Description
Recently, I have been learning a couse called "Numerical Analysis". The fixed point iteration theory
was introducted to solve the the approximate root of the equations
in this course.
Fixed point iteration theory: $x=Ψ(x) \Rightarrow x_{k+1}=Ψ(x_k)$
Example
Solving the approximate root of $f(x)=x^3+4x^2-10 = 0$
So I give three styles of $x=Ψ(x)$, shown as below:
$Ψ_1(x)=\frac{1}{2}\sqrt{10-x^3}$
$Ψ_2(x)=\sqrt{\frac{10}{x+4}}$
$Ψ_3(x)=x-\frac{f(x)}{f'(x)}=x-\frac{x^3+4x^2-10}{3x^2+8x}$
Using Mathematica to implement it and verify the results of textbook
My trial: I set the precision to 10
result1 = FixedPointList[N[1/2 Sqrt[10 - #^3] &, 10], 1.5`10]
{1.500000000, 1.286953768, 1.402540804, 1.345458374, 1.375170253, 1.360094193, 1.367846968, 1.363887004, 1.365916733, 1.364878217, 1.365410061, 1.365137821, 1.365277209, 1.365205850, 1.365242384, 1.365223680, 1.365233256, 1.365228353, 1.365230863, 1.365229578, 1.365230236, 1.365229899, 1.365230072, 1.365229984, 1.365230029, 1.365230006, 1.365230017, 1.365230011, 1.365230014, 1.365230013, 1.365230014}
result2 = FixedPointList[N[Sqrt[10/(# + 4)] &, 10], 1.5`10]
{1.500000000, 1.348399725, 1.367376372, 1.364957015, 1.365264748, 1.365225594, 1.365230576, 1.365229942, 1.365230023, 1.365230012, 1.365230014, 1.365230013}
result3 = FixedPointList[N[# - (#^3 + 4 #^2 - 10)/(3 #^2 + 8 #) &, 10], 1.5`10]
{1.500000000, 1.373333333, 1.365262015, 1.36523001, 1.36523001}
However, the last two results of the result3
is 1.36523001, 1.36523001
, whose precision is 9
rather than 10
TableForm[
Flatten[{result1, result2, result3}, {{2}, {1}}],
TableHeadings -> {None, Style[#, 15, Red] & /@ {"\[Phi]1 steps", "\[Phi]2 steps",
"\[Phi]3 steps"}}]
As the picture shown, the result2
ia same as the result of textbook
, however, the result1
and result3
are a little different from the result of textbook
. In addition, the precison of result3
is 9
, not 10
.
Question
Can someone give me a explalation about this trial? For me, I cannot understand this result.