UE Stochastic Processes - Beispiel 18

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Man zeige: Wenn 

c_j = \sum_{k=1}^j c_{j-1} c_{j-k}
c_0 = c_1=1 \,

und 

\alpha(u) = \sum_{j=0}^\infty c_j u^j \,

so ist 

\alpha^2 (u) = \frac{\alpha(u) - 1}{u} \,

d.h.

\alpha(u) = \frac{1 - \sqrt{1 - 4u}}{2u} \,



\begin{align}
c_2 & = c_0 c_1 + c_1 c_0 \\
c_3 & = c_0 c_2 + c_1 c_1 + c_2 c_0 \\
c_4 & = c_0 c_3 + c_1 c_2 + c_2 c_1 + c_3 c_0 \\
\vdots &
\end{align}

\alpha(u) = \sum_{j=0}^{\infty} c_j u^j = c_0 u^0 + c_1 u^1 + c_2 u^2 + \ldots 

\begin{align}
\alpha^2(u) 
& = (c_0 u^0 + c_1 u^1 + c_2 u^2 + \ldots) (c_0 u^0 + c_1 u^1 + c_2 u^2 + \ldots) \\
& = c_0^2 + c_0 c1 u + c_0 c_2 u^2 + \ldots + c_1 c_0 u + c_1^2 u^2 + c_1 c_2 u^3 + \ldots \\
& = c_0^2 + u (c_0 c_1 + c_1 c_0) + u^2 (c_0 c_2 + c_1 c_1 + c_2 c_0) + u^3 (c_0 c_3 + c_1 c_2 + c_2 c_1 + c_3 c_0) + \ldots\\
& = 1 + u c_2 + u^2 c_3 + u^3 c_4 + \ldots
\end{align}

\begin{align}
\frac{\alpha(u) - 1}{u} 
& = \frac{(c_0 u^0 + c_1 u^1 + c_2 u^2 + \ldots) - 1}{u} \\
& = \frac{(1 + c_1 u^1 + c_2 u^2 + \ldots) - 1}{u} \\
& = \frac{u c_1 + u^2 c_2 + u^3 c_2 + \ldots}{u} \\
& = 1 + u c_2 + u^2 c_3 + u^3 c_4 + \ldots \\
& = \alpha^2(u)
\end{align}
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