I want to calculate the Taylor polynomial of order $n$ for the funktion $ f(x) = \frac{1}{ 1−x}$ for $x_0=0$ and $0 < x < 1$ and the remainder $R_n$.

I have found that \begin{equation*}P_{0,n}(x)=\sum_{k=0}^n\frac{f^{(k)}(0)\cdot x^k}{k!}=\sum_{k=0}^n x^k \end{equation*}

Is the remainder \begin{equation*}R_n=\frac{\frac{(n+1)!}{(1-\xi)^{n+2}}}{(n+1)!}\cdot (x-0)^n=\frac{x^n}{(1-\xi)^{n+2}}\end{equation*} ? Or do we have to use the fomrula with the integral?

In some notes I read that in general the remainder doesn't converge to $0$ for all $\xi\in (0,x)$. Can you give me an example for that?

  • $\begingroup$ The remainder of " basic " function like logarith, exponential does converge to $0$. $\endgroup$ – Atmos Jan 21 '18 at 20:26
  • $\begingroup$ Ah ok! When it is asked to calculate the remainder do I have to use the formula I used in the initial post? Becayuse I have seen that there is also a formula with integral. @Atmos $\endgroup$ – Mary Star Jan 21 '18 at 20:29
  • $\begingroup$ When we consider the function $g(x)=e^x$, we get the remainder \begin{equation*}R_n=\frac{g^{(n+1)}(\xi )}{(n+1)!}\cdot x^n=\frac{e^{\xi}}{(n+1)!}\cdot x^n\end{equation*} right? We have to calculate the limit as $n\rightarrow \infty$, right? @Atmos $\endgroup$ – Mary Star Jan 21 '18 at 20:32
  • $\begingroup$ Was speaking about integrals form of the remainder $\endgroup$ – Atmos Jan 21 '18 at 20:52
  • $\begingroup$ Ah! So, does this only hold for the integral form, or are there other examples where the remainder in the form I used does not converge to $0$ ? @Atmos $\endgroup$ – Mary Star Jan 21 '18 at 20:53

The remainder for Taylor's theorem isn't the one you found.

The Lagrange form of the remainder is $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} = \frac{x^{n+1}}{(n+1)(1-\xi)^{n+1}}$

For this particular $f$, the remainder is easy to compute exactly as $$f(x)-\sum_{k=0}^n x^k=\frac{1}{1-x}-\sum_{k=0}^n x^k=\frac{x^{n+1}}{1-x}.$$

Therefore $R_n(x)=\frac{x^{n+1}}{1-x}$.

  • $\begingroup$ Ah!! So, we don't need the formula that I used? $\endgroup$ – Mary Star Jan 21 '18 at 20:33
  • $\begingroup$ @MaryStar I think that you need to precise us a specific form of remainder that you want to have. $\endgroup$ – mathcounterexamples.net Jan 21 '18 at 20:35
  • $\begingroup$ It isn't specified at the statement. If I use the formula of the inital post, is the result correct? $\endgroup$ – Mary Star Jan 21 '18 at 20:40
  • $\begingroup$ Some precision in my answer. $\endgroup$ – mathcounterexamples.net Jan 21 '18 at 20:57

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