Fixed Point Iteration
Univ.-Prof. Dr.-Ing. habil. Josef BETTEN
RWTH Aachen University
Mathematical Models in Materials Science and Continuum Mechanics
Augustinerbach 4-20
D-52056 A a c h e n , Germany
<[email protected]>
Abstract
This worksheet is concerned with finding numerical solutions of non-linear equations
in a single unknown. Using MAPLE 12 the fixed-point iteration has been applied to
some examples.
Keywords: zero form and fixed point form; LIPPSCHITZ constant; a-priori and
a-posteriori error estimation; BANACH 's fixed-point theorem
Introduction
A value x = p is called a fixed-point for a given function g(x) if g(p) = p. In finding the
solution x = p for f(x) = 0 one can define functions g(x) with a fixed-point at x = p in
several ways, for example, as g(x) = x - f(x) or as g(x) = x - h(x)*f(x) , where h(x) is a
continuous function not equal to zero within an interval [a, b] considered.
The iteration process is expressed by
> restart:
> x[n+1]:=g(x[n]); # n = 0,1,2,...
xn + 1 := g( xn )
with a selected initial value for n = 0 in the neighbourhood of x = p.
BANACH's Fixed-Point Theorem
Let g(x) be a continuous function in [a, b]. Assume, in addition, that g'(x) exists
on (a, b) and that a constant L = [0, 1) exists with
> restart:
> abs(diff(g(x),x))<=L;
1
d
g( x ) ≤ L
dx
for all x in [a, b]. Then, for any selected initial value in [a, b] the sequence defined by
> x[n+1]:=g(x[n]); # n = 0,1,2,..
xn + 1 := g( xn )
converges to the unique fixed-point p in [a, b]. The constant L is known as LIPPSCHITZ
constant. Based upon the mean value theorem we arrive from the above assumtion at
> abs(g(x)-g(xi))<=L*abs(x-xi);
g( x ) − g( ξ ) ≤ L x − ξ
for all x and xi in [a, b]. The BANACH fixed-point theorem is sometimes called the
contraction mapping principle.
From the fixed-point theorem one can proof the following error estimates:
a-priori error estimate:
> abs(x[k]-p)<=(L^k/(1-L))*abs(x[1]-x[0]); alpha:=rhs(%);
−xk + p ≤
α :=
L k −x1 + x 0
1−L
L k −x1 + x0
1−L
>
The rate of convergence depends on the factor L^k . The smaller the value of L , the faster
the convergence, which is very slow if the LIPPSCHITZ constant L is close to one.
The necessary number of iterations for a given error "epsilon" can be calculated by
the following formula [see equation (8.36) in: BETTEN, J.: Finite Elemente für Ingenieure 2,
zweite Auflage, 2004, Springer-Verlag, Berlin / Heidelberg / New York]:
> iterations[epsilon]>=ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L);
⎛ (1 − L) ε ⎞
⎟
ln⎜⎜
⎟
⎝ −x1 + x0 ⎠
≤ iterationsε
ln( L )
>
a-posteriori error estimate:
> restart:
> abs(x[k]-p)<=(L/(1-L))*abs(x[k]-x[k-1]);
−xk + p ≤
β :=
beta:=rhs(%);
L xk − xk − 1
1−L
L xk − xk − 1
1−L
where
> restart:
2
> alpha>=beta;
β≤α
>
Examples
In the following some examples of fixed point iterations should be discussed.
The first example is concerned with finding the solution f(x) = 0, where
> restart:
> f(x):=x-cos(x);
f( x ) := x − cos( x )
Using the MAPLE command "fsolve" we immediately arrive at the solution:
> Digits:=4; p:=fsolve(f(x)=0);
Digits := 4
p := 0.7391
The fixed point form is x = g(x) with g(x) = cos(x). Thus, one can read the fixed point from the
folowing Figure:
> alias(H=Heaviside,th=thickness,sc=scaling,co=color):
> p[1]:=plot({x,cos(x)},x=0..1,sc=constrained,th=3,co=black):
> p[2]:=plot(0.7391*H(x-0.7391),x=0.7390..0.7392,co=black):
> p[3]:=plot({1,H(x-1)},x=0..1.01,co=black,
title="Fixed Point at x = 0.7391"):
> p[4]:=plot([[0.7391,0.7391]],style=point,symbol=circle,
symbolsize=30):
> plots[display](seq(p[k],k=1..4));
>
The graph y = g(x) is contained in the square { (x , y) | x = [0, 1], y = [0, 1] },
id est: " g " mapps the interval [0, 1] to itself. The fixed point iteration is given as follows:
> starting_point:=x[0]=0.5; x[1]:=evalf(subs(x=0.5,cos(x)));
starting_point := x0 = 0.5
x1 := 0.8776
3
> for i from 2 to 23
do x[i]:=evalf(subs(x=%,cos(x))) od;
x2 := 0.6390
x3 := 0.8027
x4 := 0.6948
x5 := 0.7682
x6 := 0.7192
x7 := 0.7523
x8 := 0.7301
x9 := 0.7451
x10 := 0.7350
x11 := 0.7418
x12 := 0.7373
x13 := 0.7403
x14 := 0.7383
x15 := 0.7396
x16 := 0.7387
x17 := 0.7393
x18 := 0.7389
x19 := 0.7392
x20 := 0.7390
x21 := 0.7391
x22 := 0.7391
x23 := 0.7391
>
Using the starting point x[0] = 0.5 we need 21 iterations in order to arrive at the
solution p = 0.7391. If we choose x[0] = 0 or x[0] = 1 we arrive at the fixed point
after 23 or 22 iterations, respectively.
