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anonymous

  • 5 years ago

Use Newton's method to find an approximate solution for the equation x^2+x=3 Start with x0=1 and iterate Newton's method twice to find x2.

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  1. anonymous
    • 5 years ago
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    f' = 2x + 1 f(0) = -3 x1 = x0 - f(0) / f'(0) shoot - i'm not sure thats right - i'll have to check

  2. anonymous
    • 5 years ago
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    I am also not sure, if you find anything about it pls post it

  3. amistre64
    • 5 years ago
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    newtons method is finding the equation of the tangent line; getting its root for a new x value to repeat the iteration right?

  4. amistre64
    • 5 years ago
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    Xn = f'(x)(x-x0)+y0 = 0

  5. anonymous
    • 5 years ago
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    I think so

  6. amistre64
    • 5 years ago
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    (1,2) Xn = (2(1)+1)(x-1)+2

  7. amistre64
    • 5 years ago
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    Xn = 3x - 1; Xn = 1/3 then right?

  8. anonymous
    • 5 years ago
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    what do you label as Xn?

  9. amistre64
    • 5 years ago
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    new X value; Xn just seemed appropriate

  10. anonymous
    • 5 years ago
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    ok

  11. amistre64
    • 5 years ago
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    I coulda went with Larry, but.... i know a cucumber named Larry lol

  12. anonymous
    • 5 years ago
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    :)

  13. anonymous
    • 5 years ago
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    sorry still not clear, I am just working it out, but I dont see how you get y0=2

  14. amistre64
    • 5 years ago
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    x^2 +x = 3 is that a typo?

  15. amistre64
    • 5 years ago
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    x^2 +x +3 maybe?

  16. anonymous
    • 5 years ago
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    that is it

  17. anonymous
    • 5 years ago
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    first one

  18. amistre64
    • 5 years ago
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    x^2 +x = 3 ; when x=1 this is simply a false statement then 1+1 = 3???

  19. anonymous
    • 5 years ago
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    yep but the method will tend to the right answer

  20. amistre64
    • 5 years ago
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    x^2 +x -3 = 0 then

  21. anonymous
    • 5 years ago
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    that is the same

  22. amistre64
    • 5 years ago
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    then remove the 0 to find some solutions

  23. amistre64
    • 5 years ago
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    when x = 1; y = ? (1,-1) then right?

  24. anonymous
    • 5 years ago
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    -1 right

  25. amistre64
    • 5 years ago
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    use this point for your tangent equation that follows: y-y0 = m(x-x0) y = m(x-x0)+y0

  26. amistre64
    • 5 years ago
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    m = f'(x)

  27. amistre64
    • 5 years ago
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    y = (2(1)+1)(x-1)-1

  28. anonymous
    • 5 years ago
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    clear

  29. amistre64
    • 5 years ago
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    y = 3x-4; the root is x = 4/3 then right?

  30. anonymous
    • 5 years ago
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    that is why i did not see how that 2 appeared

  31. amistre64
    • 5 years ago
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    :) right concept, wrong thought lol

  32. anonymous
    • 5 years ago
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    right and I should do this again to get the result, thanks for the help!

  33. amistre64
    • 5 years ago
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    youre welcome :)

  34. anonymous
    • 5 years ago
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    A mathematica solution with comments. Refer to the pdf attachment.

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  35. anonymous
    • 5 years ago
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    andras:what/where are you studying?

  36. anonymous
    • 5 years ago
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    pure maths, just finished 1st year

  37. anonymous
    • 5 years ago
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    are you on facebook? pardon me if posting this here is a breach of ettiquet

  38. anonymous
    • 5 years ago
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    :) I dont mind. Yes I am on facebook

  39. anonymous
    • 5 years ago
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    you can add me if you wish but I dont use facebook that often. Andras Honyek

  40. anonymous
    • 5 years ago
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    will do...I just sent you a request...neat...most people i know are glued to their fb accounts.

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