Question : Find Taylor series expansion of
\[log(1-x^2) \] about x=0
Solution :
We find Taylor series expansion of Log(1-x^2) about x =0 as follows Let \[g\left(x\right)=Log{\left(1-x^2\right)}\] \[g\left(0\right)\ =0\] \[g^\prime\left(x\right)=-\frac{2x}{1-x^2}\ \] \[g^\prime\left(0\right)=0\] \[g^{\prime\prime}\left(x\right)=-\frac{4x^2}{\left(1-x^2\right)^2}-\frac{2}{1-x^2}\] \[g^{\prime\prime}\left(0\right)=-2\] \[g^{\prime\prime\prime}\left(x\right)=-\frac{16x^3}{\left(1-x^2\right)^3}-\frac{12x}{\left(1-x^2\right)^2}\ \] \[g^{\prime\prime\prime}\left(0\right)=0\] \[g^4\left(x\right)=-\frac{96x^4}{\left(1-x^2\right)^4}-\frac{96x^2}{\left(1-x^2\right)^3}-\frac{12}{\left(1-x^2\right)^2}\] \[g^4\left(0\right)=-12\] \[g^5\left(x\right)=-\frac{768x^5}{\left(1-x^2\right)^5}-\frac{960x^3}{\left(1-x^2\right)^4}-\frac{240x}{\left(1-x^2\right)^3}\] \[g^5\left(0\right)=0\] \[g^{\left(6\right)}\left(x\right)=-\frac{7680x^6}{\left(1-x^2\right)^6}-\frac{11520x^4}{\left(1-x^2\right)^5}-\frac{4320x^2}{\left(1-x^2\right)^4}-\frac{240}{\left(1-x^2\right)^3}\] \[g^{\left(6\right)}\left(0\right)=-240\] and so on Thus Taylor series expansion of Log(1-x^2 ) at about x =0 \[Log{\left(1-x^2\right)}=0+\frac{0}{1!}x+\frac{-2}{2!}x^2+\frac{0}{3!}x^3+\frac{-12}{4!}x^4+\frac{0}{5!}x^5+\frac{-240}{6!}x^6+\ldots\] \[Log{\left(1-x^2\right)}=-x^2-\frac{1}{2}x^4-\frac{1}{3}x^6-\ldots.\]