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anonymous
 3 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?
anonymous
 3 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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KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Are you sure that's the right equation? That has no roots in the interval [0,1].

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, 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
 3 years ago
Best ResponseYou've already chosen the best response.1If 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well the original function was f(x)=2x(1x^2+x)ln(x)=x^21, not sure if that changes things?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Ah. Then you should get \[2x(1x^2+x)\ln(x)x^2\large{\bf +1}.\] That might work better

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Let me see what I get.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I get x1 = 0.5  (0.116434)/(0.232868), which is 0.5  0.5 = 0

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it specifically says use x0=0.5. I just think my professor didn't double check his own work.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That'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
 3 years ago
Best ResponseYou've already chosen the best response.0yeah 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.
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