急~高等微積分

Question

Use functional iteration with a computer or a calculator to compute
the fixed point x*(λ) of exercise 1 for λ=1.6 , 1.8 , 2 .
What happens when you try functional iteration to compute the fixed
point when λ=2.5?


exercise1.
Let g(x,λ)=λ乘sinx. We assume that λ>=π/2

*是在x的右下角

Answer 1

本題有無限個fixed point啊 !
應該是求第1個正數固定點吧!

2007-12-06 16:00:55 補充：
可能有很多個fixed point!

2007-12-06 17:12:51 補充：
本題應該是要討論第一個正數fixed point(也才有意義)
設第一個固定點為x*
g(x)=λsinx, g'(x)=λcosx
x*在π/2~π之間,就設x0=2吧!

(1)λ=1.6時
x*=1.59934789085317
g'(2)=-0.665835, g'( x*)=-0.0456763均為負(振盪),且|g'(x)|<1(收歛)
By Excel:
x0=2.00000000000000 , x1=1.45487588292109
x2=1.58926199301176 ,x3=1.59972722308804 , ...
x10=1.59934789085301,x11=1.59934789085318
x12=1.59934789085317 ,x13=1.59934789085317
振盪且收歛,13次iteration得14位精確度

(2)λ=1.8時
x*=1.76586266416983
g'(2)=-0.7491, g'( x*)=-0.3489均為負(振盪),且|g'(x)|<1(收歛)
|g'(x)|<1但比λ=1.6時大,故收歛較慢
By Excel:
x0=2.00000000000000 ,x1=1.63673536828623
x2=1.79608825617589 ,x3=1.75451208161463 , ...
x10=1.76586965168191 ,x20=1.76586266435660
x30=1.76586266416984 ,x31=1.76586266416983
,x32=1.76586266416983
振盪且收歛,32次iteration得14位精確度

(3)λ=2時
x*=1.89549426703398
g'(2)=-0.8323, g'( x*)=-0.638均為負(振盪),且|g'(x)|<1(收歛)
|g'(x)|<1但比λ=1.8時大,故收歛更慢
By Excel:
x0=2.00000000000000 ,x1=1.81859485365136
x2=1.93890945306856 ,x3=1.86601601636051 ,...
x50=1.89549426705324 ,x60=1.89549426703420
x70=1.89549426703398 ,x71=1.89549426703398
振盪且收歛,71次iteration得14位精確度

(4)λ=2.5時
x*大約2.13
g'(2)=-1.040367, g'( x*)=-1.3263均為負(振盪),且|g'(x)|>1(發散)
By Excel:
x0=2.00000000000000 ,x1=2.27324356706420
x2=1.90815835769679 ,x3=2.35907778223717 ,...
x50=1.50650500188341 ,x51=2.49483506134627
x60=1.50650500188298 ,x61=2.49483506134620
x62=1.50650500188298 ,x63=2.49483506134620
振盪且不收歛,63次iteration得14位精確度兩個值