Newton's method convergence criteria To use Newton's method on interval $[a,b]$ we need to guarantee that


*

*$f(a)f(b)<0$ on the interval which is true for $[0,1]$. 

*$f'(x)$ and $f''(x)$ are continuous on the interval $[a,b]$ (which they are)

*$f'(x) \neq 0$ on $[a,b]$ and $f''(x)$ does not change signs on $[a,b]$


Then if $f(x_{0})f''(x_{0})>0$ the method will converge.
I have the following exercise in my textbook and I don't understand a part of the solution:
Using Newton's method and starting approximation $x_{0} = 1$ find the root of the equation $$f(x)=e^{x}-\sin(x)-1.5$$
Given solution:
$$f'(x) = e^{x}-\cos(x)$$
$$f''(x) = e^{x}+\sin(x)$$
$f'(x)$ and $f''(x)$ are both positive on the given interval, we can therefore use it.
We verify that $f(0)f''(0)>0$ and then start the algorithm itself.
My questions for this part of the solution:


*

*How can I determine what are the function values for a function like $f'(x)$ or $f''(x)$? Is it possible to graph them by hand?

*$f'(x)=0$ for $x=0$ which seems to be in contradiction with the convergence criteria. Is the solution right?


Thanks.
 A: I don't understand what you mean by your first question. If you know the explicit form, simply compute it. Otherwise, note that $e^x > 1$ for $x > 0$ and observe that $\cos x$ is bounded above by $1$. Therefore, $f'(x)$ must be positive over the positive reals. The same argument holds for $f''(x)$.
For your second question, since $f(0) = -0.5$, and $f(x)$ is continuous, we know that there is a neighborhood around $0$ where $f$ does not have a root. So consider an interval $[\epsilon, 1]$. That elides the problem in your first derivative criterion.
A: For your first question, you will certainly need (for your particular functions) some method to numerically approximate the values of $f'$ and $f''$ at the intermediate values --- a calculator or computer algebra system.
For your second question, all that is really required for guaranteed convergence is that $f'(x)\ne 0$ in some neighborhood of the root; in your case, since $f(0) \ne 0$ and $f$ is continuous, that is guaranteed because $f'(x) \ne 0$ in $(0,1]$.
