# Understanding the proof of Taylor's theorem

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:

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

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

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.
• +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! =) – jjepsuomi Jul 16 '14 at 8:57
• 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 =) – jjepsuomi Jul 16 '14 at 8:59
• 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. – user71352 Jul 16 '14 at 9:04