Question 22.34: Use the binomial theorem to expand (1-1/z)^-3 up to the term......
Use the binomial theorem to expand \left(1-\frac{1}{z}\right)^{-3} \text { up to the term } \frac{1}{z^4}. Hence find the sequence with z transform F(z)=\frac{z^3}{(z-1)^3}.
Step-by-Step
Using the binomial theorem, we have
\begin{aligned} \left(1-\frac{1}{z}\right)^{-3}= & 1+(-3)\left(-\frac{1}{z}\right)+\frac{(-3)(-4)}{2 !}\left(-\frac{1}{z}\right)^2 \\ & +\frac{(-3)(-4)(-5)}{3 !}\left(-\frac{1}{z}\right)^3+\frac{(-3)(-4)(-5)(-6)}{4 !}\left(-\frac{1}{z}\right)^4+\cdots \\ = & 1+\frac{3}{z}+\frac{6}{z^2}+\frac{10}{z^3}+\frac{15}{z^4}+\cdots \end{aligned}provided |z| > 1. Since
F(z)=\frac{z^3}{(z-1)^3}=\left(\frac{z-1}{z}\right)^{-3}=\left(1-\frac{1}{z}\right)^{-3}we have
F(z)=1+\frac{3}{z}+\frac{6}{z^2}+\frac{10}{z^3}+\frac{15}{z^4}+\cdotsThus F(z) can be inverted directly to give
f[k]=1,3,6,10,15, \ldots \quad \text { that is, } f[k]=\frac{(k+2)(k+1)}{2} \quad k \geqslant 0Related Answered Questions
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