Question

How do you write the taylor series for `f(x)=sqrt(x)` at `a=16` and find the radius of convergence.?

Answer 1

`f(x)=4+1/8(x-16)-1/512(x-16)^2+1/16384(x-16)^3+cdots`

and radius of convergence when

`L=abs(1-x/16)<1`

Explanation:

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`