mathematics-physics-wiki/docs/en/mathematics/set-theory/recursion-induction.md

# 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

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

> *Principle* **- Strong 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 $k \in \mathbb{Z}$ we have that $P(b), P(b+1), \dots, P(k-1)$ and $P(k)$ together imply $P(k+1)$. 
> 
> Then $P(n)$ is true for all $n \geq b$. 

For example, we claim for the recursion

$$
\begin{align*}
    &a_1 = 1, \\
    &a_2 = 3, \\
    &a_n = a_{n-2} + 2 a_{n-1}
\end{align*}
$$

that $a_n$ is odd $\forall n \in \mathbb{N}$. 

We first check the claim for for $n=1$ and $n=2$, from the definition of the recursion it may be observed that the it is true.

Now suppose that for some $i \in \{1, \dots, k\}$ $a_i$ is odd.

Then by assumption

$$
\begin{align*}
    a_{k+1} &= a_{k-1} + 2 a_k, \\
            &= a_{k-1} + 2 a_{k} + 2(a_{k-2} + 2a_{k-1}), \\
            &= 2 (a_k + a_{k-2} + 2 a_{k-1}) + a_{k-1},
\end{align*}
$$

so $a_{k+1}$ is odd.