1. Muller’s Derivation

2. Muller’s Method
quadratic
y = f(x) x x1 x2 x3

3. Muller’s Method
Start with three points (x1,y1), (x2,y2), (x3,y3) and the quadratic in the form of
  y = d1(x-x2)(x-x3) + c2(x-x3) + y3
y1 – y2
Define c1 =
x1 – x2
Substitute (x2,y2) into the above quadratic ( ), we get
y2 = d1(x2-x2)( x2-x3) + c2(x2-x3) + y3
y2 - y3
= c2
x2 - x3

4. Continue Substitute (x1,y1) into the above quadratic ( ), we get
y1 = d1(x1-x2)( x1-x3) + c2(x1-x3) + y3   
y1 - y3
= d1(x1-x2) + c2
x1 - x3
y1 – y2 + y2 - y3
c2 = d1(x1-x2)
y1 – y2 + y2 - y c2
= d1
(x1 - x3)(x1-x2) (x1-x2)

5. Continue y1 – y2 y2 - y3 (x1-x3)c2
= d1
(x1-x3)(x1-x2) (x1-x3)(x1-x2) (x1-x2)(x1-x3)
(y2 - y3)( x2 - x3)
c x2 - x (x1-x3)c2
= d1
(x1-x3) (x1-x3)(x1-x2) (x1-x2)(x1-x3)

6. Continue c1 c2 (x2-x3) (x1-x3)c2
= d1
(x1-x3) (x1-x3)(x1-x2) (x1-x2)(x1-x3)
c c2 (x2-x1)
= d1
(x1-x3) (x1-x3)(x1-x2)
c c2
= d1
(x1-x3)

7. y = d1(x-x3)(x-x3)+ d1(x3-x2)(x-x3)+ c2(x-x3) + y3
Continue
Consider the quadratic ( )again
y = d1(x-x2)(x-x3) + c2(x-x3) + y3
  y = d1(x-x3+x3-x2)(x-x3) + c2(x-x3) + y3
y = d1(x-x3)(x-x3)+ d1(x3-x2)(x-x3)+ c2(x-x3) + y3
  y = d1(x-x3)2 + d1(x3-x2)(x-x3) + c2(x-x3) + y3
Let s = d1(x3-x2) + c2
then
y = d1(x-x3)2 + s(x-x3) + y3
Solve for a root:
0 = d1(x-x3)2 + s(x-x3) + y3

8. Continue To find the root closest to x3, find x so that z = 1/(x - x3)
is as large as possible.
So solve for z in 0 = d1 + sz + y3z2
__________
z = -s   s2 – 4y3d1
2 y3
_________
z = -s - sign(s) s2 – 4y3d1
So,
x = x
___________
s + sign(s) s2 – 4y3d1

9. Muller’s Method

10. Try Some Examples Solve x3 – x + 2 = 0 >> roots([1 0 -1 2])
ans = i i
>> mullerquick(inline('x^3-x+2'),0,-.5, -1)
x(1) = i
x(2) = i
x(3) = i
x(4) = i
x(5) = i
x(6) = i
x(7) = i

11. Continue >> mullerquick(inline('x^3-x+2'),.5+1i,.6+.9i,.7+.8i)…
x(87) = i
x(88) = i
x(89) = i
x(90) = i
x(91) = i
x(92) = i
x(93) = i