Solving $x=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}$ for $y$

Question

Do you know how to solve the following equation to make $y$ the subject: $$x=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}$$

My attempts: I multiplied both sides by $\ln(2)(y+1)^2$ to get $$x\ln\left(2\right)\left(y+1\right)^{2}\left(y+2\right)=\ln\left(2\right)\left(y+1\right)e^{\ln\left(2\right)\left(y+1\right)}$$ The reasoning behind this was to get an expression of the form $xe^x$ to use the Lambert W function. However I was unable to make it work. I also tried considering the more general function $$x=\frac{z^{1+y}}{\left(y+1\right)\left(y+2\right)}$$ Differentiating both sides with respect to z cancels the $(y+1)$ term in the denominator which I thought could make it easier to work with. However again I was unable to find a solution.

Any help is appreciated. Thanks

Answer 1

$$x=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}$$

Let $t=(y+1)$ and rewrite it as $$e^{-t \log(2)}=\frac{1}{t (t+1) x}$$ which has a formal solution in terms of the generalized Lambert function (have a look at equation $(4)$). The only problem is that the solution will be given as an infinite sum of Laguerre polynomials.

Back to the original problem, assuming $x>0$, consider that we look for the zeros of function $$f(y)=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}-x$$ which shows a minimum value since $$f'(y)=\frac{2^{y+1}}{\left(y^2+3 y+2\right)^2} \Big( y^2 \log (2)+y (3\log (2)-2)+(2\log (2)-3)\Big)$$ has a positive root around $y_*=1.46958$ where $$f(y_*)=\frac{e^{1+\frac{1}{2} \sqrt{4+\log ^2(2)}} \log ^2(2)}{\sqrt{2} \left(2+\sqrt{4+\log ^2(2)}\right)}-x$$ So,

• if $x< 0.646424 $, no solution
• if $0.646424 \leq x \leq 1$, two solutions
• if $x >1$, one solution.

Considering the last case, notice that looking instead for the zero of function $$g(y)=\log \left(\frac{2^{y+1}}{(y+1) (y+2)}\right)-\log(x)$$ is easier since it is close to a straight line and this is very good for any root-finding method.

If the problem was $$\frac{2^{y+\frac{3}{2}}}{\left(y+\frac{3}{2}\right)^2}=x \sqrt 2$$ it would be very simple. This can be justified by the fact that $y$ will always be larger than $4$ and that $$\frac{1}{(y+1) (y+2)}-\frac{1}{\left(y+\frac{3}{2}\right)^2}=\frac{1}{4 y^4}+O\left(\frac{1}{y^5}\right)$$

This provides a quite good estimate $$y_0=-\frac 3 2-\frac 2{\log(2)}W_{-1}\left(-\frac{\log (2)}{2 \sqrt[4]{2} \sqrt{x}}\right)$$ Using a couple of Newton iterations to polish the root will be more than sufficient.

A table of results for $x=10^k$

$$\left( \begin{array}{cccc} k & y_0 & y_1 & \text{solution} \\ 0 & 3.8262288 & 3.7981910 & 3.7979745 \\ 1 & 9.1465586 & 9.1421901 & 9.1421849 \\ 2 & 13.447526 & 13.445525 & 13.445524 \\ 3 & 17.454761 & 17.453577 & 17.453576 \\ 4 & 21.311032 & 21.310238 & 21.310238 \\ 5 & 25.073470 & 25.072897 & 25.072897 \\ 6 & 28.771329 & 28.770894 & 28.770894 \\ 7 & 32.421776 & 32.421434 & 32.421434 \\ 8 & 36.035817 & 36.035539 & 36.035539 \\ 9 & 39.620957 & 39.620727 & 39.620727 \\ 10 & 43.182561 & 43.182368 & 43.182368 \\ 11 & 46.724604 & 46.724439 & 46.724439 \\ 12 & 50.250118 & 50.249975 & 50.249975 \\ 13 & 53.761469 & 53.761345 & 53.761345 \\ 14 & 57.260545 & 57.260435 & 57.260435 \\ 15 & 60.748873 & 60.748775 & 60.748775 \\ \end{array} \right)$$

Answer 2

$$x=\frac{2^{1+y}}{(y+1)(y+2)}$$ $$x=\frac{e^{\ln(2)(1+y)}}{(y+1)(y+2)}$$ $y\to t-1$: $$\frac{1}{t(t+1)}e^{\ln(2)t}=x$$

We cannot read from the equation how to invert it by elementary functions:
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?

