Why is it not possible to generate an explicit formula for Newton's method? Going through the recursive formula for approximating roots every time is extraordinarily tedious, so I was wondering why there was no formula that computed the $n$th iteration of Newton's method.
 A: First, note that such a formula would violate the "too good to be true" test: it would let you find the minima of any (sufficiently smooth, strictly) convex function in a single step, no matter how complicated!
For intuition, consider what Newton's method is doing: you are starting from an initial guess, and taking a sequence of "downhill steps" until you reach the minimum, where each step uses local information (a quadratic local approximation) about your neighborhood to compute the next destination. Imagine a hiker climbing down a mountain, who first hikes to the lowest point she can see, then reorients herself and goes to the lowest point she can see from her new location, etc.
The only way to make a direct beeline to the lowest point ($n\to\infty$) is if the hiker somehow knows the entire shape of the landscape in advance. In mathematical terms, this means your formula would have to use either:


*

*information about the value of your function at all points, or equivalently (for nicely-behaved functions)

*the values of all derivatives of all orders at your starting point.


And of course neither is practical in practice. Note though that sometimes you do have such complete information, e.g. when your function is quadratic, so that all higher-order derivatives are zero. And then you can easily make a beeline straight to the solution.
Also, you can write down a formula for the $n$th iteration, but as stated in the other answers, this is not useful: it gets messier and messier the larger $n$ gets (as it must, as explained above), doesn't converge to anything nice as $n\to \infty$, and amounts to nothing more than... running Newton's method for $n$ consecutive steps.
A: A tentative explanation:
As Newton's iterations converge to the same root, starting from many initial values, the function that associates the initial value to the iterate is very "flat".
Her is an illustration with the square root function:

As the expression of the iterates is a rational fraction, the degree of the numerator and denominator must increase to achieve this flatness. I have no explanation why the number of terms increases, though.
A: You wouldn't really gain much with a formula for the $n$th iteration.
The appeal of the Newton-Raphson method is that a single step:


*

*is conceptually easy to understand, 

*is fast to compute, and

*can be checked from step to step for closeness to a stationary point.


Nesting $n$ of these computations into a single formula works against all three of these things.
A: Confining the task to the square root X function, there are is a shortcut you can take.   In most computer arithmetic, division is equal in work to about three multiplications.   So once you have the square root to two or three decimal places, take half its reciprocal.  Adjust that number to K nearby that is a fast multiply (at most two internal one bits or two internal zero bits).   Then adjust your guess G <= G-K(G^2-X).  This no longer has quadratic convergence, but each step requires only one nasty multiplication and one simple multiplication.  
Working with the TI model 30 business analyst (the only calculator permitted on the actuary exams), I could implement this easily.  I stored the 1/2f'(X) in one of the several memories and started with a small value.   Anything in the ballpark will give you convergence.   If you get divergence, just reverse the sign. If the result undershoots, increase the magnitude, if the result oscillates above and below some value, decrease the magnitude of the stored correction factor.
Newton's method fails in some notable cases.   For example Ln(X) = 1, which goes nuts if your initial guess is greater than e.  (This is related to the diode equation, which we now solve by other means without itera tion).
