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johnnnboyyy

  • 2 years ago

Fine the root of the equation f(x)=2x(1−x^2+x)ln(x)-x^2+1 in the interval [0, 1] by Newton’s method where x0=0.5... The problem I am getting is that when I calculate x1 = x0 - f(x)/f'(x), I get 0, but the root is at 0.32... ideas?

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  1. KingGeorge
    • 2 years ago
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    Are you sure that's the right equation? That has no roots in the interval [0,1].

  2. johnnnboyyy
    • 2 years ago
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    Yep, that is what is on my assignment. The original ? was to write a computer program that iterates through 5 times and calculates the root using Newton's method, but my program works for everything except this function.

  3. KingGeorge
    • 2 years ago
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    If the equation is\[2x(1−x^2+x)\ln(x)-x^2−1,\] there are no roots. If it's \[2x(1−x^2+x)(\ln(x)-x^2−1),\]there is a root.

  4. johnnnboyyy
    • 2 years ago
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    well the original function was f(x)=2x(1-x^2+x)ln(x)=x^2-1, not sure if that changes things?

  5. KingGeorge
    • 2 years ago
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    Ah. Then you should get \[2x(1-x^2+x)\ln(x)-x^2\large{\bf +1}.\] That might work better

  6. johnnnboyyy
    • 2 years ago
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    yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0

  7. KingGeorge
    • 2 years ago
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    Let me see what I get.

  8. johnnnboyyy
    • 2 years ago
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    I get x1 = 0.5 - (-0.116434)/(-0.232868), which is 0.5 - 0.5 = 0

  9. KingGeorge
    • 2 years ago
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    Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.

  10. johnnnboyyy
    • 2 years ago
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    it specifically says use x0=0.5. I just think my professor didn't double check his own work.

  11. KingGeorge
    • 2 years ago
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    That's a likely guess. The sequence also doesn't converge to anything is \(x_0>.5\), so you might try something just below .5 after noting that \(x_0=.5\) sent you straight to 0.

  12. johnnnboyyy
    • 2 years ago
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    yeah choosing x0 = 0.4 works nicely. Hopefully he acknowledges his mistake and gives me credit for it. Thanks for taking the time to look at the problem though.

  13. KingGeorge
    • 2 years ago
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    No problem.

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