estimate roots newtons linearization

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estimate roots newtons linearization

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

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  • phi
If you define a function f(x)= x^2-5 it will have a zero when x is sqrt(5) You can use Newton's method to find that zero (x value) \[ x_{n+1}= x_n - \frac{f(x_n)}{f'(x_n)} \]
  • phi
can you find f'(x) ?
but i need the estimation of root5 i think ur r thinking other one
  • phi
Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(x-a)
hmm yup
  • phi
I would pick a=4 because you know the square root of 4
exactly the same
ohk
  • phi
here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{-1/2} \] evaluate this at x= a = 4
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  • phi
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
  • phi
you start with f(x) ~ f(a) + f'(a) (x-a) f(x)= sqrt(x) with a=4, f(a) is sqrt(4)
ok
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  • phi
the question asks for sqrt(5) , so plug in 5 for x into your estimation equation
2.25
  • phi
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
  • phi
pick the closer one. 31 is in the middle, so either end will not be very accurate.
root30 but this is not perfect square
  • phi
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
not to butt in but newton's method in this case is identical to mechanics rule
  • phi
I never heard of Professor Mechanic
you can use \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]
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 \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{a}{x_n}\right)\]
  • phi
es, for f(x)= x^2 -a and f'(x)= 2x newton's x- f(x)/f'(x) is x - (x^2-a)/2x = (2x^2 -x^2 +a)/2x which becomes mechanic with a few more steps
thanks all

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