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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?

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    $\begingroup$ To restate, an equation can be rewritten in various ways to express the solution as the root (zero) of a function. You ask about Newton's method, but note that it is a special case of finding a fixed point of a function by iteration. So already Newton's method can be said to involve rewriting an equation to expedite convergence to a root. $\endgroup$ – hardmath Sep 16 at 14:44
  • $\begingroup$ Related to your question, see this post for a proof of a simple condition that guarantees the desired quadratic convergence of Newton's method to a root $r$ of $f$: $f'(r) ≠ 0$ and there is some constant $m$ such that $|f''(x)/f'(r)| ≤ m$ for every $x∈dom(f)$, and your initial guess is at most $1/(3m)$ away from $r$. $\endgroup$ – user21820 Sep 17 at 12:41
  • $\begingroup$ math.stackexchange.com/questions/3078167/… is a rather good example of what I have been doing for 60+ years. $\endgroup$ – Claude Leibovici Sep 17 at 15:13
  • $\begingroup$ To give you an idea of the impact. Almost 25 years ago, I spent something like three months of work (full time) trying to make a function "more" linear. I saved $0.001$ seconds per run (CPU time). Big achievement, isn't it ? The only problem is that a single simulation was solving that equation $\sim 10 ^7$ times.which means that we saved almost $3$ hours per simulation. $\endgroup$ – Claude Leibovici Sep 18 at 9:42
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For your curiosity.

Asking this far-from-innocent question, you are almost asking me what I have been doing during the last sixty years.

My answer is : what is the transform of $f(x)$ which makes it the most linear around the solution ? If you find it, you will save a lot of iterations.

One case I enjoy for demonstration is $$f(x)=\sum_{i=1}^n a_i^x- b^x$$ where $1<a_1<a_2< \cdots < a_n$ and $b > a_n$; on this site, you could find many problems of this kind.

As written, this function is very bad since it varies very fast and it is very nonlinear. Moreover, it goes through a maximum (not easy to identify) and you must start on the right of it to converge.

Now, consider the transform $$g(x)=\log\left(\sum_{i=1}^n a_i^x \right)- x\log(b)$$ It is almost a straight line ! This means that you can start from almost anywhere and converge fast.

For illustration purposes, trying with $n=6$, $a_i=p_i$ and $b=p_{n+1}$ ($p_i$ being the $i^{th}$ prime number). Being very lazy and using $x_0=0$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 1.607120621 \\ 2 & 2.430003204 \\ 3 & 2.545372693 \\ 4 & 2.546847896 \\ 5 & 2.546848123 \end{array} \right)$$ Using $f(x)$, you must start iterating with $x_0 > 2.14$ to have convergence (big work to know it !). Let us try with $x_0=2.2$ to get as successive iterates

$$\left( \begin{array}{cc} n & x_n \\ 0 & 2.200000000 \\ 1 & 4.561765400 \\ 2 & 4.241750505 \\ 3 & 3.929819520 \\ 4 & 3.629031502 \\ 5 & 3.344096332 \\ 6 & 3.082467015 \\ 7 & 2.856023930 \\ 8 & 2.682559774 \\ 9 & 2.581375720 \\ 10 & 2.549595979 \\ 11 & 2.546866878 \\ 12 & 2.546848124 \\ 13 & 2.546848123 \end{array} \right)$$

The last point I would like to mention : even if you have a good guess $x_0$ of the solution, search for a transform $g(x)$ such that $g(x_0)\,g''(x_0) >0$ (this is Darboux theorem) in order to avoid any overshoot of the solution during the path to it.

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  • $\begingroup$ Nice (+1)! In your example, I think the notation $p_i$ means the $i$th prime? So for $n=6$, you'd have $(a_1, a_2, \ldots, a_6) = (2, 3, \ldots, 13)$, and $b=17$? $\endgroup$ – mathmandan Sep 16 at 16:52
  • $\begingroup$ @mathmandan. This is correct. Thanks & cheers $\endgroup$ – Claude Leibovici Sep 16 at 17:18
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    $\begingroup$ I think you meant to write $$f(x)=\frac{\sum_{i=1}^n a_i^x}{b^x}$$ (with division rather than subtraction). Otherwise your "transform" to $g(x)$ gives a more-or-less unrelated function; no?) $\endgroup$ – ruakh Sep 17 at 18:49
  • $\begingroup$ @ruakh. Not at all. What you suggest is better that the original function but is almost hyperbolic while $g(x)$ is almost linear. $\endgroup$ – Claude Leibovici Sep 18 at 3:23
  • $\begingroup$ @ClaudeLeibovici: Oh, wait, I see. You never mention what equation you were solving, and I made a bad assumption: I thought you were solving the equation $f(x) = y_0$ for some (arbitrary) $y_0$, and I didn't see what was gained by "transforming" that to $g(x) = z_0$ unless there was some relationship between $g$ and $f$ (such as $g(x) = \log f(x)$). I now realize that you were solving the equation $f(x) = 0$, so your "transformation" involves converting that to $f_A(x) = f_B(x)$, then to $g_A(x) = g_B(x)$, then to $g(x) = 0$, where $g_A(x) = \log f_A(x)$ and $g_B(x) = \log f_B(x)$. $\endgroup$ – ruakh Sep 18 at 4:20
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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)

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    $\begingroup$ There are also functions, for which Newton's method diverges everywhere, also near the root. See e.g. $$y=\sqrt {\lvert x\rvert}$$ or $$y=\operatorname{sgn}(x)\cdot\sqrt{ \lvert x\rvert}$$ for which Newton's method jumps between $a$ and $-a$ for any starting point $a\ne 0.$ $\endgroup$ – CiaPan Sep 16 at 14:38
  • $\begingroup$ If $f$ has a root of multiplicity strictly greater than one, then Newton's method converges linearly. A method of order three or higher is worthwhile when very high accuracy is sort, i.e., thousands of significant figures. $\endgroup$ – Carl Christian Sep 16 at 19:48
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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.

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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:

enter image description here

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 $0.3708102352\ldots$

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