# Can anyone help me in this proving?

• Mar 31st 2008, 05:54 AM
chris_panda85
Can anyone help me in this proving?
Given

x_(n+1) = x_n - f(x)/f'(x_n) (1)

x_(n+1) = { x_n + -f'(x_n) ± √[f'^2(x_n) - 2f(x_n)f"(y_n)] } / f''(y_n) (2)
n=1,2,...

(2) can be write as,
x_(n+1) = x_n + { [2f(x_n)f'(x_n)] / [2f'^2(x_n) - f''(y_n)f(x_n)] } (3)

f(x_n) = f'(r) [ e_n + c_2 (e_n)^2 + c_3 (e_n)^3 + ... ] (4)
f'(x_n) = f'(r) [ 1 + 2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...] (5)

From (1) and (2),

f(x_n) / f'(x_n) = e_n - c_2 (e_n)^2 + 2[ (c_2)^2 - c_3] (e_n)^3 + ... (6)

How do I get eqn (6) using (4) and (5)? Please ...

Thanks !!!

P/s: _ represent subscript while ^ represent superscript / power
• Mar 31st 2008, 10:57 PM
CaptainBlack
Quote:

Originally Posted by chris_panda85
Given

x_(n+1) = x_n - f(x)/f'(x_n) (1)

x_(n+1) = { x_n + -f'(x_n) ± √[f'^2(x_n) - 2f(x_n)f"(y_n)] } / f''(y_n) (2)
n=1,2,...

(2) can be write as,
x_(n+1) = x_n + { [2f(x_n)f'(x_n)] / [2f'^2(x_n) - f''(y_n)f(x_n)] } (3)

f(x_n) = f'(r) [ e_n + c_2 (e_n)^2 + c_3 (e_n)^3 + ... ] (4)
f'(x_n) = f'(r) [ 1 + 2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...] (5)

From (1) and (2),

f(x_n) / f'(x_n) = e_n - c_2 (e_n)^2 + 2[ (c_2)^2 - c_3] (e_n)^3 + ... (6)

How do I get eqn (6) using (4) and (5)? Please ...

Thanks !!!

P/s: _ represent subscript while ^ represent superscript / power

Assume the errors are small so:

$\frac{1}{f'(x_n)} = \frac{1}{f'(r)} ~\frac{1}{ 1 + [2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]}$

Now expand the second term as a power series:

$\frac{1}{f'(x_n)} =$ $\frac{1}{f'(r)} [1 - [2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]+$ $[2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]^2 - ... ]$

etc.

RonL