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Exercises of Chapter 1 of gfology

  1. Find the ordinary power series generating functions of each of the following sequences, in simple, closed-form. In each case, the sequence is defined for all \(n\ge 0\).
  • \(a_n=n\).

sol:

\[\begin{aligned} f(z)&=\sum_{n\ge 0} nz^n=z\sum_{n\ge 1}nz^{n-1}\\ &=z\sum_{n\ge 1}(z^n)'=z\left(\sum_{n\ge 1}z^n\right)'\\ &=\dfrac{z}{(1-z)^2} \end{aligned} \]

  • \(a_n=\alpha n+\beta\).

sol:

\[\begin{aligned} f(z)&=\sum_{n\ge 0} (\alpha n+\beta)z^n\\ &=\alpha\dfrac{\alpha}{(1-z)^2}+\dfrac{1}{(1-z)} \end{aligned} \]

  • \(a_n=3^n\)

sol:

\[\begin{aligned} f(z)=\sum_{n\ge 0} 3^nz^n=\sum_{n\ge 0} (3z)^n = \dfrac{1}{1-3z} \end{aligned} \]

  1. For each of the sequences given in part 1, find the exponential generating function of the sequence in simple, closed-form.
  • \(a_n=n\).

sol:

\[\begin{aligned} f(z)=\sum_{n\ge 1}\dfrac{z^{n}}{(n-1)!}=z\sum_{n\ge 0}\dfrac{z^n}{n!}=ze^z \end{aligned} \]

  • \(a_n=\alpha n+\beta\)

sol:

\[f(z)=\alpha\sum_{n\ge 0} n\dfrac{z^n}{n!}+\beta\sum_{n\ge 0} \dfrac{z^n}{n!}=(\alpha z+\beta)e^z \]

  • \(a_n=3^n\)

sol:

\[f(z)=\sum_{n\ge 0}\dfrac{(3z)^n}{n!}=e^{3z} \]

  1. If \(f(z)\) is the ordinary power series generating function of the sequence \(\{a_n\}_{n\ge 0}\), then express simply in terms of \(f(z)\), the ordinary power series generating functions of the following sequences. In each case the range of \(n\) is \(0,1,2,\dots\)
  • \(\{a_n+c\}\)
    sol:

\[g(z)=\sum_{n\ge 0}(a_n+c)z^n=f(z)+\dfrac{c}{(1-z)} \]

  • \(\{na_n\}\)

sol:

\[g(z)=\sum_{n\ge 0} na_nz^n=z\sum_{n\ge 1}na_nz^{n-1}=\sum_{n\ge 1}a_n(z^n)'=(f(z)-a_0)'=f'(z) \]

  • \(\{a_{n+h}\}\), \(h\) is a given constant.

\[g_z=\sum_{n\ge 0}a_{n+h}z^n=\dfrac{f(z)-\sum_{j=0}^{h-1} a_jz^j}{z^h} \]

  1. Find
  • \([z^n] e^{2z}\)

sol:

\[[z^n] e^{2z}=\dfrac{2^n}{n!} \]

  • \(\left[\dfrac{z^n}{n!}\right]e^{\alpha z}\)

sol:

\[\left[\dfrac{z^n}{n!}\right]e^{\alpha z}=\alpha^n \]

  • \([z^n/n!]\sin z\)

sol:

\[( [z^n/n!]\sin z=[z^n/n!]\dfrac{e^{iz}-e^{-iz}}{2i}=\dfrac{[z^n/n!](e^{iz}-e^{-iz})}{2i}=\dfrac{1+(-1)^{n+1}}{2} \]

posted @ 2022-03-28 10:30  feicheng  阅读(10)  评论(0编辑  收藏  举报