Differentiating closed formulas in formal power series In Herbert Wilfs' book gfology, the generating function is defined "formally" as 

If $\displaystyle f = \sum_{i \geq 0} a_i x^i$, and $\displaystyle g = \sum_{i \geq 0} b_i x^i $, we define 
  
  
*
  
*$\displaystyle f+g := \sum_{i \geq 0} (a_i + b_i) x^i$ 
  
*$\displaystyle fg := \sum_{i \geq 0}(\sum_{m+n = i} a_mb_n)x^i$. 
  
*$\displaystyle f' := \sum_{i \geq 0} ia_{i+1} x^i$ (Call this $\star$)
  

This is fine and I understand this and have no problem with this. But when we say something like this in a formal setting:

$\displaystyle e^x = \sum_{i \geq 0} \frac{x^n}{n!}$

Or, 

$\displaystyle \frac{1}{\sqrt{1-4x}} = \sum_{k \geq 0} \binom{2k}{k} x^k 
$ (Call this equation $\spadesuit$)

My question is: How is $e^x$ or $\displaystyle \frac{1}{\sqrt{1-4x}}$ defined in a formal setting ? 
Are they defined as the generating functions of the sequences I mentioned ? 
If it's defined like that, then how can you proceed to "differentiate" both sides of $\spadesuit$ to obtain 

$\displaystyle \frac{2}{\sqrt{(1-4x)^3}} = \sum_{k \geq 0} k \binom{2(k+1)}{k+1} x^k $

I understand how you formally differntiate the right hand side of $\spadesuit$ since that's defined in $\star$, but how the formal differentiation of $\frac{1}{\sqrt{1-4x}}$ defined (especially when limits may not make much sense working on some rings)? 
 A: There are essentially two ways to define $e^x$:

*

*The lazy way is to define $e^x$ as $\sum_{k=0}^{\infty} \dfrac{x^k}{k!}$. This is easy, but then you have to explain what $e^{x+y}$ and $e^{-x}$ and $e^{e^x-1}$ are, and whatever other exponentials you need.


*The optimal way (to me) is to define $e^a$ as $\sum_{k=0}^{\infty} \dfrac{a^k}{k!}$ whenever $a$ is an element of a topological $\mathbb{Q}$-algebra for which the sum $\sum_{k=0}^{\infty} \dfrac{a^k}{k!}$ converges. This defines the power series $e^x$, $e^{x+y}$, $e^{-x}$, $e^{e^x-1}$ and many others simultaneously and also $e^a$ when $a$ is a nilpotent element of a $\mathbb{Q}$-algebra (here, the sum converges because all but its first few terms are zero).
That said, the difference between these two ways is merely aesthetic: If you have defined $e^x$ in the first way, you can then define $e^a$ as the result of evaluating $e^x$ at $x=a$ whenever this evaluation is well-defined. Thus, the definitions boil down to the same.
Now, what about $\sqrt{1-4x}$? If $p$ is a formal power series with coefficient $1$ over a $\mathbb{Q}$-algebra $\mathbb{K}$, then I see three ways to define $\sqrt{p}$:

*

*The most natural way is to define $\sqrt{p}$ to be the unique power series with constant term $1$ whose square is $p$. It takes a bit of work to check that this power series actually does exist and is unique (this is why we require that it has constant term $1$; otherwise it would not be unique). Then, it takes a bit more work to prove basic identities, like $\sqrt{p}\cdot\sqrt{q} = \sqrt{pq}$ for arbitrary $p$ and $q$ when $\mathbb{K}$ is commutative. Ultimately this is still fairly fast and easy. The downside of this approach is that it doesn't generalize very far: It lets you define $\sqrt{p}$ and $\sqrt[m]{p}$ for positive integers $m$, but not (for example) $p^{\sqrt{2}}$.


