# `f(x)=sqrt(1+x^3)` Use the binomial series to find the Maclaurin series for the function.

A binomial series is an example of infinite series. It is a series that is only convergent when we have `|x|lt1` and with a sum of `(1+x)^k `  where k is any number. To apply binomial series in determining the Maclaurin series of a given function `f(x) = (1+x)^k` , we may apply 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) = sqrt(1+x^3)` , we may apply the radical property: `sqrt(x)= x^(1/2)` . The function becomes:

`f(x) =(1+x^3)^(1/2)`

or

`f(x) =(1+x^3)^0.5`

To apply the aforementioned formula for binomial series, we may replace "`x` " with "`x^3` " and "k" with "`0.5` ". We let:

`(1+x^3)^0.5 = sum_(n=0)^oo (0.5(0.5-1)(0.5-2)...(0.5-n+1))/(n!) (x^3) ^n`

`=sum_(n=0)^oo (0.5(-0.5)(-1.5)...(0.5-n+1))/(n!) x^(3n)`

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

`=1+0.5x^3-0.25/(1*2)x^6+0.375/(1*2*3)x^9-0.9375/(1*2*3*4)x^(12)+...`

`=1+0.5x^3-0.25/2x^6+0.375/6x^9-0.9375/24x^(12)+...`

`=1+x^3/2-x^6/8+x^9/16-(5x^(12))/128+...`

Then, the Maclaurin series for the `f(x)=sqrt(1+x^3) ` can be expressed as:

`sqrt(1+x^3)=1+x^3/2-x^6/8+x^9/16-(5x^(12))/128+...`