Combinatorial Numbers

Let there be a recurrence relation \({ P }_{ a }\left( n \right) =n\cdot { P }_{ a }\left( n-1 \right) +{ a }^{ n }\) , where \({ P }_{ a }\left( 0 \right) =1\). Show that \({ P }_{ a }\left( n \right) \sim { e }^{ a }n!\) for large \(n\).

Solution

We first show that the above recurrence relation constructs the sum \[{P}_{a}(n) = n!\left( \sum _{ k=0 }^{ n }{ \frac { { a }^{ k } }{ k! } } \right).\]

Now we prove by induction.

When \(k=1\)

\[1! \left(\frac{{a}^{0}}{0!} + \frac{{a}^{1}}{1!}\right) = 1(1)+a \]

\[1!\left( \sum _{ k=0 }^{ 1 }{ \frac { { a }^{ k } }{ k! } } \right) = 1\cdot {P}_{a}(0) + {a}^{1} = {P}_{a}(1).\]

When \(k=n\)

\[n! \left(\frac{{a}^{0}}{0!} + \frac{{a}^{1}}{1!} + ...+ \frac{{a}^{n}}{n!}\right) = n(n-1)! \left(\frac{{a}^{0}}{0!} + \frac{{a}^{1}}{1!} + ...+ \frac{{a}^{n-1}}{(n-1)!}\right) + {a}^{n} \]

\[n!\left( \sum _{ k=0 }^{ n }{ \frac { { a }^{ k } }{ k! } } \right) = n\cdot {P}_{a}(n-1) + {a}^{n-1} = {P}_{a}(n).\]

When \(k=n+1\)

\[(n+1)! \left(\frac{{a}^{0}}{0!} + \frac{{a}^{1}}{1!} + ...+ \frac{{a}^{n+1}}{(n+1)!}\right) = (n+1)(n)! \left(\frac{{a}^{0}}{0!} + \frac{{a}^{1}}{1!} + ...+ \frac{{a}^{n}}{n!}\right) + {a}^{n+1} \]

\[(n+1)!\left( \sum _{ k=0 }^{ n+1 }{ \frac { { a }^{ k } }{ k! } } \right) = (n+1)\cdot {P}_{a}(n) + {a}^{n+1} = {P}_{a}(n+1)\]

Notice that the sum \( \sum _{ k=0 }^{ n }{ \frac { { a }^{ k } }{ k! } } \) approaches \({e}^{a}\) for large \(n\).

Hence, \[{ P }_{ a }\left( n \right) \sim { e }^{ a }n!\] for large \(n\).

Note: the way I found this recurrence relation is by investigating the integral \[{ P }_{ a }(n)=\int _{ a }^{ \infty }{ { e }^{ a-x } } { x }^{ n }dx .\] As an exercise, prove that this integral is equivalent to the defined recurrence equation.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
3 years, 12 months ago

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