Taylor theorem with general remainder formula I came across Taylor theorem as following:
Let the function $f(x)$ have $n+1$ derivatives in $(a- \delta , a+\delta )$ of the point $a$ and $p>0$.Then for every $x \in (a- \delta , a+\delta )$ there exist $c \in (a,x)$ so that:
$$\begin{align}f(x) =  &\sum_{k=0}^{n} \frac{(x-a)^k}{k!}f^{(k)}(a) \\&+ \left(\frac{x-a}{x-c}\right)^p\frac{(x-c)^{n+1}}{n!p}f^{(n+1)}(c)  \end{align}  $$
I am curious , what is the proof of this theorem?
if you plug in for example $p=n+1$ you get the Lagrange remainder.But I didn't quite find anywhere a proof with $p$ where the remainder is expressed like that. Can someone share full proof of this theorem or link where this is proved ?
 A: Corollary of the Mean Value Theorem
Assume that $g(x)\gt0$ for $a\lt x\lt b$ and define
$$
h(x)=\int_a^xf(t)g(t)\,\mathrm{d}t
$$
and
$$
k(x)=\int_a^xg(t)\,\mathrm{d}t
$$
Then there is a $c$ between $a$ and $b$ so that
$$
\bbox[5px,border:2px solid #C0A000]{\frac{\int_a^bf(t)g(t)\,\mathrm{d}t}{\int_a^bg(t)\,\mathrm{d}t}
=\frac{h(b)-h(a)}{k(b)-k(a)}=\frac{h'(c)}{k'(c)}=\frac{f(c)g(c)}{g(c)}=f(c)}
$$

The Taylor's Formula with the Integral form of the Remainder is
$$
f(x)=\sum_{k=0}^n\frac{(x-a)^k}{k!}f^{(k)}(a)+\int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)\,\mathrm{d}t
$$
The corollary above then says,
$$
\begin{align}
\int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)\,\mathrm{d}t
&=\int_a^x\frac{(x-t)^{p-1}\color{#C00}{(x-t)^{n-p+1}}}{\color{#C00}{n!}}\color{#C00}{f^{(n+1)}(t)}\,\mathrm{d}t\\
&=\color{#C00}{\frac{(x-c)^{n-p+1}f^{n+1}(c)}{n!}}\int_a^x(x-t)^{p-1}\,\mathrm{d}t\\
&=\frac{(x-c)^{n-p+1}f^{n+1}(c)}{n!}\frac{(x-a)^p}p\\
&=\left(\frac{x-a}{x-c}\right)^p\frac{(x-c)^{n+1}}{n!p}f^{(n+1)}(c)
\end{align}
$$
A: I have a proof for you. It's based on Lagrange's approach to proving the "usual" Lagrange form of the remainder formula. (See, for example, problem 19 in Chapter 20 of Spivak's Calculus, 4th edition.) His trick is to let the point $a$ vary and consider
$$f(x) = P_{n,t}(x) + R_{n,t}(x).$$
Here $P_{n,t}(x) = \sum\limits_{k=0}^n \dfrac{f^{(k)}(t)}{k!}(x-t)^k$ and $R_{n,t}$ is the remainder. For ease of notation, since we will fix $n$ and $x$, let's write $R(t)=R_{n,t}(x)$.
Note that $R(x)-R(a)=0-R_{n,a}(x) = -(f(x)-P_{n,a}(x))$. Let $g(t)=(x-t)^p$. Then $g(x)-g(a) = -(x-a)^p$.  It looks very promising to apply the Cauchy Mean Value Theorem:
$$\frac{R(x)-R(a)}{g(x)-g(a)} = \frac{f(x)-P_{n,a}(x)}{(x-a)^p} = \frac{R'(c)}{g'(c)}.$$
Since $g'(c)=-p(x-c)^{p-1}$, so things are starting to look good. The critical step will be to compute $R'(t)$.
We have $R(t) = f(x) - \sum\limits_{k=0}^n \dfrac{f^{(k)}(t)}{k!}(x-t)^k$, so
\begin{align*}
R'(t) &= -\sum_{k=0}^n \left(\frac{f^{(k+1)}(t)}{k!}(x-t)^k -\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\right) \\
&= -\sum_{k=0}^n \frac{f^{(k+1)}(t)}{k!}(x-t)^k + \sum_{k=0}^{n-1} \frac{f^{(k+1}(t)}{k!}(x-t)^k \\
&= -\frac{f^{(n+1)}(t)}{k!}(x-t)^n.
\end{align*}
And now it all falls in place:
\begin{align*}
\frac{f(x)-P_{n,a}(x)}{(x-a)^p} &= \frac{R'(c)}{g'(c)}\\
&=\frac{-\frac{f^{(n+1)}(c)}{k!}(x-c)^n}{-p(x-c)^{p-1}} \\ &= \frac{f^{(n+1)}(c)}{k!p} (x-c)^{n+1-p},
\end{align*}
from which we get
\begin{align*}
f(x)-P_{n,a}(x) &= \frac{f^{(n+1)}(c)}{k!p} (x-c)^{n+1-p} (x-a)^p \\ &= \left(\frac{x-a}{x-c}\right)^p \frac{f^{(n+1)}(c)}{k!p} (x-c)^{n+1},
\end{align*}
as desired.