The next example is concerned with finding the solution f(x) = 0, where
> restart:
> f(x):=x-(1/2)*(sin(x)+cos(x));
1
1
sin( x ) − cos( x )
2
2
Using the MAPLE command "fsolve" we immediately arrive at the solution:
> p:=fsolve(f(x)=0);
f( x ) := x −
p := 0.7048120020
The fixed point form is x = g(x) with g(x) = [sin(x) + cos(x)]/2. Thus, one can read the
fixed point from the following Figure:
> alias(H=Heaviside,th=thickness,sc=scaling,co=color):
4
> p[1]:=plot({x,(sin(x)+cos(x))/2},x=0..1,
sc=constrained,th=3,co=black):
> p[2]:=plot({1,H(x-1),-H(x-1.001),0.7048*H(x-0.7048),
0.7048*H(x-0.7058)},x=0..1.001,co=black,
title="Fixed Point at x = 0.7048"):
> p[3]:=plot([[0.7048,0.7048]],style=point,symbol=circle,
symbolsize=30):
> plots[display](seq(p[k],k=1..3));
>
The fixed point iteration is given as follows:
> g(x):=(sin(x)+cos(x))/2;
> x[0]:=0; x[1]:=evalf(subs(x=0,g(x)),2);
>
g( x ) :=
1
1
sin( x ) + cos( x )
2
2
x0 := 0
x1 := 0.50
> for i from 2 to 10 do x[i]:=evalf(subs(x=%,g(x))) od;
x2 := 0.6785040503
x3 := 0.7030708012
x4 := 0.7047118221
x5 := 0.7048062961
x6 := 0.7048116773
x7 := 0.7048119834
x8 := 0.7048120009
x9 := 0.7048120019
5
x10 := 0.7048120020
We see, the 10th iteration x[10] is identical to the above MAPLE solution.
The following steps are concerned with both the a-priori and the a-posteriori error estimate.
At first we determine the LIPPSCHITZ constant from the absolute derivative of the function g(x):
> absolute_derivative:=abs(Diff(g(xi),xi))=abs(diff(g(x),x));
d
1
1
g( ξ ) = cos( x ) − sin( x )
dξ
2
2
> p[1]:=plot(abs(diff(g(x),x)),x=0..1,sc=constrained,
th=3,co=black):
> p[2]:=plot({H(x),H(x-1),-H(x-1.001)},x=0..1.001,co=black,
title="Absolute Derivative of g(x)"):
> plots[display](p[1],p[2]);
absolute_derivative :=
The greatest value of the absolute derivative in the neighbourhood of the expected fixed point,
here in the interval x = [0, 1], may be assumed to be as the LIPPSCHITZ constant:
> L:=evalf(subs(x=0,abs(diff(g(x),x))),2);
L := 0.50
Assuming " i " iterations, then the a-priori error estimate is given by:
> for i from 1 by 3 to 10 do
a_priori_estimate[i]:=evalf((L^i/(1-L))*abs(x[1]-x[0]),5) od;
a_priori_estimate1 := 0.50000
a_priori_estimate4 := 0.062500
a_priori_estimate7 := 0.0078125
a_priori_estimate10 := 0.00097655
>
The necessary number of iterations for a given error " epsilon " can be calculated by
the formula on page 2:
> iterations[epsilon]>=ln((1-L_)*epsilon/abs(X[1]-X[0]))/ln(L_);
⎛ ( 1 − L_ ) ε ⎞
⎟
ln⎜⎜
⎟
⎝ −X1 + X0 ⎠
≤ iterationsε
ln( L_ )
> iterations[epsilon]:=
evalf(ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L),3);
6
iterationsε := −1.44 ln( 1.00 ε )
> for i from 1 to 5 do iterations[epsilon=10^(-i)]:=
evalf(ln((1-L)*10^(-i)/abs(x[1]-x[0]))/ln(L),2) od;
iterationsε = 1 / 10 := 3.3
iterationsε = 1 / 100 := 6.7
iterationsε = 1 / 1000 := 10.
iterationsε = 1 / 10000 := 13.
iterationsε = 1 / 100000 := 17.
>
The a-posteriori error estimate after the 10th iteration is given by:
> a_posteriori_estimate[k]:=(L_/(1-L_))*abs(x[k]-x[k-1]);
a_posteriori_estimatek :=
L_ xk − xk − 1
1 − L_
> x[k]:=x[10]; x[k-1]:=x[9];
xk := 0.7048120020
xk − 1 := 0.7048120019
> a_posteriori_estimate[10]:=
subs({k=10,L_=L},a_posteriori_estimate[k]);
a_posteriori_estimate10 := 0.1000000000 10-9
>
The 10th iteration is identical to the MAPLE solution: x[10] = p. The necessary
a-priori-iterations with the above a-posteriori error = 0.1*10^(-9) is given by:
> a_priori_iterations[epsilon]:=-1.44*ln(1.00*epsilon);
a_priori_iterationsε := −1.44 ln( 1.00 ε )
> a_priori_iterations[evalf(0.1*10^(-9),2)]:=
evalf(subs(epsilon=0.1*10^(-9),%),3);
a_priori_iterations
0.10 10
-9
:= 33.1
>
The necessary a-priori-iterations can alternatively expressed as follows:
> for k in [0.1,0.01,0.001,0.0001,0.000001,0.0000000001] do
a_priori_iterations[epsilon=k]:=evalf(-1.44*ln(k),2) od;
a_priori_iterationsε = 0.1 := 3.2
a_priori_iterationsε = 0.01 := 6.4
a_priori_iterationsε = 0.001 := 9.7
a_priori_iterationsε = 0.0001 := 13.
a_priori_iterations
a_priori_iterations
ε = 0.1 10
-5
:= 20.
ε = 0.1 10
-9
:= 32.
>
Another example is concerned with the zero form f(x) = 0, where
7
> restart:
> f(x):=1+cosh(x)*cos(x);
f( x ) := 1 + cosh( x ) cos( x )
The MAPLE command "fsolve" immediately furnishes the solution:
> p:=fsolve(f(x)=0);
p := 1.875104069
The fixed point form is x = g(x) with g(x) = x - h(x)*f(x). In this example we have
to introduce a continuous function h(x) according to BANACH's theorem. A suitable
function is given by:
> h(x):=-exp(-x);
h( x ) := −e
( −x )
> g(x):=x-h(x)*f(x);
( −x )
g( x ) := x + e
( 1 + cosh( x ) cos( x ) )
Thus, one can read the fixed point from the following Figure:
> alias(H=Heaviside,sc=scaling,th=thickness,co=color):
> p[1]:=plot({x,g(x)},x=1..3,sc=constrained,th=3,
xtickmarks=5,ytickmarks=4,co=black):
> p[2]:=plot({3,3*H(x-3)},x=1..3.001,1..3,co=black,
title="Fixed Point at x = 1.8751"):
> p[3]:=plot({2+H(x-1.5)-1.5*H(x-1.5)},x=1.49..1.5001,1.5..2,
linestyle=4,th=2,co=black):
> p[4]:=plot({2+H(x-2)-1.5*H(x-2)},x=1.99..2.001,1.5..2,
linestyle=4,th=2,co=black):
> p[5]:=plot({1.5,2},x=1.5..2,linestyle=4,th=2,co=black):
> p[6]:=plot([[1,1],[3,1],[3,3],[1,3],
[1.5,1.5],[2,1.5],[2,2],[1.5,2],[1.8751,1.8751]],
style=point,symbol=circle,symbolsize=30):
> plots[display](seq(p[k],k=1..6));
8
>
In this Figure we have drawn two squares, in which the fixed point x = 1.8751 is contained.
