# “Multi-stage logarithm” series expansion (e.g. $a^x+b^x+c^x=d$)

As everyone knows, the solution to $a^x=b$ is $x=\log_a{b}$. (Edit: Corrected from $x=\log_b{a}$.)

But what about $a^x+b^x=c$?

Let's define a "multilogarithm" function as:

$a_0^x+a_1^x+...+a_n^x=b$

$x=\text{Lg}\left(b\mid a_0, a_1,...,a_n\right)$.

For example, $3^x=40000$ gives $x=\log_3\left(40000\right)$, and $2^x+e^x+3^x+10^x=20000$ would give $x=\text{Lg}\left(20000\mid 2, e, 3, 10\right)$.

First, is there an infinite series for a logarithm that converges for all positive numbers?

And what would be an appropriate series for the multilogarithm?

Is there already a name for this type of function?

• When you say "series" do you mean "power series"? – Milo Brandt May 12 '18 at 15:18
• Not necessarily. Any function possible – El Ectric May 12 '18 at 15:20
• The ordinary logarithm has an singularity at $x=0$, so no entire series can have an infinite radius of convergence. – Yves Daoust May 12 '18 at 17:59
• I considered this a while back, I wrote a few notes but didn't get very far: authorea.com/users/5445/articles/… You might even consider an infinite sequence as a base $$\text{Lg}(\zeta(s)|1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots) = s$$ – Benedict W. J. Irwin Jun 1 '18 at 14:45

May be an idea for a sequence (not a series).

Let us suppose (for the time being) that $1 <a_0 <a_1< \cdots < a_n$, $b >n+1$ and that a solution exists.

So, we want to solve $$\log(a_n^x)=\log\left(b-\sum_{i=0}^{n-1} a_i^x\right)$$ that is to say to find the zero of function $$f(x)=x \log(a_n)-\log\left(b-\sum_{i=0}^{n-1} a_i^x\right)$$ Newton iterates could be a way to build a converging sequence (being lazy, let us start with $x_0=0$ (knowing in advance that we shall face one overshoot of the solution since $f(0)\times f''(0) < 0$ - Darboux-Fourier theorem).

Let me try with $a_n=p_{n+1}$ and $n=10$ and $b=10^6$

Newton iterates would be $$\left( \begin{array}{cc} k & x_k \\ 0 & 0.000000000 \\ 1 & 4.102824161 \\ 2 & 3.988842463 \\ 3 & 3.935316697 \\ 4 & 3.930965343 \\ 5 & 3.930944175 \end{array} \right)$$

I am sure that we can build a better starting point. Using for axample $x_0=\frac{\log(b)}{\log(a_n)}$, Newton iterates would be almost the same $$\left( \begin{array}{cc} k & x_k \\ 0 & 4.102850256 \\ 1 & 3.988866252 \\ 2 & 3.935320622 \\ 3 & 3.930965381 \\ 4 & 3.930944175 \end{array} \right)$$

For the inverse function, define the notation $\;E(x | {\bf a}) = E(x | a_1, a_2,\dots,a_n) := \sum_{i=1}^n e^{a_ix}.\;$ Define the power sums $\;p_k = p_k({\bf a}) := \sum_{i=1}^n a_i^k \;$ which are used in the power series expansion of our function $\;E(x| {\bf a}) = \sum_{k=0}^\infty p_k({\bf a})\frac {x^k}{k!} = n + (a_1 + \dots + a_n)x + \cdots.\;$ The inverse function $\;L(x|{\bf a})\;$ is implicitly defined by $\;x = L( E( x | {\bf a}) | {\bf a}) = E( L( x | {\bf a}) | {\bf a}).\;$ Its power series expansion is given by $$\;L( x | {\bf a}) = \frac{1}{p_1} \frac{(x-n)^1}{1!} - \frac {p_2} {p_1^3} \frac{(x-n)^2}{2!} + \frac {3p_2^2-p_1p_3} {p_1^5} \frac{(x-n)^3}{3!} - \cdots .$$ It has a finite radius of convergence just like $\;\ln(x) = \frac{(x-1)^1}{1} - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots.\;$