Find the Maclaurin series for f(x)= 1/(1+x^2)^2

Question

I could really use some help please!

Answer 1

The Maclaurin Series Generated By f(x) is given by : f(0)+f'(0)x¹/1!+f''(0)x²/2!+f'''(0)x³/3!+f^(IV)(0)x⁴/4! +...+

f⁽ⁿ⁾(0)xⁿ/n!+...

For f(x) equal to 1/(1+x²)² or (1+x²)^-2, obtain f'(x)=-4x(1+x²)^-3; f''(x)=24x²(1+x²)^-4;

f'''(x)=-192x³(1+x²)^-5; f^(IV)(x)=1920x⁴(1+x²)^-6...

The Maclaurin Series Generated By f(x)=1/(1+x²)² can then be written as 1/(1+0²)²+[-4(0)(1+0²)^-3/1!]x¹+

[24(0)²(1+0²)^-4/2!]x²+[-192(0)³(1+0²)^-5/3!]x³+[1920(0)⁴(1+0²)^-6/4!]x⁴+...

Each ratio term in bold type contains a power of 0 in the numerator which evaporates every term of the Maclaurin Series Expansion except for f(0) equal to 1/(1+0²)² or 1, which is in accord with 1/(1+x²)² at x=0 being equal to 1/(1+0²)² or 1.