Question

How do you find the first five terms of the Taylor series for `f(x)=x^8+x^4+3` at `x=1`?

Answer 1

The Taylor series is a particular way to approximate a function with a polynomial in the neighbourhood of a generic point `(x_0,f(x_0))`.

`f(x)=f(x_0)+f'(x_0)(x-x_0)^1/(1!)+f''(x_0)(x-x_0)^2/(2!)+...`

or, with the series notation:

`f(x)=sum_(i=0)^(+oo)f^((i))(x_0)(x-x_0)^i/(i!)`.

So, let's calculate all the derivatives we need:

`f(1)=5`

`f^((1))(x)=8x^7+4x^3` and `f^((1))(1)=12`

`f^((2))(x)=56x^6+12x^2` and `f^((2))(1)=68`

`f^((3))(x)=336x^5+24x` and `f^((3))(1)=360`

`f^((4))(x)=1680x^4+24` and `f^((4))(1)=1704`

so:

`f(x)=5+12(x-1)^1/(1!)+68(x-1)^2/(2!)+360(x-1)^3/(3!)+1704(x-1)^4/(4!)`

It's easy to understand that if you would calculate the Taylor's series until the 8th order you will obtain the function itself, because it is a polynomial!