`f(x)=1/sqrt(4+x^2)` Use the binomial series to find the Maclaurin series for the function.

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marizi eNotes educator| Certified Educator

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) = 1/sqrt(4+x^2)` , we may apply `4+x^2=4(1+x^2/4)` .

The function becomes:

`f(x) =1/sqrt(4(1+x^2/4))`

`f(x) =1/(2sqrt(1+x^2/4))`

Apply radical property: `sqrt(x) = x^(1/2)` . The function becomes:

`f(x) =1/(2(1+x^2/4)^(1/2))`

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

`f(x) =1/2(1+x^2/4)^(-1/2)`

or `f(x)=1/2(1+x^2/4)^(-0.5)`

Apply the aforementioned formula on `(1+x^2/4)^(-0.5)` by letting:

`x=x^2/4 ` and `k =-0.5.`

`(1+x^2/4)^(-0.5) = sum_(n=0)^oo (-0.5(-0.5-1)(-0.5-2) ...(-0.5-n+1))/(n!) (x^2/4)^n`

`= sum_(n=0)^oo (-0.5(-1.5)(-2.5) ...(-0.5-n+1))/(n!) x^(2n)/4^n`

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

`= 1 -0.5x^2/4+ 0.75/2* x^4/16-1.875/6x^6/64 +6.5625/24x^8/256+...`

`= 1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+...`

Applying `(1+x^2/4)^(-0.5) =1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+...` , we get:

`1/2(1+x^2/4)^(-0.5) = 1/2[1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+...]`

`=1/2 -x^2/16+ (3x^4)/256-(5x^6)/2048+(35x^8)/65536+...`

Therefore, the Maclaurin series for the function `f(x) =1/sqrt(4+x^2)` can be expressed as:

`1/sqrt(4+x^2)=1/2 -x^2/16+ (3x^4)/256-(5x^6)/2048+(35x^8)/65536+...`