is that the recursive algorithm is useful when writing computer programs to execute the interpolation.


Once you get the values how do you put it into integer form for the polynomial
Reply 5
Avatar for joostan
joostan
13
Original post by Mihael_Keehl
Once you get the values how do you put it into integer form for the polynomial


What do you mean by integer form?
Once you compute the values of the coefficients of each approximant p(x)=A0+∑j=1n(Aj∏i=0j−1(x−xi))p(x)=A_0+\displaystyle\sum_{j=1}^n \left( A_j \displaystyle\prod_{i=0}^{j-1}(x-x_i)\right).
This is itself the polynomial you're looking for. If you so wished, you could expand these out to obtain something of the form: p(x)=∑i=0naixip(x)=\displaystyle\sum_{i=0}^n a_i x^i.
(edited 8 years ago)

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