- #1
math199711
- 1
- 0
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[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - [itex]\frac{tanhx}{sech^{2}x}[/itex]
I simplified it to:
x[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - ([itex]\frac{1}{2}[/itex]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:
[itex]\frac{1}{2}[/itex]ln(2[itex]\pm \sqrt{3}[/itex])
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.
So far I set up the algorithm:
x[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - [itex]\frac{tanhx}{sech^{2}x}[/itex]
I simplified it to:
x[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - ([itex]\frac{1}{2}[/itex]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:
[itex]\frac{1}{2}[/itex]ln(2[itex]\pm \sqrt{3}[/itex])
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.