Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the two equations (Taylor's theorem):

$f(x+h) = f(x) + f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}h^{n}f^{(n)}(x)+R_{n+1}$


$f(x-h) = f(x) - f'(x)\cdot h + \dfrac{1}{2}h^{2}f''(x)+\dots+\dfrac{1}{n!}(-h)^{n}f^{(n)}(x)+R_{n+1}$

Are these two remainder-terms equal?

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.