Cfrgtkky

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

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

Is this derivative correct?

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

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

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

Baz

$$frac{(1-e^{-lambda tau})}{lambda tau}-e^{-lambda tau}$$
We see, the function depends on $lambdatau$. Set $lambdatau=x$ therefore.
$$frac{(1-e^{-x})}{x}-e^{-x}$$
Calculating the extrema:
$$frac{d}{dx}frac{(1-e^{-x})}{x}-e^{-x}stackrel{!}{=}0$$
$$frac{x^2e^{-x}+xe^{-x}+e^{-x}-1}{x^2}stackrel{!}{=}0$$
$$curvearrowright xneq0$$
We see, the function in the equation depends on $x$ and $e^{-x}$. Both are algebraically independent of each other. Unfortunately, the equation cannot have 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

$$lambdatau=1.793282132900..., f(lambdatau)=0.2984256075256...$$

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

As said in comments and answers, using $lambdatau=x$, you want to maximize
$$f(x)=frac{(1-e^{-x})}{x}-e^{-x}$$ As already said, the derivative is
$$f'(x)=frac{e^{-x} left(x^2+x+1-e^xright)}{x^2}$$ 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
$$g(x)=g(x_*)+frac 12 g''(x_*)(x-x_*)^2+Oleft((x-x_*)^3right)$$ 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
$$g(x)simeq frac{7-e^2+frac{left(-36+11 e^2-e^4right) }{2
left(e^2-5right)}(x-2)}{1+frac{left(2-e^2right) }{2
left(e^2-5right)}(x-2)}$$ 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
$$left(
begin{array}{cccc}
n & x_{(n)} & x_{(n)} approx & f(x_{(n)}) \
0 & frac{-3+e^2}{-5+e^2} & 1.8371507 & 0.29835622 \
1 & frac{2+2 e^2}{36-11 e^2+e^4} & 1.8005100 & 0.29842369 \
2 & frac{192-98 e^2+28 e^4-2 e^6}{-660+290 e^2-40 e^4+2 e^6} & 1.7943170 &
0.29842557 \
3 & frac{-12528+8412 e^2-1860 e^4+228 e^6-12 e^8}{21456-11934 e^2+2490 e^4-210
e^6+6 e^8} & 1.7934191 & 0.29842561
end{array}
right)$$