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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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KingGeorgeBest ResponseYou've already chosen the best response.1
Are you sure that's the right equation? That has no roots in the interval [0,1].
 one year ago

johnnnboyyyBest ResponseYou've already chosen the best response.0
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

johnnnboyyyBest ResponseYou've already chosen the best response.0
yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Let me see what I get.
 one year ago

johnnnboyyyBest ResponseYou've already chosen the best response.0
I get x1 = 0.5  (0.116434)/(0.232868), which is 0.5  0.5 = 0
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.
 one year ago

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

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

johnnnboyyyBest ResponseYou've already chosen the best response.0
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.
 one year ago
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