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 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+...