How to smoothly bound Newton's iteration within known interval (not the interval Newton method) Let's say that we have a $C^\infty$ differentiable function , $f(x)$ and we either know apriori, or make the ansatz/postulate that a root which we are seeking is known to lie in some interval [m-1,m+1] where m is the midpoint. 
What are some ways of modifying the function $f(x)$ so that the Newton iteration
$N_f (t) = t - \frac{f (t)}{\dot{f} (t)}$
cannot go outside of the interval if started from a point within the interval (which is generally different from the midpoint)
Using min/max, or piecewise functions, is not a solution because these functions are not differentiable, and the iteration would simply reach the boundary of the interval and stay there.
Squashing $\frac{f (t)}{\dot{f} (t)}$ with $\tanh \left( \frac{f (t)}{\dot{f} (t)} \right)$ will also not work in this case, since it could still escape the interval by making multiple hops even though each hop $x_{n+1}$ would move no greater than 1 from $x_n$
I thought about introducing a repelling fixed point at the ends of the interval but I'm not quite sure how to do that.
I'm not looking for the interval Newton method, that's something different.
thanks in advance 
based on the comment below , the mapping function 
$$f(t) = a+{(b-a) \left( 1+{{\rm e}^{-{\frac {t-a}{b-a}}}} \right) ^{-1}}$$
can be used, here, with $a=308.36982554421727540$  and $b=310.36982554421727540$
the inverse function is 
$${\it invf} \left( t \right) =a\ln  \left( -{\frac {b-t}{-t+a}}
 \right) -\ln  \left( -{\frac {b-t}{-t+a}} \right) b+a$$

and the function I am trying to find the roots of, the Hardy Z function in this case, is 

where the root of the modified function isn't even found in the interval.. it seems that I need the mapping function $f(t)$ to map to a and b much faster than it is, since indeed the root is there, but not in the interval. If I go ahead and find that root and invert the mapping, thats not the root either.
The function Z(f(t))  has  its root at 312.16854074213271881 so it cannot be inverted to a real number since it lies outside the interval, it evaluates to 309.86500608950761935-6.28318530717958647*I 
 A: You cannot quite do what you ask for, because if there is any point $p$ within the interval that has a very small derivative of $f$ and a nonzero value of $f$ such that $f/'f$ then $p$ will take the next guess outside the interval, regardless of whether you add a "barrier" at the end of the interval.
However, you can do what you want as follows:
Consider the function 
$$
g(x) = f\left(\frac2\pi \tan^{-1}x\right)
$$
Use the ordinary Newton's method to solve $g(x) = 0$.  The set of guesses  cannot leave the interval of $(-1,1)$ in the argument of $f(t)$ since the arctan is always between $-\frac\pi 2$ and $\frac\pi 2$.  When you have your solution $x_0$, then the solution of $f(t)=0$ is at
$$
t=\frac2\pi \tan^{-1}x_0
$$
A: You know that the root $x$ you are looking for is such that $a<x<b$ and your solver is not able to handle bounds.
What you can do is to define $$x=a+\frac{b-a}{1+e^{-t}}$$ which then gives an infinite domain for $t$.
Start Newton at $t=0$ which corresponds to $x=\frac{a+b}2$.
A: Time traveller is here. Perhaps you're looking for a Interval Newton's method wiki not wiki. The idea is to iterate intervals $[a_k..b_k]$. Defining a point on each interval, say $x_k\equiv (a_k+b_k)/2$ we

*

*get new interval

$$
[a_{k+1}..b_{k+1}] = x_k - f(x_k)/f'([a_k..b_k]),
$$
Note that left side of equation is not simply $[x_k - f(x_k)/f'(a_k) \,\, .. \,\, x_k - f(x_k)/f'(b_k)]$, see here


*find its intersection with the previous

$$
[a_{k+1}..b_{k+1}] := [a_{k+1}..b_{k+1}] \cap [a_k..b_k].
$$


*repeat.

