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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1353960049993:dw
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Is this what you mean http://en.wikipedia.org/wiki/Newton's_method
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
If you define a function f(x)= x^25 it will have a zero when x is sqrt(5) You can use Newton's method to find that zero (x value) \[ x_{n+1}= x_n  \frac{f(x_n)}{f'(x_n)} \]
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
can you find f'(x) ?
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
but i need the estimation of root5 i think ur r thinking other one
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(xa)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I would pick a=4 because you know the square root of 4
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
exactly the same
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{1/2} \] evaluate this at x= a = 4
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1353961871007:dw
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, so your approximate function is f(x) = 2 + (1/4)*(x4) reasonable for x values near 4
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
how did u get 2 u said 4
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
you start with f(x) ~ f(a) + f'(a) (xa) f(x)= sqrt(x) with a=4, f(a) is sqrt(4)
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1353962194869:dw
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the question asks for sqrt(5) , so plug in 5 for x into your estimation equation
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, and 2.25^2 is 5.0625 which is pretty close to 5
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
got u one small doubt what if it is root31 shall i pick root36 or root25
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
pick the closer one. 31 is in the middle, so either end will not be very accurate.
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
root30 but this is not perfect square
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
30 is not a perfect square. so you want to pick a nearby "a" where you know the root. 25 is the closest square. 36 is almost as close. Using 25 is probably better by a little bit that using 36
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
not to butt in but newton's method in this case is identical to mechanics rule
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I never heard of Professor Mechanic
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you can use \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
mechanic predated newton in fact i think mechanic was babylonian, maybe circa 3000 BC it is an old method for finding roots, but if you do the algebra with newtons method for square roots it ends up being the same. the recursion to find \(\sqrt{a}\) is \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{a}{x_n}\right)\]
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
es, for f(x)= x^2 a and f'(x)= 2x newton's x f(x)/f'(x) is x  (x^2a)/2x = (2x^2 x^2 +a)/2x which becomes mechanic with a few more steps
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
thanks all
 2 years ago
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