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

Mar 1, 2015

The Taylor series is a particular way to approximate a function with a polynomial in the neighbourhood of a generic point $\left({x}_{0} , f \left({x}_{0}\right)\right)$.

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:

$f \left(1\right) = 5$

${f}^{\left(1\right)} \left(x\right) = 8 {x}^{7} + 4 {x}^{3}$ and ${f}^{\left(1\right)} \left(1\right) = 12$

${f}^{\left(2\right)} \left(x\right) = 56 {x}^{6} + 12 {x}^{2}$ and ${f}^{\left(2\right)} \left(1\right) = 68$

${f}^{\left(3\right)} \left(x\right) = 336 {x}^{5} + 24 x$ and ${f}^{\left(3\right)} \left(1\right) = 360$

${f}^{\left(4\right)} \left(x\right) = 1680 {x}^{4} + 24$ and ${f}^{\left(4\right)} \left(1\right) = 1704$

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!