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.