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5.3.2.1 Golden Section Method
Assume that, by examining the graph of the function under consideration,
you have established the local minimum x min ∈ [a, b]. This means that the
curve of the function is strictly decreasing in the interval [a, x min ] and is
strictly increasing in the interval [x min , b]. Next, choose a number r < 1/2, but
whose precise value will be determined later, and define the internal points c
and d such that:

c = a + r(b – a) (5.22)

d = a + (1 – r)(b – a) (5.23)

and such that a < c < d < b. Next, evaluate the values of the function at c and
d. If we find that f(c) ≥ f(d), we can assert that x min ∈ [c, b]; that is, we narrowed
the external bounds of the interval. (If the inequality was in the other sense,
we could have instead narrowed the outer limit from the right.) If in the sec-
ond iteration, we fix the new internal points such that the new value of c is the
old value of d, then all we have to compute at this step is the new value of d.
If we repeat the same iteration k-times, until the separation between c and d is
smaller than the desired tolerance, then at that point we can assert that:

ck() + dk()
x = (5.24)
min
2
Now, let us determine the value of r that will allow the above iteration to
proceed as described. Translating the above statements into equations, we
desire that:

c()2 = d( )1
(5.25)
⇒ c()2 = a()2 + r b( ( )2 − a( ))2 = a( ) (1 + 1− r b)( ( )1 − a( ))1

a( )2 = c()1 = a()1 + r b( ( )1 − a( ))1 (5.26)

b()2 = b( )1 (5.27)

Now, replacing the values of a(2) and b(2) from Eqs. (5.26) and (5.27) into Eq.
(5.25), we are led to a second-degree equation in r:

2
r – 3r + 1 = 0 (5.28)

The desired root is the value of the Golden ratio:

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