Can $e^x$ be expressed as a linear combination of $(1 + \frac x n)^n$? In other words, does there exist an infinite sequence $(a_k)_{k \in \mathbb N_0}$ such that $$e^x = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$$ for all $x \in \mathbb R$? Call the series on the right $s(x)$.

I can answer the question in the negative when the series is absolutely convergent. In the conditionally convergent case, I'm not so sure. My thoughts were to use the fact that:

$$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x}$$

and use the lower bound when $a_k$ is negative, and the upper bound when $a_k$ is positive. This gets stuck because it's not always the case that if some $b_k$ is a decaying sequence then $\sum_{k} \frac{b_k}{k}$ is convergent.

The strengthened inequality $$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x - \frac{x^2}{2k} + \frac{x^3}{3k^2}}$$ looks like it might make more progress...

[EDIT 2019/08/14 14:00 GMT]

This is the solution in the absolutely convergent case, given by lemmas 1 and 2:

Definition: Let $s(x) = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$.

Lemmas and proofs follow:

Lemma 1: If $s(x)$ converges absolutely for some $x\geq 0$, then $s(x)$ converges absolutely for all $x \geq 0$.


Pick an $x_0 \geq 0$ for which $s(x_0)$ converges absolutely.

By the condition stated in the lemma, the series $\sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x_0} k\right|^k$ must converge. We also observe that $|a_k| \leq |a_k| \left|1 + \frac {x_0} k\right|^k$ is true for all $k$. So by the Direct Comparison Test, the series $\sum_{0 \leq k < \infty} |a_k|$ must also converge. In other words, $s(0)$ is absolutely convergent.

Consider now any $x \geq 0$. The series $\sum_{0 \leq k < \infty} |a_k| e^{x}$ converges because it is equal to $e^{x} \sum_{0 \leq k < \infty} |a_k|$, which we proved to be convergent in the previous paragraph. We observe that $|a_k| \left|1 + \frac {x} k\right|^k \leq |a_k| e^{x}$ is true for all $k$. So by the Direct Comparison Test, the series $|a_0| + \sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x} k\right|^k$ must also converge. So by the definition of absolute convergence, we have that $a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k=s(x)$ converges absolutely, where $x \geq 0$ was arbitrary.


Lemma 2: If $s(x)$ converges absolutely when $x \geq 0$, then for large enough $x$ we have that $e^x > s(x)$.


Let $z_n(x) = |a_0| + \sum_{1 \leq k < n} |a_k| \left(1 + \frac x k\right)^k$.

Pick some $\epsilon < \frac 1 2$.

Observe that there must be a large enough $n$ such that $z_\infty(0) - z_n(0) \leq \epsilon$.

Using the triangle inequality, we have that: $$\begin{aligned} |s(x)| &\leq z_\infty(x)\\ &\leq z_n(x) + (z_\infty(x) - z_n(x))\\ \end{aligned}$$

Since $z_n(x)$ is a polynomial, there is a large enough $X$ such that all $x \geq X$ it's true $z_n(x) < \epsilon \cdot e^x$. So we have that $$\begin{aligned} |s(x)| &<\epsilon\cdot e^x + (z_\infty(x) - z_n(x))\\ &\leq \epsilon\cdot e^x + (z_\infty(0) - z_n(0)) e^x\\ &\leq \epsilon\cdot e^x + \epsilon\cdot e^x\\ & = 2\epsilon \cdot e^x\\ &< e^x. \end{aligned}$$ The claim above that $z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x$ follows from $$\begin{aligned} &|a_k| \left(1 + \frac x k\right)^k \leq |a_k| e^x\\ \implies &\sum_{k \geq {n+1}}\left(1 + \frac x k\right)^k \leq \sum_{k \geq {n+1}}|a_k| e^x\\ \implies & z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x \end{aligned}$$

We are done.


