Asymptotic Error Analysis for Different Sequences in Numerical Analysis, Assignments of Mathematical Methods for Numerical Analysis and Optimization

Instructions for homework #2 in math 128a, focusing on asymptotic error analysis for three different sequences in numerical analysis. Students are required to compute the sequence pn for each sequence and plot pn to determine the limit. The sequences involve calculating the limit of (1 + 1/n)^n, the difference between consecutive terms of xn = xn-1 - (xn-1/xn), and the recursive sequence xn+1 = 1/2xn + 1/xn. The document also encourages experimenting with various starting values for the third sequence to observe the convergence to √2.

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Pre 2010

Uploaded on 10/01/2009

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MATH 128A
Homework #2
Due (in class) Tuesday, June 29.
All numbered exercises are from the textbook Numerical Analysis by Bur-
den and Faires, 7th ed.
a) Section 1.3 6(a,b),7(d),8,10,15,16
b) Section 2.1 10,12,14,16
If xnxthen let us define a new seqence pn=|xn+1x|
|xnx|α.
For α= 1 and pnλ < 1, we say that xnxlinearly; if λ= 1, the
convergence is sublinear. (What happens if λ > 1?)
For α2 if pnλ(notice, no restriction on λ), we call this convergence
of order α.
For the following sequences perform the asymptotic error analysis, i.e.
numerically compute pnfor n= 1, . . . , 1000:
a) xn= (1 + 1
n)n; here x=eand α= 1
b) xn+1 =xnx2
n
x2
n+x2
n1
;x0= 20, x1= 15; here x= 0 and α= 1 and
c) xn+1 = 1/2xn+1
xn,x0= 2; here x=2, try α= 1,2
Plot pnthat you are getting and guess what the limit is in each case.
For the last sequence, experiment with various starting values and see for
what range of starting values the original sequence xnwill still converge to
2.

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MATH 128A

Homework # Due (in class) Tuesday, June 29.

All numbered exercises are from the textbook Numerical Analysis by Bur- den and Faires, 7th ed.

a) Section 1.3 6(a,b),7(d),8,10,15, b) Section 2.1 10,12,14,

If xn → x∗^ then let us define a new seqence pn = |xn+1−x

∗| |xn−x∗|α^. For α = 1 and pn → λ < 1, we say that xn → x∗^ linearly; if λ = 1, the convergence is sublinear. (What happens if λ > 1?) For α ≥ 2 if pn → λ (notice, no restriction on λ), we call this convergence of order α. For the following sequences perform the asymptotic error analysis, i.e. numerically compute pn for n = 1,... , 1000: a) xn = (1 + (^1) n )n; here x∗^ = e and α = 1

b) xn+1 = xn − x

(^2) n x^2 n+x^2 n− 1 ;^ x^0 = 20, x^1 = 15; here^ x

∗ (^) = 0 and α = 1 and

c) xn+1 = 1/ 2 xn + (^) x^1 n , x 0 = 2; here x∗^ =

2, try α = 1, 2

Plot pn that you are getting and guess what the limit is in each case. For the last sequence, experiment with various starting values and see for what range of starting values the original sequence√ xn will still converge to