Both x and y = g(x) are elements of [a, b] = [1, 3] or, alternatively, of [a, b] = [1.5, 2].
That is: the operator " g " maps an interval [a, b] into itself, valid for both squares.
In the following the fixed point iteration has been discussed based upon the two squares.
We will see that the rate of convergence is faster by considering the smaller square. Firstly,
let us discuss the iteration based upon the greater square. After that we compare these results
with the rate of convergence based upon the smaller square.
The LIPPSCHITZ constant can be determined from the absolute derivative of the function g(x):
> absolute_derivative:=abs(Diff(g(xi),xi))=abs(diff(g(x),x)):
> simplify(%):
> subs({sinh(x)=(exp(x)-exp(-x))/2,cosh(x)=(exp(x)+exp(-x))/2},%):
> simplify(%);
>
>
>
>
>
d
1
1 ( −2 x )
( −x )
( −2 x )
g( ξ ) = −1 + e
+e
cos( x ) + sin( x ) + e
sin( x )
dξ
2
2
p[1]:=plot(rhs(%),x=1..3,th=3,co=black):
p[2]:=plot({1,H(x-3)},x=1..3.001,co=black,
title="Absolute Derivative of g(x)"):
p[3]:=plot({0.3651,0.3651*H(x-1.8751)},x=1..1.8752,
linestyle=4,co=black):
p[4]:=plot([[1.8751,0.3651]],style=point,
symbol=circle,symbolsize=30):
plots[display](seq(p[k],k=1..4));
>
The greatest value of the derivative in this Figure can be considered as the
LIPPSCHITZ constant:
> L:=evalf(subs(x=3,abs(diff(g(x),x))),4);
L := 0.8824
Assuming the starting point x[0] = 2, we arrive at the following fixed point iteration:
> x[0]:=2; x[1]:=evalf(subs(x=2,g(x)));
x0 := 2
x1 := 1.923450867
> for i from 2 to 21 do x[i]:=evalf(subs(x=%,g(x))) od;
x2 := 1.893171371
x3 := 1.881762275
9
x4 := 1.877544885
x5 := 1.875997102
x6 := 1.875430574
x7 := 1.875223412
x8 := 1.875147687
x9 := 1.875120010
x10 := 1.875109895
x11 := 1.875106198
x12 := 1.875104847
x13 := 1.875104353
x14 := 1.875104173
x15 := 1.875104107
x16 := 1.875104083
x17 := 1.875104074
x18 := 1.875104071
x19 := 1.875104070
x20 := 1.875104069
x21 := 1.875104069
> alpha:=Lambda^k*abs(xi[1]-xi[0])/(1-Lambda);
Λ −ξ 1 + ξ 0
k
α :=
1−Λ
> alpha:=evalf(subs({k=20,Lambda=L,xi[0]=x[0],xi[1]=x[1]},%),4);
α := 0.05363
We see, the 20th iteration x[20] is identical to the MAPLE solution p = 1.875104069.
The following steps illustrade both the a-priori and the a-posteriori error estimate.
Assuming " i " iterations, then the a-prori estimate is given by:
> for i in [1,5,10,15,20,30,40,50] do
a_priori_estimate[i]:=evalf(L^i*abs(x[1]-x[0])/(1-L),4) od;
a_priori_estimate1 := 0.5777
a_priori_estimate5 := 0.3503
a_priori_estimate10 := 0.1874
a_priori_estimate15 := 0.1003
a_priori_estimate20 := 0.05363
a_priori_estimate30 := 0.01535
a_priori_estimate40 := 0.004392
a_priori_estimate50 := 0.001257
10
>
The necessary number of iterations for a given error " epsilon " can be calculated by
the formula on page 2:
> iterations[epsilon]:=
evalf(ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L),4);
iterationsε := −7.994 ln( 1.527 ε )
> for i in
[0.5777,0.3503,0.1003,0.05363,0.01535,0.004392,0.001257] do
iterations[epsilon=i]:=
evalf(subs(epsilon=i,iterations[epsilon]),2) od;
iterationsε = 0.5777 := 1.1
iterationsε = 0.3503 := 5.2
iterationsε = 0.1003 := 15.
iterationsε = 0.05363 := 20.
iterationsε = 0.01535 := 30.
iterationsε = 0.004392 := 40.
iterationsε = 0.001257 := 50.
>
The a-posteriori error estimate after the 20th iteration is given by:
> x[20]:=1.875104069;
x[19]:=1.875104070;
x20 := 1.875104069
x19 := 1.875104070
> beta:=a_posteriori_estimate[20]=L*abs(x[20]-x[19])/(1-L);
β := a_posteriori_estimate20 = 0.7503401361 10-8
>
Iserting this value into the a-priori estimate, we find the following number of necessary
iterations:
> a_priori_iterations[epsilon=evalf(0.75*10^(-8),2)]:=
evalf(subs(epsilon=0.75*10^(-8),iterations[epsilon]),3);
a_priori_iterations
ε = 0.75 10
-8
:= 146.
