What is maclaurin-ln(1-x^2) ?

The answer to maclaurin-ln(1-x^2) is x^2+1/2 x^4+1/3 x^6+1/4 x^8+1/5 x^{10}+\ldots

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maclaurin-ln(1-x^2)

maclaurin

- ln (1 - x^{2})

x^{2} + \frac{1}{2} x^{4} + \frac{1}{3} x^{6} + \frac{1}{4} x^{8} + \frac{1}{5} x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( - ln ( 1 - x ^{2} ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( - ln ( 1 - x ^{2} ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( - ln ( 1 - x ^{2} ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 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} + \frac{0}{7 !} x^{7} + \frac{10080}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{725760}{10 !} x^{10} + \dots

Refine

= x^{2} + \frac{1}{2} x^{4} + \frac{1}{3} x^{6} + \frac{1}{4} x^{8} + \frac{1}{5} x^{10} + \dots

\frac{d}{d x} (\frac{x}{753600 x ^{2} + 900})

y = 4 x + 4, y = 5 x + 5, x = 2

\int \frac{2 x - 1}{x ^{2} ( x + 3 )} d x

x \to 0 lim (\frac{( 2. 7 ^{x} - 1 )}{x})

\frac{d}{d t} (12 t - 40 π sin (π t))

What is maclaurin-ln(1-x^2) ?
The answer to maclaurin-ln(1-x^2) is x^2+1/2 x^4+1/3 x^6+1/4 x^8+1/5 x^{10}+\ldots