Need help for Recurrence!

Hello! Any hints for this? (Characteristic eqn doesn't seem to work for me here...)

$a_n = (n-1) a_{n-1} +n$ .

Thanks! :) Please help reshare this too!

#Algebra #RecurrenceRelations #HelpMe! #Mathematics

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

You have to do plug and chug

This is called plug :

$a_n = (n-1)((n-2)a_{n-2}+n-1) + n$

This is called chug

$a_n = (n-1)(n-2)a_{n-2}+(n-1)^{2}+n$

Continue doing this until you arrive at a base case....You can see the general form after a few chugs. which you can prove by induction....

Eddie The Head - 7 years ago

I don't think there is an easy way to do this. Clearly there is no characteristic polynomial as the coefficients are variable. I plugged it into wolfram alpha after working on it to no avail and its general form is $a_n=2e\Gamma(n,1)-1$ where $a_1=1$ .

$\Gamma(a,x)$ is the Incomplete Gamma Function which I have no clue of its definition.

Daniel Liu - 7 years ago

My original strategy was to substitute $b_n=a_{n+1}-c_1a_n$ for some constant $c$ such that everything cancels out neatly, but it got really messy and I ended up having $c_1=\dfrac{-n+\sqrt{n^2-4}}{2}$ (which is almost certainly wrong) then I stopped.

Daniel Liu - 7 years ago

Yes me too! I got that too, but it doesn't work the same way as typical characteristic eqn with variables...

Happy Melodies - 7 years ago

Divide through by $(n-1)!$ to get $\frac{a_n}{(n-1)!} = \frac{a_{n-2}}{(n-2)!} + \frac{n}{(n-1)!}$ . Then if we let $b_n = \frac{a_n}{(n-1)!}$ , we should have $b_1 = a_1$ and $b_n = b_{n-1} + \frac{n}{(n-1)!}$ , so $b_n = a_1 + \sum_{k=2}^{n} \frac{k}{(k-1)!}$ . Then $a_n = a_1(n-1)! + \sum_{k=2}^n \frac{k(n-1)!}{(k-1)!}$ . Perhaps that sum has a nice simple closed form?

Patrick Corn - 7 years ago

Is there a nice way to express the summation in a closed form? Summations in general do not count as a closed form. $\displaystyle\sum_{i=1}^ni$ is not a closed form, but $\dfrac{n(n+1)}{2}$ is.

Daniel Liu - 7 years ago

No, I don't think so. Although $\sum \frac{(k-2)}{(k-1)!}$ is telescoping, this still leaves $\sum \frac2{(k-1)!}$ , twice the partial sum for $e$ .

Patrick Corn - 7 years ago

I would use generating functions.

Sean Roberson - 7 years ago

a1=1,a2=3,a3=9,a4=31

Sai Satti - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$