>
In the above discussion we have considered a range x = [1, 3] , where the LIPPSCHITZ constant
is L = [0, 1) . Besides the fixed point p = 1.875104069 the function f(x) = 1 + cosh(x)*cos(x)
has
the following zeros:
> for i from 1 to 8 do ZERO[i-1..i]:=fsolve(f(x)=0,x,i-1..i) od;
ZERO0 .. 1 := fsolve( 1 + cosh( x ) cos( x ) = 0, x, 0 .. 1 )
ZERO1 .. 2 := 1.875104069
ZERO2 .. 3 := fsolve( 1 + cosh( x ) cos( x ) = 0, x, 2 .. 3 )
11
ZERO3 .. 4 := fsolve( 1 + cosh( x ) cos( x ) = 0, x, 3 .. 4 )
ZERO4 .. 5 := 4.694091133
ZERO5 .. 6 := fsolve( 1 + cosh( x ) cos( x ) = 0, x, 5 .. 6 )
ZERO6 .. 7 := fsolve( 1 + cosh( x ) cos( x ) = 0, x, 6 .. 7 )
ZERO7 .. 8 := 7.854757438
> p[1]:=plot(f(x),x=0..5,-20..10,th=3,co=black):
> p[2]:=plot({-20,10,10*H(x-5),-20*H(x-5)},x=0..5.001,co=black,
title="f(x) := 1 + cosh(x)*cos(x)"):
> plots[display](seq(p[k],k=1..2));
>
The absolute derivative of the function g(x) in the interval x = [0, 5] is illustrated in
the following Figure:
> p[1]:=plot(abs(diff(g(x),x)),x=0..5,0..1.5,th=3,co=black):
> p[2]:=plot({1,1.5,1.5*H(x-5),1.5*H(x-3.22)},x=0..5.001,co=black,
title="Absolute Derivative of g(x)"):
> plots[display](p[1],p[2]);
> x[derivative=1]:=evalf(fsolve(abs(diff(g(x),x))=1,x,-1..5),3);
xderivative = 1 := 3.22
>
We see, in the above considered interval x = [1, 3] the absolute derivative of the function g(x) is
smaller than one. In this range we find the fixed point p = 1.875104069. Note, instead of the
interval x = [1, 3] it would be possible and more comfortable to consider the range x = [1.5, 2] ,
12
for instance. Then we arrive at the following detals, in contrast to the foregoing results:
> restart:
> f(x):=1+cosh(x)*cos(x);
f( x ) := 1 + cosh( x ) cos( x )
> p:=fsolve(f(x)=0);
p := 1.875104069
> h(x):=-exp(-x);
h( x ) := −e
( −x )
> g(x):=x-h(x)*f(x);
( −x )
>
>
>
>
>
>
g( x ) := x + e
( 1 + cosh( x ) cos( x ) )
alias(H=Heaviside,sc=scaling,th=thickness,co=color):
p[1]:=plot({x,g(x)},x=1.5..2,1.5..2,
sc=constrained,th=3,co=black,
title="Fixed Point at x = 1.8751"):
p[2]:=plot({2,2*H(x-2)},x=1.5..2.001,1.5..2,
xtickmarks=5,ytickmarks=4,co=black):
p[3]:=plot(1.8751*H(x-1.8751),x=1.8750..1.8752,1.5..1.8751,
co=black):
p[4]:=plot([[1.8751,1.8751]],style=point,symbol=circle,
symbolsize=30):
plots[display](seq(p[k],k=1..4));
>
The LIPPSCHITZ constant can be determined from the absolute derivative | g'(x) | .
> absolute_derivative:=abs(Diff(g(xi),xi))=
simplify(abs(diff(g(x),x))):
> simplify(subs({sinh(x)=(exp(x)-exp(-x))/2,
cosh(x)=(exp(x)+exp(-x))/2},%));
d
1
1 ( −2 x )
( −x )
( −2 x )
g( ξ ) = −1 + e
cos( x ) + sin( x ) + e
sin( x )
+e
dξ
2
2
> p[1]:=plot(rhs(%),x=1.5..2,0..0.5,th=3,co=black):
13
> p[2]:=plot({0.5,H(x-2)},x=1.5..2.001,co=black,
title="Absolute Derivative | g'(x) |"):
> p[3]:=plot(0.3651*H(x-1.8751),x=1.8750..1.8752,0..0.3651,
linestyle=4,co=black):
> p[4]:=plot([[1.8751,0.3651]],style=point,symbol=circle,
symbolsize=30):
> p[5]:=plot(0.3651,x=1.5..1.8751,linestyle=4,co=black):
> plots[display](seq(p[k],k=1..5));
>
The greatest value of the derivative in this Figure can be considered as LIPPSCHITZ' s constant:
> L:=evalf(subs(x=2,abs(diff(g(x),x))),4);
L := 0.4090
in contrast to the foregoing value L = 0.8824. Inserting the fixed point p = 1.8751 into the
absolute derivative, one arrives at an optimal value:
> L[opt]:=evalf(subs(x=1.8751,abs(diff(g(x),x))),4);
Lopt := 0.3651
>
However, the fixed point " p " is "not yet" known. Thus, the following results have been based
upon the value L = 0.4090.
From the foregoing fixed point iteration we know the following data:
> x[0]:=2;
x[1]:=evalf(subs(x=2,g(x)));
x[19]:=1.875104070;
x[20]:=1.875104069; L:=0.4090;
x0 := 2
x1 := 1.923450867
x19 := 1.875104070
x20 := 1.875104069
L := 0.4090
>
We see, the 20th iteration x[20] is identical to the MAPLE solution finding by using the MAPLE
command fsolve. With the improved LIPPSCHITZ constant L = 0.4090 in contrast to L = 0.8824
we calculate:
> alpha:=Lambda^k*abs(xi[1]-xi[0])/(1-Lambda);
14
Λ −ξ 1 + ξ 0
k
α :=
1−Λ
> alpha:=evalf(subs({k=20,Lambda=L,xi[0]=x[0],xi[1]=x[1]},%),2);
α := 0.30 10-8
in contrast to alpha = 0.05363 of the foregoing calculation based upon L = 0.8824.
The following steps illustrade both the a-priori and the a-posteriori error estimate.
Assuming " i " iterations, then the a-priori estimate is given by:
> for i in [1,5,10,15,20] do
a_priori_estimate[i]:=evalf(L^i*abs(x[1]-x[0])/(1-L),2) od;
a_priori_estimate1 := 0.069
a_priori_estimate5 := 0.0020
a_priori_estimate10 := 0.000022
a_priori_estimate15 := 0.27 10-6
a_priori_estimate20 := 0.30 10-8
>
The last value is identical to alpha.
