5.11 Cubic Interpolation Method 285Iteration 1
To find the value ofλ ̃∗and to test the convergence criteria, we first computeZandQ
as
Z=3 ( 5. 0 − 113. 0 )
3. 2
− 20. 0 + 350. 688 = 229. 588
Q=[229. 5882 0 +( 2. 0 )( 350. 688 )]^1 /^2 = 442. 0
Hence
λ ̃∗= 3. 2(
− 20. 0 + 229. 588 ± 244. 0
− 20. 0 + 350. 688 + 459. 176
)
= 1 .84 or − 0. 1396By discarding the negative value, we have
λ ̃∗= 1. 84Convergence criterion: Ifλ ̃∗ is close to the true minimum,λ∗, thenf′(λ ̃∗)=
df (λ ̃∗)/dλshouldbe approximately zero. Sincef′= 5 λ^4 − 51 λ^2 − 0, 2
f′(λ ̃∗) = 5 ( 1. 84 )^4 − 51 ( 1. 84 )^2 − 02 = − 13. 0Since this is not small, we go to the next iteration or refitting. Asf′( ̃λ∗) < 0,we take
A=λ ̃∗and
fA= f(λ ̃∗) =( 1. 84 )^5 − 5 ( 1. 84 )^3 − 02 ( 1. 84 )+ 5 = − 41. 70Thus
A= 1. 84 , fA= − 41. 70 , fA′= − 13. 0B= 3. 2 , fB= 131. 0 , fB′= 503. 688A <λ ̃∗< BIteration 2
Z=
3 (− 41. 7 − 113. 0 )
3. 20 − 1. 84
− 13. 0 + 350. 688 = − 3. 312
Q=[(− 3. 312 )^2 + 3 ( 1. 0 )( 350. 688 )]^1 /^2 = 76. 5
Hence
λ ̃∗= 1. 84 + −^13.^0 −^3.^312 ±^67.^5
− 13. 0 + 350. 688 − 6. 624( 3. 2 − 1. 84 )= 2. 05
Convergence criterion:f′(λ ̃∗) = 5. 0 ( 2. 05 )^4 − 51. 0 ( 2. 05 )^2 − 02. 0 = 5. 35Since this value is large, we go the next iteration withB=λ ̃∗= 2. 0 5 [asf′(λ ̃∗) ]> 0
and
fB= ( 2. 05 )^5 − 5. 0 ( 2. 05 )^3 − 02. 0 ( 2. 05 )+ 5. 0 = − 42. 90