Asked by seth
n(x)=x^5+xe^x, find the real zeroes using newton-raphson methodology.
Answers
Answered by
drwls
The method is explained here:
http://en.wikipedia.org/wiki/Newton's_method
You will need the derivative of n(x) to use the method. In your case it is
n'(x) = 5x^4 + e^x + x e^x
One of the zeroes is obviously x = 0.
Fox x<0, n(x)<0
For x>0, n(x)>0
It seems to me like x=0 is the only solution, and that you don't need to use the method at all to see that. You could assume a value of 0.1, for example, and use the Newton method to get a better approximation.
http://en.wikipedia.org/wiki/Newton's_method
You will need the derivative of n(x) to use the method. In your case it is
n'(x) = 5x^4 + e^x + x e^x
One of the zeroes is obviously x = 0.
Fox x<0, n(x)<0
For x>0, n(x)>0
It seems to me like x=0 is the only solution, and that you don't need to use the method at all to see that. You could assume a value of 0.1, for example, and use the Newton method to get a better approximation.
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