ksaimouli
  • ksaimouli
estimate roots newtons linearization
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ksaimouli
  • ksaimouli
|dw:1353960049993:dw|
phi
  • phi
Is this what you mean http://en.wikipedia.org/wiki/Newton's_method
ksaimouli
  • ksaimouli
yes

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phi
  • 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
  • phi
can you find f'(x) ?
ksaimouli
  • ksaimouli
but i need the estimation of root5 i think ur r thinking other one
phi
  • phi
Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(x-a)
ksaimouli
  • ksaimouli
hmm yup
phi
  • phi
I would pick a=4 because you know the square root of 4
ksaimouli
  • ksaimouli
exactly the same
ksaimouli
  • ksaimouli
ohk
phi
  • phi
here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{-1/2} \] evaluate this at x= a = 4
ksaimouli
  • ksaimouli
|dw:1353961871007:dw|
phi
  • phi
yes, so your approximate function is f(x) = 2 + (1/4)*(x-4) reasonable for x values near 4
ksaimouli
  • ksaimouli
how did u get 2 u said 4
phi
  • phi
you start with f(x) ~ f(a) + f'(a) (x-a) f(x)= sqrt(x) with a=4, f(a) is sqrt(4)
ksaimouli
  • ksaimouli
ok
ksaimouli
  • ksaimouli
|dw:1353962194869:dw|
phi
  • phi
the question asks for sqrt(5) , so plug in 5 for x into your estimation equation
ksaimouli
  • ksaimouli
2.25
phi
  • phi
yes, and 2.25^2 is 5.0625 which is pretty close to 5
ksaimouli
  • ksaimouli
got u one small doubt what if it is root31 shall i pick root36 or root25
phi
  • phi
pick the closer one. 31 is in the middle, so either end will not be very accurate.
ksaimouli
  • ksaimouli
root30 but this is not perfect square
phi
  • 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
anonymous
  • anonymous
not to butt in but newton's method in this case is identical to mechanics rule
phi
  • phi
I never heard of Professor Mechanic
anonymous
  • anonymous
you can use \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]
anonymous
  • anonymous
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
  • 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
ksaimouli
  • ksaimouli
thanks all

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