## 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?

1. KingGeorge

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

2. johnnnboyyy

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

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

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

Ah. Then you should get $2x(1-x^2+x)\ln(x)-x^2\large{\bf +1}.$ That might work better

6. johnnnboyyy

yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0

7. KingGeorge

Let me see what I get.

8. johnnnboyyy

I get x1 = 0.5 - (-0.116434)/(-0.232868), which is 0.5 - 0.5 = 0

9. KingGeorge

Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.

10. johnnnboyyy

it specifically says use x0=0.5. I just think my professor didn't double check his own work.

11. 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.

12. johnnnboyyy

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

No problem.