The exponential function has the well-known power series representation/definition:

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

And the natural logarithm has the less well-known power series representation/definition:

$\ln(x) = \sum_{k=1}^\infty \frac{(x - 1)^k}{k}(-1)^{k + 1}$

Let's for the purpose of this question pretend we have no previous concept of exponentials and logarithms: all we have are two power series called $e^x$ and $\ln(x)$ and, just for the fun of it, we want to see if they are the inverse of one another. So we put one power series into the other:

$\ln(e^x) = \sum_{k=1}^\infty \frac{(\sum_{n=0}^\infty \frac{x^n}{n!} - 1)^k}{k}(-1)^{k + 1}$

How does one prove, using only the above power series, any algebraic manipulation of the expression and any theorems pertaining to power series that do not require the anterior knowledge of $e^x$ or $\ln(x)$, that this is x?


3 Answers 3


I enjoyed Christian's answer, and if you want the shortest proof starting from the series definitions, then that is the way to go. But, like you, I was curious about this question because I like to manipulate power series, and because this problem ended up being harder than I thought it would be.

Here is a solution that meets your requirements. The only thing I have omitted is proving that exchanging the order of summation is valid when I do it below, which is probably not the part of the proof you are interested in.

A few lemmas are needed, which I will list first. The first two are well-known combinatorial identities, and I will omit the proofs. I'll follow your approach, but I'll delay the expansion of $e^x$ a little bit longer than you did, because otherwise you get sums over an arbitrarily large number of variables, which end up being difficult to handle.

Lemma 1: For $m,n \in \mathbb{Z}$ with $0 \le m < n$, $$ \sum_{k=0}^n (-1)^k \binom{n}{k} k^m = 0 $$ The easiest way to prove this is with calculus, but it can be proven by induction on $m$ from the result that for $n \ge 1$, $$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0 $$ which itself can be proven by induction on $n$. So, no calculus required.

Lemma 2: $$ \sum_{n=k}^m \binom{n}{k} = \binom{m+1}{k+1} $$ This can also be proven by induction.

Lemma 3: $ e^x e^y = e^{x+y} $

Proof: This one is a lot easier than the main result. Expand the right hand side using the series definition and the binomial theorem to get $$ e^{x+y} = \sum_{m=0}^\infty \frac{(x + y)^m}{m!} = \sum_{m=0}^\infty \sum_{i=0}^m \frac{1}{m!} \binom{m}{i} x^i y^{m-i} = \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} $$ Let $j=m-i$ and the sums become independent: $$ \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} = \sum_{i=0}^\infty \sum_{j=0}^\infty \frac{x^i y^j}{i!j!} = \left( \sum_{i=0}^\infty \frac{x^i}{i!} \right) \left( \sum_{j=0}^\infty \frac{x^j}{j!} \right) = e^x e^y $$

Lemma 4: $ (e^x)^y = e^{xy} $ when $y$ is a nonnegative integer.

Proof: This is obvious using Lemma 3.

Main result: $ \log(e^x) = x $

Proof: Apply the power series for the logarithm, then a binomial expansion, and then the power series for the exponential.

$$ \begin{align} \log(e^x) &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} ( e^x - 1 )^n \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} e^{kx} \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \sum_{m=0}^\infty \frac{k^m x^m}{m!} \\ &= -\sum_{m=0}^\infty \frac{x^m}{m!} \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \\ &= -\sum_{m=0}^\infty \frac{x^m}{m!} a_m \tag{1} \label{1} \end{align} $$

where $a_m$ has been defined as

$$ a_m = \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$

Split the sum over $n$ into two parts, from $1$ to $m$ and from $m+1$ to $\infty$.

$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m + \sum_{n=m+1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$

On the right, the summation over $k$ is zero by Lemma 1 because $n > m$. This gives us

$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \label{2} \tag{2} $$

In this expression, when $k \ge 1$, we can apply the identity

$$ \frac{1}{n} \binom{n}{k} = \frac{1}{k} \binom{n-1}{k-1} \label{3} \tag{3} $$

Fortunately, the $k=0$ terms in \eqref{2} are all zero because of the $k^m$ term (since we are only considering $a_m$ where $m \ge 1$). Dropping the $k=0$ terms, moving the summation over $n$ to the right, and applying \eqref{3} in \eqref{2}, we get

$$ \begin{align} a_m &= \sum_{k=1}^m (-1)^k k^m \sum_{n=k}^m \frac{1}{k} \binom{n-1}{k-1} \\ &= \sum_{k=1}^m (-1)^k k^{m-1} \sum_{n=k}^m \binom{n-1}{k-1} \end{align} $$

Applying Lemma 2 to the summation over $n$ gives $$ a_m = \sum_{k=1}^m (-1)^k k^{m-1} \binom{m}{k} $$

This can be simplified by again applying Lemma 1. However, note that Lemma 1 requires the $k=0$ terms, which we dropped earlier so that we could apply \eqref{3}. Now the $k=0$ term in the sum must be replaced by including $k=0$ in the sum and subtracting it outside the sum.

