Use the binomial theorem to expand (1-1/z)^-3 up to the term 1/z^4. Hence find the sequence with z transform F(z) = z³/(z-1)³.

Question

Use the binomial theorem to expand [latex]\left(1-\frac{1}{z} ight)^{-3} ext { up to the term } \frac{1}{z^4}[/latex]. Hence find the sequence with z transform [latex]F(z)=\frac{z^3}{(z-1)^3}[/latex].

Accepted Answer

Using the binomial theorem, we have

[latex]\begin{aligned} \left(1-\frac{1}{z} ight)^{-3}= & 1+(-3)\left(-\frac{1}{z} ight)+\frac{(-3)(-4)}{2 !}\left(-\frac{1}{z} ight)^2 \ & +\frac{(-3)(-4)(-5)}{3 !}\left(-\frac{1}{z} ight)^3+\frac{(-3)(-4)(-5)(-6)}{4 !}\left(-\frac{1}{z} ight)^4+\cdots \ = & 1+\frac{3}{z}+\frac{6}{z^2}+\frac{10}{z^3}+\frac{15}{z^4}+\cdots \end{aligned}[/latex]

provided |z| > 1. Since

[latex]F(z)=\frac{z^3}{(z-1)^3}=\left(\frac{z-1}{z} ight)^{-3}=\left(1-\frac{1}{z} ight)^{-3}[/latex]

we have

[latex]F(z)=1+\frac{3}{z}+\frac{6}{z^2}+\frac{10}{z^3}+\frac{15}{z^4}+\cdots[/latex]

Thus F(z) can be inverted directly to give

[latex]f[k]=1,3,6,10,15, \ldots \quad ext { that is, } f[k]=\frac{(k+2)(k+1)}{2} \quad k \geqslant 0[/latex]

Question 22.34: Use the binomial theorem to expand (1-1/z)^-3 up to the term......

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