# Recursion and induction

## Recursion

A recursively defined function $f$ needs two ingredients:

* a *base*, where the function value $f(n)$ is defined, for some value of $n$.
* a *recursion*, in which the computation of the function in $n$ is explained with the help of the previous values smaller than $n$.

For example, the sum

$$
    \begin{align*}&\sum_{i=1}^1 i = 1,\\ &\sum_{i=1}^{n+1} i = (n + 1) + \sum_{i=1}^{n} i.\end{align*}
$$

Or the product

$$
    \begin{align*}&\prod_{i=0}^0 i = 1,\\ &\prod_{i=0}^{n+1} i = (n+1) \cdot \prod_{i=0}^{n} i.\end{align*}
$$

## Induction

> *Principle* **- Natural induction**: suppose $P(n)$ is a predicate for $n \in \mathbb{Z}$, let $b \in \mathbb{Z}$. If the following holds
>
> * $P(b)$ is true,
> * for all $k \in \mathbb{Z}$, $k \geq b$ we have that $P(k)$ implies $P(k+1)$. 
>
> Then $P(n)$ is true for all $n \geq b$.

For example, we claim that $\forall n \in \mathbb{N}$ we have

$$
    \sum_{i=1}^n i = \frac{n}{2} (n+1).
$$

We first check the claim for $n=1$:

$$
    \sum_{i=1}^1 i = \frac{1}{2} (1+1) = 1.
$$

Now suppose that for some $k \in \mathbb{N}$ 

$$
    \sum_{i=1}^k i = \frac{k}{2} (k+1).
$$

Then by assumption

$$
\begin{align*}
    \sum_{i=1}^{k+1} i &= \sum_{i=1}^k i + (k+1), \\
                       &= \frac{k}{2}(k+1) + (k+1), \\
                       &= \frac{k+1}{2}(k+2).
\end{align*}
$$

Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that $\forall n \in \mathbb{N}$ we have

$$
    \sum_{i=1}^n i = \frac{n}{2}(n+1).
$$