Question #cf120 Calculus Power Series Constructing a Taylor Series 1 Answer Narad T. Aug 23, 2017 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+......# Answer link Related questions How do you find the Taylor series of #f(x)=1/x# ? How do you find the Taylor series of #f(x)=cos(x)# ? How do you find the Taylor series of #f(x)=e^x# ? How do you find the Taylor series of #f(x)=ln(x)# ? How do you find the Taylor series of #f(x)=sin(x)# ? How do you use a Taylor series to find the derivative of a function? How do you use a Taylor series to prove Euler's formula? How do you use a Taylor series to solve differential equations? What is the Taylor series of #f(x)=arctan(x)#? What is the linear approximation of #g(x)=sqrt(1+x)^(1/5)# at a =0? See all questions in Constructing a Taylor Series Impact of this question 1468 views around the world You can reuse this answer Creative Commons License