# 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?

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yes. But I think you need just to consider the difference of sign + or - – 89085731 Mar 3 '12 at 12:39
@Gingerjin: ??? – Did Mar 3 '12 at 13:45

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: $$R_{n}(0, h) = e^{-1/h^2}$$ $$S_{n}(0, -h) = 0$$

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I think for the function $$f(x) = \begin{cases} 0 & x \leq 0 \\ e^{-1/x^2} & x > 0 \end{cases}$$, $$R_{n}(0, h) = p(h)e^{-1/h^2}$$,p is a polynomial – 89085731 Mar 4 '12 at 1:02