The necessary number of iterations for a tolerated error " epsilon " can be calculated
by the formula on page 2:
> iterations[epsilon]:=
evalf(ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L),4);
iterationsε := −1.119 ln( 7.675 ε )
> for i in [0.069,0.002,0.000022,evalf(0.27*10^(-6),2),
evalf(0.3*10^(-8),2)] do iterations[epsilon=i]:=
evalf(subs(epsilon=i,iterations[epsilon]),2) od;
iterationsε = 0.069 := 0.69
iterationsε = 0.002 := 4.6
iterationsε = 0.000022 := 9.6
iterations
iterations
ε = 0.27 10
-6
:= 14.
ε = 0.30 10
-8
:= 20.
>
The a-posteriori error estimate after the 20th iteration is given by:
> beta:=Lambda*abs(xi[k]-xi[k-1])/(1-Lambda);
β :=
Λ ξk − ξ k − 1
1−Λ
> beta:=
simplify(subs({k=20,Lambda=L,xi[k]=x[20],xi[k-1]=x[19]},%));
β := 0.6920473773 10-9
> beta:=evalf(%,2);
β := 0.69 10
> Q:=Alpha/Beta=evalf(alpha/beta,3);
-9
15
Q :=
Α
= 4.35
Β
>
Inserting the value of beta = 0.692*10^(-9) into the a-priori estimate, we find again the
necessary iterations:
> a_priori_iterations[epsilon=evalf(0.69*10^(-9),2)]:=
evalf(subs(epsilon=0.69*10^(-9),iterations[epsilon]),2);
a_priori_iterations
ε = 0.69 10
-9
:= 21.
>
Comparing the above results, based upon L = 0.8824 and L = 0.4090, we see that the rate
of convergence essentially depends on the factor L^k in the formula for alpha. The smaller the
value of L , the faster the convergence, which is very slow if the LIPPSCHITZ constant L is
close to one.
The next example is concerned with the zero form f(x) = x - exp(x^2-2) = 0.
> restart:
> f(x):=x-exp(x^2-2); g(x):=x-f(x);
f( x ) := x − e
2
(x − 2)
2
(x − 2)
g( x ) := e
The MAPLE command "fsolve" immediately furnishes the solution:
> p:=fsolve(f(x)=0);
p := 0.1379348256
The fixed point form is x = g(x) with g(x) = exp(x^2-2). Thus, one can read the fixed point
from the following Figure:
> alias(H=Heaviside,sc=scaling,th=thickness,co=color):
> p[1]:=plot({x,g(x)},x=0.1..0.2,0.1..0.2,
sc=constrained,th=3,co=black):
> p[2]:=plot(0.1379*H(x-0.1379),x=0.137..0.138,co=black):
> p[3]:=plot({0.2,0.2*H(x-0.2)},x=0.1..0.2001,co=black,
title="Fixed Point at x = 0.1379"):
> p[4]:=plot([[0.1379,0.1379]],style=point,symbol=circle,
symbolsize=30):
> plots[display](seq(p[k],k=1..4));
16
>
The fixed point iteration is given as follows:
> x[0]:=0.12;
x[1]:=evalf(subs(x=0.12,g(x)));
x0 := 0.12
x1 := 0.1372982105
> for i from 2 to 7 do x[i]:=evalf(subs(x=%,g(x))) od;
x2 := 0.1379106591
x3 := 0.1379339061
x4 := 0.1379347905
x5 := 0.1379348242
x6 := 0.1379348256
x7 := 0.1379348256
We see, the 6th iteration x[6] is identical to the above MAPLE solution.
The following steps are concerned with both the a-priori and the a-posteriori error estimate.
At first we determine the LIPPSCHITZ constant from the absolute derivative | g'(x) | .
> absolute_derivative:=abs(Diff(g(xi),xi))=diff(g(x),x);
>
>
>
>
>
>
2
d
(x − 2)
absolute_derivative :=
g( ξ ) = 2 x e
dξ
p[1]:=plot(diff(g(x),x),x=0.1..0.2,0..0.06,th=4,co=black):
p[2]:=plot({0.06,0.06*H(x-0.2)},x=0.1..0.2001,co=black,
title="Absolute Derivative | g'(x) |"):
p[3]:=plot([[0.1379,0.038]],style=point,symbol=circle,
symbolsize=30):
p[4]:=plot(0.038,x=0.1..0.1379,linestyle=4,co=black):
p[5]:=plot(0.038*H(x-0.1379),x=0.137..0.138,linestyle=4,
co=black):
plots[display](seq(p[k],k=1..5));
17
The greatest value of the absolute derivative in the neighbourhood of the expected fixed point,
here in the interval x = [0.1, 0.2], may be assumed to be as the LIPPSCHITZ constant:
> L:=evalf(subs(x=0.2,diff(g(x),x)),2);
L := 0.056
Assuming " i " iterations, then the a-priori error estimate is given by:
> for i from 1 to 6 do a_priori_estimate[i]:=
evalf((L^i/(1-L))*abs(x[1]-x[0]),2) od;
a_priori_estimate1 := 0.0012
a_priori_estimate2 := 0.000066
a_priori_estimate3 := 0.38 10-5
a_priori_estimate4 := 0.20 10-6
a_priori_estimate5 := 0.12 10-7
a_priori_estimate6 := 0.66 10-9
>
The necessary number of iterations for a given error "epsilon" can be calculated by the formula
on page 2:
> iterations[epsilon]>=
ln((1-Lambda)*epsilon/abs(xi[1]-xi[0]))/ln(Lambda);
⎛ (1 − Λ) ε ⎞
⎟
ln⎜⎜
⎟
−
ξ
+
ξ
⎝
1
0 ⎠
≤ iterationsε
ln( Λ )
> iterations[epsilon]:=
evalf(subs({Lambda=L,xi[1]=x[1],xi[0]=x[0]},lhs(%)),4);
iterationsε := −0.3470 ln( 54.56 ε )
> for i in[0.0012,0.000066,evalf(0.38*10^(-5),2),
evalf(0.20*10^(-6),2),evalf(0.12*10^(-7),2),
