can we express $x=f(y)$, if $y=\exp(x/1.45)(1-\exp(-x))$?

Question

Given $y=\exp(x/1.45)(1-\exp(-x))$, I need to express $x = f(y)$.

I tried taking log of both side but didnt get any fruitful expression.

Is it possible? If yes, how?

Answer 1

If $t = \exp(x)$, this can be written as $t^{9/29} y = t - 1$, which is a polynomial equation in $t^{1/29}$. Of course since the degree is greater than $4$, we don't expect a solution involving radicals. There is, however, a nice series solution in powers of $y$, obtained from the Lagrange Inversion Formula: if $b = 9/29$,

$$ \eqalign{t &= 1 + y + b y^2 + \dfrac{b(3b-1)}{2!} y^3 + \dfrac{b(4b-1)(4b-2)}{3!} y^4 _+ \ldots\cr &= 1 + \sum_{n=1}^\infty \left( \prod_{j=0}^{n-2} (nb-j)\right) \frac{y^n}{n!}}$$

Answer 2

You will not find any explicit solution if you want $x$ such that $$y=e^{20 x/29} \left(1-e^{-x}\right)$$ Setting $e^x=z$ the equation would write $$y=z^{20/29}(1-\frac 1 z)=\frac{P_{29}(z)}{z^9}$$ with which you would have a lot of fun (just kidding !).

If $y$ is small, you could approximate the function by its Taylor expansion around $x=0$ which will give $$e^{20 x/29} \left(1-e^{-x}\right)=x+\frac{11 x^2}{58}+\frac{301 x^3}{5046}+O\left(x^4\right)$$ and using more or less terms, you could find an approximation of $x$ as a function of $y$. But this works for a very limited range of $y$.

For the most general case, the only way is to consider a method such as Newton which, starting at a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For your case $$f(x)=e^{20 x/29} \left(1-e^{-x}\right)-y$$ $$f'(x)=\frac{1}{29} e^{-9 x/29} \left(20 e^x+9\right)$$ For illustration purposes, let us consider two cases.

First case : $y=12345$. A look at the plot of the function shows a solution around $x=15$. So, let us apply the method using $x_0=15$. The successive iterates will then be $$\left( \begin{array}{cc} n & x_n \\ 1 & 14.12564931 \\ 2 & 13.72770282 \\ 3 & 13.66199628 \\ 4 & 13.66046179 \\ 5 & 13.66046098 \end{array} \right)$$

Second case : $y=-12345$. A look at the plot of the function shows a solution around $x=-30$. So, let us apply the method using $x_0=-30$. The successive iterates will then be $$\left( \begin{array}{cc} n & x_n \\ 1 & -30.37705418 \\ 2 & -30.35664112 \\ 3 & -30.35657618 \end{array} \right)$$

Answer 3

Here are the details of the Lagrange Inversion computation. We have

$$y = \exp(bx)(1-\exp(-x)).$$

where $b=20/29.$ Putting $\exp(x) = t$ so that $x = \log t$ we get

$$y = t^b (1-1/t).$$

We seek to compute

$$[y^n] t(y) = \frac{1}{2\pi i} \int_{|y|=\epsilon} \frac{1}{y^{n+1}} t(y) dy.$$

Using the definition we put with $w$ being complex

$$y = w^b (1-1/w) = w^{b-1} (w-1)$$

so that

$$dy = (b w^{b-1} - (b-1) w^{b-2}) \; dw.$$

We get for the integral

$$\frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{w^{(b-1)(n+1)}} \frac{1}{(w-1)^{n+1}} \\ \times w \times w^{b-2} (bw - (b-1)) \; dw.$$

This has two pieces, the first is

$$\frac{b}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{w^{(b-1)n-1}} \frac{1}{(w-1)^{n+1}} \; dw \\ = \frac{b}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(1+(w-1))^{(b-1)n-1}} \frac{1}{(w-1)^{n+1}} \; dw $$

and the second is

$$\frac{b-1}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{w^{(b-1)n}} \frac{1}{(w-1)^{n+1}} \; dw \\ = \frac{b-1}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(1+(w-1))^{(b-1)n}} \frac{1}{(w-1)^{n+1}} \; dw.$$

Extracting coefficients yields

$$(-1)^n \left(b {(b-1)n-2+n\choose n} - (b-1) {(b-1)n-1+n\choose n}\right) \\ = (-1)^n \left(b {bn-2\choose n} - (b-1) {bn-1\choose n}\right).$$

This is one when $n=0.$ Continuing we have

$$\frac{(-1)^n}{n!} (b (bn-2)^{\underline{n}} - (b-1) (bn-1)^{\underline{n}}) \\ = \frac{(-1)^n}{n!} (bn-1)^{\underline{n}} \left(b \frac{bn-(n+1)}{bn-1} - (b-1)\right) \\ = \frac{(-1)^{n+1}}{n!} (bn-1)^{\underline{n}} \frac{1}{bn-1} \\ = \frac{(-1)^{n+1}}{n!} (bn-2)^{\underline{n-1}}.$$

We thus obtain

$$t = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1} \prod_{j=0}^{n-2} (nb-2-j) \\ = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1} \prod_{j=2}^{n} (nb-j).$$

Remark. Observe that the accepted answer and this post solve slightly different equations which both yield the correct result.