Solving $k\left(e^{a(2k+1)}-1\right)=1$ for k?

Question

I've been studying the structure of partial sums of the Dirichlet eta function and noticed that certain critical points occur when the summing from $n=1$ to $n\approx N(k)$ for $k\in\mathbb{N}$ where $$N(k)=\frac{1}{e^{ak}-1},$$ and $a$ is some constant determined by the input of the eta function.

There is one special critical point which happens when $k=N(2k+1)$, which I'd like to isolate (i.e; solve for $k$). Thus,

How do we solve (or at least approximate) for $k$ when $k\left(e^{a(2k+1)}-1\right)=1$?

Note: Trivially, this is generally unsolvable for $k\in\mathbb{N}$, however solving it for $k\in\mathbb{R}$ will still yield useful results for my purpose.

I have been fumbling with this equation for quite some time to convert it into a form suitable to apply the Lambert W function to no avail. At this point I'm not sure if this is solvable with such a method. Even WolframAlpha cannot solve it.

I presume this has no nice elementary solution, and probably involves some transcendental function / infinite series; which I am content with. Any advice?

Answer 1

The solution of the equation $$k\left(e^{a(2k+1)}-1\right)-1=0$$ is the inverse of $$a=\frac{\log \left(1+\frac{1}{k}\right)}{2 k+1}$$

Expand the rhs as series for lage values of $k$ and use series reversion to obtain with $\color{red}{b=\sqrt{\frac a 2}}$ (to avoid fractional exponents)

$$\color{red}{k=\frac 1 {2b}\Big[1-b+\frac 16 b^2\sum_{n=0}^\infty c_n\,b^{2n}\Big]}\tag 1$$ the $c_n$ forming the sequence $$\left\{1,\frac{11}{60},\frac{17}{840},\frac{-281}{100800},\frac{-44029}{19958400}, \frac{-12147139}{21794572800},\frac{-2030761}{130767436800},\frac{13609741669}{29 6406190080000}\right\}$$

Using this limited truncated series (which is $O(b^{17})$) $$k\left(e^{a(2k+1)}-1\right)-1=-\frac{397228981889 }{31141145704660992000}a^9+O\left(a^{19/2}\right)$$which is not much.

For a bit less accuracy, we can replace the summation by a $[2n,2n]$ Padé approximant and write $$\color{blue}{k=\frac 1 {2b}\Big[1-b+\frac 16 b^2\, \frac {c_1+(c_1^2-c_2)b^2 } {c_1-c_2b^2 }\Big]}\tag 2$$ which is $O(b^{7})$.

Now, using $(1)$, some results for $a=2^{-n}$

$$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 0 & \color{red}{0.27170231}440411687350110886003 & 0.27170231920910422397935487003 \\ 1 & \color{red}{0.5436268955}7861786499425948417 & 0.54362689559153720894863155701 \\ 2 & \color{red}{0.9443606736303}0013554101955879 & 0.94436067363033521102732030538 \\ 3 & \color{red}{1.521073680700180}52565415902743 & 1.52107368070018062163224524534 \\ 4 & \color{red}{2.343243204480015204}11186436522 & 2.34324320448001520437568154439 \\ 5 & \color{red}{3.51044655743237831905}233206469 & 3.51044655743237831905305896211 \\ 6 & \color{red}{5.16423050404273624607050}413428 & 5.16423050404273624607050613956 \\ 7 & \color{red}{7.5052120648670598212721246}4712 & 7.50521206486705982127212465266 \\ 8 & \color{red}{10.8173926658157477667511730966} & 10.8173926658157477667511730966 \\ 9 & \color{red}{15.5026046329576986633373698117} & 15.5026046329576986633373698117 \\ 10 & \color{red}{22.1292585867287527840972383970} & 22.1292585867287527840972383970 \end{array} \right)$$

For $n=50$, the absolute difference between the estimate and the solution is $5.025\times 10^{-127}$ (the solution itself being $k=23726566.$).

