Find the sequence $(y_{k})_{k}$ that gives rise to the generating function $Y(s) = (3 - s)^{n}$, $n \in \mathbb{N}$.
Someone can help me? Thank you in advance.
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1$\begingroup$ Given a sequence, do you know how to write down its generating function? $\endgroup$rogerl– rogerl2019-11-27 20:52:14 +00:00Commented Nov 27, 2019 at 20:52
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$\begingroup$ What have you tried? People will be happy to help you once they know you've put in some effort too. $\endgroup$Chris Grossack– Chris Grossack2019-11-27 20:52:42 +00:00Commented Nov 27, 2019 at 20:52
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2$\begingroup$ Hint: Write $Y(s) = 3^{1/n}\left(1-\frac13 s\right)^n$. Also note that $Y(s)$ is a polynomial of degree $n$, so the coefficients of the sequence $y_k$ will be zero for $k>n$. $\endgroup$Math1000– Math10002019-11-27 21:03:16 +00:00Commented Nov 27, 2019 at 21:03
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1$\begingroup$ Do you know of a theorem that gives a formula to easily expand the binomial power $(x+y)^n$? $\endgroup$Brian Moehring– Brian Moehring2019-11-27 21:06:26 +00:00Commented Nov 27, 2019 at 21:06
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2$\begingroup$ @BrianMoehring Yes, I meant $3^n\left(1-\frac13 s\right)^n$. Moreover, $$ \left(1-\frac13s\right)^n = \sum_{k=0}^n \binom nk \left(-\frac13s\right)^k $$ by the binomial theorem. $\endgroup$Math1000– Math10002019-11-28 02:31:23 +00:00Commented Nov 28, 2019 at 2:31
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1 Answer
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We observe $Y(s)=(3-s)^n$ is a polynomial in $s$ of degree $n$. This implies the polynomial admits a representation \begin{align*} Y(s)=\sum_{k=0}^ny_k s^k \end{align*} Application of the binomial theorem to $Y(s)=(3-s)^n$ gives \begin{align*} Y(s)&=\sum_{k=0}^n\binom{n}{k}(-s)^k3^{n-k}\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k3^{n-k}s^k \end{align*} from which we deduce the sequence $(y_k)_k$ is the finite sequence \begin{align*} (y_k)_{0\leq k\leq n}=\color{blue}{\left(\binom{n}{k}(-1)^k3^{n-k}\right)_{0\leq k\leq n}} \end{align*}
