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I have been given the following recurrence relation:

$b_{n} = (−1)^{n}(n + 1)a_{0} + (−1)^{n−1}(n)a_1 + · · · + (−1)(2)a_{n−1} + a_n $

I am trying to find a generating function for $b_{n}$ in terms of the generating function for $a_{n}$

I get the pattern for the sequence is $-1^{n}(n+1),-1^{n-1}(n), -1^{n-2}(n-1) \dots -1^{n-(n-1)}(2), -1^{n-n}(1)$

However, the thing that confuses me is how I would index this sequence. Is there a way of indexing 'backwards', such that the sum goes from n to zero. For example:

$\sum_{k=n}^{0} = ((k+1)(-1)^{k})a_k\ x^{n-k}$

However, I have no idea how I would be able to express this in closed form, let alone whether it is possible. and as such I was wondering whether someone could explain what direction I could take.

I have also thought of modifying the sequence to make it so that:

$\frac{b_{n}}{g(n)} = a_{n}$ where $g(n) = \frac{1}{1+n} + \frac{2}{1-n}$ which I believe is the closed form expression for the sequence 1,-1,1,-1,1.

However, this still leaves me with the problem of finding the sequence for $a_{n}$ which would still be $(n+1)a_{0} + (n)a_{1} + \dots + (2)a_{n-1} + (1)a_{n}$

I am confused as to whether the above sequence is finite or infinite, as such I was wondering whether someone could explain what direction I could take from here.

Thanks in advance.

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  • $\begingroup$ The suggested writing $\sum_{k=n}^0$ is not helpful, since a sum is defined to be zero when the upper limit is less than the lower limit. A way to cope with it is addressed in my answer at (2). $\endgroup$ – Markus Scheuer Apr 29 '18 at 13:58
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We consider a generating function $A(z)=\sum_{n=0}^\infty a_n z^n$ and are looking for a generating function $B(z)=\sum_{n=0}^\infty b_n z^n$ with \begin{align*} b_n&=(-1)^n(n+1)a_0+(-1)^{n-1}na_1+\cdots -2a_{n-1}+a_n\\ &=\sum_{k=0}^n(-1)^{n-k}(n+1-k) a_k\tag{1} \end{align*}

We apply the Cauchy product formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^n]A(z)=a_n \end{align*}

We obtain from (1): \begin{align*} \color{blue}{b_n} &=\sum_{k=0}^n(-1)^{n-k}(n+1-k) a_k\\ &=\sum_{k=0}^n(-1)^{k}(k+1) a_{n-k}\tag{2}\\ &=\sum_{k=0}^n\left([z^k]\frac{1}{(1+z)^2}\right)\left([z^{n-k}]A(z)\right)\tag{3}\\ &\,\,\color{blue}{=[z^n]\frac{1}{(1+z)^{2}}A(z)} \end{align*} and conclude \begin{align*} \color{blue}{B(z)=\frac{A(z)}{(1+z)^2}} \end{align*}

Comment:

  • In (2) we change the order of summation by replacing the index $k$ with $n-k$.

  • In (3) we use the series representation \begin{align*} \sum_{k=0}^\infty(-1)^k(k+1)z^k=\frac{1}{(1+z)^2} \end{align*} and apply the Cauchy product formula \begin{align*} Q(z)&=\sum_{k=0}^\infty q_kz^k,\qquad R(z)=\sum_{j=0}^\infty r_jz^j\\ Q(t)R(t)&=\sum_{n=0}^\infty\left(\sum_{{k+j=n}\atop{k,j\geq 0}}q_kr_j\right)t^n=\sum_{n=0}^\infty \left(\sum_{k=0}^n q_k r_{n-k}\right)t^n\\ &=\sum_{n=0}^\infty\sum_{k=0}^n \left([z^k]Q(z)\right)\left([z^{n-k}]R(z)\right)t^n \end{align*}

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