Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the coefficient of $x^n$ in the expansion of $$\left(1 + \frac{x}{1!} + \frac{x^2}{2!}+\cdots +\frac{x^n}{n!} \right)^2$$

How do you even start this problem? Do you use multinomial theorem or binomial theorem? Could anyone please help? I found this in a textbook of mine. What I feel hard is what to do with the factorials?

share|cite|improve this question
Start by writing out some of the terms in the square of that sum. What do they look like? – abiessu Apr 21 '14 at 14:41
up vote 4 down vote accepted

To do this, you need to look at every combination of terms that gives $x^n = x^j \cdot x^{n-j}$. We can do this using the multinomial theorem, or just multiply it out.

\begin{align*}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)^2 &= \left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\\ &= 1\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right) + \frac{x}{1!}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right) + \ldots\\ &+ \frac{x^n}{n!}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\\ &= 1 + 2\frac{x}{1!} + \frac{x^2}{2!} + \left(\frac{x}{1!}\right)^2 + \frac{x^2}{2!} + \ldots \end{align*} So the $1$ coefficient is $1$, the $x$ term is $\frac{2}{1!}$, the $x^2$ term is $\frac{1}{2!} + \frac{1}{1!}\frac{1}{1!} + \frac{1}{2!}$, the $x^3$ term is $\frac{1}{3!} + \frac{1}{1!}\frac{1}{2!} + \frac{1}{2!}\frac{1}{1!} + \frac{1}{3!}$. Continuing in this manner, we get that the $x^n$ term is (using the first few terms to predict the pattern):

$$\frac{1}{n!} + \frac{1}{1!}\frac{1}{(n-1)!} + \frac{1}{2!}\frac{1}{(n-2)!} + \cdots + \frac{1}{(n-2)!}\frac{1}{2!} + \frac{1}{(n-1)!}\frac{1}{1!} + \frac{1}{n!}$$ or rewriting as a sum, we could rewrite this as \begin{align*}\sum_{j=0}^n \frac{1}{j!}\frac{1}{(n-j)!}&= \frac{1}{n!}\sum_{j=0}^n \frac{n!}{j!(n-j)!}\\ &= \frac{1}{n!}\sum_{j=0}^n \left(\begin{array}{c}n\\j\end{array}\right) 1^j 1^{n-j}\\ &= \frac{1}{n!}(1+1)^n\\ &= \frac{1}{n!}2^n \end{align*} (the second-to-last equality is the binomial theorem)

(To make sense of this when $j=0$ or $j=n$, we use the convention $0! = 1$)

share|cite|improve this answer
+1. Delete the square in the RHS at the first line. – Tunk-Fey Apr 21 '14 at 15:04
Thanks for the heads up. Fixed now. – Nicholas Stull Apr 21 '14 at 15:06
Superb answer, really clever thinking!!!! – user34304 Apr 21 '14 at 15:20

Hint From this expression $$ \left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}\right)\cdot\left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}\right), $$ you can see, that you will get every element in the final sum by taking one element from the first bracket, lets say $\frac{x^i}{i!}$ and one element from the second bracket e.g. $\frac{x^j}{j!}$, where $0\leq i, j \leq n, i, j \in \mathbb{N}$. Now you can see, that $x^n$ will come from $$ \frac{x^i}{i!} \cdot \frac{x^j}{j!} = \frac{1}{i!\cdot j!}x^{i+j}, $$ where $i + j = n$.

Lets assume $n = 2k, k \in \mathbb{N}$. Then you get

$$ 2\cdot\frac{x^n}{n!} + 2\cdot\frac{x^{n-1}}{(n-1)!}\frac{x}{1}+ \cdots +2\cdot \frac{x^{n-\frac{n}{2}-1}}{\left(n-\frac{n}{2}-1\right)!}+ \frac{x^{n-\frac{n}{2}}}{\left(n-\frac{n}{2}\right)!}\frac{x^{n-\frac{n}{2}}}{\left(n-\frac{n}{2}\right)!} $$

The $2$'s comes from this fact: for $i\neq j $ we have two choices from which bracket we take $\frac{x^i}{i!}$ (and then $\frac{x^j}{j!}$ from the other). But for the last element - since we assume, that $n$ is even we have to take $\cfrac{x^{n-\frac{n}{2}}}{(n-\frac{n}{2})!}$ from both.

For $n= 2k+1, k \in \mathbb{N}$ it is quite similar.

share|cite|improve this answer

Note that $$1+\frac x{1!}+\frac{x^2}{2!} +\ldots +\frac{x^n}{n!}=e^x+O(x^{n+1}) $$ hence $$ \left(1+\frac x{1!}+\frac{x^2}{2!} +\ldots +\frac{x^n}{n!}\right)^2=e^{2x}+O(x^{n+1})$$

share|cite|improve this answer
I'm not sure I understand how the $O(x^{n+1})$ is preserved from the first line to the second... – abiessu Apr 21 '14 at 14:43
What is O(x^n+1)? – user34304 Apr 21 '14 at 14:48


$$\left(\sum_{i=0}^n \frac {x^i}{i!}\right)^2$$

The first $\dfrac{x^n}{n!}$ term comes from $1\cdot\dfrac{x^n}{n!}$. The second term is $\dfrac{x^1}{1!}\cdot\dfrac{x^{n-1}}{(n-1)!}=\dfrac{x^n}{(n-1)!}$, and the $k$th term is $\dfrac{x^k}{k!}\cdot\dfrac{x^{n-k}}{(n-k)!}=\dfrac{x^n}{k!(n-k)!}$ for a full sum of

$$S_n = x^n\sum_{i=0}^n{1\over i!(n-i)!}$$

This sum is very much like that of binomial sums, but it is missing the $n!$ term in the numerator to make it a simple binomial sum. Of course, this only means that it could be added like so:

$$S_n = \frac{x^n}{n!}\sum_{i=0}^n{n!\over i!(n-i)!}=\frac{x^n}{n!}\sum_{i=0}^n{n\choose i}$$

But the sum over all the terms of a binomial base $n$ is $2^n$, so we have

$$S_n = 2^n\frac{x^n}{n!}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.