What is the Range of Convergence for Newton's Method on tanh(x)?

[math199711]

I was hoping someone could help me out with this problem. I need to find the values, that Newton's method converges for tanh(x).

So far I set up the algorithm:

x$\tiny_{k+1}$ = x$_{k}$ - $\frac{tanhx}{sech^{2}x}$

I simplified it to:

x$\tiny_{k+1}$ = x$_{k}$ - ($\frac{1}{2}$sinh(2x))

And then I took the derivative to see when the absolute value of the derivative was less than one.

|1 - cosh(2x)| < 1

I did all of this and came up with the interval of convergence as being:

$\frac{1}{2}$ln(2$\pm \sqrt{3}$)

These numbers are approximately .-658 and .658 respectively.

However when I test out 1, it converges using Newton's method, yet it falls out of the range that I determined.

Any help on the problem would be appreciated.

 

[Adrmay]

The range of convergence for Newton's method for tanh(x) is actually [-0.881, 0.881]. This is because the derivative of tanh(x) is sech^2(x) which has an absolute value of 1 at x= +/-0.881.