[高微]help數學問題!!

Question

Let f(x) = logx - x + 2

a)Show that f(x) = 0 has two roots in x > 0.

b)Show that for Newton's method to converge to the root closer to zero, we must take 0 < Xo < e(-1).

c)Using a Newton's code, approximate both roots to within 10(-12).

e(-1) 是 e 的 -1 次方
10(-12) 是 10 的 -12 次方

請解釋一下題目和解答
謝謝大大

Answer 1

(a)證明x>0時f(x)=0有兩根
(b)使用Newton法求f(x)=0與0較近之根,必須設x0在1~1/e之間
(c)用Newton法求f(x)=0兩根的近似值,誤差<10^(-12)

作法:
http://tw.knowledge.yahoo.com/question/question?qid=1607120410787

Answer 2

A)
(1)f(x) is continuous and differentiable on [0.1, ∞), and f'(x)=1/x-1
(2)f(0.1)<0, f(1)=1>0, f(4)<0
(3)
f'(x)>0 in (0.1, 1)==> f is increasing in (0.1, 1)
f'(x)<0 in (1, ∞)==> f is decreasing in (1, ∞)
So that, f(x) has exact two roots in x>0

(B)
To find the root in [0.1, 1), let 0 Newtons method: x1=x0-f(x0)/f'(x0)
=>x1=x0-[ln(x0)-x0+2]/(1/x0-1) must be positive (in domain of f(x))
=>x0>[ln(x0)-x0+2]/(1/x0-1)
=>x0(1/x0-1)>ln(x0)-x0+2 (since 1/x0-1>0)
=>-1>ln(x0)
=>0
(C)By Excel
x0=0.2
x1=0.1523594781085250
x2=0.1584478213815620
x3=0.1585942591145130
x4=0.1585943395630150
x5=0.1585943395630390
x6=0.1585943395630390
x7=0.1585943395630390
x8=0.1585943395630390
x9=0.1585943395630390
=>the first root ≒ 0.158594339563

x0=2.5
x1=3.1938178864569300
x2=3.1463567017224600
x3=3.1461932225994200
x4=3.1461932206205800
x5=3.1461932206205800
x6=3.1461932206205800
x7=3.1461932206205800
x8=3.1461932206205800
x9=3.1461932206205800
=>the second root ≒3.1461932206206