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

*print*Print*list*Cite

** Binomial series** is an example of an infinite series. When it is convergent at `|x|lt1` , we may follow the sum of the binomial series as `(1+x)^k` where `k` is any number. We may follow the formula:

`(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)` , we may apply radical property: `sqrt(x) = x^(1/2)` . The function becomes:

`f(x) = 1/ (1-x)^(1/2)`

Apply Law of Exponents: `1/x^n = x^(-n)` to rewrite the function as:

`f(x) = (1-x)^(-1/2)`

or ` f(x)= (1 -x)^(-0.5)`

This now resembles `(1+x)^k` form. By comparing "`(1+x)^k` " with "`(1 -x)^(-0.5) or (1+(-x))^(-0.5)` ”, we have the corresponding values:

`x=-x` and `k =-0.5` .

Plug-in the values on the aforementioned formula for the binomial series, we get:

`(1-x)^(-0.5) =sum_(n=0)^oo (-0.5(-0.5-1)(-0.5-2)...(-0.5-n+1))/(n!)(-x)^n`

`=1 + (-0.5)(-x) + (-0.5(-0.5-1))/(2!) (-x)^2 + (-0.5(-0.5-1)(-0.5-2))/(3!)(-x)^3 +(-0.5(-0.5-1)(-0.5-2)(-0.5-3))/(4!)(-x)^4+...`

`=1 + 0.5x + (-0.5(-1.5))/(1*2) (-1)^2x^2 + (-0.5(-1.5)(-2.5))/(1*2*3) (-1)^3x^3 +(-0.5(-1.5)(-2.5)(-3.5))/(1*2*3*4)(-1)^4x^4+...`

`=1 + 0.5x + 0.75/2 (1)x^2 + (-1.875)/6 (-1)x^3 +(6.5625)/24(1)x^4+...`

`=1 + 1/2x + (3x^2)/8 + (5x^3)/16 +(35x^4)/128+...`

Therefore, the** Maclaurin series** for the function `f(x) =1/sqrt(1-x)` can be expressed as:

`1/sqrt(1-x)=1 + x/2 + (3x^2)/8 + (5x^3)/16 +(35x^4)/128+...`