Question

Taylor polynomial of degree 4 of the function f(x)=arctan(x) at a=0

Answer 1

`f(x)=\arctan x`

`(\mathrm df)/(\mathrm dx)=\frac1{1+x^2}`

`(\mathrm d^2f)/(\mathrm dx^2)=-(2x)/((1+x^2)^2)`

`(\mathrm d^3f)/(\mathrm dx^3)=(6x^2-2)/((1+x^2)^3)`

`(\mathrm d^4f)/(\mathrm dx^4)=(24x-24x^3)/((1+x^2)^4)`

The Taylor polynomial takes the form


`P_4(x)=f(0)+(f'(0))/(1!)(x-0)+(f''(0))/(2!)(x-0)^2+(f'''(0))/(3!)(x-0)^3+(f^((4))(0))/(4!)(x-0)^4`

`P_4(x)=\frac11x-\frac26x^3=x-\frac13x^3`

Notice that the fourth degree term vanishes, since
`f^((4))(0)=0`, so you can stop at the third degree.