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Let $g:I \to \mathbb{R}$ be a $C^{k+1}$ function, $a,x \in I$, then: $$g(x)=g(a)+g'(a)(x-a)+ \ldots +\frac{1}{k!}g^{(k)}(a)(x-a)^k+ \quad \frac{1}{k!}\int_a^x(x-s)^kg^{(k+1)}(s)\,ds $$ Proof: Taking $x$ fixed, the RHS of the equation above is a function of $a$ and its derivative wrt $a$ is zero (that can be easily verified, so I will not write the computations, for the last term simply use the Fundamental Theorem of Calculus), so it remains constant when varying $a$. In particular when it is evaluated at $a=x$ its value is $g(x)$ as wanted.
Since the function is constant (when varying $a$) then the RHS equals $g(x)$ for all $a \in I$ and since $x$ was arbitrary the equation holds for all $x \in I$

Can anyone confirm this proof is correct? It seems tricky and so simple that makes me suspect

BTW this is taken from the apendix of Hubbard's book Vector calculus, linear algebra and differential forms

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    @mixedmath I don't understand why it would require $k+2$ differentiability of $g$. I also don't realize of the problem with $h$. I mean, I could have stated the theorem without any mention to $h$, using only $x$ as variable. I will edit, to make it clearer – 2017-02-10
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    @chilangoincomprendido why is that I can't conclude the derivative wrt $a$ of the RHS is zero, $h$ is not in the equation anymore after the change of variable – 2017-02-10
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    Yes, I see what you did now. Looks fine. You verified the formula. But of course, verifying a formula without seeing "where it comes from" is not very instructive. Anyway, I will delete my previous comments. – 2017-02-10
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    No, it's fine. I agree it's not very instructive, it is completly artificial and the procedure seems kind of tricky, but on the other hand it's very simple. BTW I still don't "trust" in that proof, that's why I'm asking for a confirmation of correctness – 2017-02-10
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    Your proof is correct. But notice that using integrals you need the continuity of $f^{(n+1)}$. The more general theorem holds with just existence of $f^{(n+1)}$ and it avoids integrals. – 2017-02-10
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    @ParamanandSingh actually no. The general case (which I know) assumes $f^n$ is absolutely continuous. The rhs is obviously AC in $a$. Now, the derivative is $0$ almost everywhere. A AC function on an interval is constant iff its derivative is $0$ almost everywhere. So referring to the right version of FTC the proof still works in the general case. – 2017-02-10
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    @user251257: You are referring to the case when the integral in question is Lebesgue integral, right? – 2017-02-10
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    @ParamanandSingh yes – 2017-02-10

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