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songoku
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Homework Statement
a. Using equation x2 – a = 0, show that [tex]x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)[/tex]
b. Given that Xn = √a + e, where e is a small error. Show that:
[tex]x_{n+1}=a^{1/2}+\frac{1}{2}\left(\frac{e^2}{a^{1/2}+e}\right)[/tex]
c. If x0 = 4 and a = 7, where √7 = 2.646, find x3. By using the value of x3, find the error for X4 (1 significant figure)
Homework Equations
Newton Raphson
The Attempt at a Solution
I am able to do (a) and (b) and having trouble with (c). I've tried to find x3 using the formulas (a) and (b) and I got different answer.
Using formula (a) :
[tex]x_{1}=\frac{1}{2}\left(x_0+\frac{a}{x_0}\right)=2.875[/tex]
[tex]x_{2}=\frac{1}{2}\left(x_1+\frac{a}{x_1}\right)=2.65489[/tex]
[tex]x_{3}=\frac{1}{2}\left(x_2+\frac{a}{x_2}\right)=2.64577[/tex]
[tex]x_{4}=\frac{1}{2}\left(x_3+\frac{a}{x_3}\right)=2.64575[/tex]
So :
X4 = √a + e4
e4 = -2.5 x 10-4
Using formula (b) :
I think the error e is not a constant value because if it's constant, the value Xn = √a + e will be constant. So I assume [tex]e\rightarrow e_n[/tex] ----> just guessing (is it possible [tex]e\rightarrow e_{n-1} ?[/tex] )
Then :
X0 = √a + e0
e0 = 4 - 2.646 = 1.354
Using the value of e0 :
[tex]x_{1}=a^{1/2}+\frac{1}{2}\left(\frac{(e_0)^2}{a^{1/2}+e_0}\right)=2.875[/tex]
X1 = √a + e1
e1 = 2.875 - 2.646 = 0.229
[tex]x_{2}=a^{1/2}+\frac{1}{2}\left(\frac{(e_1)^2}{a^{1/2}+e_1}\right)=2.65512[/tex]
X2 = √a + e2
e2 = 2.65512 - 2.646 = 9.12 x 10-3
[tex]x_{3}=a^{1/2}+\frac{1}{2}\left(\frac{(e_2)^2}{a^{1/2}+e_2}\right)=2.646015663[/tex]
X3 = √a + e3
e3 = 2.646015663 - 2.646 = 1.5663 x 10-3
[tex]x_{4}=a^{1/2}+\frac{1}{2}\left(\frac{(e_3)^2}{a^{1/2}+e_3}\right)=2.646000461[/tex]
X4 = √a + e4
e4 = 2.646000461 - 2.646 = 4.6136 x 10-7
Which one is right? I think the positive one is right because from the graph of y = x2 - 7 and x0=4, the approximation will be overestimate, so the error should be positive.
But I'm not sure..
Thanks
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