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関数のべき級数展開

■問題

次の関数をべき級数展開（マクローリン展開）をせよ．

${\displaystyle log\left(1-x\right)}$

■ヒント

${\displaystyle f\left(x\right)=log\left(1-x\right)}$ とおいて微分を5回行い ${\displaystyle f^{\prime }\left(x\right)}$ ～ ${\displaystyle f^{(5)}\left(x\right)}$ を求める．

それぞれに ${\displaystyle x=0}$ を代入して ${\displaystyle f\left(0\right)}$ ～ ${\displaystyle f^{(5)}\left(0\right)}$ を求める．

最後に ${\displaystyle f\left(0\right)}$ ～ ${\displaystyle f^{(5)}\left(0\right)}$ 値を マクローリン展開の式に代入する．

■答

${\displaystyle f\left(x\right)=log\left(1-x\right)}$ とおいて， ${\displaystyle f^{\prime }\left(x\right)}$ から ${\displaystyle f^{(5)}\left(x\right)}$ を求める．

${\displaystyle f^{\prime }\left(x\right)}$${\displaystyle =\frac{-1}{1-x}}$

${\displaystyle f^{″}\left(x\right)}$${\displaystyle ={\left(\frac{-1}{1-x}\right)}^{\prime }}$${\displaystyle ={\left\{\left(-1\right){\left(1-x\right)}^{-1}\right\}}^{\prime }}$${\displaystyle =\left(-1\right)\left\{\left(-1\right){\left(1-x\right)}^{-2}\times \left(-1\right)\right\}}$${\displaystyle =\left(-1\right){\left(1-x\right)}^{-2}}$${\displaystyle =\frac{-1}{{\left(1-x\right)}^2}}$

${\displaystyle f^{‴}\left(x\right)}$${\displaystyle ={\left\{\frac{-1}{{\left(1-x\right)}^2}\right\}}^{\prime }}$${\displaystyle ={\left\{\left(-1\right){\left(1-x\right)}^{-2}\right\}}^{\prime }}$${\displaystyle =\left(-1\right)\left\{\left(-2\right){\left(1-x\right)}^{-3}\times \left(-1\right)\right\}}$${\displaystyle =\left(-2\right){\left(1-x\right)}^{-3}}$${\displaystyle =\frac{-2}{{\left(1-x\right)}^3}}$

${\displaystyle f^{(4)}\left(x\right)}$${\displaystyle ={\left\{\frac{-2}{{\left(1-x\right)}^3}\right\}}^{\prime }}$${\displaystyle ={\left\{\left(-2\right){\left(1-x\right)}^{-3}\right\}}^{\prime }}$${\displaystyle =\left(-2\right)\left\{\left(-3\right){\left(1-x\right)}^{-4}\times \left(-1\right)\right\}}$${\displaystyle =\left(-6\right){\left(1-x\right)}^{-4}}$${\displaystyle =\frac{-6}{{\left(1-x\right)}^4}}$

${\displaystyle f^{(5)}\left(x\right)}$${\displaystyle ={\left\{\frac{-6}{{\left(1-x\right)}^4}\right\}}^{\prime }}$${\displaystyle ={\left\{\left(-6\right){\left(1-x\right)}^{-4}\right\}}^{\prime }}$${\displaystyle =\left(-6\right)\left\{\left(-4\right){\left(1-x\right)}^{-5}\times \left(-1\right)\right\}}$${\displaystyle =\left(-24\right){\left(1-x\right)}^{-5}}$${\displaystyle =\frac{-24}{{\left(1-x\right)}^5}}$

${\displaystyle x=0}$ のとき ${\displaystyle f\left(x\right)}$ から ${\displaystyle f^{(5)}\left(x\right)}$ の値は

となる．マクローリン展開の式に ${\displaystyle f\left(0\right)}$ から ${\displaystyle f^{(5)}\left(0\right)}$ の値を代入する．

${\displaystyle log\left(1-x\right)}$${\displaystyle =f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}\left(0\right)}{2!}{x}^2}$${\displaystyle +\frac{f^{‴}\left(0\right)}{3!}{x}^3}$${\displaystyle +\cdots }$${\displaystyle +\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n}$${\displaystyle +\cdots }$

${\displaystyle =0+\left(-1\right)x+\frac{-1}{2}{x}^2+\frac{-2}{6}{x}^3}$${\displaystyle +\frac{-6}{24}{x}^4}$${\displaystyle +\frac{-24}{120}{x}^5+\cdots }$

${\displaystyle =-x-\frac{1}{2}{x}^2-\frac{1}{3}{x}^3-\frac{1}{4}{x}^4}$${\displaystyle -\frac{1}{5}{x}^5+\cdots }$

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最終更新日： 2024年5月28日

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