## anonymous one year ago $a_{n+2}-(2^{n+1}+2^{-n}) a_{n+1}+a_n=0$$$a_1=1, a_2=3$$. Find $$a_n$$.

1. anonymous

What have you tried?

2. ganeshie8

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...

3. anonymous

thanks gane :)

4. Empty

Here's my thoughts so far: The ordinary power series generating function satisfies this equation: $f(x)-\frac{2}{x} f(\frac{x}{2})-2f(2x)-\frac{1}{x}=0$

5. anonymous

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

6. Empty

I study everything I can get my hands on lol

7. anonymous

good :) btw, how you got that?

8. Empty

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

9. anonymous

10. anonymous

$\begin{split} a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\ &= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\ &= \left(b^2-1\right)a_{n-2}-a_{n-3} \\ &= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\ &= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\ &= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\ &= \left(b^{k+1}-\frac{1-b^{k}}{1-b}\right)a_{n-k-1}-a_{n-k-2} \\ \end{split}$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 $$b=2^{n+1}+2^{-n}$$.

11. anonymous

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 :)

12. anonymous

$\begin{split} a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\ &= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\ &= \left(b^2-1\right)a_{n-2}-a_{n-3} \\ &= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\ &= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\ &= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\ &= \left(b^{k}-\frac{1-b^{k-1}}{1-b}\right)a_{n-k}-a_{n-k-1} \\ \end{split}$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 $$b=2^{n+1}+2^{-n}$$.

13. anonymous

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

14. anonymous

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

15. anonymous

thanks

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