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As $e$ is transcendental, there is no polynomial equation with integer coefficients having $e$ as a root.

Are there classical equations of the form

$$\sum_{i=0}^{\infty} a_ix^i =1$$

that have $e$ or $1/e$ as root, with $a_i\in \mathbb{Z}$ for each $i$?

For $1/e$, is it possible to have $a_i\ge 0$ for each $i$?

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  • $\begingroup$ The Taylor series for $\ln(1+x)$ has fractional coefficients, so it doesn't seem to help...but I don't see an obvious argument that settles this. $\endgroup$
    – user44143
    Commented Jul 29, 2019 at 20:01
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    $\begingroup$ To people voting to close: is the answer so obvious? Let's discard the smart-aleck solution $a_0 = 1, a_i = 0$ for $i \geq 1$. $\endgroup$
    – Todd Trimble
    Commented Jul 29, 2019 at 20:04
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    $\begingroup$ @ToddTrimble, isn't it obvious that we can cook up such a series for any positive number smaller than $1$? Just take $a_0:=0$ and inductively $a_i$ to be the largest positive integer such that the partial sum is still smaller than 1? $\endgroup$
    – Kostya_I
    Commented Jul 29, 2019 at 20:28
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    $\begingroup$ @Kostya_I, the question uses the word "classical", which could be charitably interpreted here as "coming from a function satisfying a differential equation with rational coefficients". That may make the question more interesting. $\endgroup$
    – user44143
    Commented Jul 29, 2019 at 20:39
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    $\begingroup$ @Kostya_I Yes, thanks very much, but let's perhaps try to be generous as Matt F. suggests. $\endgroup$
    – Todd Trimble
    Commented Jul 29, 2019 at 20:50

1 Answer 1

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If $a_i=0$ for all large enough $i$, then $e$, being transcendental, cannot be a solution to your equation, unless $a_i=0$ for all $i\ge1$. Otherwise, if $a_i\ne0$ for infinitely many $i$, then the series for $e$ cannot converge, given that the $a_i$'s are integers, because then $|a_ie^i|\ge e^i\not\to0$ if $a_i\ne0$.

However, this can be done for $1/e$, as follows: let $a_0:=0$, and then let $a_j:=\max\big\{k\in\{0,1,\dots\}\colon ke^{-j}+S_{j-1}\le1\big\}$ recurrently for $j=1,2,\dots$, where $S_{j-1}:=\sum_{i=0}^{j-1}a_ie^{-i}$. Indeed, then for $j=1,2,\dots$ we have $S_j=a_je^{-j}+S_{j-1}\le1$, whereas $S_j+e^{-j}=(a_j+1)e^{-j}+S_{j-1}>1$, so that $1-e^{-j}<S_j\le1$, whence $S_j\to1$ as $j\to\infty$, as desired.

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  • $\begingroup$ Any idea for $1/e$? $\endgroup$
    – Todd Trimble
    Commented Jul 29, 2019 at 20:24
  • $\begingroup$ @ToddTrimble : I have added the positive answer for $1/e$. $\endgroup$
    – Iosif Pinelis
    Commented Jul 29, 2019 at 20:48

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