Update

For the case where $a$ is large, $k$ is small and one iteration of the next Newton-like iterative method after Fouseholder's method (with $k_0=0$) gives as a reasonable estimate $$\color{red}{k_1=\frac {3A^3+3(4a-3)A^2+3(2 a^2-4 a+3)A-3} {3A^4+6 (3 a-2)A^3+6 \left(4 a^2-6 a+3\right)A^2+\left(4 a^3-12 a^2+18 a-12\right)A+3}}$$ with $\color{red}{A=e^a}$

$$\left( \begin{array}{ccc} a & \text{estimate} & \text{solution} \\ 1 & \color{red}{0.2}6847689395796102422 & 0.27170231920910422398 \\ 2 & \color{red}{0.09}974834418269860608 & 0.09983932012886691696 \\ 3 & \color{red}{0.04060}235654552477461 & 0.04060622503324579480 \\ 4 & \color{red}{0.016334}39070827734458 & 0.01633453474367618202 \\ 5 & \color{red}{0.00636280}472550865216 & 0.00636280836365866688 \\ 6 & \color{red}{0.00241379}594072638865 & 0.00241379600562880126 \\ 7 & \color{red}{0.00090125998}824875719 & 0.00090125998916343416 \\ 8 & \color{red}{0.0003337872099}8329316 & 0.00033378720999441105 \\ 9 & \color{red}{0.000123151705248}29952 & 0.00012315170524842250 \\ 10 &\color{red}{0.00004536081839098}537 & 0.00004536081839098664 \\ 11 &\color{red}{0.00001669584598485734} & 0.00001669584598485734 \end{array} \right)$$

Answer 2

I prefer to add a second answer since based on a different approach.

Let $x=2a k$ and $A=e^a$ which makes the equation $$A e^x x=2a+x \implies e^{-x}= A \, \frac{ x}{x+2 a}$$ which has an explicit solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper). This is interesting but, unfortunately, not very practical.

Considering the function (I assume $a>0$) $$f(x)=A e^x x-x-2a$$ its solution is the inverse of $$a=-\frac{x}{2} -W\left(-\frac{x}{2} e^{\frac x2} \right)\implies x < 2 W\left(\frac{1}{e}\right)$$

If $x$ is "small", we have $$a=\frac {x^2}2 \Bigg[1+\sum_{n=1}^\infty \alpha_n\,x^n\Bigg]$$ the first coefficients being $$\left\{1,\frac{7}{6},\frac{3}{2},\frac{41}{20},\frac{263}{90},\frac{5413}{1260},\frac{3259}{504},\frac{89953}{9072},\frac{129653}{8400} ,\frac{8092127}{332640},\frac {20712491}{534600},\frac{216331303}{3474900}\right\}$$ and series reversion would give $$x=b \Bigg[1-\frac 12 b+\frac 1{24}b^2\sum_{n=0}^\infty \beta_n\,b^{2n}\Bigg]\qquad \text{with} \qquad b=\sqrt{2a}$$ the first coefficients being $$\left\{1,\frac{11}{240},\frac{17}{13440},-\frac{281}{6451200},-\frac{44029}{5109350 400},-\frac{12147139}{22317642547200}\right\}$$ More tedious is the expansion around $x= 2 W\left(\frac{1}{e}\right)$.

$$a=-\left(\frac{t}{2}+T\right)-\frac{2 (t+1) T+t}{2 t (T+1)}(x-t)-\frac{T (t-2 T) (t+2 T+4)}{8 t^2 (T+1)^3}(x-t)^2+O\left(\left(x-t\right)^{3}\right)$$ where $$t=2 W\left(\frac{1}{e}\right)\qquad \text{and} \qquad T=W\left(-\frac{t}{2} e^{\frac t2} \right)$$ which can easily be inversed since $$a=\gamma_0+\gamma_1(x-t)+\gamma_2(x-t)^2++O\left((x-t)^3\right)$$ leads to $$x=t+\frac{1}{\gamma_1}(a-\gamma_0)-\frac{\gamma_2 }{\gamma_1 ^3}(a-\gamma_0)^2+O\left((a-\gamma_0)^3\right)$$