anonymous
  • anonymous
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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KingGeorge
  • KingGeorge
Are you sure that's the right equation? That has no roots in the interval [0,1].
anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
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.

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anonymous
  • anonymous
well the original function was f(x)=2x(1-x^2+x)ln(x)=x^2-1, not sure if that changes things?
KingGeorge
  • KingGeorge
Ah. Then you should get \[2x(1-x^2+x)\ln(x)-x^2\large{\bf +1}.\] That might work better
anonymous
  • anonymous
yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0
KingGeorge
  • KingGeorge
Let me see what I get.
anonymous
  • anonymous
I get x1 = 0.5 - (-0.116434)/(-0.232868), which is 0.5 - 0.5 = 0
KingGeorge
  • KingGeorge
Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.
anonymous
  • anonymous
it specifically says use x0=0.5. I just think my professor didn't double check his own work.
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
No problem.

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