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

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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)`

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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+...`

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