Question

F1 If, in the Fibonacci search algorithm. when ,f(p) /(q) we let p = b-- (instead of b-2c), prove that p = q Likewise, if when f(p) > f(q) we let q - a+(a) (instead of a +2e), prove that p q. | (b-a) F1 F2

Answer 1

The expressions of and are developed for this version of the algorithm and it will be seen that the same value is obtained for and .

When $f(p)\leq f(q)$ , we let $p=b-\frac{F_{1}}{F_{2}}(b-a)$ and knowing that $F_{1}=1$ and $F_{2}=2$ are numbers from the Fibonacci series, you have:

$p=b-\frac{F_{1}}{F_{2}}(b-a)=b-\frac{1}{2}(b-a)=b-\frac{b}{2}+\frac{a}{2}$ $=\frac{b}{2}+\frac{a}{2}=\frac{b+a}{2}\ (1)$

$q=a+\frac{F_{1}}{F_{2}}(b-a)=a+\frac{1}{2}(b-a)=a+\frac{b}{2}-\frac{a}{2}$ $=\frac{a}{2}+\frac{b}{2}=\frac{a+b}{2}=\frac{b+a}{2}\ (2)$

Then, of (1) and (2), it is concluded that   $p=q$ .

When , we let $q=a+\frac{F_{1}}{F_{2}}(b-a)$ and knowing that $F_{1}=1$ and $F_{2}=2$ are numbers from the Fibonacci series, you have:

$q=a+\frac{F_{1}}{F_{2}}(b-a)=a+\frac{1}{2}(b-a)=a+\frac{b}{2}-\frac{a}{2}$ $=\frac{a}{2}+\frac{b}{2}=\frac{a+b}{2}=\frac{b+a}{2}\ (3)$ $p=b-\frac{F_{1}}{F_{2}}(b-a)=b-\frac{1}{2}(b-a)=b-\frac{b}{2}+\frac{a}{2}$ $=\frac{b}{2}+\frac{a}{2}=\frac{b+a}{2}\ (4)$

Then, of (3) and (4), it is concluded that   $p=q$ .