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Lily@pie
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Homework Statement
let x0, x1,... be the approximations of pi from the Newton's Method. Use Mean Value theorem to show that
|pi-xj+1|=|tan2cj||pi-xj|
for some cj between xj and pi
Homework Equations
pi is defined as smallest positive number r when sin r =0
The Attempt at a Solution
I have tried to let f(x) = sin x and the bound to be [x,pi].
By using the mean value theorem, there exist some c such that
f'(c) = (sin pi - sin x) / (pi - x)
cos c = - sin x / (pi-x)
But I couldn't get the form, especially the |tan2cj| part.
I have also tried to let f(x) = tan x - x in [x,pi]
so, f'(c)=sec2c + 1 = (tan pi + pi - tan x - x)/(pi-x)
tan2 c = (tan pi - tan x + pi - x)/(pi-x)
But tan pi is undefined...
Am I on the correct path?