Vrftsjtryk

Question:

a) Find the coefficient of $x^n$ at the following generating function: $ne^{2x}$

b) Find the coefficient of $frac{x^n}{n!}$ at the following generating function: $n cos x$

My Approach:

It has been proved that $e^x = sum_{n=0}^{infty} frac{x^n}{n!}$ and that $cos x = sum_{n=0}^{infty} (-1)^nfrac{x^{2n}}{(2n)!}$.

Therefore, for (a) we have:
$$
e^{2x}=sum_{n=0}^{infty} frac{(2x)^n}{n!}
$$
Taking the derivative with respect to $x$:
$$
2e^{2x}=sum_{n=0}^{infty} nfrac{2^ncdot x^{n-1}}{n!}
$$
Multiplying by $x$:
$$
2xe^{2x}=sum_{n=0}^{infty} nfrac{2^ncdot x^n}{n!} = sum_{n=0}^{infty}frac{2^n cdot x^n}{(n-1)!}
$$
Therefore:
$$
frac{2^n}{(n-1)!}
$$
is the coefficient... But that's the coefficient for the function $2xe^{2x}$. I don't know if I am missing something, but it's not clear to me what $n e^{2x}$ is... I did the same thing for (b), but I'll hide the details.

Maybe that $n$ is just a constant with different index, and therefore we're actually looking for:
$$
n e^{2x} = n (sum frac{2^k x^k}{k!}) underbrace{rightarrow}_{text{Coef. of }x^k} frac{n 2^k}{k!}
$$
... But I can't think about that because we're asked to find the coefficient of $x^n$, and therefore the index of the infinite series that represent $e^x$ is $n$... What a MESS!

Can someone please clarify to me what is going on with that $n$ multiplying the functions in terms of $x$?

Thank you!

It is sufficient to clarify a.) since we can use the same arguments for b.). It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

At first we look at a specific case, for instance $n=5$:

• 
  

  Find the coefficient of $x^5$ at the following generating function $5e^{2x}$.

  
  
  

  This is easy to calculate, since we have no notational difficult situation.
  begin{align*}
  color{blue}{[x^5]left(5e^{2x}right)}&=5[x^5]sum_{k=0}^infty frac{(2x)^{k}}{k!}\
  &=5cdotfrac{2^5}{5!}\
  &,,color{blue}{=frac{4}{3}}
  end{align*}

  
  

Now the problem a.)

• 
  

  Find the coefficient of $x^n$ at the following generating function $ne^{2x}$.

  
  
  

  Note that $n$ is a variable independent of any series representation of $e^{2x}$. When expanding the series we conveniently use an index variable which is different to $n$.

  
  

begin{align*}
color{blue}{[x^n]left(ne^{2x}right)}&=n[x^n]sum_{k=0}^infty frac{(2x)^{k}}{k!}tag{1}\
&=ncdotfrac{2^n}{n!}\
&,,color{blue}{=frac{2^n}{(n-1)!}}
end{align*}

Note: The index variable $k$ in (1) is a so-called bound variable. This means that the scope (i.e. range of validity) is determined by the sigma-operator $sum$ and the operator precedence rules.

In fact it is even correct to write
begin{align*}
color{blue}{n}e^{2x}&=color{blue}{n}sum_{n=0}^infty frac{(2x)^n}{n!}tag{2}
end{align*}
where the factor $color{blue}{n}$ outside the sum is a free variable different to the bound index variable $n$ of the sigma-operator $sum$. But of course this is a misuse of notation and strictly to avoid in order to support readability of the text.

Side note: A great example of a notational misuse is given by Richard P. Stanley in this MO post.