Find $a$ in terms of $b$ if $b = ({a \over 5})^{a+1}$

Question

Find $a$ in terms of $b$ if $b = ({a \over 5})^{a+1}$.

Firstly,

is this possible (within elementary functions)

and secondly,

if so, how is it done?

also, can someone help me tag this, I don't really now which tags to use.

Answer 1

Even with Lambert function, there is no explicit solution because of the $+1$ in the exponent.

I guess that numerical methods would be needed. For starting, let $a=5c$ and consider that you look for the zero's of function $$f(c)=c^{5 c+1}-b$$ or, much better $$g(c)=(5c+1)\log(c)-\log(b)$$ Computing the derivatives $$g'(c)=\frac{1}{c}+5 \log (c)+5\qquad \text{and} \qquad g''(c)=\frac{5 c-1}{c^2}$$

The first derivative cancels at $$c_*=-\frac{1}{5 W\left(-\frac{e}{5}\right)}$$ which is not a real. So, the derivative is positive and $g(c)$ is an increasing function with a positive second derivative as soon as $c > \frac 15$.

Newton method will not face any problem; the only thing, foor saving iterations, is that we need a "reasonable" guess $c_0$ to start with.

Ignoring the $+1$, we could that start using $$c_0=\frac{\log (b)}{5 W\left(\frac{\log (b)}{5}\right)}$$ which will be an overestimate of the solution since $g(c_0) > 0$.

Let us try for $b=123456789$. The iterates would be $$\left( \begin{array}{cc} n & c_n \\ 0 & 3.201940306 \\ 1 & 3.097390221 \\ 2 & 3.096653690 \\ 3 & 3.096653653 \end{array} \right)$$ then $a=15.483268263$.

Trying for $b=123456789\times 10^{100}$, the iterates would be $$\left( \begin{array}{cc} n & c_n \\ 0 & 17.41954887 \\ 1 & 17.27183432 \\ 2 & 17.27167350 \end{array} \right)$$

Edit

As I wrote, if you want to save iterations, a "good" estimate $c_0$ is required.

Let me take the second case with $c_0=1$. The method will provide the following iterates $$\left( \begin{array}{cc} n & c_n \\ 0 & 1 \\ 1 & 42.48165184 \\ 2 & 19.29197737 \\ 3 & 17.29900434 \\ 4 & 17.27167903 \\ 5 & 17.27167350 \end{array} \right)$$ All of that can very easily be done with Excel (or whatever).

Update

Assuming that you would accept the proposed expression for $c_0$ you could have a better approximation using the simple $[1,1]$ Padé approximant of $g(c)$ built at $c=c_0$. This would give $$c_1=c_0+\frac{2 \,g(c_0)\, g'(c_0)}{g(c_0)\, g''(c_0)-2\, g'(c_0)^2}$$

For the worked examples, this would give $c_1=3.09667$ and $c_1=17.2717$. This does not seem to be too bad. Be sure that we can still improve it using $[1,n]$ Padé approximant.