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 \[g\left(x\right)=Log{\left(1-x^2\right)}\] \[g\left(0\right)\ =0\] \[g^\prime\left(x\right)=-\frac{2x}{1-x^2}\ \] \[g^\prime\left(0\right)=0\] \[g^{\prime\prime}\left(x\right)=-\frac{4x^2}{\left(1-x^2\right)^2}-\frac{2}{1-x^2}\] \[g^{\prime\prime}\left(0\right)=-2\] \[g^{\prime\prime\prime}\left(x\right)=-\frac{16x^3}{\left(1-x^2\right)^3}-\frac{12x}{\left(1-x^2\right)^2}\ \] \[g^{\prime\prime\prime}\left(0\right)=0\] \[g^4\left(x\right)=-\frac{96x^4}{\left(1-x^2\right)^4}-\frac{96x^2}{\left(1-x^2\right)^3}-\frac{12}{\left(1-x^2\right)^2}\] \[g^4\left(0\right)=-12\] \[g^5\left(x\right)=-\frac{768x^5}{\left(1-x^2\right)^5}-\frac{960x^3}{\left(1-x^2\right)^4}-\frac{240x}{\left(1-x^2\right)^3}\] \[g^5\left(0\right)=0\]…
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