evalf(0.66*10^(-9),2)] do iterations[epsilon=i]:=
evalf(subs(epsilon=i,-0.347*ln(54.56*epsilon)),2) od;
iterationsε = 0.0012 := 0.94
iterationsε = 0.000066 := 2.0
18
iterations
iterations
iterations
iterations
ε = 0.38 10
-5
:= 3.0
ε = 0.20 10
-6
:= 3.8
ε = 0.12 10
-7
:= 4.9
ε = 0.66 10
-9
:= 6.0
>
The a-posteriori error estimate after the 6th iteriation is given by:
> x[5]:=0.1379348242; x[6]:=0.1379348256; L:=0.056;
x5 := 0.1379348242
x6 := 0.1379348256
L := 0.056
> beta:=a_posteriori_estimate[6]=L*abs(x[6]-x[5])/(1-L);
β := a_posteriori_estimate6 = 0.8305084746 10-10
> alpha:=Lambda^k*abs(xi[1]-xi[0])/(1-Lambda);
Λ −ξ 1 + ξ 0
k
α :=
> x[0]:=0.12;
1−Λ
x[1]:=0.1372982105; L:=0.056;
x0 := 0.12
x1 := 0.1372982105
L := 0.056
> alpha:=a_priori_error_estimate[6]=L^6*abs(x[1]-x[0])/(1-L);
α := a_priori_error_estimate6 = 0.5651416893 10-9
> Q:=Alpha/Beta=evalf(0.565142*10^(-9)/(0.83051*10^(-10)),4);
Q :=
Α
= 6.805
Β
>
Inserting the value of beta = 0.8305*10^(-10) into the a-priori estimate, we find again the
necessary iterations:
> a_priori_iterations[epsilon=evalf(0.805*10^(-10),2)]:=
evalf(subs(epsilon=0.8305*10^(-10),-0.347*ln(54.56*epsilon)),1);
a_priori_iterations
ε = 0.80 10
-10
:= 6.
>
In the last example the convergence is very fast, since the LIPPSCHITZ constant L = 0.056 is very
small.
>
LEGENDRE Polynomials
Evaluating integrals numerically the GAUSS - LEGENDRE quadrature is very convenient.
In order to apply this method one need the so called GAUSS points, which are identical to
the zeros of the LEGENDRE polynomials [BETTEN, J.: Finite Elemente für Ingenieure 2,
19
zweite Auflage, Springer-Verlag, Berlin / Heidelberg / New York 2004].
Using the MAPLE package orthopoly we immediately find the LEGENDRE polynomials:
> restart:
> with(orthopoly):
> Legendre[n]:=P(n,x);
Legendren := P( n, x )
> for i in [0,1,4,5,8] do Legendre[i]:=P(i,x) od;
Legendre0 := 1
Legendre1 := x
3 35 4 15 2
x −
x
+
8 8
4
63 5 35 3 15
Legendre5 :=
x −
x +
x
8
4
8
Legendre4 :=
Legendre8 :=
35 6435 8 3003 6 3465 4 315 2
+
x −
x +
x −
x
128 128
32
64
32
>
the zeros of which in the interval (0, 1) are given as follows:
> for i in [1,4,5,8] do ZERO[i][0..1]:=
fsolve(P(i,x)=0,x,0..1) od;
ZERO1
:= 0.
0 .. 1
ZERO4
:= 0.3399810436, 0.8611363116
0 .. 1
ZERO5
:= 0., 0.5384693101, 0.9061798459
0 .. 1
ZERO8
:= 0.1834346425, 0.5255324099, 0.7966664774, 0.9602898565
0 .. 1
>
The roots of the LEGENDRE polynomials lie in the interval (-1, 1) and have a symmetry with
respect to the origin as has been illustrated in the following Figure:
> alias(H=Heaviside,sc=scaling,th=thickness,co=color):
> p[1]:=plot({P(1,x),P(3,x),P(4,x),P(5,x)},x=-1..1,-1..1,
sc=constrained,th=3,co=black,xtickmarks=4,ytickmarks=4,
title="LEGENDRE Polynomials P(n,x)"):
> p[2]:=plot({-1,1,H(x-1),-H(x-1)},x=0..1.001,co=black):
> p[3]:=plot({H(x+1),-H(x+1)},x=-1.001..0,co=black):
> plots[display](seq(p[k],k=1..3));
20
>
Instead of using the MAPLE command fsolve in finding the zeros of the LEGENDRE
polinomials we are going to use the fixed point iteration in the following.
In the first example we consider the polynomial P(4,x):
> restart:
> with(orthopoly):
> Legendre[P][4]:=P(4,x);
LegendreP :=
4
3 35 4 15 2
x −
x
+
8 8
4
> f(x):=8*%;
f( x ) := 3 + 35 x4 − 30 x2
> ZERO[0..1]:=fsolve(f(x)=0,x,0..1);
ZERO0 .. 1 := 0.3399810436, 0.8611363116
>
From the polynomial f(x) we read the following fixed point forms:
> g[1](x):=x+f(x);
g1( x ) := x + 3 + 35 x4 − 30 x2
> g[2](x):=sqrt(1/10+7*x^4/6);
g2( x ) :=
90 + 1050 x4
30
> g[4](x):=(6*x^2/7-3/35)^(1/4);
⎛ 6 x2 3 ⎞
g4( x ) := ⎜⎜
− ⎟⎟
⎝ 7
35 ⎠
(1 / 4)
>
One can show that only g[2] is compatible with BANACH's fixed-point theorem.
> g(x):=g[2](x);
g( x ) :=
90 + 1050 x4
30
> alias(H=Heaviside,sc=scaling,th=thickness,co=color):
> p[1]:=plot({x,g(x)},x=0.3..0.4,0.3..0.4,sc=constrained,
21
>
>
>
>
xtickmarks=4,ytickmarks=4,th=3,co=black,
title="Fixed Point at x = 0.34"):
p[2]:=plot({0.4,0.4*H(x-0.4)},x=0.3..0.4001,co=black):
p[3]:=plot(0.34*H(x-0.34),x=0.339..0.3401,co=black):
p[4]:=plot([[0.34,0.34]],style=point,symbol=circle,
symbolsize=30):
plots[display](seq(p[k],k=1..4));
>
This Figure shows that the operator " g " maps the interval x = [0.3, 0.4] into itself :
The graph y = g(x) is contained in the square Q = { (x, y) | x = [0.3, 0.4], y = [0.3, 0.4] }.
