Could anyone help me solve this equation in terms of $x$ using the Lambert $W$ function or even other methods:
$$x^{-a}e^{ – kx}
=c(n-x)^{-a}e^{ – k(n-x)}$$
Where $k$, $c$, and $n$ are constants, and $a$ is the path loss exponent of the system with a value (2.2), I am trying to change the frequency in order to have an integer $a$.
What I have tried is :
$$x^{-a}e^{ – kx}
=c(n-x)^{-a}e^{ – kn} e^{ kx}$$
Then, let $b=ce^{ – kn}$ and dividing both sides by $e^{ kx}$, we get:
$$x^{-a}e^{ -2 kx}
=b(n-x)^{-a}$$
I am stuck here afterward.
Thanks in advance
Best Answer
$$x^{-a}e^{-kx}=c(n-x)^{-a}e^{-k(n-x)}$$
We see, this equation can be rearranged to a polynomial equation of more than one algebraically independent monomials. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.
$$x^{-a}=ce^{-kn}(n-x)^{-a}e^{2kx}$$
$x\to \frac{1}{2k}t\ $ *: $$\left(\frac{1}{2k}t\right)^{-a}=ce^{-kn}\left(n-\frac{1}{2k}t\right)^{-a}e^t$$ $$\frac{c\left(\frac{2kn-t}{2k}\right)^{-a}}{e^{kn}\left(\frac{1}{2k}t\right)^{-a}}e^t=1$$
for $a\neq 0$:
for $(a\in\mathbb{Q})\lor(k,n,t\in\mathbb{R})$:
$$\frac{c^\frac{1}{a}t}{e^\frac{kn}{a}(2kn-t)}e^\frac{t}{a}=1$$
$t\to au\ $ *: $$\frac{ac^\frac{1}{a}u}{e^\frac{kn}{a}(2kn-au)}e^u=1$$ $$\frac{au}{2kn-au}e^u=c^{-\frac{1}{a}}e^{\frac{kn}{a}}$$ $$\frac{au}{au-2kn}e^u=-c^{-\frac{1}{a}}e^{\frac{kn}{a}}$$ $$\frac{u}{\frac{au-2kn}{a}}e^u=-c^{-\frac{1}{a}}e^{\frac{kn}{a}}$$ $$\frac{u}{u-\frac{2kn}{a}}e^u=-c^\frac{-1}{a}e^\frac{kn}{a}$$
We see, we cannot solve this equation in terms of Lambert W, but in terms of Generalized Lambert W.
$$u=W\left(^{\ \ 0}_\frac{2kn}{a};-c^\frac{-1}{a}e^\frac{kn}{a}\right)$$ $$t=aW\left(^{\ \ 0}_\frac{2kn}{a};-c^\frac{-1}{a}e^\frac{kn}{a}\right)$$ $$x=\frac{a}{2k}W\left(^{\ \ 0}_\frac{2kn}{a};-c^\frac{-1}{a}e^\frac{kn}{a}\right)$$
So we have closed forms for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.
*) Because I don't want to use the transformation after equation (5) in [Mező/Baricz 2017] where I possibly would make mistakes, I want to transform the exponent of $e$ that contains $x$ to an exponent that is only a variable. Clearly you can proceed without this substitution.
$\ $
[Maignan/Scott 2016] Maignan, A.; Scott, T. C.: Fleshing out the Generalized Lambert W Function. ACM Communications in Computer Algebra 50 (2016) (2) 45-60
[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mező/Baricz 2017] Mező, I.; Baricz, Á.: 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