MULLER’S METHOD

f(x)=0.5x³-4x²+6x-2

f(5)=62.5-100+30-2=-9.5

f(6)=108-144+36-2=-2

f(7)=171.5-196+42-2=15.5

We need a quadratic that fits these points: (5,-9.5), (6,-2), (7,15.5).

Let the quadratic be ax²+bx+c, then

(1) 25a+5b+c=-9.5

(2) 36a+6b+c=-2

(3) 49a+7b+c=15.5

(4): (2)-(1): 11a+b=7.5, 22a+2b=15

(5): (3)-(1): 24a+2b=25

(6): (5)-(4): 2a=10, a=5, so b=7.5-11a=7.5-55=-47.5, c=-9.5-(25a+5b)=103.

Quadratic is: 5x²-47.5x+103.

The zeroes are (47.5±√(47.5²-2060))/10.

The two values for x are X₁=6.15089 and X₂=3.34911 approx.

f(X₁)=-0.0737, f(X₂)=-7.9888. So f(X₁) is closer to zero than f(X₂) which means that X₁ is closer to a zero of f(x) than X₂. Muller’s Method converts this process into a formula, and I use x₃ iteration as an example:

x₃=x₂-(x₂-x₁)(2c/max(b±√(b²-4ac)),

where:

a=qf(x₂)-q(1+q)f(x₁)+q²f(x₀),

b=(2q+1)f(x₂)-(1+q)²f(x₁)+q²f(x₀),

c=(1+q)f(x₂), and where:

q=(x₂-x₁)/(x₁-x₀).

The same formula applies to x₄ and subsequent iterations of x: just add 1 to the iteration number in the expression.

So x₀=5, x₁=6, x₂=7; q=(7-6)/(6-5)=1.

a=15.5-2(-2)+(-9.5)=15.5+4-9.5=10,

b=3(15.5)+8-9.5=45,

c=2(15.5)=31.

b²-4ac=2025-1240=785, √785=28.01785 approx.

max(b±√(b²-4ac))=45+28.01785=73.01785.

x₃=7-62/73.01785=6.15089 approx and f(x₃)=-0.07373 approx.

The operation is repeated to find the next iteration.

The final value using this method to 4 decimal places is x=6.1563.