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

1 Answer
Mar 1, 2015

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!