How do you determine which representation of a function to use for Newton's method? Take the equation:
$$x^2 + 2x = \frac{1}{1+x^2}$$ 
I subtracted the right term over to form $~f_1(x)~$:
$$x^2 + 2x - \frac{1}{1+x^2} = 0$$ 
I wanted to take the derivative, so I rearranged things to make it a bit easier, call it $~f_2(x)~$:
$$x^4 + 2x^3 + x^2 + 2x - 1 = 0$$
I noticed when I graphed $~f_1(x)~$ and $~f_2(x)~$ that their plots were different $($although they shared the same solution for $~x)~$.
Newton's method iterates down the graph line, so I'd imagine that Newton's method for these two equations are not equivalent. They'd find the same solution, but they would get there different ways. In that case, is there a way to decide which equation to use for Newton's to obtain the best/quickest result?
 A: It depends on the derivative of your function at the zeros of your function.
The formula suggests that  $$x_{n+1} = x_n-\frac {f(x_n)}{f'(x_n)}$$
Thus you get a better (faster) result if the derivative of your function is higher at the zeros of $f(x)$. 
The process becomes very chaotic if your derivative at a 
 staritng point is close to zero.
As a very interesting case you may consider $$y=\sin x$$ and try to find $x=\pi$ starting at a point close to $\pi/2$ 
Without knowing the zeros of $f(x)$  making a decision is not an easy task. 
A: It is not difficult to show that there are just two real solutions, since


*

*$x^2+2x$ is decreasing from $+\infty$ to $-1$ on $(-\infty,-1]$, while $\frac{1}{x^2+1}$ is increasing from $0$ to $\frac{1}{2}$;

*over $[-1,0]$ we have $x^2+2x\leq 0$ but $\frac{1}{x^2+1}>0$;

*$x^2+2x$ is increasing from $0$ to $+\infty$ on $\mathbb{R}^+$, while $\frac{1}{x^2+1}$ is decreasing from $1$ to $0$.


Let's say we are interested in finding the positive root, which lies in $I=\left[0,\frac{1}{2}\right]$.
Over such interval $a(x)=x^2+2x$ is increasing and convex, $b(x)=\frac{1}{x^2+1}$ is decreasing and concave.
This suggests a tweak of the Newton-Raphson iteration:

We are interested in the intersection between the blue and orange curves, which lies on the left of the intersection between the sienna and the purple tangent lines. By solving 
$$ 3x-\frac{1}{4} = \frac{28}{25}-\frac{16}{25}x $$
we find that $\frac{137}{364}$ is already a good approximation of the solution in $[0,1]$. Newton's method (applied to $a(x)-b(x)$) with such starting point is granted to converge quadratically (and in a monotonic way) to the actual solution $0.3708102352\ldots$
A: Newton's method is very robust once you are near a root - exactly what function you chose to run it on is not of that much consequence; it will converge quickly no matter what. You might as well just use whatever is easiest.
To be more precise, let $f$ be the function we wish to find a root of and define $g(x)=x-\frac{f(x)}{f'(x)}$ to be the function performing a single step of Newton's method. First: it should be clear that things go badly if $f'$ and $f$ share a root, so let's assume they don't - and we'll assume that $f$ is smooth near the root, for convenience.
The idea is that $g$ pulls everything near a root to the root - and to quantify how fast, we can take a look at its Taylor series at a root. Let $x_0$ be a root of $f$. If we differentiate $g$ then simplify at $x_0$ by noting that $f(x_0)=0$, we can see the following:
$$g(x_0)=x_0$$
$$g'(x_0)=0$$
$$g''(x_0)=\frac{f''(x_0)}{f'(x_0)}$$
In particular, this tells us that, once we are close enough to $x_0$, if the current amount of error is $e$, one more application of $g$ makes the error something more like $g''(x_0)e^2$ - this is pretty fantastic irrespective of what $g''(x_0)$ is - if you take a small number and square it over and over and over, it gets really small really fast - which is what we want to happen to the error. I suppose if you were really eager, you could try to make $g''(x_0)$ small by choosing $f$ carefully - but realistically, it does not matter, since it tends to be the case that running an additional step of the method completely dwarfs any gains from fine-tuning $f$.

This said, Newton's method can behave pretty badly on a global scale - even for polynomials it can do really nasty things*. Choosing your function carefully isn't likely to help you much here either, since when functions have really small derivatives, the method is unstable, but when they have rapidly growing derivatives (like near an asymptote), the method converges very very slowly - basically, this isn't a good method to look for a root without having any idea of where one might be, and fine tuning $f$ won't help you there either.
(*Example: Look up "Newton's method basins of attraction" on Google - you'll get these amazing plots of where the method converges if you run it on polynomials as simple as $x^3-1$ starting at various points in the complex plane. Basically, when it's not near a root, anything can happen)
