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|dw:1353960049993:dw|

Is this what you mean
http://en.wikipedia.org/wiki/Newton's_method

yes

can you find f'(x) ?

but i need the estimation of root5 i think ur r thinking other one

Maybe you mean a taylor series approximation?
f(x) ~ f(a) + f'(a)(x-a)

hmm yup

I would pick a=4 because you know the square root of 4

exactly the same

ohk

here
f(x)= \(x^{\frac{1}{2}}\)
\[ f'(x)= \frac{1}{2}x^{-1/2} \]
evaluate this at x= a = 4

|dw:1353961871007:dw|

yes, so your approximate function is
f(x) = 2 + (1/4)*(x-4)
reasonable for x values near 4

how did u get 2 u said 4

you start with
f(x) ~ f(a) + f'(a) (x-a)
f(x)= sqrt(x)
with a=4, f(a) is sqrt(4)

ok

|dw:1353962194869:dw|

the question asks for sqrt(5) , so plug in 5 for x into your estimation equation

2.25

yes, and 2.25^2 is 5.0625 which is pretty close to 5

got u one small doubt what if it is root31 shall i pick root36 or root25

pick the closer one. 31 is in the middle, so either end will not be very accurate.

root30 but this is not perfect square

not to butt in but newton's method in this case is identical to mechanics rule

I never heard of Professor Mechanic

you can use
\[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]

thanks all