> abs(Diff(g(xi),xi))=abs(diff(g(x),x));
d
g( ξ ) = 70
dξ
>
>
>
>
>
x3
90 + 1050 x4
p[1]:=plot(rhs(%),x=0.3..0.4,th=3,co=black,
xtickmarks=3,ytickmarks=3,
title="Absolute Derivative | g'(x) |"):
p[2]:=plot({0.6,0.6*H(x-0.4)},x=0.3..0.4001,0..0.6,co=black):
p[3]:=plot([[0.34,subs(x=0.34,abs(diff(g(x),x)))]],
style=point,symbol=circle,symbolsize=30):
p[4]:=plot((subs(x=0.34,abs(diff(g(x),x)))*H(x-0.34),
x=0.3..0.34001,co=black)):
plots[display](seq(p[k],k=1..4));
>
We see, in the neighbourhood of the expected fixed point x = p the absolute derivative is
less than one.
The LIPPSCHITZ constant can be assumed to be the greatest value in (p-delta, p+delta):
22
> L:=(p-delta,p+delta)[max];
L := ( p − δ, p + δ )
max
> L:=evalf(subs(x=0.4,abs(diff(g(x),x))),2);
L := 0.41
>
With the starting point x[0] = 0.3 we get the following fixed point iteration:
> x[0]:=0.3; x[1]:=subs(x=0.3,g(x));
x0 := 0.3
x1 := 0.3308322838
> for i from 2 to 15 do x[i]:=subs(x=%,g(x)) od;
x2 := 0.3376030997
x3 := 0.3393458083
x4 := 0.3398101553
x5 := 0.3399349860
x6 := 0.3399686240
x7 := 0.3399776943
x8 := 0.3399801403
x9 := 0.3399808000
x10 := 0.3399809780
x11 := 0.3399810260
x12 := 0.3399810390
x13 := 0.3399810423
x14 := 0.3399810433
x15 := 0.3399810437
>
The 15th iteration x[15] is "identical" to the MAPLE solution based upon the command fsolve.
Assuming " i " iterations, then the a-priori error estimate is given by:
> for i from 3 by 4 to 15 do a_priori_estimate[i]:=
evalf((L^i/(1-L))*abs(x[1]-x[0]),2) od;
a_priori_estimate3 := 0.0036
a_priori_estimate7 := 0.000096
a_priori_estimate11 := 0.28 10-5
a_priori_estimate15 := 0.81 10-7
>
The necessary number of iterations for a given error " epsilon " can be calculated by
the formula on page 2:
> iterations[epsilon]:=
23
evalf(ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L),4);
iterationsε := −1.122 ln( 19.16 ε )
> for i in [0.0036,0.000096,evalf(0.28*10^(-5),2),
evalf(0.81*10^(-7),2)] do iterations[epsilon=i]:=
evalf(subs(epsilon=i,iterations[epsilon]),2) od;
iterationsε = 0.0036 := 3.0
iterationsε = 0.000096 := 6.9
iterations
iterations
ε = 0.28 10
-5
:= 11.
ε = 0.81 10
-7
:= 14.
> beta:=a_posteriori_estimate[15]=L*abs(x[15]-x[14])/(1-L);
β := a_posteriori_estimate15 = 0.2779661017 10-9
> beta:=evalf(%,2);
β := a_posteriori_estimate15 = 0.28 10-9
> alpha:=a-priori-estimate[15]=a_priori_estimate[15];
α := a − priori − estimate15 = 0.81 10
> Q:=A/B=evalf(0.81*10^(-7)/(0.69*10^(-9)),4);
-7
Q :=
A
= 117.4
B
>
Inserting the value of beta = 0.28*10^(-9) into the a-priori estimate, we find the
necessary a-priori iterations of i = 21 instead of 15 iterations with alpha = 0.81*10^(-7) :
> a_priori_iterations[epsilon=evalf(0.28*10^(-9),2)]:=
evalf(subs(epsilon=0.28*10^(-9),iterations[epsilon]),2);
a_priori_iterations
ε = 0.28 10
-9
:= 21.
>
The next example is concerned with finding the zero of the LEGENDRE polynom P(5, x) in the
interval x = [0.5, 0.6]. The results are briefly listed in the following:
> restart:
> with(orthopoly):
> Legendre[p][5]:=P(5,x);
Legendrep :=
5
63 5 35 3 15
x −
x +
x
8
4
8
> f(x):=8*%;
f( x ) := 63 x5 − 70 x3 + 15 x
> ZERO[0.5..0.6]:=evalf(fsolve(f(x)=0,x,0.5..0.6),8);
ZERO0.5 .. 0.6 := 0.53846931
>
From the polynomial f(x) we read the following fixed point forms:
> g[0](x):=x+f(x);
g0( x ) := 16 x + 63 x5 − 70 x3
24
> g[1](x):=(70*x^3-63*x^5)/15;
14 3 21 5
x −
x
3
5
> g[3](x):=((15*x+63*x^5)/70)^(1/3);
g1( x ) :=
9 5⎞
⎛3
x ⎟⎟
g3( x ) := ⎜⎜ x +
10 ⎠
⎝ 14
> g[5](x):=((70*x^3-15*x)/63)^(1/5);
5 ⎞
⎛ 10
g5( x ) := ⎜⎜ x3 −
x ⎟⎟
21 ⎠
⎝9
(1 / 3)
(1 / 5)
>
One can show that only g[3] is compatible with BANACH's fixed- point theorem,
since g[3] maps the interval x = [0.5, 0.6] to itself and the absolute derivative
| g'[3] | is smaller than one.
