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$$\sum_{n=0}^∞ a_n D^n = 1 / \left(D^2-3D+2\right)$$ $$\left(2-3D+D^2\right)\sum_{n=0}^∞ a_n D^n = 1$$ $$\begin{align} 2&\sum_{n=0}^∞ a_n D^n&-3D&\sum_{n=0}^∞ a_n D^n&+D^2&\sum_{n=0}^∞ a_n D^n &= 1\\ &\sum_{n=0}^∞ 2a_n D^n&-&\sum_{n=0}^∞ 3a_n D^{n+1}&+&\sum_{n=0}^∞ a_n D^{n+2} &= 1\\ &\sum_{n=0}^∞ 2a_n D^n&-&\sum_{n=1}^∞ 3a_{n-1} D^{n}&+&\sum_{n=2}^∞ a_{n-2} D^{n} &= 1\\ 2a_0+2a_1D+&\sum_{n=2}^∞ 2a_n D^n&-3a_0D-&\sum_{n=2}^∞ 3a_{n-1} D^{n}&+&\sum_{n=2}^∞ a_{n-2} D^{n} &= 1 \end{align}$$ $$2a_0+\left(2a_1-3a_0\right)D+\sum_{n=2}^∞\left( 2a_n- 3a_{n-1}+ a_{n-2}\right)D^n= 1$$ $$\begin{align} a_0 &= \frac{1}{2}\\ a_1 &= \frac{3a_0}{2}=\frac{3}{4}\\ a_n &= \frac{3a_{n-1}-a_{n-2}}{2} \end{align}$$ $$\frac{1}{D^2−3D+2}=\frac{1}{2}+\frac{3}{4}D+\frac{7}{8}D^2+\frac{15}{16}D^3+\cdots+\left(1-\frac{1}{2^{n+1}}\right)D^n+\cdots$$
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Mon, 19 Apr 2021 15:09 GMT