`f(x)=root(4)(1+x)` Use the binomial series to find the Maclaurin series for the function.

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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. The formula will be:

`(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!)...

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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. The formula will be:

`(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) = root(4)(1+x)` , we may apply radical property: `root(n)(x) = x^(1/n)` . The function becomes:

`f(x)= (1+x)^(1/4)`

or   `f(x) =(1+x)^(0.25)`

By comparing "`(1+x)^k` " with "`(1+x)^(0.25)` ”, we have the corresponding values:

`x=x` and `k =0.25`

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

`(1+x)^(0.25) =sum_(n=0)^oo (0.25(0.25-1)(0.25-2)...(0.25-n+1))/(n!)x^n`

`=sum_(n=0)^oo (0.25(-0.75)(-1.75)...(0.25-n+1))/(n!)x^n`

`=1 + 0.25x + (0.25(-0.75))/(2!) x^2 + (0.25(-0.75)(-1.75))/(3!)x^3 +(0.25(-0.75)(-1.75)(-2.75))/(4!)x^4+...`

`=1 + 0.25x + (-0.1875)/(1*2) x^2 + (0.328125)/(1*2*3)x^3+(-0.90234375)/(1*2*3*4)x^4+...`

`=1 + 0.25x -0.1875/2 x^2 + 0.328125/6x^3 -0.90234375/24x^4+...`

`=1 + x/4 -(3x^2)/32 +(7x^3)/128 -(77x^4)/2048 +...`

Therefore, the Maclaurin series  for  the function `f(x) =root(4)(1+x)` can be expressed as:

`root(4)(1+x)=1 + x/4 -(3x^2)/32 +(7x^3)/128 -(77x^4)/2048 +...` 

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