5.10 Quadratic Interpolation Method 275
provided that
c=
fC+fA− 2 fB
2 t^2
> 0 (5.41)
The inequality (5.41) can be satisfied if
fA+fC
2
>fB (5.42)
(i.e., the function valuefBshould be smaller than the average value offAandfC).
This can be satisfied iffBlies below the line joiningfAandfCas shown in Fig. 5.12.
Thefollowing procedure can be used not only to satisfy the inequality (5.42) but
also to ensure that the minimumλ ̃∗lies in the interval 0< ̃λ∗< 2 t.
1 .Assuming thatfA= f(λ=0) and the initial step sizet 0 are known, evaluate
the functionf atλ=t 0 and obtainf 1 = f(λ=t 0 ) The possible outcomes are.
shown in Fig. 5.13.
2.Iff 1 >fAis realized (Fig. 5.13c), setfC=f 1 and evaluate the functionf at
λ=t 0 /2 andλ ̃∗using Eq. (5.40) witht=t 0 /2.
3 .Iff 1 ≤fAis realized (Fig. 5.13aorb), setfB=f 1 , and evaluate the functionf
atλ= 2 t 0 to findf 2 = f(λ= 2 t 0 ) This may result in any one of the situations.
shown in Fig. 5.14.
4.Iff 2 turns out to be greater thanf 1 (Fig. 5.14borc), setfC=f 2 and compute
λ ̃∗according to Eq. (5.40) witht=t 0.
5 .Iff 2 turns out to be smaller thanf 1 , set newf 1 =f 2 andt 0 = 2 t 0 , and repeat
steps 2 to 4 until we are able to findλ ̃∗.
Stage 3. Theλ ̃∗found in stage 2 is the minimum of the approximating quadratic
h(λ)and we have to make sure that thisλ ̃∗is sufficiently close to the true minimumλ∗
of f(λ)before takingλ∗≃λ ̃∗. Several tests are possible to ascertain this. One possible
test is to comparef (λ ̃∗) ithw h(λ ̃∗) nd considera λ ̃∗a sufficiently good approximation
f(l)
l
Figure 5.12 fBsmaller than (fA+fC)/2.