A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function #f(x)# about a point #x=a# is given by
#f(x)=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f^(3)(a))/(3!)(x-a)^3+cdots+(f^(n)(a))/(n!)(x-a)^n+cdots#
We are given
#f(x)=sqrt(x)# and #a=16#
Here are the first four derivatives
#f'(x)=1/(2sqrt(x))#
#f''(x)=-(1)/(4x^(3//2))#
#f^((3))(x)=(3)/(8x^(5//2))#
#f^((4))(x)=-(15)/(16x^(7//2))#
Plug these derivatives into the Taylor series expansion above
#f(x)=sqrt(a)+1/(2sqrt(a))(x-a)-(1)/(2!4a^(3//2))(x-a)^2+(3)/(3!8a^(5//2))(x-a)^3-(15)/(4!16a^(7//2))(x-a)+cdots#
Plug in #a=16#
#f(x)=sqrt(16)+1/(2sqrt(16))(x-16)-(1)/(2!4(16)^(3//2))(x-16)^2+(3)/(3!8(16)^(5//2))(x-16)^3-(15)/(4!16(16)^(7//2))(x-16)+cdots#
Simplify #sqrt(16)=4#
#f(x)=4+1/(2(4))(x-16)-(1)/(2!4(4)^(3))(x-16)^2+(3)/(3!8(4)^(5))(x-16)^3-(15)/(4!16(4)^(7))(x-16)+cdots#
Simplify the fractions
#f(x)=4+1/8(x-16)-1/512(x-16)^2+1/16384(x-16)^3-1/2097152(x-16)^4+cdots#
Finally, the radius of convergence is found when
#L=abs(1-x/16)<1#
