Find the first five terms of the Taylor series for x^8+ x^4 +3 " at " x = 0?

Find the first five terms of the Taylor series for ${x}^{8} + {x}^{4} + 3 \text{ at } x = 0$?

May 30, 2016

First nonzero terms:
$3 + {x}^{4} + {x}^{8}$

Series evaluated from $n = 0$ to $n = 4$:
$3 + 0 + 0 + 0 + {x}^{4}$

Explanation:

As it is not entirely clear whether the question is asking for the first five nonzero terms, or the first five terms produced from the sigma notation of a Taylor series, both will be given.

The Taylor series centered at $c$ for a function $f \left(x\right)$ is given by

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

where ${f}^{\left(n\right)} \left(c\right)$ is the ${n}^{\text{th}}$ derivative of $f$, evaluated at $c$.

In our case, we have $f \left(x\right) = {x}^{8} + {x}^{4} + 3$

As $f$ is a polynomial, though, we can tell that ${f}^{\left(n\right)} \left(0\right)$ will be nonzero only when ${f}^{\left(n\right)} \left(x\right)$ contains a constant term, that is, for $n \in \left\{0 , 4 , 8\right\}$. For all other derivatives, every term will have a factor of $x$, and thus will be $0$ when evaluated at $0$.

As such, the Taylor series for $f \left(x\right)$ will have only three terms:

f(0)/(0!)(x-0)^0 = (0^8+0^4+3)/1*1 = 3

f^((4))(0)/(4!)(x-0)^4 = (8*7*6*5*0^4+4*3*2*1)/(4*3*2*1)x^4 = x^4

f^((8))(0)/(8!)(x-0)^8 = (8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1)x^8 = x^8

Therefore, the Taylor series for $f \left(x\right)$ is $3 + {x}^{4} + {x}^{8}$, which is just $f \left(x\right)$. We should not be surprised by this result, as a Taylor series is, in effect, a polynomial representation of a function. When the function is a polynomial, then, they should be identical.

As for the terms produced from $n = 0$ to $n = 4$ in the series, we have, by the above reasoning, that f^((n))(0)/(n!)(x-0)^n = 0 for $n \in \left\{1 , 2 , 3\right\}$. Thus the first five terms the series produces are $3 , 0 , 0 , 0 , {x}^{4}$.