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Is it possible to solve for x this kind of equation? Since 1,2,3 are not multiples to each other I see a priori no possibility. $$1^x+2^x+3^x=0; x?$$

Computing this on Wolfram Alpha, for example, returns two complex solutions for $x$. Wolfram Alpha usually returns a formal and algebraical expression to represent the solution, but this time no formal one is returned, instead, just a long, infinite and complex number with seems to be the result of an iterating computation rather than a formal way to solve it.

You can check the results with Newton-Rhapson's method in my GeoGebra's plot.

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    $\begingroup$ I've voted this up as it leads to an interesting project. $\endgroup$
    – Martin Hansen
    Commented Mar 9, 2019 at 8:29
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    $\begingroup$ For the complex zeros $|2^z+3^z| < 1 $ for $ \Re(z) > 1$ and $|3^{-z}+(3/2)^{-z}| < 1$ for $\Re(x) < -1$ so the zeros are located in a strip $\Re(x) \in [-1,1]$. There is at least one zero at $-0.454397 \pm 3.59817i$ and by almost periodicity there are infinitely many of them, almost linear spaced vertically. $\endgroup$
    – reuns
    Commented Mar 9, 2019 at 9:02
  • $\begingroup$ I guess one of the problems is I pretend to find an analytical solution for a transcendental equation... $\endgroup$
    – Javier
    Commented Mar 9, 2019 at 16:45
  • $\begingroup$ @Fco. Javier López I've added an image of the zero at -0.45+3.59 plus a link to the free webpage complex grapher used to obtain it $\endgroup$
    – Martin Hansen
    Commented Mar 10, 2019 at 10:15

2 Answers 2

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It certainly can't have any real value solutions since, $$1^x+2^x+3^x=0$$ $$1+2^x+3^x=0$$ $$2^x+3^x=-1$$ And $a^x>0$ for all real $x$ and all real $a>0$.

So it makes sense that if you want a solution, the search needs the extended number system of complex numbers.

If you ask Wolfram Alpha to solve $$2^x+3^x=-1$$ it will give you the numerical solutions $$x=-0.454397 \pm 3.59817i$$

If you want to try and obtain these by hand you could try applying the numerical Newton Raphson method to the equation; it works with complex numbers as well as the reals, but the arithmetic may be daunting especially if you are interested in solutions to $$\sum_{k=1}^{n} k^x=0$$

To explore this properly would need software that can apply the Newton Raphson method in complex numbers and show you what it's doing.

A quick google search makes me think Maxima may do it for free : http://maxima.sourceforge.net/ or GNU Octave : http://www.gnu.org/software/octave/ although I have no experience of either, or even if they can cope with complex numbers.

The irony is, of course, that you ideally want to start near a solution. I see this is becoming an interesting project; be good to graph the modulus of the complex function for small values of k, as often done for Zeta function.

Update: There's a free interactive Complex Grapher embedded within a webpage which I've used to look at the zero at -0.45+3.59i which is shown at the black centre of the coloured circle.

The zero at -0.45+3.59i

The webpage is here : https://www.complexgrapher.com/

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  • $\begingroup$ I know. But how do you solve it? $\endgroup$
    – Javier
    Commented Mar 8, 2019 at 22:15
  • $\begingroup$ I've added the numerical answers to my post, but I agree with your original question which said these look like solutions obtained by iteration rather than symbolic manipulation $\endgroup$
    – Martin Hansen
    Commented Mar 8, 2019 at 23:21
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    $\begingroup$ The key problem is the initial guess. Suppose that we have the equation $\sum_{k=2}^n k^x=-1$ where $n$ is given, any idea for $x_0=a+b i$ ? $\endgroup$
    – Claude Leibovici
    Commented Mar 9, 2019 at 4:16
  • $\begingroup$ @Claude Leibovici I've added a reply to my answer as the limit on the length of comment prevented it going here. $\endgroup$
    – Martin Hansen
    Commented Mar 9, 2019 at 8:20
  • $\begingroup$ Keep in mind that this is just ONE solution. Depending on the branch cut, there are many other solutions. $\endgroup$
    – Fomalhaut
    Commented Mar 15, 2019 at 17:45
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Too long for a comment.

After @Martin Hansen's answer, I had a look to he zero of equation $$f_n(x)=\sum_{k=1}^{n} k^x$$ (give a lot of attention to @reuns's comment).

I used Newton method. For the base case $(n=3)$ starting with $x_0=-1+\pi\,i$ the iterates are

$$\left( \begin{array}{cc} 1 & -0.501820 \,+ \,3.86512 \,i \\ 2 & -0.487049 \,+ \,3.58176\,i \\ 3 & -0.454116 \,+ \,3.59875 \,i \\ 4 & -0.454397 \,+ \,3.59817\, i \end{array} \right)$$

Below are given some results which could be a good start (I hope) for a deeper exploration (as @Martin Hansen suggested). $$\left( \begin{array}{cc} 3 & -0.454397\, + \,3.59817 \, i \\ 4 & -0.625971\, + \,3.12712 \, i \\ 5 & -0.714285\, + \,2.83349 \, i \\ 6 & -0.767633\, + \,2.62901 \, i \\ 7 & -0.803209\, + \,2.47644 \, i \\ 8 & -0.828584\, + \,2.35711 \, i \\ 9 & -0.847585\, + \,2.26049 \, i \\ 10 & -0.862348\, + \,2.18022 \, i \\ 11 & -0.874153\, + \,2.11212 \, i \\ 12 & -0.883812\, + \,2.05341 \, i \\ 13 & -0.891868\, + \,2.00209 \, i \\ 14 & -0.898692\, + \,1.95672 \, i \\ 15 & -0.904551\, + \,1.91623 \, i \\ 16 & -0.909640\, + \,1.87978 \, i \\ 17 & -0.914103\, + \,1.84675 \, i \\ 18 & -0.918051\, + \,1.81661 \, i \\ 19 & -0.921571\, + \,1.78897 \, i \\ 20 & -0.924730\, + \,1.76349 \, i \end{array} \right)$$

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  • $\begingroup$ How long did it take to you to converge to the root? (How many iterations) $\endgroup$
    – Javier
    Commented Mar 9, 2019 at 15:39
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    $\begingroup$ @Fco.JavierLópez. It is not bad. I shall add details tomorrow morning. By the way, if I may ask, where are you located ? $\endgroup$
    – Claude Leibovici
    Commented Mar 9, 2019 at 17:15
  • $\begingroup$ @ClaudeLeivici I'm located in Spain. $\endgroup$
    – Javier
    Commented Mar 9, 2019 at 17:25
  • $\begingroup$ @Fco.JavierLópez. Pues, somos vecinos, amigo ! $\endgroup$
    – Claude Leibovici
    Commented Mar 10, 2019 at 4:24

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