Question

Question #cf120

Answer 1

The answer is `=1+1/2x^2-1/8x^4+1/16x^6+......`

Explanation:

The Taylor series is

`(1+x)^n=1+n/(1!)x+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+...`

Therefore,

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

`=1+1/2x^2-1/8x^4+1/16x^6+......`