- dan815

expand x/(1-x-x^2) as a power series about the origin

- schrodinger

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- ganeshie8

looks suspiciously related to fibonacci

- anonymous

yeah, that is

- ganeshie8

https://d2m81yxg6b9itc.cloudfront.net/flex-sequence-1/processed/fibonacci-formula.60d0194eb1ccc3a725abd73435bc877e/full/540p/index.mp4

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## More answers

- dan815

lol... someone needs to teach him how to hold a pen

- dan815

|dw:1435870607209:dw|

- Empty

let's say \[f(x)=\sum_{n=0}^\infty a_n x^n\] then this means the difference quotient\[\frac{f(x)-f(0)}{x}= \frac{f(x)-a_0}{x}=\sum_{n=0}^\infty a_{n+1}x^n\] This is taught at the very beginning of chapter 2 in generatingfunctionology which I'm glad you've just started reading, I'll stop here because I think this is sufficiently interesting. ;P

- DanJS

hmm, free book , thanks

- dan815

hmm interseting

- dan815

u can carry this process for getting powerseries with index shifting higher and higher

- dan815

|dw:1435871132463:dw|

- dan815

oh my

- dan815

did u get can equation for a powerseiries with k index shifted

- dan815

interms of f(x) and f(0)s

- Empty

hint: \( g(0)=a_1 \)

- Empty

remember you have to dived the whole difference by x, not just the second part

- dan815

ya just noticed

- dan815

|dw:1435871684516:dw|

- dan815

|dw:1435871700596:dw|

- dan815

i didnt work out that pattern, just thinking itd be nice if it was this

- dan815

its gotta be this right

- dan815

oh ya its kinda trivial hehe

- dan815

so what can we do with this

- Empty

Well guess you gotta read the book ;P

- dan815

will come into stuff like descrete methods of representations for the derivatives

- dan815

now that f(0) is popping up all over

- dan815

cause for the contiuos version L{f'}=s*L{f}-f(0)

- dan815

tell me or i will suspend u

- anonymous

I would try separating the denominator into partial fractions, then rewriting until the components can be expressed as geometric sums.\[1-x-x^2=-\left(x+\frac{1+\sqrt5}{2}\right)\left(x+\frac{1-\sqrt5}{2}\right)\]
then
\[\frac{x}{1-x-x^2}=\frac{1-\sqrt5}{\sqrt5\left(2x+1-\sqrt5\right)}-\frac{1+\sqrt5}{\sqrt5\left(2x+1+\sqrt5\right)}\]
or something like that...

- anonymous

Actually, those \(\sqrt5\)s should be \(5\)s in the denominator.

- anonymous

Just to add some more detail, the next step would be to do the following:
\[\frac{1}{ax-b}=-\frac{1}{\frac{1}{b}\left(1-abx\right)}=-b\sum_{k=0}^\infty (abx)^k\]

- anonymous

$$f(x)=\frac{x}{1-x-x^2}\\(1-x-x^2)f(x)=x$$let \(f(x)=\sum_{n=0}^\infty a_n x^n\) so $$f(x)-xf(x)-x^2 f(x)=x\\\sum_{n=0}^\infty a_n x^n-\sum_{n=1}^\infty a_{n-1} x^n-\sum_{n=2}^\infty a_{n-2} x^n=x\\a_0+a_1 x-a_0 x+\sum_{n=2}^\infty (a_n-a_{n-1}-a_{n-2}) x^n=x$$so it follows \(a_n=a_{n-1}+a_{n-2};\quad a_0=0, a_1=a_0+1=1\)

- anonymous

ergo \(a_n=F_n\) where \(F_n\) is the n-th Fibonacci number where \(F_0=0\) and so $$\frac{x}{1-x-x^2}=\sum_{n=0}^\infty F_n x^n=x+x^2+2x^3+3x^4+5x^5+\dots$$

- anonymous

^^ by the way, this general rule works for all series whose generating functions are rational \(P(x)/Q(x)\) -- they are defined by a linear recurrence

- anonymous

and generating functions are great (especially in DSP when it comes to z-transforms and analyzing the behavior of systems on digital sequences of data)

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