Simple Derivative plus tricky algebraic expression to simplify?

Question

I need to find the maximum of:

$$\frac{(1-e^{-\lambda \tau})}{\lambda \tau}-e^{-\lambda \tau}$$

apply quotient rule to the fraction term

$$\frac{(e^{-\lambda \tau}\lambda \tau-\tau(1-e^{-\lambda \tau}))}{(\lambda \tau)^2}+e^{-\lambda \tau}\tau$$

give common denominator

$$\frac{(e^{-\lambda \tau}\lambda \tau-\tau(1-e^{-\lambda \tau}))+e^{-\lambda \tau}\tau^3\lambda^2}{(\lambda \tau)^2}$$

Is this derivative correct?

I now need to set this equal to zero and solve for lambda!

$$(e^{-\lambda \tau}\lambda \tau-\tau(1-e^{-\lambda \tau}))+e^{-\lambda \tau}\tau^3\lambda^2=0$$

Can this be solved analytically? Is so could I please have a hint as to how to approach the problem?

Baz

Answer 1

$$\frac{(1-e^{-\lambda \tau})}{\lambda \tau}-e^{-\lambda \tau}$$ We see, the function depends on $\lambda\tau$. Set $\lambda\tau=x$ therefore. $$\frac{(1-e^{-x})}{x}-e^{-x}$$ $$\curvearrowright x\neq0$$ Calculating the extrema: $$\frac{d}{dx}\frac{(1-e^{-x})}{x}-e^{-x}\stackrel{!}{=}0$$ $$\frac{x^2e^{-x}+xe^{-x}+e^{-x}-1}{x^2}\stackrel{!}{=}0$$ $$x^2e^{-x}+xe^{-x}+e^{-x}-1\stackrel{!}{=}0$$ We see, the function on the left-hand side of the equation is an algebraic equation that depends on $x$ and $e^{-x}$. Both are algebraically independent of each other for all $x$ except $0$. Unfortunately, the equation cannot have therefore a solution that is an elementary function (see Wikipedia: Elementary function for the definition of the elementary functions).

A known special function for inverting functions of $x$ and $e^x$ is the Lambert W function. It is the inverse of $xe^x$. But the function in the equation cannot be represented in the form of equation (1) of my answer at Algebraic solution to natural logarithm equations like 1−x+xln(−x)=0. Unfortunately, the equation cannot be solved with help of Lambert W therefore.

If a function in dependence of a variable in polynomials and in exponential functions cannot be solved with Lambert W, some generalizations of Lambert W function are available. See the references in Wikipedia: Lambert W function - Generalizations as an entry. Ask if you need further literature references.

The numerical solution for the maximum is

$$\lambda\tau=1.793282132900...,\ f(\lambda\tau)=0.2984256075256...$$

$$\lambda=\tau=1.339134844928...,\ f(\lambda\tau)=0.2984256075256...$$

Answer 2

As said in comments and answers, using $\lambda\tau=x$, you want to maximize $$f(x)=\frac{(1-e^{-x})}{x}-e^{-x}$$ As already said, the derivative is $$f'(x)=\frac{e^{-x} \left(x^2+x+1-e^x\right)}{x^2}$$ and you then need to find the zero of function $$g(x)=x^2+x+1-e^x$$

May be, this could be done using the generalization of the Lambert function with complex coeddicients but I am afraid that this could be a nightmare.

However, we can easily obtain rather good approximations.

The derivative $$g'(x)=2x+1-e^x$$ cancels for $$x_*=-\frac{1}{2}-W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)$$ where appears the second branch of Lambert function. At this point (where the first derivative is zero, we can build a Taylor series and get $$g(x)=g(x_*)+\frac 12 g''(x_*)(x-x_*)^2+O\left((x-x_*)^3\right)$$ Neglecting the higher order terms, the solutions are then $$x_\pm=x_*\pm \sqrt{-\frac{2g(x_*)}{g''(x_*)}}$$ In the considered problem, we need to consider the largest solution which is $x_-$.

For more simplicity in notations, let $t=W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right) \approx-1.75643$ and then

$$x_-=-\frac{2 t^2+3 t+1+\sqrt{-4 t^3-12 t^2-11 t-3}}{2 (t+1)}\approx 1.90907$$ making $f(x_-)\approx 0.297958$ which is not too bad when compared to the exact solution @IV_ gave in his/her answer.

Could we do better ? Graphing $f(x)$ we can notice that the maximum is "around" $x=2$. So, using the simplest Padé approximant,w ehave $$g(x)\simeq \frac{7-e^2+\frac{\left(-36+11 e^2-e^4\right) }{2 \left(e^2-5\right)}(x-2)}{1+\frac{\left(2-e^2\right) }{2 \left(e^2-5\right)}(x-2)}$$ which cancel at $$x=\frac{2+2 e^2}{36-11 e^2+e^4} \approx 1.80051 \implies f(x) \approx 0.298424$$ which is better.

We could continue that way builing around $x=2$ the $[1,n]$ Padé approximants which will write $$g(x)=\frac{7-e^2+a_1^{(n)}(x-2)}{1+\sum_{k=1}^n b_k (x-2)^n}\implies x_{(n)}=2+\frac{e^2-7}{ a_1^{(n)}}$$ and get the following results $$\left( \begin{array}{cccc} n & x_{(n)} & x_{(n)} \approx & f(x_{(n)}) \\ 0 & \frac{-3+e^2}{-5+e^2} & 1.8371507 & 0.29835622 \\ 1 & \frac{2+2 e^2}{36-11 e^2+e^4} & 1.8005100 & 0.29842369 \\ 2 & \frac{192-98 e^2+28 e^4-2 e^6}{-660+290 e^2-40 e^4+2 e^6} & 1.7943170 & 0.29842557 \\ 3 & \frac{-12528+8412 e^2-1860 e^4+228 e^6-12 e^8}{21456-11934 e^2+2490 e^4-210 e^6+6 e^8} & 1.7934191 & 0.29842561 \end{array} \right)$$