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ksaimouli

  • 2 years ago

estimate roots newtons linearization

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  1. ksaimouli
    • 2 years ago
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    |dw:1353960049993:dw|

  2. phi
    • 2 years ago
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    Is this what you mean http://en.wikipedia.org/wiki/Newton's_method

  3. ksaimouli
    • 2 years ago
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    yes

  4. phi
    • 2 years ago
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    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
    • 2 years ago
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    can you find f'(x) ?

  6. ksaimouli
    • 2 years ago
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    but i need the estimation of root5 i think ur r thinking other one

  7. phi
    • 2 years ago
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    Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(x-a)

  8. ksaimouli
    • 2 years ago
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    hmm yup

  9. phi
    • 2 years ago
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    I would pick a=4 because you know the square root of 4

  10. ksaimouli
    • 2 years ago
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    exactly the same

  11. ksaimouli
    • 2 years ago
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    ohk

  12. phi
    • 2 years ago
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    here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{-1/2} \] evaluate this at x= a = 4

  13. ksaimouli
    • 2 years ago
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    |dw:1353961871007:dw|

  14. phi
    • 2 years ago
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    yes, so your approximate function is f(x) = 2 + (1/4)*(x-4) reasonable for x values near 4

  15. ksaimouli
    • 2 years ago
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    how did u get 2 u said 4

  16. phi
    • 2 years ago
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    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
    • 2 years ago
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    ok

  18. ksaimouli
    • 2 years ago
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    |dw:1353962194869:dw|

  19. phi
    • 2 years ago
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    the question asks for sqrt(5) , so plug in 5 for x into your estimation equation

  20. ksaimouli
    • 2 years ago
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    2.25

  21. phi
    • 2 years ago
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    yes, and 2.25^2 is 5.0625 which is pretty close to 5

  22. ksaimouli
    • 2 years ago
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    got u one small doubt what if it is root31 shall i pick root36 or root25

  23. phi
    • 2 years ago
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    pick the closer one. 31 is in the middle, so either end will not be very accurate.

  24. ksaimouli
    • 2 years ago
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    root30 but this is not perfect square

  25. phi
    • 2 years ago
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    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
    • 2 years ago
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    not to butt in but newton's method in this case is identical to mechanics rule

  27. phi
    • 2 years ago
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    I never heard of Professor Mechanic

  28. satellite73
    • 2 years ago
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    you can use \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]

  29. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    thanks all

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