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ksaimouli Group Title

estimate roots newtons linearization

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    |dw:1353960049993:dw|

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

    • one year ago
  3. ksaimouli Group Title
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    yes

    • one year ago
  4. phi Group Title
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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) \[ x_{n+1}= x_n - \frac{f(x_n)}{f'(x_n)} \]

    • one year ago
  5. phi Group Title
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    can you find f'(x) ?

    • one year ago
  6. ksaimouli Group Title
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    but i need the estimation of root5 i think ur r thinking other one

    • one year ago
  7. phi Group Title
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    Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(x-a)

    • one year ago
  8. ksaimouli Group Title
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    hmm yup

    • one year ago
  9. phi Group Title
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    I would pick a=4 because you know the square root of 4

    • one year ago
  10. ksaimouli Group Title
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    exactly the same

    • one year ago
  11. ksaimouli Group Title
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    ohk

    • one year ago
  12. phi Group Title
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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

    • one year ago
  13. ksaimouli Group Title
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    |dw:1353961871007:dw|

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

    • one year ago
  15. ksaimouli Group Title
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    how did u get 2 u said 4

    • one year ago
  16. phi Group Title
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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)

    • one year ago
  17. ksaimouli Group Title
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    ok

    • one year ago
  18. ksaimouli Group Title
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    |dw:1353962194869:dw|

    • one year ago
  19. phi Group Title
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    the question asks for sqrt(5) , so plug in 5 for x into your estimation equation

    • one year ago
  20. ksaimouli Group Title
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    2.25

    • one year ago
  21. phi Group Title
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    yes, and 2.25^2 is 5.0625 which is pretty close to 5

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

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

    • one year ago
  24. ksaimouli Group Title
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    root30 but this is not perfect square

    • one year ago
  25. phi Group Title
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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

    • one year ago
  26. satellite73 Group Title
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    not to butt in but newton's method in this case is identical to mechanics rule

    • one year ago
  27. phi Group Title
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    I never heard of Professor Mechanic

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

    • one year ago
  29. satellite73 Group Title
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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 \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{a}{x_n}\right)\]

    • one year ago
  30. phi Group Title
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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

    • one year ago
  31. ksaimouli Group Title
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    thanks all

    • one year ago
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