  • 2
    $\begingroup$ One approach: Weierstrass M-test can be used to establish uniform convergence of the series (and of those of the derivatives) on some interval with $x=0$ as its upper bound thus one should be able to term-by-term differentiate the series and evaluate at $x=0$. Doing this $n$ times you should get something like $\sum_{k=1}^\infty a_k = 1$ and $\sum_{k=1}^\infty a_k/k^n = 1$ for $n=2,3,4,\ldots$ which allows you to deduce the $a_k$'s and get a contradiction. $\endgroup$ – Winther Aug 14 '19 at 12:49
  • $\begingroup$ @Winther Your approach only works for the absolutely convergent case (which I had already solved, but you've provided another way of doing). It leaves the conditionally convergent case. $\endgroup$ – jkabrg Aug 14 '19 at 13:05
  • $\begingroup$ Yes I implicitly assumed the coefficients had to be positive. Another possible way to look at it is to use that $(1+x/k)^k$ is basically $e^x$ for large $k$. Thus around $x=0$ we have $e^x = \sum_{k=0}^N a_k (1+x/k)^k + e^x \sum_{k=N+1}^\infty a_k + \epsilon$ where $\epsilon$ is some tiny correction (which likely also has tiny derivatives). Solving for $e^x$ and analyze the resulting series might be easier (if you can control $\epsilon$ well enough). Don't know how easy it is to make this precise though. $\endgroup$ – Winther Aug 14 '19 at 13:25
  • $\begingroup$ Not all coefficients are positive. In order to get the $x^2$ term, we need some coefficients negative. (In order to get the $x$ term, we need $a_0 = 0$.) $\endgroup$ – GEdgar Aug 14 '19 at 13:30
  • $\begingroup$ Is there any question left about my answer ? --- Maybe you are looking for another type of explanation but then please add that to your question. $\endgroup$ – user90369 Aug 16 '19 at 20:41

A reason, why $a_k$ cannot be evaluated:

Let's use the power tower function here written as $h(z)$ .

We have $~\displaystyle h(z)=z^{h(z)}~$ for $~e^{-1/e}<z<e^{1/e}~$ .

And we have (please have a look at the second section) :

$$\frac{ h(e^{z/e})^x}{1-\ln h(e^{z/e})} = \sum\limits_{k=0}^\infty\frac{(zk/e)^k}{k!}\left(1+\frac{x}{k}\right)^k ~~ , ~~~~ |z|<1 $$

Now we see:

$$z \uparrow 1 ~ \Rightarrow ~ h\left(e^{z/e}\right) \uparrow e ~ \Rightarrow ~ \frac{ h(e^{z/e})^x}{1-\ln h(e^{z/e})} \uparrow \frac{e^x}{0}$$

And $~\displaystyle \frac{e^x}{0}~$ is not defined. $~$ ( It's interesting, that e.g. for $2^x$ your series exists. )


Let $~\displaystyle a_k(z):= (1-\ln h(e^{z/e}))\frac{(zk/e)^k}{k!}~$ and $~\displaystyle p_k(x) := \left(1+\frac{x}{k}\right)^k~$ (polynomial of $k^{th}$ degree),

it's $~\displaystyle h(e^{z/e})^x = \sum\limits_{k=0}^\infty a_k(z)\,p_k(x) = \sum\limits_{v=0}^\infty\frac{(xz/e)^v}{v!}\sum\limits_{k=0}^\infty a_k(z)\,p_k(v)~$ using $~\displaystyle\ln h(e^{z/e})=\frac{z}{e}h(e^{z/e})~$ .

It follows $~\displaystyle e^x = \sup h(e^{z/e})^x = \lim\limits_{z\uparrow 1}\sum\limits_{k=0}^\infty a_k(z)\,p_k(x) = \sum\limits_{v=0}^\infty\frac{x^v}{v!}~$ .

Here is a summary of calculations for a possible proof for the formula above.

$\displaystyle \frac{d}{dz}h(z) = \frac{h(z)^2}{z(1-\ln h(z))}~$ which comes from $~\displaystyle (x^{1/x})'=x^{1/x}\frac{1-\ln x}{x^2}~$

With $~\displaystyle \frac{\partial^k}{\partial z^k}h(e^z)^x|_{z=0} =x(x+k)^{k-1}~$ for $~k\in\mathbb{N}_0 , ~ x\in\mathbb{C} ~$ we get a Taylor series at $~z=0~$ :

$$h(e^z)^x = \sum\limits_{k=0}^\infty \frac{z^k}{k!}x(x+k)^{k-1}~, ~~~x(x+k)^{k-1}|_{k=0}:=1$$

The proof for this formula can be done by using $~\displaystyle \frac{\partial}{\partial x}h(e^z)^x = z h(e^z)^{x+1}$ .

It follows $~\displaystyle \frac{z^x h(e^z)^x}{1-\ln h(e^z)} = \frac{z}{x}\frac{\partial}{\partial z}(z^x h(e^z)^x) = z^x\sum\limits_{k=0}^\infty\frac{z^k}{k!}(x+k)^k ~$

and from that directly the formula above.


