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In trying to find complex zeros of equation $3^z+4^z=5^z$, we could divide both sides by $(\sqrt{15})^z$ to transform it into $(\frac{4}{\sqrt{15}})^z=(\sqrt{\frac53})^z-(\sqrt{\frac35})^z$. Let's write $a=\log\sqrt{\frac53}\approx 0.255, b=\log\frac4{\sqrt{15}}\approx 0.032$, we could transform the equation into $f(z)=\sinh(az)-\frac12\exp(bz)=0$.

Let $u_0=-1.785...$ which is real root of $3^x=4^x+5^x$, for any $u<u_0, v>2$ lines $\Re(z)=u, \Re(z)=v, \Im(z)=\frac{(k+\frac12)\pi i}a$ forms some rectangles and it is easy to show that in edges of those rectangle, $|\sinh(az)|\gt |\frac12\exp(bz)|$. By applying Rouché's Theorem, we could get that f(z) has exact one zero inside each rectangle, same as function $\sinh(az)=0$.

Next, starting from each zero of function $\sinh(az)=0$, or $z_0=\frac{k\pi i}a$, and using Newton's method that $z_{h+1}=z_h-\frac{f(z)}{f'(z)}$, computer shows that it will finally converges to the zero of f(z) in the rectangle.

Another very interesting result is that after sorting all zeros of f(z) by imaginary part, the difference of neighbour zeros are very close to an ellipse $\frac{x^2}{3.73^2}+\frac{(y-\frac{\pi}a)^2}{3.96^2}=1$.

Another example is polynomail $z^{n+m}-2z^n+1=0$, we could transform it into $z^m-2+1/z^{-n}=0$ and shows that there're exactly m roots with absolute value more than 1. Similarly, computer shows that starting from any root of $z^m-2=0$, we could reach different root of the equation by using Newton's method.

Is there any special mathematics relationship between the Rouché's Theorem and Newton's method?

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  • $\begingroup$ What sort of relationship are you looking for? Newton's Method helps you algorithmically find roots, and Rouche's Theorem tells us about the distribution of roots. $\endgroup$ Sep 25, 2021 at 3:10
  • $\begingroup$ Yes. But Newton's Method does not tell which root the iteration will end. How could we make sure that starting for $k\pi i$, all roots could be found (and no two of them will generate same root?) $\endgroup$
    – Zhaohui Du
    Sep 25, 2021 at 5:22
  • $\begingroup$ In theory, once you find a root, you can remove it. Suppose you have some meromorphic function $f$. Every time you find a root $z_0$ of multiplicity $k$ for $f$, you modify $f$ to be $(z-z_0)^{-k}f(z)$ so that $f$ is some nonzero constant about $z_0$, and then run Newton's Method again. Concerning where Newton's Method will land, we actually get a fractal.... it is night impossible to guess where Newton's method will land if on the edge of the fractal, so I don't imagine you'll get a very explicit relationship between Rouche's Theorem and Newton's Method $\endgroup$ Sep 25, 2021 at 6:19
  • $\begingroup$ For more on the fractal we get, see the Wikipedia page on Newton's fractal. On the plus side, we know that Newton's Method converges quadratically in a small enough neighborhood around a root, so I suppose one connection is that you could use Rouche's theorem repeatedly to narrow in on the roots, namely applying it enough to find a small enough neighborhood about some root to get quadratic convergence. $\endgroup$ Sep 25, 2021 at 6:19
  • $\begingroup$ Yes. Now I have found that if we start from a neighborhood whose distance to the root is no more than 0.8, it will converge to the root because $|f(z)f''(z)|\lt |f'(z)^2|$. But the distance from $\frac{k\pi i}a$ to the root is around 2. $\endgroup$
    – Zhaohui Du
    Sep 25, 2021 at 14:00

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We could find that for function $f(z)=z\exp(-2.9z)$ and $g(z)=(\frac13z^2+z+\frac13)\exp(-2.9z)$.

In circle $|z|=1$, we have $|f(z)-g(z)|\le |f(z)|$ But if we started unique root (z=0) of f(z) inside the circle and using Newton's method for g(z), finally, it converges to root of g(z) outside the circle instead of the one inside circle.

It looks like a special exception around z=0:

Newton's method result for g(z) where <span class=$-5\le \Re z\le 5, -5\le \Im z\le 5$" />

Points in same color family means they will finally converge to same root and points with darker color converges faster.

Newton's method result for g(z) where <span class=$-0.5\le \Re z\le 0.5, -0.5\le \Im z\le 0.5$" />

We could find a small green region around z=0.

Below is corresponding result for $sh(az)=\frac12 exp(bz)$

Newton's method result for <span class=$sh(az)=\frac12 exp(bz)$ where $-10\le \Re z \le 10, -100\le \Im z\le 100$" />

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