Note to grader: the software used to make my homework legible does not do any calculation. — David Wagner 2007/09/13 16:14//

Problem 1

$u/u_{mean} = ({3n+1}/{n+1})(1-{r/R}^{{n+1}/n})$ = $0.9 = ({3n+1}/{n+1})(1-0.8^{{n+1}/n})$

→ $({3n+1}/{n+1})(1-0.8^{{n+1}/n})-0.9 =0$ = $({3n+1}/{n+1})-({3n+1}/{n+1})0.8^{{n+1}/n}-0.9 =0$

= $1-0.8^{{n+1}/n}-0.9({n+1}/{3n+1})=0$ = $0.9({n+1}/{3n+1})+0.8^{{n+1}/n}-1=0$

n ≈ 0.309161

Problem 2

ε=0.001

f(x)=x^3-0.2589x^2+0.02262x-0.001122=0

f^prime(x)=3x^2-0.5178x+0.02262

x_1=0 ; f(0)= -0.001122 ; f^prime(0)= 0.022620

$x_2 = x_1-{f(x_1)}/{f^prime(x_1)}=0-{{-0.001122}/0.02262}=0.0496020$ ; f(0.049602)= -0.000515 ; f^prime(0.049602)= 0.004317

|f(0.049602)| = 0.000515 < 0.001 → |f(x)| < ε

x ≈ 0.049602

However, ε is probably too large. See the convergence below.

i	x	f(x)	f'(x)
1	0.000000	-0.00112200	0.02262000
2	0.049602	-0.00051495	0.00431713
3	0.168883	0.00013071	0.02073666
4	0.162579	0.00000959	0.01773255
5	0.162038	0.00000007	0.01748583
6	0.162035	0.00000000	0.01748409