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

A binomial series is an example of infinite series. When we have a function of `f(x) = (1+x)^k` such that k is any number and convergent to `|x| lt1` , we may apply the sum of series as the value of `(1+x)^k` . This can be expressed in a 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))/(2!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...`

To evaluate the given function `f(x) = sqrt(1+x^2)` , we express it in term...

A binomial series is an example of infinite series. When we have a function of `f(x) = (1+x)^k` such that k is any number and convergent to `|x| lt1` , we may apply the sum of series as the value of `(1+x)^k` . This can be expressed in a 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))/(2!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...`

To evaluate the given function `f(x) = sqrt(1+x^2)` , we express it in term of fractional exponent:

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

or

`f(x) =(1+x^2)^0.5`

To apply the aforementioned formula for binomial series, we may replace "`x` " with "`x^2` " and "`k` " with "`0.5` ". We let:

`(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(0.5-1)(0.5-2)...(0.5-n+1))/(n!) x^(2n)`

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

`=1+0.5x^2-0.25/(1*2)x^4+0.375/(1*2*3)x^6-0.9375/(1*2*3*4)x^8+...`

`=1+0.5x^2-0.25/2x^4+0.375/6x^6-0.9375/24x^8+...`

`=1+x^2/2-x^4/8+x^6/16-(5x^8)/128+...`

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

`sqrt(1+x^2)=1+x^2/2-x^4/8+x^6/16-(5x^8)/128+...`