$$ a_m = \left( \sum_{k=0}^m (-1)^k k^{m-1} \binom{m}{k} \right) - 0^{m-1} $$

Obviously this term on the right is zero when $m>1$, but the $m=1$ case results in $0^0$. Here the appropriate convention is that $0^0=1$. (For example, when we expanded the exponential as $e^{kx} = \sum_m k^m x^m/m!$, the $m=0$ term is interpreted as $1$ even when $k=0$.) Since the $0^{m-1}$ term yields 0 for $m \ne 1$ and 1 for $m=1$, we can write this using the Kronecker delta as $\delta_{m,1}$.

$$ a_m = \left( \sum_{k=0}^m \binom{m}{k} (-1)^k k^{m-1} \right) - \delta_{m,1} $$

Now Lemma 1 can be applied again. (Note that the conditions of the lemma are satisfied because $m-1 < m$.) This gives $ a_m = -\delta_{m,1} $. Inserting this into $\eqref{1}$ gives the final result,

$$ \log(e^x) = x $$

  • 3
    $\begingroup$ An elegant slog. $\endgroup$ Oct 9, 2016 at 6:43
  • $\begingroup$ This is pretty excellent, but the $-\delta_{m,1}$ mystifies me. $\endgroup$
    – brothir
    Oct 10, 2016 at 21:01
  • $\begingroup$ @brothir: Sorry about that. I will update the derivation to make it clearer. $\endgroup$
    – sasquires
    Oct 12, 2016 at 21:50
  • $\begingroup$ @brothir: I just got a chance to do the update. I hope this helps. I feel like there must be a simpler way to do this derivation without dropping and re-adding the $k=0$ terms, but every time I tried to write it up that way, it just became more complicated. If anyone manages to make it simpler, please feel free to post another answer. $\endgroup$
    – sasquires
    Oct 12, 2016 at 23:58
  • $\begingroup$ Accepted your answer. For people not familiar with the Kronecker delta: en.wikipedia.org/wiki/Kronecker_delta. $\endgroup$
    – brothir
    Oct 14, 2016 at 11:04

Consider the two functions $$f(x):=\sum_{k=1}^\infty{(-1)^{k-1}\over k}x^k\qquad\bigl(=\log(1+x)\bigr)$$ and $$g(y):=\sum_{j=1}^\infty{1\over j!}y^j\qquad\bigl(=e^y-1\bigr)\ .$$ Both are defined in a neighborhood of the orgin, one has $f(0)=g(0)=0$, and termwise differentiation of the two series reveils that $$\eqalign{f'(x)&=\sum_{k=1}^\infty(-1)^{k-1}x^{k-1}={1\over 1+x}\ ,\cr g'(y)&=\sum_{j=1}^\infty{j\over j!}y^{j-1}=\sum_{j'=0}^\infty{1\over j'!}y^{j'}=1+g(y)\ .\cr}$$ We now consider the composition $$p(y):=f\bigl(g(y)\bigr)$$ of these two functions. According to the chain rule (no further series manipulation required!) we obtain $$p'(y)=f'\bigl(g(y)\bigr)\>g'(y)={1\over 1+g(y)}\bigl(1+g(y)\bigr)\equiv 1\ .$$ As $p(0)=0$ we can conclude that $p(y)\equiv y$, which proves that $f$ and $g$ are indeed inverses of each other.

  • 3
    $\begingroup$ Technically and pedantically speaking this answer fulfills the criteria, but I was hoping for an answer not using any calculus. $\endgroup$
    – brothir
    Mar 21, 2015 at 10:51
  • $\begingroup$ I think this is nice. How will anything involving an infinite series not involve calculus?! $\endgroup$
    – user21820
    Mar 21, 2015 at 12:35
  • $\begingroup$ @brothir: Unless you're hoping to avoid termwise differentiation? $\endgroup$
    – user21820
    Mar 21, 2015 at 12:35
  • $\begingroup$ To be more specific, operations of calculus (in this case differentiation). If you see the question, I asked to use only theorems pertaining to power series as well as any algebraic manipulation required to mold the expression in such a way that any such theorems might be used. Now, differentiation is TECHNICALLY an algebraic operation, but that was not the "spirit" of the question: I was more thinking of stuff like splitting the sums up into separate parts and similar. $\endgroup$
    – brothir
    Mar 21, 2015 at 12:39
  • $\begingroup$ Also, I specifically wanted that the final power series could be shown to equate specifically to $x$, not just that they are inverse. $\endgroup$
    – brothir
    Mar 21, 2015 at 12:42

I like the series approaches above, quite interesting.

I had a much more mundane approach that involves using the chain rule and inverse functions. Suppose that g(x) is the inverse of f(x), whereby


Take the derivative of both sides (remember the chain rule):


Take the special case that f(x)=exp(x).

(dg/df)exp(x)= (dg/df)f = 1 or dg/df = 1/f

Integrating both sides, you'll get that g(f) is whatever the integral of 1/f comes to. Calling that the logarithm,


That works with x too: g(x)=ln(x) which is the inverse of f(x)=exp(x)

  • 1
    $\begingroup$ I think that this is a reasonable approach to the problem in general. But you need to read the OP's question carefully. He wants to do a proof that uses power series alone ("using only the above power series") without relying on any additional concepts from calculus, such as derivatives or integrals. $\endgroup$
    – sasquires
    Feb 10, 2020 at 23:39
  • $\begingroup$ For formal power series, derivatives and integrals can hardly be considered non-algebraic. $\endgroup$ Oct 14, 2021 at 5:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .