Example 3. Find the Newton polynomial of degree  n=3  that passes through the four points [Graphics:Images/BackSubstitutionMod_gr_142.gif].  

Solution 3.

Set up the four equations to be solved.

[Graphics:../Images/BackSubstitutionMod_gr_143.gif]


[Graphics:../Images/BackSubstitutionMod_gr_144.gif]

[Graphics:../Images/BackSubstitutionMod_gr_145.gif]

[Graphics:../Images/BackSubstitutionMod_gr_146.gif]

[Graphics:../Images/BackSubstitutionMod_gr_147.gif]

 

Solve these equations for the coefficients  [Graphics:../Images/BackSubstitutionMod_gr_148.gif].

[Graphics:../Images/BackSubstitutionMod_gr_149.gif]


[Graphics:../Images/BackSubstitutionMod_gr_150.gif]

If the matrix method  AX = B  is preferred, then we can have Mathematica convert the 4 equations to a 4⨯4 matrix and a column vector.

[Graphics:../Images/BackSubstitutionMod_gr_151.gif]

[Graphics:../Images/BackSubstitutionMod_gr_152.gif]

The first part of this data structure is the matrix A and the second part is the column vector B.

[Graphics:../Images/BackSubstitutionMod_gr_153.gif]




[Graphics:../Images/BackSubstitutionMod_gr_154.gif]

[Graphics:../Images/BackSubstitutionMod_gr_155.gif]

[Graphics:../Images/BackSubstitutionMod_gr_156.gif]

[Graphics:../Images/BackSubstitutionMod_gr_157.gif]

The solution can be found by using our forward substitution subroutine.

[Graphics:../Images/BackSubstitutionMod_gr_158.gif]



[Graphics:../Images/BackSubstitutionMod_gr_159.gif]

[Graphics:../Images/BackSubstitutionMod_gr_160.gif]

[Graphics:../Images/BackSubstitutionMod_gr_161.gif]
[Graphics:../Images/BackSubstitutionMod_gr_162.gif]
[Graphics:../Images/BackSubstitutionMod_gr_163.gif]

If you prefer, you could even let Mathematica do it with the built in procedure LinearSolve.

[Graphics:../Images/BackSubstitutionMod_gr_164.gif]

[Graphics:../Images/BackSubstitutionMod_gr_165.gif]

In any event, we have determined the coefficients of the desired Newton polynomial.

[Graphics:../Images/BackSubstitutionMod_gr_166.gif]



[Graphics:../Images/BackSubstitutionMod_gr_167.gif]

[Graphics:../Images/BackSubstitutionMod_gr_168.gif]

[Graphics:../Images/BackSubstitutionMod_gr_169.gif]
[Graphics:../Images/BackSubstitutionMod_gr_170.gif]
[Graphics:../Images/BackSubstitutionMod_gr_171.gif]
[Graphics:../Images/BackSubstitutionMod_gr_172.gif]

We see that  P[t]  passes through the given points [Graphics:../Images/BackSubstitutionMod_gr_173.gif].

We are done.

We can graph the polynomial, just for fun!

[Graphics:../Images/BackSubstitutionMod_gr_174.gif]


[Graphics:../Images/BackSubstitutionMod_gr_175.gif]

[Graphics:../Images/BackSubstitutionMod_gr_176.gif]
[Graphics:../Images/BackSubstitutionMod_gr_177.gif]
[Graphics:../Images/BackSubstitutionMod_gr_178.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004