# Formal (series/sum/derivative…)

I have come across a lot of cases where terms such as formal sum rather than simply sum is used, similarly in case of derivatives/infinite series/power series.

As I understand in case of series/sum, the term formal is used when the notion of convergence is not clear.

I would appreciate any precise definition or explanation of where formal is used. I also cannot get how it relates to derivatives. Also, where else is "formal" used?

• The case of differentiation is a good one because things can get a bit weird once you start looking at distributions. Take for instance the Heaviside step function. It is not differentiable in the usual sense at $0$ however distributionally, it has a Dirac delta as a derivative. So you could say "formally $H'(x) = \delta(x)$" or something along those lines. – Cameron Williams Jul 27 '13 at 21:13

## 2 Answers

A formal sum is where we write something using a $+$ symbol, or other way normally used for sums, even when there may be no actual operation defined.

An example. Someone may have defined a quaternion as a formal sum of a scalar and a (3-dimensional) vector, for example. Before this definition, at least in that book, there was no defined "sum" of a scalar plus a vector.

After that definition, our author may tell us what is means for two quaternions to be equal $$\lambda + \mathbf{x} = \mu + \mathbf{y} \quad \Longleftrightarrow \quad\text{??}$$ how to add quaterntions $$(\lambda + \mathbf{x}) + (\mu + \mathbf{y}) = \text{??}$$ how to multiply quaterntions $$(\lambda + \mathbf{x}) \cdot (\mu + \mathbf{y}) = \text{??}$$ and so on.

The formal power series over a commutative ring are basically just the sequences in that ring, with addition and multiplication defined to match. So $\sum_{n\in\Bbb N}a_n x^n$ is basically equivalent to the sequence $(a_n)_{n\in \Bbb N}$, but there are rules for addition, subtraction, and multiplication of the former: $$\sum_{n\in \Bbb N}a_n x^n + \sum_{n\in N}b_n x^n := \sum_{n\in\Bbb N}(a_n+b_n)x^n,$$ $$-\sum_{n\in\Bbb N}a_n x^n = \sum_{n\in\Bbb N}(-a_n)x^n,$$ and $$\Bigl(\sum_{n\in \Bbb N}a_nx^n\Bigr)\Bigl(\sum_{m\in\Bbb N}b_nx^m\Bigr):=\sum_{n\in \Bbb N}\Bigl(\sum_{j+k=n}a_jb_k\Bigr)x^n.$$