## johnnnboyyy Group Title 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

1. KingGeorge Group Title

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

2. johnnnboyyy Group Title

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 Group Title

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 Group Title

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 Group Title

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

6. johnnnboyyy Group Title

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

7. KingGeorge Group Title

Let me see what I get.

8. johnnnboyyy Group Title

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

9. KingGeorge Group Title

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

10. johnnnboyyy Group Title

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

11. KingGeorge Group Title

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 Group Title

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 Group Title

No problem.