I'm consider $e^{-x}$ with $x >> 0$. I want to express it in a series of the form

$$ e^{-x} = \sum_{i=1}^{\infty} \frac{a_i}{x^k}$$

But analysis seems to resist it:


$ \frac{1}{x^k}$ rapidly decays to 0, as does $e^{-x}$ so I'm thinking a sum of such terms should intuitively yield a series.

Work So Far:

My first idea was that substitution $ u = \frac{1}{x}$ yield that

$$ e^{-x} = e^{-\frac{1}{u}}$$

Now I could look to express this as a Taylor series around $u = \gamma$, observing that

$$ \frac{d}{du} e^{ - \frac{1}{u}} = \frac{1}{u^2} e^{-\frac{1}{u}} $$

Deriving again we have

$$ \frac{d^2}{du^2} e^{ - \frac{1}{u}} = \left(- \frac{2}{u^3} + \frac{1}{u^4} \right)e^{-\frac{1}{u}} $$


$$ \frac{d^3}{du^3} e^{ - \frac{1}{u}} = \left( \frac{2\times 3}{u^4} - \frac{4}{u^5} + \frac{1}{u^6}\right)e^{-\frac{1}{u}} $$

And In General

$$ \frac{d^k}{du^k} e^{ - \frac{1}{u}}= $$

$$ \left((-1)^{k-1} \frac{2 \times 3 \times ... k}{u^{k+1}} + (-1)^{k-2} \frac{4 \times 5 \times 6 ... \times (k+1)}{u^{k+2}} + (-1)^{k-3} \frac{6 \times 7 \times ... (k+2)}{u^{k+3}}... +\frac{1}{u^{2k}} \right) e^{-\frac{1}{u}} $$

$$ = e^{- \frac{1}{u}} \sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} \frac{(-1)^{-i}}{u^{2k-i}} = \frac{e^{-\frac{1}{u}}}{u^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} u^{i} $$

Then it follows that we can define $e^{-\frac{1}{u}}$ around a point $u = \gamma$ as

$$ e^{- \frac{1}{u}} = e^{-\frac{1}{\gamma}} + \sum_{k=1}^{\infty} \left(\frac{e^{-\frac{1}{\gamma}}}{k!\gamma^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} \gamma^{i} \right) ( u - \gamma)^{k} $$

And therefore:

$$ e^{-x} = e^{-\frac{1}{\gamma}} + \sum_{k=1}^{\infty} \left(\frac{e^{-\frac{1}{\gamma}}}{k!\gamma^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} \gamma^{i} \right) \left( \frac{1}{x} - \gamma \right)^{k} $$

But this explodes if $\gamma = 0$, so the mechanics of building THAT converging series are going to be very different than this (namely NO convergence around $x < 1$ and then something nice...)


There is an error in my derivation, namely:

$$ \frac{d^3}{du^3} e^{ - \frac{1}{u}} = \left( \frac{2\times 3}{u^4} - \frac{6}{u^5} + \frac{1}{u^6}\right)e^{-\frac{1}{u}} $$.

The remaining series derivation is also incorrect for the same mistake not caught.

  • 1
    $\begingroup$ I am not sure, but I think the laurent-series will not converge fast enough towards zero to represent the function $e^{-x}$ $\endgroup$
    – Peter
    Mar 31 '16 at 9:11
  • 2
    $\begingroup$ Such a series development cannot exist because the (complex) exponential function has an essential singularity at infinity. $\endgroup$
    – Martin R
    Mar 31 '16 at 9:20
  • $\begingroup$ If you want a practical solution, apart from using the FPU-provided function, find $n$ with $2^n>|x|$ and compute as $(e^{-2^{-n}x})^{2^n}$ where the outer power is the result of $n$ squarings. $\endgroup$ Mar 31 '16 at 11:55

$$\frac{\partial}{\partial x} \exp(-x)=-\exp(-x)\\ \frac{\partial}{\partial x}\sum_{i=1}^{\infty} a_i x^{-i}=\sum_{i=1}^{\infty} \frac{a_i}{-i} x^{-i-1}$$ We would like $$\sum_{i=1}^{\infty} a_i x^{-i}=\sum_{i=1}^{\infty} \frac{a_i}{i} x^{-i-1}$$ If the coefficients in $x^{-1}$ are equal, then $a_1$ must be 0, because the rhs. has no term in $x^{-1}$. By induction we see that all $a_i$ must be 0. Thus, there is no such series that approximates $\exp(-x)$.

You may be interested in Pade approximations.


You wish to have an asymptotic expansion of $e^{-x}$ as $x \to \infty$. Allowing $x$ to be complex, and writing $z = 1/x$, we have $e^{-1/z}$ for which we want a Taylor series expansion around $z = 0$. Like Wouter has pointed, this is impossible because $e^{-1/z}$ has an essential singularity at $z = 0$. You may write a Laurent series, but I expect this to be identical to the Taylor series of $e^{-x}$.

This also shows that $e^{-x}$ is a term that is "beyond all orders" when it comes to asymptotics. It decays faster than every possible power law.


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