We see, as already stated in the other answers, this equation isn't solvable by elementary functions and Lambert W. But it is solvable by elementary functions together with Generalized Lambert W.

$$t=\frac{1}{\ln(2)}W\left(^{\ \ \ \ \ \ -}_{0,-\ln(2)};\frac{x}{\ln^2(2)}\right)$$ $$y=\frac{1}{\ln(2)}W\left(^{\ \ \ \ \ \ -}_{0,-\ln(2)};\frac{x}{\ln^2(2)}\right)-1$$

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 3

I prefer to make a second answer for the generalization of the problem.

We want to solve for $y$ the general equation $$\color{red}{\large x=\frac{z^{a+y}}{\prod_{k=1}^n\left(y+b_k\right)}}$$ where $x$ is supposed to be large.

As in my first answer, the rhs will be approximated as $$\frac{z^{a+y}}{(y+c)^n}=z^{a-c} \frac{z^{y+c}}{(y+c)^n}\quad \implies \quad x\,z^{c-a}=\frac{z^{y+c}}{(y+c)^n}$$ which gives, as an approximation, $$\color{blue}{\large y_0=-c-\frac{n}{\log (z)}\,W_{-1}\left(-\frac{\log (z)}{n\,z^{\frac{c-a}{n}}}\,\,\frac 1{\sqrt[n]x }\right)}$$

As usual, to stay in the real domain, the requirement is $$\frac{\log (z)}{n\,z^{\frac{c-a}{n}}}\,\,\frac 1{\sqrt[n]x } \leq \frac 1 e$$

The only problem left is to decide about $c$. In the first answer was used the mean of the $b_k$'s; this is justified by the fact that, at least for large $y$, $$\frac{1}{\prod_{k=1}^n\left(y+b_i\right)}-\frac{1}{(y+c)^n}=\frac {n\,c-\sum_{k=1}^n b_k}{y^{n+1}}+O\left(\frac{1}{y^{n+2}}\right)$$

However, may be, it could be possible to have a better approximation of $c$ and a function of the $b_k$ (to be explored later).

An example for illustration

Trying for $x=10^{20}$, $z=\pi$, $a=e$ and $b_k$ being the $k^{\text{th}}$ prime number, some results as a function of $n$

$$\left( \begin{array}{ccc} n & y_0 & \text{solution} \\ 2 & 44.2276 & 44.2275 \\ 3 & 47.8232 & 47.8224 \\ 4 & 51.5652 & 51.5630 \\ 5 & 55.4718 & 55.4655 \\ 6 & 59.4969 & 59.4866 \\ 7 & 63.6580 & 63.6411 \\ 8 & 67.9169 & 67.8943 \\ 9 & 72.2904 & 72.2598 \\ 10 & 76.7898 & 76.7464 \\ 11 & 81.3617 & 81.3080 \\ 12 & 86.0407 & 85.9725 \\ 13 & 90.7989 & 90.7162 \\ 14 & 95.6140 & 95.5195 \\ 15 & 100.500 & 100.393 \\ \end{array} \right)$$

Again, a couple of Newton iterations to polish the root will be more than sufficient.

For a maximum numerical efficiency, the function to be solved write $$f(y)=(y+a)\log(z)-\sum_{k=1}^n \log(y+b_k)-\log(x)$$ $$f'(y)=\log(z)-\sum_{k=1}^n \frac 1{y+b_k}$$

For the working case with $n=15$, $y_0=100.499980$, $y_1=100.393061$ while the solution is $y=100.393055$.

Update

Is seems that the approximation is not very sensitive to the value of $c$ since $$\frac {\partial\, y_0}{\partial c}=\frac{n}{(y_0+c) \log (z)-n}$$

For example, in the last worked case for which $c=\frac{328}{15}$, at half this value $y=99.1089$ while doubling it gives $y=102.868$ in agreement with the theoretical slope of $0.12$.

I think that we can avoid the search of a "better" definition of $c$.