I'm trying to understand the proof of Taylor's theorem from here:

I already made a question about the remainder part of the theorem and got an answer for it here: Remainder term in Taylor's theorem

My question is again about the remainder part of the theorem. In the wikipedia page it is stated that:

$$\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\;dt = -\left[ \frac{f^{(n+1)}(t)}{(n+1)n!}(x-t)^{n+1}\right]_a^x + \int_a^x \frac{f^{(n+2)}(t)}{(n+1)n!}(x-t)^{n+1}\;dt$$

$$=\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}+ \int_a^x \frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1}\;dt,$$

The last integral can be solved immediately, which leads to

$$R_n = f^{(n+1)}(\xi)\frac{(x-a)^{n+1}}{(n+1)!}$$

where $f$ is an $n+1$ times differentiable function on the open interval $(a,b), x\in(a,b)$.

The answer I was given used the Second mean-value theorem for the integral:


This is how the remainder term was solved by the user who answered my question. I'm satisfied with this answer, but I'm still wondering about the wikipedia's proof...it had a different approach and I'm interested about the phrase:

The last integral can be solved immediately, which leads to

My question is: How? Why does:

$$\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}+ \int_a^x \frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1}\;dt = f^{(n+1)}(\xi)\frac{(x-a)^{n+1}}{(n+1)!}=:R_n$$

Another question: Why does this statement involve the $f^{(n+2)}(t)$? I thought $f$ was supposed to be only $n+1$ times differentiable...or does $f^{(n+2)}(t)=0$? If yes, then does this mean that:

$$\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} ?$$

Hope my questions is clear =) Thank you for any help! Please let me know if you need more information. All the details are found on the wikipedia page

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The author of the Taylor's Theorem proof was proving the integral form of the remainder as stated at the beginning. The author proceeded by induction on the degree of smoothness, $n$. This is why the $n+2$ appears because in demonstrating the inductive case he assume that $f$ is $(n+1)+1$ times differentiable. I'm afraid I can't clear up the comment on why the last integral can be solved immediately. If you are willing to add the hypothesis that $f^{(k+1)}$ is continuous then you can use, what I know as, the First Mean Value Theorem of Integration to obtain the last portion of the proof you read (http://en.wikipedia.org/wiki/Mean_value_theorem). In the other answer you received this was known as the Second Mean Value Theorem of Integration.

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  • $\begingroup$ +1 Thank you for your help! =) The proof is confusing to me, because first it is stated that $f$ is $n+1$ times differentiable and then right after it is assumed that it is $n+2$ times differentiable... :/ Appreciate your help! =) $\endgroup$ – jjepsuomi Jul 16 '14 at 8:57
  • $\begingroup$ P.S. can by any chance explain to me (or show me the steps), why: $\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}+ \int_a^x \frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1}\;dt = f^{(n+1)}(\xi)\frac{(x-a)^{n+1}}{(n+1)!}=:R_n$? This is what interests me the most =) $\endgroup$ – jjepsuomi Jul 16 '14 at 8:59
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    $\begingroup$ Remember that the $n+2$ only arose due to the induction argument. Assuming $n+1$ derivatives (and continuity of the $n+1$ derivative) then you just apply the First Mean value Theorem of Integration to $\int_{a}^{x}f^{(n+1)}(t)\frac{(x-t)^{n}}{n!}dt$. So take $f^{(n+1)}(x)$ as your continuous function and $\frac{(x-a)^{n}}{n!}$ as your non-negative function and apply the First Mean Value Theorem of Integration. $\endgroup$ – user71352 Jul 16 '14 at 9:04
  • $\begingroup$ +1 Thank you =) I think that cleared it up, the wiki-proof is still a bit in the dark x) But I got it =) $\endgroup$ – jjepsuomi Jul 16 '14 at 9:06
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    $\begingroup$ You're welcome. $\endgroup$ – user71352 Jul 16 '14 at 9:06

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