f(x)=1/sqrt(1-x^2) Use the binomial series to find the Maclaurin series for the function.

Recall binomial series  that is convergent when |x|lt1 follows: 
(1+x)^k=sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!)x^n
or
(1+x)^k = 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4+...
To evaluate the given function f(x) = 1/sqrt(1-x^2) , we may apply radical property: sqrt(x) = x^(1/2) . The function becomes:
f(x) = 1/ (1-x^2)^(1/2)
Apply Law of Exponents: 1/x^n = x^(-n) to rewrite  the function as:
f(x) = (1-x^2)^(-1/2)
or f(x)= (1 -x^2)^(-0.5)
 This now resembles (1+x)^k form. By comparing "(1+x)^k " with "(1 -x^2)^(-0.5) or (1+(-x^2))^(-0.5) ”, we have the corresponding values:
x=-x^2 and k = -0.5 .
Plug-in the values on the aforementioned formula for the binomial series, we get:
(1-x^2)^(-0.5) =sum_(n=0)^oo (-0.5(-0.5-1)(-0.5-2)...(-0.5-n+1))/(n!)(-x^2)^n
=sum_(n=0)^oo (-0.5(-1.5)(-2.5)...(-0.5-n+1))/(n!)(-1)^nx^(2n)
=1 + (-0.5)(-1)^1x^(2*1) + (-0.5(-1.5))/(2!) (-1)^2x^(2*2)+ (-0.5(-1.5)(-2.5))/(3!)(-1)^3x^(2*3) +(-0.5(-1.5)(-2.5)(-3.5))/(4!)(-1)^4x^(2*4)+...
=1 + (-0.5)(-1)x^2 + (-0.5(-1.5))/(1*2) (1)x^4 + (-0.5(-1.5)(-2.5))/(1*2*3) (-1)x^6 +(-0.5(-1.5)(-2.5)(-3.5))/(1*2*3*4)(1)x^8+...
=1 +0.5x^2 + 0.75/2x^4 + 1.875/6x^6 +6.5625/24x^8+...
=1 + x^2/2+ (3x^4)/8 + (5x^6)/16 +(35x^8)/128+...
Therefore, the Maclaurin series for the function f(x) =1/sqrt(1-x^2) can be expressed as:
1/sqrt(1-x^2)=1 + x^2/2+ (3x^4)/8 + (5x^6)/16 +(35x^8)/128+...