# Analytical approximate solution to a trascendental equation

I have the following equation to solve $$z+e^{z^2}\operatorname{erfc}(z)=0$$ being $$\operatorname{erfc}(z)=1-\frac{2}{\sqrt{\pi}}\int_0^ze^{-s^2}ds.$$ I solved it numerically and appears to have an infinity of complex solutions. I am in need to find an approximate analytical approximation for the first root. I made several attempts, e.g. using either Taylor or asymptotic expansions, but to no avail. I was not able to recover the first root that seems to be $$-.3378723369\ldots-.9499598705\ldots i$$.

My question: is it possible to find a successful approximation scheme to get an analytical expression for the roots of the given equation?

Edit: There appear to be infinite roots in $$\mathbb{C}$$. This can be seen straightforwardly in Fig. 1 where both the real and imaginary part are given of $$-z$$ and $$e^{z^2}\operatorname{erfc}(z)$$

Orange is $$Re(-z)$$, green is $$Im(-z)$$. There is a common line between this twos crossing the part with the erfc showing a possible infinite set of roots.

I was able to find several roots by choosing a different starting point for the search algorithm. E.g. I have got, listing just a few along with their complex conjugates implied,

$$-1.90593 - 1.82833 i$$

$$-3.19368 - 3.06637 i$$

$$-4.07604 - 3.94595 i$$

$$-6.71931 - 6.60297 i$$

$$-8.00551 - 7.89692 i$$

$$-10.2496 - 10.1527 i$$

The aforementioned first root appears a recurring one in the search.

• Please, have a look to my edit. I do not think that the equations has many roots. Did you find other solutions ? If you did, please tell me. Cheers. – Claude Leibovici Feb 13 at 4:39
• @ClaudeLeibovici From the plot I made with Mathematica, it seemed to me that the number of roots was large. E.g. I also get $-5.114532877548192 + 4.9884679626263475i$ and finding or not another root depends on the starting point in FindRoot.. I think you can easily check by yourself that the one I showed you is another zero. – Jon Feb 13 at 7:09
• Have a look to my edit. Using the contour plot of $\Phi(x,y)=1$, I had to zoom crazy to "see" the solution you reported in your last comment (in fact, I did not see it - my wife did it for me !). I wonder if we could face some accuracy problems here with $\Phi(x,y)$, – Claude Leibovici Feb 13 at 8:01
• Well, I plotted the imaginary and real part in a 3d plot and drawn the x and y plans. It is seen in this way that there seems to be a multitude of roots. Maybe, a contour plot could be not so easy to read. I used FindRoot starting from $-5-5i$ and the root come out. – Jon Feb 13 at 9:04
• Good to know that ! It would be good you list in the post some of the roots. I am sure that you know that if $(a+ib)$ is a root $(a-ib)$ is another one. – Claude Leibovici Feb 13 at 9:42

You can obtain the first root using Newton method.

Using for example $$z_0=1+i$$, it would give the following iterates $$\left( \begin{array}{cc} n & z_n \\ 0 & 1+ i \\ 1 & -0.570752+0.455687 i \\ 2 & -0.217277+0.691623 i \\ 3 & -0.298698+1.012267 i \\ 4 & -0.336117+0.946645 i \\ 5 & -0.337868+0.949970 i \\ 6 & -0.337872+0.949960 i \end{array} \right)$$

Using $$z_0=1-i$$, it would give the following iterates $$\left( \begin{array}{cc} n & z_n \\ 0 & 1-i \\ 1 & -0.570752-0.455687 i \\ 2 & -0.217277-0.691623 i \\ 3 & -0.298698-1.012267 i \\ 4 & -0.336117-0.946645 i \\ 5 & -0.337867-0.949968 i \\ 6 & -0.337872-0.949960 i \end{array} \right)$$

Starting with $$z_0=i$$ seems to converge faster $$\left( \begin{array}{cc} n & z_n \\ 0 & i \\ 1 & -0.400169+0.909374 i \\ 2 & -0.339614+0.946181 i \\ 3 & -0.337864+0.949950 i \\ 4 & -0.337872+0.949960 i \end{array} \right)$$

The same starting with $$z_0=-i$$ $$\left( \begin{array}{cc} n & z_n \\ 0 & - i \\ 1 & -0.400169-0.909374 i \\ 2 & -0.339614-0.946181 i \\ 3 & -0.337864-0.949950 i \\ 4 & -0.337872-0.949960 i \end{array} \right)$$

Concerning approximations, I am stuck. The only thing I really tried is a series expansion around $$z=i$$ and got $$\frac{-i \text{erfi}(1)+1+i e}{e}+\frac{\left(2 \text{erfi}(1)+2 i+e-\frac{2 e}{\sqrt{\pi }}\right) (z-i)}{e}+\frac{\left(i \text{erfi}(1)-1-\frac{2 i e}{\sqrt{\pi }}\right) (z-i)^2}{e}+O\left((z-i)^3\right)$$ Ignoring the higher order terms, solving the quadratic in $$(z-i)$$ and evaluating leads to $$z\approx -0.351584 + 0.947053 i$$ Doing the same around $$z=-i$$ gives the other root.

Edit

In terms of approximation of the solution, we can do better using the simplest $$[1,1]$$ Padé approximant built at $$z_\pm=\pm i$$. For example, $$z_+=-\frac{i \left(-4 e \left(\pi -2 \sqrt{\pi }\right) (\text{erfi}(1)+i)-5 \pi (\text{erfi}(1)+i)^2+4 e^2 \left(\sqrt{\pi }-1\right)\right)}{e \left(5 \pi -6 \sqrt{\pi }\right) (\text{erfi}(1)+i)+3 \pi (\text{erfi}(1)+i)^2+e^2 \left(4-6 \sqrt{\pi }+\pi \right)}$$ which is $$-0.347274 + 0.955038 i$$ much better than the previous one.

We could continue with $$[1,n]$$ Padé approximant to get explicit approximations (the expressions are really messy). The values are reporteds below $$\left( \begin{array}{cc} n & z^{(n)}_+\\ 0 & -0.400169+0.909374 i \\ 1 & -0.347274+0.955038 i \\ 2 & -0.337395+0.951302 i \\ 3 & -0.337591+0.949901 i \\ 4 & -0.337859+0.949907 i \end{array} \right)$$

Update

Looking at the table of the roots given in Jon' edit, they really look like $$z_k \sim -k \pm ki$$. Using the $$[1,1]$$ Padé approximant built at $$z=-k-ik$$ (the formula is too messy to be reported here) an approximation of the root can be obtained.

Below are listed some values (to be compared to Jon's) $$\left( \begin{array}{cc} k & \text{approximation} \\ 1 & -0.34995-0.95803 i \\ 2 & -1.90291-1.81163 i \\ 3 & -3.23300-3.00433 i \\ 4 & -4.06884-3.94940 i \\ 5 & -5.10001-4.96837 i \\ 6 & -6.05119-5.85692 i \\ 7 & -7.09451-6.88121 i \\ 8 & -8.03695-7.91822 i \\ 9 & -9.07841-8.95282 i \\ 10 & -10.0648-9.96230 i \\ 11 & -11.0449-10.9348 i \\ 12 & -12.0518-11.9671 i \\ 13 & -13.0490-12.9660 i \\ 14 & -14.0285-13.9565 i \\ 15 & -15.0394-14.9589 i \end{array} \right)$$

• Thanks a lot. Very nice. – Jon Feb 12 at 13:14
• Very neat answer. This is what I needed. Thank you very much. – Jon Feb 13 at 7:03