Σ[k=1,n-1]p^ksinkxなどの等式

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^{n-1} p^k \sin kx=\frac{p \sin x-p^n \sin nx+p^{n+1} \sin (n-1)x}{1-2p \cos x+p^2}\\
&(2)  \displaystyle\sum_{k=0}^{n-1} p^k \cos kx=\frac{1-p \cos x-p^n \cos nx+p^{n+1} \cos (n-1)x}{1-2p \cos x+p^2}\\
\end{alignat}










<証明>

どちらも、等級数列の和の公式を用います

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^{n-1} p^k \sin kx=\displaystyle\sum_{k=1}^{n-1} p^k \cdot \frac{e^{ikx}-e^{-ikx}}{2i}\\
&=\frac{1}{2i}\left\{\displaystyle\sum_{k=1}^{n-1} (pe^{ix})^k-\displaystyle\sum_{k=1}^{n-1} (pe^{-ix})^k\right\}\\
&=\frac{1}{2i}\left[\frac{pe^{ix}\{1-p^{n-1}e^{i(n-1)x}\}}{1-pe^{ix}}-\frac{pe^{-ix}\{1-p^{n-1}e^{-i(n-1)x}\}}{1-pe^{-ix}}\right]\\
&=\frac{1}{2i}\left(\frac{pe^{ix}-p^n e^{inx}}{1-pe^{ix}}-\frac{pe^{-ix}-p^n e^{-inx}}{1-pe^{-ix}}\right)\\
&=\frac{1}{2i}\cdot \frac{(1-pe^{-ix})(pe^{ix}-p^ne^{inx})-(1-pe^{ix})(pe^{-ix}-p^ne^{-inx})}{(1-pe^{ix})(1-pe^{-ix})}\\
&=\frac{1}{2i}\cdot \frac{pe^{ix}-p^ne^{inx}-p^2+p^{n+1}e^{i(n-1)x}-pe^{-ix}+p^ne^{-inx}+p^2-p^{n+1}e^{-i(n-1)x}}{1-pe^{ix}-pe^{-ix}+p^2}\\
&=\frac{1}{2i} \cdot \frac{p(e^{ix}-e^{-ix})-p^n(e^{inx}-e^{-inx})+p^{n+1}\{e^{i(n-1)x}-e^{i(n-1)x)}\}}{1-p(e^{ix}-e^{-ix})+p^2}\\
&=\frac{p \sin x-p^n \sin nx+p^{n+1} \sin (n-1)x}{1-2p \cos x+p^2}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^{n-1} p^k \sin kx=\frac{p \sin x-p^n \sin nx+p^{n+1} \sin (n-1)x}{1-2p \cos x+p^2}$$










\begin{alignat}{2}
&(2)  \displaystyle\sum_{k=0}^{n-1} p^k \cos kx=\displaystyle\sum_{k=0}^{n-1} p^k \cdot \frac{e^{ikx}+e^{-ikx}}{2}\\
&=\frac{1}{2}\left\{\displaystyle\sum_{k=0}^{n-1} (pe^{ix})^k+\displaystyle\sum_{k=0}^{n-1} (pe^{-ix})^k\right\}\\
&=\frac{1}{2}\left(\frac{1-p^n e^{inx}}{1-pe^{ix}}+\frac{1-p^n e^{-inx}}{1-pe^{-ix}}\right)\\
&=\frac{1}{2}\cdot \frac{(1-pe^{-ix})(1-p^n e^{inx})+(1-pe^{ix})(1-p^n e^{-inx})}{(1-pe^{ix})(1-pe^{-ix})}\\
&=\frac{1}{2}\cdot \frac{1-p^ne^{inx}-pe^{-ix}+p^{n+1}e^{i(n-1)x}+1-p^ne^{-inx}-pe^{ix}+p^{n+1}e^{-i(n-1)x}}{1-pe^{ix}-pe^{-ix}+p^2}\\
&=\frac{1}{2} \cdot \frac{2-p(e^{ix}+e^{-ix})-p^n(e^{inx}+e^{-inx})+p^{n+1}\{e^{i(n-1)x}+e^{i(n-1)x}\}}{1-p(e^{ix}+e^{-ix})+p^2}\\
&=\frac{1-p \cos x-p^n \cos nx+p^{n+1} \cos (n-1)x}{1-2p \cos x+p^2}
\end{alignat}以上より$$\displaystyle\sum_{k=0}^{n-1} p^k \cos kx=\frac{1-p \cos x-p^n \cos nx+p^{n+1} \cos (n-1)x}{1-2p \cos x+p^2}$$

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