First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 125
There are eight equations and eight unknowns. However, a 0 = c 0 =0, so that reduces the
number of equations and unknowns to six. We rewrite the equations below, substituting
a 0 = c 0 =0, and simplifying when possible:
b 0 + d 0 =1
a 1 + b 1 + c 1 + d 1 =1
a 1 +2b 1 +4c 1 +8d 1 =0
b 0 +3d 0 = b 1 +2c 1 +3d 1
3d 0 = c 1 +3d 1
c 1 +6d 1 =0
We will use Julia to solve this system of equations. To do that, we first rewrite the system
of equations as a matrix equation
Ax = v
where
⎡
⎤ ⎡ ⎤ ⎡ ⎤
1 1 0 0 0 0 b 0 1
0 0 1 1 1 1
d 0
1
0 0 1 2 4 8
a 1
0
A =
,x=
,v =
.
1 3 0 −1 −2 −3
b 1
0
⎢
⎣0 3 0 0 −1 −3
⎥ ⎢
⎦ ⎣c ⎥ ⎢
1 ⎦ ⎣0
⎥
⎦
0 0 0 0 1 6 d 1 0
We enter the matrices A, v in Julia and solve the equation Ax = v using the command
A\v.
In [1]: A=[1 1 0 0 0 0 ; 0 0 1 1 1 1; 0 0 1 2 4 8; 1 3 0 -1 -2 -3;
0 3 0 0 -1 -3; 0 0 0 0 1 6]
Out[1]: 6×6 Array{Int64,2}:
1 1 0 0 0 0
0 0 1 1 1 1
0 0 1 2 4 8
1 3 0 -1 -2 -3
0 3 0 0 -1 -3
0 0 0 0 1 6