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CSE/MATH 451: Numerical Computations, spring 2005
Answers to home work 5
1 Bisection method
For the bisection method, if e is the error tolerance, the number of iteration must satisfy (see text book)
n ≥ log(b − a) − log(2e) log(2)
In our case we have e = 10−^8 , a = 0 and b = 1, set these in we get n ≥ 25 .57. So we need at least 26 iterations.
2 Fixed point iteration
a). One can easily check that there is a root in the interval [1. 1 , 1 .6], because f (1.1) = − 0 .1207 and f (1.6) = 0.2311. Since f (x) is continuous, then it must cross 0 at some point between [1. 1 , 1 .6].
b). One can rewrite it as x = g(x) = f (x) + x = e−x^ − cos(x) + x. The iteration scheme will then be: xn+1 = g(xn) = e−xn^ − cos(xn) + xn.
Start with x 0 = 1.6, and do 4 iterations, and we get x 1 = 1. 8311 , x 2 = 2. 2487 , x 3 =
- 9814 , x 4 = 4.0193. It is not converging to the root we are looking for, because g′(x) = −e−x^ + sin(x) + 1 is strictly bigger than 1 on the interval [1. 1 , 1 .6].
c). If one rewrite it as x = g(x) = −f (x) + x = −e−x^ + cos(x) + x. The iteration scheme will then be: xn+1 = g(xn) = −e−xn^ + cos(xn) + xn. Start with x 0 = 1.6, and do 4 iterations, and we get x 1 = 1. 3689 , x 2 = 1. 3150 , x 3 =
- 2996 , x 4 = 1.2948. It looks like it is converging. In fact, if we compute 5 more steps, we will get x 9 = x 8 = 1.2927. The convergence is guaranteed because g′(x) = e−x^ − sin(x) + 1 satisfies 0 < g′(x) < 1 on the interval [1. 1 , 1 .6]. Therefore by the theorem it converges.
d). The limit value of the iteration is a fixed point for x = 12 x + (^1) x. By simply calculation we find x = ±
2, if the method converges. To see the convergence with x 0 = 1, we first write x = g(x) = 12 x + (^1) x , and then we have g′(x) = 12 − (^) x^12. One easily see that |g′(x)| < 1 on the interval [1, 2] which contains the root
- Therefore the method converges.
3 Newton’s method
a). Using Newton’s method, with f (x) = x^2 − R og f ′(x) = 2x we get
xn+1 = xn − f (xn) f ′(xn)
= xn − x^2 n − R 2 xn
(xn + R/xn).
There we have
x^2 n+1 − R =
(xn + R/xn)^2 − R
4 x^2 n
[
x^4 n + 2x^2 nR + R^2 − 4 x^2 nR
]
4 x^2 n
[x^2 n − R]^2.
This means that the residual f (xn) = x^2 n − R has quadratic convergence.
f (xn+1) =
4 x^2 n
f (xn)^2. (1)
We want to show the same with the error, en := x − xn. From (1) we get
(xn+1 −
R)(xn+1 +
R) =
4 x^2 n
(xn −
R)^2 (xn +
R)^2
and xn+1 −
R =
4 x^2 n
(xn +
R)^2
xn+1 +
R
(xn −
R)^2
where (xn +
R)^2
xn+1 +
R
(xn +
R)^2
(x^2 n + R + 2xn
R)/ 2 xn
= 2xn,
and xn+1 −
R =
2 xn (xn −
(R))^2
which means that en+1 = −
2 xn
e^2 n
and therefore the mathod converges quadratically.
b). For R = 10, the iteration becomes:
xn+1 =
xn +
xn
With x 0 = 3, we get x 1 = 3. 1666667 , x 2 = 3. 1622807 , x 3 = 3. 1622776 , x 4 = 3.1622776. We see that it converges very quickly.
c). With f (x) = (x − 1)^8 and f ′(x) = 8(x − 1)^7 , Newton’s method becomes
xn+1 = xn −
f (xn) f ′(xn)
= xn −
(x − 1)^8 8(x − 1)^7
= xn − (xn − 1)/7 =
xn +
v=newton(’f’,’df’,3.0,1e-12,10) n xn f(xn) 0 3.000000 9.000000e+ 1 2.437500 2.036865e+ 2 2.213033 2.563634e- 3 2.175555 6.463361e- 4 2.174560 4.479068e- 5 2.174559 2.157385e- 6 2.174559 8.881784e- v =
In order to use Newton’s method on our function f , we need to define the files f1.m and df1.m. My f1.m looks like:
function y = f1(x) y=exp(-x)-cos(x);
and my df1.m looks like:
function y = df1(x) y=-exp(-x)+sin(x);
I get the following result:
v=newton(’f1’,’df1’,1.6,1e-12,10)
n xn f(xn) 0 1.600000 2.310960e- 1 1.310289 1.217111e- 2 1.292814 8.160649e- 3 1.292696 3.871718e- 4 1.292696 0.000000e+ v =
So after only 4 iteration the method got a very accurate approximation.