*The easiest way to define $p^a$ for any power series $p$ with constant coefficient $1$ and any $a \in \mathbb{K}$ is the following: Write $p$ as $p = 1+r$ for some formal power series $r$ with constant coefficient $0$, and set
\begin{equation}
p^a = \sum_{k=0}^{\infty} \dbinom{a}{k} r^k ,
\end{equation}
where $\dbinom{a}{k} = \dfrac{a\left(a-1\right)\cdots\left(a-k+1\right)}{k!}$ is computed in the $\mathbb{Q}$-algebra $\mathbb{K}$. This is called "Newton's binomial formula". Setting $a=1/2$, you then obtain $p^{1/2}$ (that is, $\sqrt{p}$). The downside of this approach is that the basic rules of exponents (such as $\left(p^a\right)^b = p^{ab}$ and $\left(pq\right)^a = p^a q^a$ when $\mathbb{K}$ is commutative) are hard to prove.


*Finally, the probably simplest approach is to first define $e^a$ for any formal power series $a$ with constant term $0$ (by $e^a = \sum_{k=0}^{\infty} \dfrac{a^k}{k!}$) and define $\log b$ for any formal power series $b$ with constant term $1$ (by $\log b = - \sum_{i=1}^{\infty} \dfrac{\left(1-b\right)^i}{i}$). Once these are defined, you can set $p^a = e^{a \log p}$ for any formal power series $p$ with constant term $1$ and any $a \in \mathbb{K}$. The usual properties of exponentials and logarithms then yield the basic rules of exponents.
It is an instructive exercise to show that the last two of these three definitions are equivalent, and that the first one is equivalent to them whenever it applies.
Note that the first and second definitions have an advantage that the third does not: They can be extended to more general classes of commutative rings $\mathbb{K}$. For example, $\sqrt{p}$ is defined not just when $\mathbb{K}$ is a $\mathbb{Q}$-algebra, but already when $\mathbb{K}$ is a ring in which $2$ is invertible. More generally, if $a \in \mathbb{Q}$, then $p^a$ is defined whenever the denominator of $a$ is invertible in $\mathbb{K}$. The third definition does not apply in this generality, since it requires $\mathbb{K}$ to be a $\mathbb{Q}$-algebra.
A: The exponential function, $e^x$ is defined formally as $$e^x = \sum_{n\ge 0} \frac{x^n}{n!}. $$
One can easily show that the formal derivative of $e^x$ is itself and that $e^{x + y} = e^xe^y$ and other properties a formal exponential should have.
On the other hand, $f(x)=\frac{1}{\sqrt{1-4x}}$ is a bit more subtle. Whatever this power series is, we would like it to satisfy
$$ f(x)^2 = \sum_{n \ge 0} 4^n x^n = \frac{1}{1-4x}, \; f(0) = 1. \tag{1}$$
That is, we define $f(x)$ as the unique root of $(1 - 4x)y^2 - 1$ with positive constant term.
There are now two ways to proceed. First, using complex analysis where we can speak about Taylor series and where $f(x)$ will be a power series defined in some neighbourhood of zero. With this approach, we already know what the derivative should be and then we can make use of a theorem which says that the derivative of a power series can be computed term-by-term.
The second approach is to take $(1)$ as the definition. Then, one shows that the formal derivative satisfies the usual rules:


*

*$(f + g)' = f' + g'$

*$(fg)' = f'g + g'f$

*$(f/g)' = (f'g - fg')/g^2$ whenever $g$ is invertible (i.e. $g(0) \ne 0$)


In particular, if we let $g(x) = 1 - 4x$ and differentiate both sides of
$$ f(x)^2 = \frac{1}{g(x)}, $$
using the second and third bullet points, we get
$$ 2f(x)f'(x) = -\frac{g'(x)}{g(x)^2}. $$
Now squaring both sides and using $(1)$, we have
\begin{align}
4f(x)^2f'(x)^2 &= \frac{g'(x)^2}{g(x)^4} \\
4\frac{f'(x)^2}{g(x)} &= \frac{g'(x)^2}{g(x)^4} \\
f'(x)^2 &= \frac{g'(x)^2}{4g(x)^3} = \frac{4}{(1 - 4x)^3}.
\end{align}
Therefore $f'(x)$ is the root of $(1 - 4x)^3y^2 - 4$ with positive constant term.
The point being, if you have a power series that satisfies some algebraic relation, then the derivative will satisfy another algebraic relation.
