Question

\[a_{n+2}-(2^{n+1}+2^{-n}) a_{n+1}+a_n=0\]\(a_1=1, a_2=3\).

Find \(a_n\).

Answer 1

What have you tried?

Answer 2

I saw this problem before and know a solution which is somewhat complicated. I'm going to wait for other solutions :) I think this problem is composed by mukushla, he wants us to try...

Answer 3

thanks gane :)

Answer 4

Here's my thoughts so far:

The ordinary power series generating function satisfies this equation:

\[f(x)-\frac{2}{x} f(\frac{x}{2})-2f(2x)-\frac{1}{x}=0\]

Answer 5

@Empty you study physics? That seems like a physicist approach!

Answer 6

I study everything I can get my hands on lol

Answer 7

good :) btw, how you got that?

Answer 8

Sorry that wasn't quite the link I wanted, this is the book I'm reading currently generatingfunctionology:

https://www.math.upenn.edu/~wilf/DownldGF.html

Answer 9

thanks for the link

Answer 10

\[\begin{split}
a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\
&= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\
&= \left(b^2-1\right)a_{n-2}-a_{n-3} \\
&= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\
&= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\
&= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\
&= \left(b^{k+1}-\frac{1-b^{k}}{1-b}\right)a_{n-k-1}-a_{n-k-2} \\
\end{split}\]For our base case, we plug in \(k=n-3\): \[
a_n = \left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)a_2+a_1 = 3\left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)+1
\]Then substitute the \(b=2^{n+1}+2^{-n}\).

Answer 11

It looks good, maybe you have some little mistakes, find \(a_3\) from your formula and once again directly from the relation and compare, see if the formula is right or not :)

Answer 12

\[\begin{split}
a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\
&= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\
&= \left(b^2-1\right)a_{n-2}-a_{n-3} \\
&= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\
&= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\
&= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\
&= \left(b^{k}-\frac{1-b^{k-1}}{1-b}\right)a_{n-k}-a_{n-k-1} \\
\end{split}\]For our base case, we plug in \(k=n-2\): \[
a_n = \left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)a_2+a_1 = 3\left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)+1
\]Then substitute the \(b=2^{n+1}+2^{-n}\).

Answer 13

There is a special case at \(n=3\) because for that case, we have the coefficient \(b\) rather than \(b-\frac{1}{1-b}\)

Answer 14

In fact, all the \(n<4\) cases probably will not hold, because that geometric sum assumes higher values for \(n\).

Answer 15

thanks