関数のべき級数展開

${\displaystyle log\left(1+x\right)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\cdot \cdot \cdot }$

${\displaystyle =x\left(2+3x\right)-\frac{1}{2}{\left\{x\left(2+3x\right)\right\}}^2}$${\displaystyle +\frac{1}{3}{\left\{x\left(2+3x\right)\right\}}^3}$${\displaystyle -\frac{1}{4}{\left\{x\left(2+3x\right)\right\}}^4}$${\displaystyle +\cdot \cdot \cdot }$

${\displaystyle =\left(2x+3x^2\right)-\frac{1}{2}\left(4x^2+12x^3+9x^4\right)}$${\displaystyle +\frac{1}{3}\left(8x^3+36x^4+54x^5+27x^6\right)}$${\displaystyle -\frac{1}{4}\left(16x^4+{96}^5+216x^6+216x^7+81x^8\right)+\cdot \cdot \cdot }$

${\displaystyle =2x+\left(3x^2-2x^2\right)+\left(-6x^3+\frac{8}{3}{x}^3\right)}$${\displaystyle +\left(-\frac{9}{2}{x}^4+12x^4-4x^4\right)}$${\displaystyle +\left(18x^5-24x^5\right)}$${\displaystyle +\left(9x^6-54x^6\right)}$${\displaystyle -216x^7-81x^8+\cdot \cdot \cdot }$