TITLE: How do you determine which representation of a function to use for Newton's method?
QUESTION [21 upvotes]: 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?

REPLY [2 votes]: 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 $0.3708102352\ldots$