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## The Formula for Taylor Series We have computed power series representations for some
functions, including the following. $\begin{eqnarray} All of these have radius of convergence $R=1$, which is a result
of their geometric series origins.
This says that if a function can be represented by a power
series, its coefficients must be those in Taylor's Theorem.
This formula works both ways: if we know the $n$-th derivative
evaluated at $a$, we can figure out $c_n$; if we know $c_n$, we
can figure out the $n$-th derivative evaluated at $a$. To
use this theorem, we have the conventions that Example: We consider the series representation for
$\displaystyle f(x)=\frac{1}{1-x}$ (at the top of the page).(Notice that this series is centered around $a=0$.) By Taylor's Theorem, $c_nx^n=x^n$ since $x^n$ are the terms of our series. So $c_n=1$ for all $n$. On the other hand, also by Taylor's Theorem, $c_n=\frac{f^{(n)}(0)}{n!}$, so we must have $f^{(n)}(0)=n!$ for all $n$ here. Let's see if this is true.
$\begin{array}{lllll} Warning: The coefficients $c_n$ do
not contain the variable $x$, since the derivatives
in the $c_n$ are evaluated at $a$.The video will explain why Taylor's theorem works, in general. |