63 lines
1.6 KiB
Markdown
63 lines
1.6 KiB
Markdown
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# Recursion and induction
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## Recursion
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A recursively defined function $f$ needs two ingredients:
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* a *base*, where the function value $f(n)$ is defined, for some value of $n$.
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* a *recursion*, in which the computation of the function in $n$ is explained with the help of the previous values smaller than $n$.
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For example, the sum
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$$
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\begin{align*}&\sum_{i=1}^1 i = 1,\\ &\sum_{i=1}^{n+1} i = (n + 1) + \sum_{i=1}^{n} i.\end{align*}
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$$
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Or the product
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$$
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\begin{align*}&\prod_{i=0}^0 i = 1,\\ &\prod_{i=0}^{n+1} i = (n+1) \cdot \prod_{i=0}^{n} i.\end{align*}
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$$
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## Induction
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> *Principle* **- Natural induction**: suppose $P(n)$ is a predicate for $n \in \mathbb{Z}$, let $b \in \mathbb{Z}$. If the following holds
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>
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> * $P(b)$ is true,
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> * for all $k \in \mathbb{Z}$, $k \geq b$ we have that $P(k)$ implies $P(k+1)$.
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>
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> Then $P(n)$ is true for all $n \geq b$.
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For example, we claim that $\forall n \in \mathbb{N}$ we have
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$$
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\sum_{i=1}^n i = \frac{n}{2} (n+1).
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$$
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We first check the claim for $n=1$:
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$$
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\sum_{i=1}^1 i = \frac{1}{2} (1+1) = 1.
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$$
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Now suppose that for some $k \in \mathbb{N}$
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$$
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\sum_{i=1}^k i = \frac{k}{2} (k+1).
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$$
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Then by assumption
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$$
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\begin{align*}
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\sum_{i=1}^{k+1} i &= \sum_{i=1}^k i + (k+1), \\
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&= \frac{k}{2}(k+1) + (k+1), \\
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&= \frac{k+1}{2}(k+2).
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\end{align*}
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$$
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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
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$$
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\sum_{i=1}^n i = \frac{n}{2}(n+1).
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$$
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