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

1. ksaimouli Group Title

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2. phi Group Title

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

3. ksaimouli Group Title

yes

4. phi Group Title

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 Group Title

can you find f'(x) ?

6. ksaimouli Group Title

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

7. phi Group Title

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

8. ksaimouli Group Title

hmm yup

9. phi Group Title

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

10. ksaimouli Group Title

exactly the same

11. ksaimouli Group Title

ohk

12. phi Group Title

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

13. ksaimouli Group Title

|dw:1353961871007:dw|

14. phi Group Title

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

15. ksaimouli Group Title

how did u get 2 u said 4

16. phi Group Title

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 Group Title

ok

18. ksaimouli Group Title

|dw:1353962194869:dw|

19. phi Group Title

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

20. ksaimouli Group Title

2.25

21. phi Group Title

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

22. ksaimouli Group Title

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

23. phi Group Title

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

24. ksaimouli Group Title

root30 but this is not perfect square

25. phi Group Title

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 Group Title

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

27. phi Group Title

I never heard of Professor Mechanic

28. satellite73 Group Title

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

29. satellite73 Group Title

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 Group Title

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 Group Title

thanks all