Find the sequence.

Question

I am given generating function $\ f(x)=\frac{1}{x-3} $
I have to find formula of the sequence which is generated by this function. Have to achieve this?
I find some tutorials but it was hard for me to understand this.
In some of them people where starting from generating function for $a_{n}=1$ (I mean $ \sum_{n=0}^\infty x^n $ ) and then made transformation to achieve desired result. I would be extremely grateful if someone would explain me how to do this.

score 2 · Accepted Answer · answered Dec 10 '13 at 19:07

2

Note that $$\frac{1}{x-3}=-\frac{1}{3}\cdot\frac{1}{1-x/3}.$$ You probably know that (for suitable $t$) $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.$$ Plug in $t=x/3$ and simplify a bit to get the expansion of your function.

answered Dec 10 '13 at 19:07

André Nicolas

507,029

Am i doing it right? $ \frac{1}{1-{ \frac{x}{3}} } = \sum_{n=0}^\infty \frac{x^n}{3^n} $ Then i multiply it by $ \frac{-1}{3} $ . I got $ \sum_{n=0}^\infty x^n \cdot \frac{-1}{3^{2n}} $ and my sequence formula will be $ a_{n}=\frac{-1}{3^{2n}} $ ? – Marcin Majewski Dec 10 '13 at 19:15
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When you multiply $\frac{1}{3^n}$ by $\frac{1}{3}$ you should get $\frac{1}{3^{n+1}}$, not what you wrote. – André Nicolas Dec 10 '13 at 19:32
Eh, stupid mistake. So my final sequence formula will be $ a_{n}=\frac{-1}{3^{n+1}}$ ? – Marcin Majewski Dec 10 '13 at 19:37
1

Yes, $\frac{-1}{3^{n+1}}$, or anything equivalent. – André Nicolas Dec 10 '13 at 19:39
Thank you, it helped me a lot. – Marcin Majewski Dec 10 '13 at 19:39
You are welcome. Hope this adds a tool to the toolbox! – André Nicolas Dec 10 '13 at 19:41

score 2 · Answer 2 · answered Dec 10 '13 at 19:12

You're looking for a power series $$y = \sum_{n = 0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$$ such that $y = \frac{1}{x - 3}$. Or equivalently such that $(x - 3)y = 1$. Well we have $$xy = x\sum_{n = 0}^\infty a_nx^n = \sum_{n = 0}^\infty a_nx^{n + 1} = \sum_{n = 1}^\infty a_{n - 1}x^n$$ and $$3y = 3\sum_{n = 0}^\infty a_nx^n = \sum_{n = 0}^\infty 3a_nx^n$$ so $$xy - 3y = \sum_{n = 1}^\infty a_{n - 1}x^n - \sum_{n = 0}^\infty 3a_nx^n = \sum_{n = 1}^\infty a_{n - 1}x^n - \sum_{n = 1}^\infty 3a_nx^n - 3a_0$$ $$= \sum_{n = 1}^\infty[a_{n - 1} - 3a_n]x^n - 3a_0$$ For this to equal $1$ we must have that the coefficient of $x^0$ which is $3a_0$ equal $1$, so $a_0 = \frac13$. For $n > 0$ the coefficient of $x^n$, which is $a_{n - 1} - 3a_n$ must equal $0$. So $a_n = \frac{a_{n - 1}}{3}$. Hence $a_1 = \frac{1}{3^2}$, $a_2 = \frac{1}{3^3}$, and so on. Can you figure out a formula for $a_n$?

I do not understand why we need $ 3a_{0} $ equal to $ 1 $ ? Can you elaborate on this a little bit more please. Why $a_{n - 1} - 3a_n$ should equal to 0 ? If we get this equal to zero and $3a_{0}$ equal to $1$ we will get $-1$ not $1$? — Marcin Majewski, Dec 10 '13 at 19:35
Two power series are equal if and only if their coefficients are equal. As a power series $1 = 1 + 0x + 0x^2 + \cdots$ has coefficients $(1, 0, 0, \ldots)$ and we computed that $xy - 3y$ has coefficients $(3a_0, a_0 - 3a_1, a_1 - 3a_2, \ldots)$ so we must have $1 = 3a_0$, $0 = a_0 - 3a_1$, $0 = a_1 - 3a_2$, $\ldots$ — Jim, Dec 10 '13 at 23:00

score 1 · Answer 3 · answered Dec 10 '13 at 19:05

1

To achieve this, you expand the function in a power series around the origin. The coefficients of this power series form the sequence you seek.

answered Dec 10 '13 at 19:05

Igor Rivin

25,994
1
19
40

Find the sequence.

3 Answers3