Let $R_n(z)=\sum_{1 \leq k \le n} a_k \left(1 + \frac z k\right)^k$. Assuming the hypothesis we will show:

1: $R_n(z) \to e^z-a_0$ uniformly in the disc $|z| \le \frac{1}{2}$

2: $\sum_{1}^{\infty}{\frac{a_k}{k^q}}=0, q \ge 1$ arbitrary integer

3: $a_k=0, k \ge 1$

We use that if $|z| \le \frac{1}{2}, k \ge 1$, $|(1+\frac{z}{k})^k-(1+\frac{z}{k+1})^{k+1}| \le B|\frac{1}{k^2}|, B>0$ constant, which follows from $k\log(1+\frac{z}{k})=z-\frac{z^2}{2k}+O(\frac{z^3}{k^2}), |z| \le \frac{1}{2}, k \ge 1$, so $(1+\frac{z}{k})^k=e^{z-\frac{z^2}{2k}+O(\frac{z^3}{k^2})}$ and then subtracting the relations for $k, k+1$ since the $O$ terms are at most $\frac{1}{8k^2}$ in absolute value and $|\frac{z^2}{2k}-\frac{z^2}{2k+2}| \le \frac{1}{8k^2}, |z| \le \frac{1}{2}$

We also note that $|(1+\frac{z}{k})^k| \le (1+\frac{|z|}{k})^k \le e^{|z|} \le e^{\frac{1}{2}} <e$ since the binomial coefficients are positive and the triangle inequality works

For simplicity let $b_k(z)=(1+ \frac z k)^k$, so if $|z| \le \frac{1}{2}$

$|b_k(z)-b_{k+1}(z)| \le B|\frac{1}{k^2}|$

$|b_k(z)| \le e$

Then since $\Sigma{a_k} \to 1$ by hypothesis for $x=0$, it folows that $|\sum_{N}^{M}a_k| \le A$ for all $N \le M$ and some constant $A>0$, while $|\sum_{N}^{M}a_k| \to 0, N,M \to \infty$ so if we pick arbitrary $\epsilon >0, |\sum_{N}^{M}a_k| \le \epsilon, M>N >N(\epsilon)$ and then we sum by parts:


where $A_p=\sum_{N+1}^{p}a_k, p \ge N+1$

So $|R_M(z)-R_N(z)| \le A\sum_{N+1}^{M-1}{B|\frac{1}{k^2}}|+e|A_M| \le AB\frac{1}{N}+e\epsilon, M>N > N(\epsilon)$ which shows that $R_N(z)$ is uniformly Cauchy in $|z| \le \frac{1}{2}$. But this means $R_n(z)$ converges uniformly to an analytic function $f(z)$ on the disc of radius $r=\frac{1}{2}$ and since we know by hypothesis that $f(x)=e^x-a_0$ on the $[-\frac{1}{2},\frac{1}{2}]$ segment it follows by the identity principle that $f(z)=e^z-a_0$ and this is 1 above

Now, we can differentiate term by term and get $R_n(z)' \to e^z$ uniformly on the disc of radius $\frac{1}{2}$ and then plugging $z=0$ we get $\sum_{k \ge 1}a_k=1$ hence $a_0=0$, hence $R_n(z) \to e^z$ uniformly on the above disc.

Subtracting we get $\sum_{k = 1}^n\frac{a_kz}{k}(1+\frac{z}{k})^{k-1} \to 0$ uniformly which clearly implies $\sum_{k = 1}^n\frac{a_k}{k}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k \ge 1}\frac{a_k}{k}=0$

($zf_n(z) \to 0$ uniformly on the disc of radius $r$, means that for any $\epsilon >0$ there is $N(\epsilon), |zf_n(z)| \le \epsilon, |z| \le r, n \ge N(\epsilon)$

Schwarz lemma implies $|zf_n(z)| \le \frac{|z|}{r}\epsilon, |z| \le r, n \ge N(\epsilon)$ or $|f_n(z)| \le \frac{1}{r}\epsilon, |z| \le r, n \ge N(\epsilon)$)

But now (with all convergences below being uniform) we can integrate on the straight line from $0$ to $z$, $\sum_{k = 1}^{n}\frac{a_k}{k}(1+\frac{w}{k})^{k-1} \to 0$ and get $\sum_{k = 1}^{n}\frac{a_k}{k}(1+\frac{z}{k})^{k} \to 0$.

