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\large 
\underline
{MA2160 \hspace{5mm}Calculus II with Technology \hspace{50mm} Ch. 10-2}

In the previous section, we approximated functions wit $n$-degree polynomials called Taylor polynomials.  Now we will extend that concept to actual representations of a function with a polynomial of infinite degree


\Large
Goal
\large
\begin{enumerate}
\item Define Taylor series
\item Apply Taylor series to find new representations of common functions
\item Apply Taylor series to find binomial expansions of non-positive integer exponents


\end{enumerate}
What is a Taylor series?  A Taylor series is an extension of a Taylor polynomial of degree $n\rightarrow \infty$.

From our previous study we know the following Taylor series (about $x=0$):
\begin{enumerate}
\item $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$
\item $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$
\item $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....$
\end{enumerate}

Because these functions are defined for all $x \in \mathit{R}$, this series equals the function for all values of $x$.  However, not all functions are like this.  Take the natural log function about $x=1$:
$$\ln x = (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+....$$
The natural log function is not defined for $x\leq 0$, so the series is only good for estimating values of $x>0$.  Since we are centering the polynomial at $x=1$, it makes sense that the function will approximate values of $x$ the same distance to the right as to the left.  As the series only works for a length of 1 away from the center to the left, it stands to reason that the series only works for a length of 1 to the right as well.
\newpage
We know from the binomial expansion theorem how to find $(1+x)^p$ when $p$ is a positive integer.  However, what if $p$ is not a positive integer, maybe negative or a positive non-integer?  We can use the same algebraic calculation using the $n$th derivatives to generate the following series:

\begin{small}
$$(1+x)^p=1+px+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\frac{p(p-1)(p-2)(p-3)}{4!}x^4+...$$
\end{small}



Again, because we don't know what $p$ is, the series will only work for values of $|x|<1$, because for some values of $p$, the function will not be defined for $x\leq -1$

Does this really work for a positive integer $p$?  Absolutely, for each term with degree greater than $p$, the coefficient is 0 because of the subtraction in the numerator.

How does this change if we have the function $(1-x)^p$?  Everything is the same except the terms will alternate positive and negative based on whether the degree of the term is even or odd.
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