## ksaimouli Group Title estimate roots newtons linearization one year ago one year ago

1. ksaimouli

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2. phi

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

3. ksaimouli

yes

4. 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) $x_{n+1}= x_n - \frac{f(x_n)}{f'(x_n)}$

5. phi

can you find f'(x) ?

6. ksaimouli

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

7. phi

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

8. ksaimouli

hmm yup

9. phi

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

10. ksaimouli

exactly the same

11. ksaimouli

ohk

12. phi

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

13. ksaimouli

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14. phi

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

15. ksaimouli

how did u get 2 u said 4

16. phi

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

17. ksaimouli

ok

18. ksaimouli

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19. phi

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

20. ksaimouli

2.25

21. phi

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

22. ksaimouli

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

23. phi

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

24. ksaimouli

root30 but this is not perfect square

25. 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

26. satellite73

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

27. phi

I never heard of Professor Mechanic

28. satellite73

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

29. satellite73

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 $x_{n+1}=\frac{1}{2}\left (x_n+\frac{a}{x_n}\right)$

30. 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

31. ksaimouli

thanks all