Taylor remainder f(x+h) and f(x-h) I have the two equations (Taylor's theorem):
$f(x+h) = f(x) + f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}h^{n}f^{(n)}(x)+R_{n+1}$
and
$f(x-h) = f(x) - f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}(-h)^{n}f^{(n)}(x)+R_{n+1}$
Are these two remainder-terms equal? 
 A: I assume the variable $h$ is supposed to represent a non-negative value, correct?
Then, if you define a remainder $R_{n+1}(x, h)$ by
$$f(x+h) = f(x) + f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}h^{n}f^{(n)}(x)+R_{n+1}(x, h)$$
then the domain of $R_{n+1}(u,v)$ only includes non-negative values for $v$.
If you define another remainder term $S_{n+1}(x,h)$ by
 $$f(x-h) = f(x) - f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}(-h)^{n}f^{(n)}(x)+S_{n+1}(x, -h)$$
then the domain of $S_{n+1}(u,v)$ only includes non-positive values for $v$.
Their domains aren't the same so they can't be the same. Of course, they agree where their domains overlap ($R_{n+1}(u,0) = S_{n+1}(u,0)$), so I can define a new function $T_{n+1}(u,v)$ by
$$ T_{n+1}(u,v) = \begin{cases} R_{n+1}(u,v)  & \text{when defined} \\
S_{n+1}(u,v)  & \text{when defined} \end{cases} $$
and have an equation
$$f(x+\epsilon) = f(x) + f'(x)\cdot \epsilon + \dfrac{1}{2}\epsilon^{2}f''(x)+\dots+\dfrac{1}{n!}\epsilon^{n}f^{(n)}(x)+T_{n+1}(x, \epsilon)$$
valid for all $\epsilon$: positive, negative, and zero.
The values of $R_{n+1}(x, h)$ and $S_{n+1}(x, -h)$ don't really have anything at all to do with each other either, so they're completely different in that sense too.
However, if $f(x)$ is an analytic function, then $R_{n+1}(u,v)$ and $S_{n+1}(u,v)$ are analytic continuations of each other, so in that sense they are the "same" function.
The formulas for estimating the remainder are also symmetric, of course -- so they're similar in that fashion. Of course, $R$ depends on values of $f$ for $h > 0$, and $S$ depends on values of $f$ for $h < 0$, so again they are fairly independent ideas.
One of the most famous counterexamples for dealing with Taylor series is the function
$$ f(x) = \begin{cases} 0 & x \leq 0 \\ e^{-1/x^2} & x > 0 \end{cases} $$.
Every derivative of this function at zero is zero. So, formulas for the two remainders at $x=0$ are:
$$ R_{n}(0, h) = e^{-1/h^2} $$
$$ S_{n}(0, -h) = 0 $$