Subtracting gives $\sum_{k = 1}^n\frac{a_kz}{k^2}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k = 1}^n\frac{a_k}{k^2}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k \ge 1}\frac{a_k}{k^2}=0$. A clear induction (integrate, subtract, divide the $z$) gives 2 above

3 is a trivial consequence of 2 since wlog we can assume $\sum|a_k| < \infty$ in 2 by going to $b_k=\frac{a_k}{k^2}$ which is absolutely convergent since $a_k$ is bounded and which clearly satisfies 2; then if $p \ge 1$ is the first index for which $a_p \ne 0, |a_p|=a>0$ and with $A=\max|a_k| \ge a>0$ we easily find a large $q$ s.t. $A\frac{p^q}{(p+m)^q} \le .0001\frac{a}{(p+m)^2}, m>1$ as $(1+\frac{m}{p})^{q-2} \ge (1+\frac{1}{p})^{q-2} \to \infty$ with $q$ for fixed $p$, leading to a term with absolute value $a$ plus sum that is at most $.0001a\frac{\pi^2}{6}$ in absolute value being zero and that is impossible.

So all $a_k$ must be zero if 2 is satisfied and we are finally done!

  • $\begingroup$ let $p \ge 1$ the first index $|a_p|=a>0$, $A = \sup|a_k| < \infty$.As $(1+\frac{1}{p+m})^q \ge (1+\frac{1}{p+1})^q \to \infty$ with $q$ for $m \ge 1$,there is $q$, $(1+\frac{1}{p+m})^{q-2} \ge (1+\frac{1}{p+1})^{q-2} > 10^{100}Aa^{-1}p^2$, so when $\sum_{k \ge p}{\frac{a_k}{k^q}}=0$ is multiplied by $p^q$, the first term has absolute value $a$, while the rest are bounded by $Ap^2(10^{100}Aa^{-1}p^2)^{-1}\frac{1}{(m+p)^2}=10^{-100}a\frac{1}{(m+p)^2}$, so their sum is bounded by $10^{-100}a\sum_{k \ge 1}\frac{1}{k^2}$ - contradiction with all terms (first plus the rest) adding to zero $\endgroup$ – Conrad Aug 17 '19 at 16:17
  • $\begingroup$ This is the same proof that Dirichlet series that are zero, have all coefficients zero, though here we just have that the associated Dirichlet series is zero only at all integers greater or equal to $1$, and since the coefficients are generally alternating (could be complex numbers too), we cannot conclude directly, but need to repeat the proof $\endgroup$ – Conrad Aug 17 '19 at 16:22
  • $\begingroup$ More generally, the proof above shows that for any convergent series (conditionally enough, complex numbers ok), $\sum{a_k} \to a$, the series $\sum_{k \ge 1}{a_k(1 + \frac {z}{k})^k}$ is uniformly convergent near zero to an analytic function (most likely could make it entire but the estimates are trickier when $|z| >1$ so $|z|^2$ doesn't dominate $|z|^m, m \ge 2$ and then you can essentially try and identify coefficients if the result is known on the reals like here - many times it should work, here just the special nature of $e^z$ leads to a contradiction $\endgroup$ – Conrad Aug 17 '19 at 16:32
  • $\begingroup$ The argument is above - if the first non-zero term is $|a_p|=a>0$, multiplying by $p^q$ gives $a_p+\sum_{m \ge 1}{a_{p+m}\frac{p^q}{(p+m)^q}}=0$ and then you just make $\frac{p^q}{(p+m)^q}$ very small times $\frac{1}{(m+p)^2}$, so the sum of all the other terms except $a_p$ are very, very small in terms of $a$ and that is the contradiction $0=a_p+$ very small; I put explicit numbers in the comment above but it is a standard argument $\endgroup$ – Conrad Aug 17 '19 at 21:47
  • $\begingroup$ The problem was, that you were writing too complicate. The comment is much better then the text in your answer. What you mean is: $~\displaystyle \sum\limits_{k=p}^\infty\frac{a_k}{k^q}=0~$ for all $~q\in\mathbb{N} ~~ \& ~~ |a_k|<\infty ~ \Rightarrow ~ a_p=-\sum\limits_{m=1}^\infty\frac{a_{p+m}}{\left(1+\frac{m}{p}\right)^q} ~ \to ~ 0 ~$ for $~q\to\infty~$ . $\endgroup$ – user90369 Aug 18 '19 at 7:30

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