The Maclaurin series for f left parenthesis x right parenthesis equals fraction numerator 1 over denominator 1 minus x squared end fraction is
You asked:
Find the Maclaurin series of the function \(f\left( x \right) = \frac{1}{1 - {x}^{2}}\).
MathBot Answer:
Series expansion:
\[1 + x^{2} + x^{4} + x^{6} + x^{8} + x^{10}+ \cdots\]
Series representations:
\[\sum_{n=0}^{\infty} \left(x^{2}\right)^{n}\]
\[\sum_{n=0}^{\infty} \left(-1\right)^{n} \left(- x^{2}\right)^{n}\]
Converges for \(\left|{x^{2}}\right| < 1\)
\[\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n} \log{\left(1 - x^{2} \right)}^{n}}{n!}\]
Converges for all values of \(x\)