Question

How do you find the second-degree Taylor polynomial `T_2(x)` for the function `f(x) = sqrt(3+x^2)` at the number x=1?

Answer 1

`T_2(x)=2+1/2(x-1)+3/16(x-1)^2`

Explanation:

A complete Taylor polynomial for the function `f` centered around `x=c` is given by:

`T(x)=sum_(n=0)^oo(f^((n))(c))/(n!)(x-c)^n`

So the second degree Taylor polynomial will be the sum of the terms `n=0` through `n=2`, or:

`T_2(x)=(f^((0))(c))/(0!)(x-c)^0+(f^((1))(c))/(1!)(x-c)^1+(f^((2))(c))/(2!)(x-c)^2" "`

`c=1` and we can make a couple other simplifications:

`T_2(x)=f(1)+f'(1)(x-1)+(f''(1))/2(x-1)^2" "" "color(red)star`

We see that:

`f(1)=sqrt(3+1)=2`

And:

`f'(x)=1/2(3+x^2)^(-1/2)(2x)=x/sqrt(3+x^2)`

`f'(1)=1/sqrt(3+1)=1/2`

Finding `f''` using `f'(x)=x(3+x^2)^(-1/2)`:

`f''(x)=(3+x^2)^(-1/2)+x(-1/2(3+x^2)^(-3/2))(2x)`

`f''(x)=(3+x^2)^(-1/2)-x^2(3+x^2)^(-3/2)`

`f''(x)=3/(3+x^2)^(3/2)`

`f''(1)=3/(1+3)^(3/2)=3/8`

Returning to `color(red)star`:

`T_2(x)=2+1/2(x-1)+3/16(x-1)^2`