> g(x):=g[3](x);
(1 / 3)
>
>
>
>
>
>
⎛3
9 5⎞
g( x ) := ⎜⎜ x +
x ⎟⎟
⎝ 14
10 ⎠
alias(H=Heaviside,sc=scaling,th=thickness,co=color):
p[1]:=plot({x,g(x)},x=0.5..0.6,0.5..0.6,sc=constrained,
xtickmarks=4,ytickmarks=4,th=3,co=black,
title="Fixed Point at x = 0.538"):
p[2]:=plot({0.6,0.6*H(x-0.6)},x=0.5..0.6001,co=black):
p[3]:=plot(0.538*H(x-0.538),x=0.537..0.539,co=black):
p[4]:=plot([[0.538,0.538]],style=point,symbol=circle,
symbolsize=30):
plots[display](seq(p[k],k=1..4));
>
In this Figure the graph y = g(x) is contained in the square
Q = { (x, y) | x = [0.5, 0.6], y = [0.5, 0.6] }.
The operator " g " maps the interval x = [0.5, 0.6] into itself .
> abs(Diff(g(xi),xi))=abs(diff(g(x),x));
25
3 9 x4
+
14
2
>
>
>
>
>
d
1
g( ξ ) =
(2 / 3)
dξ
3 ⎛3
9 5⎞
⎜⎜ x +
x ⎟⎟
⎝ 14
10 ⎠
p[1]:=plot(rhs(%),x=0.5..0.6,0.6..0.8,th=3,co=black,
xtickmarks=3,ytickmarks=3,
title="Absolute Derivative | g'(x) |"):
p[2]:=plot({0.8,0.8*H(x-0.6)},x=0.5..0.6001,co=black):
p[3]:=plot(0.681*H(x-0.538),x=0.537..0.539,co=black):
p[4]:=plot([[0.538,0.681]],style=point,symbol=circle,
symbolsize=30):
plots[display](seq(p[k],k=1..4));
>
We see, in the neighbourhood of the expected fixed point x = p the absolute derivative
is less than one. The LIPPSCHITZ constant can be assumed to be the greatest value
in the range (p-delta, p+delta):
> L:=(p-delta, p+delta)[max];
L := ( p − δ, p + δ )
max
> L:=evalf(subs(x=0.6,abs(diff(g(x),x))),2);
L := 0.76
>
With the starting point x[0] = 0.5 we receive the following fixed point iteration:
> x[0]:=0.5;
x[1]:=evalf(subs(x=0.5,g(x)),8);
x0 := 0.5
x1 := 0.51333184
> for i from 2 to 25 do x[i]:=evalf(subs(x=%,g(x)),8) od;
x2 := 0.52180783
x3 := 0.52732399
x4 := 0.53096887
x5 := 0.53340154
x6 := 0.53503604
26
x7 := 0.53613917
x8 := 0.53688593
x9 := 0.53739248
x10 := 0.53773657
x11 := 0.53797051
x12 := 0.53812967
x13 := 0.53823800
x14 := 0.53831176
x15 := 0.53836199
x16 := 0.53839620
x17 := 0.53841950
x18 := 0.53843538
x19 := 0.53844620
x20 := 0.53845357
x21 := 0.53845859
x22 := 0.53846200
x23 := 0.53846433
x24 := 0.53846591
x25 := 0.53846699
>
The 25th iteration x[25] is "identical" to the MAPLE solution based on the command fsolve.
Assuming " i " iterations, then the a-priori error estimate is given by:
> for i from 5 by 5 to 25 do
a_priori_estimate[i]:=evalf((L^i/(1-L))*abs(x[1]-x[0]),4) od;
>
a_priori_estimate5 := 0.01406
a_priori_estimate10 := 0.003563
a_priori_estimate15 := 0.0009033
a_priori_estimate20 := 0.0002290
a_priori_estimate25 := 0.00005808
>
The necessary number of iterations for an allowable error " epsilon " can be calculated
by the formula on page 2:
> iterations[epsilon]:=
evalf(ln((1-L)*epsilon/abs(x[1]-x[0]))/ln(L),5);
iterationsε := −3.6438 ln( 18.005 ε )
27
> for i in [0.014,0.003563,0.0009033,0.000229,0.0000581] do
iterations[epsilon=i]:=
evalf(subs(epsilon=i,iterations[epsilon]),2) od;
iterationsε = 0.014 := 5.0
iterationsε = 0.003563 := 9.7
iterationsε = 0.0009033 := 15.
iterationsε = 0.000229 := 20.
iterationsε = 0.0000581 := 25.
> beta:=a_posteriori_estimate[25]=L*abs(x[25]-x[24])/(1-L);
β := a_posteriori_estimate25 = 0.3420000000 10
> beta:=evalf(%,3);
-5
β := a_posteriori_estimate25 = 0.342 10-5
> alpha:=a_priori_estimate[25];
α := 0.00005808
> Q:=Alpha/Beta=evalf(alpha/rhs(beta),4);
Q :=
Α
= 16.98
Β
>
Inserting the value of beta = 0.342*10^(-5) into the a-priori estimate, we find the necessary
a-priori iterations of i = 35 instead of 25 iterations with alpha = 0.5808*10^(-4) :
> a_priori_iterations[epsilon=evalf(0.342*10^(-5),3)]:=
evalf(subs(epsilon=0.342*10^(-5),iterations[epsilon]),2);
a_priori_iterations
ε = 0.342 10
-5
:= 35.
>
In the same way as before one can find the zeros of the CHEBYSHEV polynomials T(n, x),
which are also orthogonal polynomials contained in the MAPLE package orthopoly.
Applications of the fixed point iteration to systems of nonlinear equations have been
discussed in more detail, for instance, by BETTEN, J.: Finite Elemente für Ingenieure 2,
zweite Auflage, Springer Verlag, Berlin / Heidelberg / New York 2004.
Instead of the fixed point iteration one can utilize the NEWTON or NEWTON-RAPHSON
iteration for solving root-finding problems. NEWTON's method is one of the most well-known
and powerful numerical methods, the convergence order of which is at least two. In contrast, the
fixed point iteration is linearly convergent. In general, a sequence with a high order of
convergence
converges more rapidly than a sequence with a lower order. Some examples have been discussed
by BETTEN, J: "NEWTON's Method in Comparison with the Fixed Point Iteration" published as
a worksheet in the MAPLE Application Center.
>
28