Intuition behind Lagrange remainder term in Taylor's Theorem Let $f:\mathbb R\to\mathbb R$ be an $n+1$-times differentiable function. Taylor's Theorem states that for each $x\in\mathbb R$,
$$
f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(\lambda)}{(n+1)!}x^{n+1} \tag{*}\label{*}
$$
where $\lambda$ is strictly in between $0$ and $x$.
Note that in the case $n=0$, this is simply a restatement of the mean value theorem. However, the proof for general $n$ found in many textbooks is a rather messy inductive proof.
My question is: is there an intuitive reason why $\eqref{*}$ holds? The mean value theorem has an appealing geometric interpretation, that at some point the instantaneous rate of change over an interval has to equal the average rate of change. I wonder if there is an analogous interpretation of $\eqref{*}$.
To clarify, it has already been discussed on MSE why we ought to have
$$
f(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n \, ,
$$
for $x$ close to $0$, and I do find this rather intuitive. The reason I find Taylor's theorem somewhat unintuitive is because it is unclear why the remainder term can be expressed in the above form.
 A: Here is one simple but not as robust approach:
Suppose $f$ is $C^{n + 1}$ on $[0, x]$ with $x > 0$. Let $M_0 = \inf_{[0, x]}f^{(n + 1)}$, $M_1 = \sup_{[0, x]}f^{(n + 1)}$, which exist since $f^{(n + 1)}$ is assumed to be continuous. For $y \in [0, x]$,
$$M_0 \leq f^{(n + 1)}(y) \leq M_1.$$
Integrating this from $0$ to $y$ gives
$$M_0y \leq f^{(n)}(y) - f^{(n)}(0) \leq M_1y.$$
Integrating from $0$ to $y$ again gives
$$\frac{M_0y^2}{2} \leq f^{(n - 1)}(y) - f^{(n - 1)}(0) - f^{(n)}(0)y \leq \frac{M_1y^2}{2}.$$
Doing this $n - 1$ more times yields
$$\frac{M_0y^{n + 1}}{(n + 1)!} \leq f(0) - f'(0)y - \frac{f''(0)y^2}{2} - \dots - \frac{f^{(n)}(0)}{n!}y^n \leq \frac{M_1y^{n + 1}}{(n + 1)!}.$$
Thus the remainder $R_n(x, 0) = f(x) - P_n(x, 0)$ satisfies
$$\frac{M_0x^{n + 1}}{(n + 1)!} \leq R_n(x, 0) \leq \frac{M_1x^{n + 1}}{(n + 1)!}.$$
Applying the intermediate value theorem yields the Lagrange remainder formula.
A: You wrote "the proof for general $n$ found in many textbooks is a rather messy inductive proof."  I agree, but it turns out there is a much simpler argument relying on nothing more than Rolle's theorem, so it's depends on nothing more than a special case of the Mean Value Theorem.
I discovered this yesterday in a comment by Pieter-Jan De Smet on the blog page by Gowers about the Lagrange form of the remainder here, where the simpler proof comes from a textbook around 40 years ago.  Or see Theorem 1.1 and its proof in Section 2 here, which is my account of two different forms of the remainder. I used to have an inductive proof there for the Lagrange form of the remainder, but I rewrote it to present the simpler argument from the blog comment.
A: It's a matter of applying an appropriate mean-value theorem (namely the Cauchy MVT) to the remainder function as you vary the base point. Explicitly, let us fix $0$ and $x>0$. For each $s\in [0,x]$, let us consider the expansion of $f$ about $s$:
\begin{align}
f(x)&=\sum_{k=0}^n\frac{f^{(k)}(s)}{k!}(x-s)^k+\rho(s),
\end{align}
i.e we are fixing $0$ and $x$ and defining $\rho(s)$ to be the difference $f(x)-\sum_{k=0}^n\frac{f^{(k)}(s)}{k!}(x-s)^k$. Since $f$ is $n+1$ times differentiable, we certainly have that $\rho$ is once differentiable. Observe also that $\rho(0):= f(x)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k$ is the remainder we are interested in, and $\rho(x)=0$ from the definition. So, $\rho(0)=\rho(0)-\rho(x)$. Ah, such a difference is exactly what the mean-value theorem, and by extension, the Cauchy MVT are intended to give us information about. So, in anticipation of this, we note that a direct calculation using the product rule and evaluating the resulting telescoping sum yields $\rho'(s)=-\frac{f^{(n+1)}(s)}{n!}(x-s)^n$.
Suppose we have another function $\phi:\Bbb{R}\to\Bbb{R}$ which is nice enough. Then, how can we express our remainder $\rho(0)$ of interest in terms of these functions? Well,
\begin{align}
\rho(0)&=\rho(0)-\rho(x)\\
&=\frac{\rho(0)-\rho(x)}{\phi(0)-\phi(x)}[\phi(0)-\phi(x)]\\
&=\frac{\rho'(\lambda)}{\phi'(\lambda)}[\phi(0)-\phi(x)]\\
&=\frac{f^{n+1}(\lambda)(x-\lambda)^n}{n!}\cdot\frac{\phi(x)-\phi(0)}{\phi'(\lambda)}
\end{align}
for some $\lambda\in (0,x)$, by the Cauchy MVT. So, we can state Taylor's remainder theorem with respect to a function $\phi$ as:
\begin{align}
f(x)&=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k+\rho(0)\\
&=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k + \frac{f^{(n+1)}(\lambda)(x-\lambda)^n}{n!}\cdot\frac{\phi(x)-\phi(0)}{\phi'(\lambda)}
\end{align}
So, by varying the base point $s$ in the Taylor polynomial, we get a hole bunch of remainders, $\rho(s)$, and this function is differentiable. The Cauchy MVT, which I find to be pretty geometric (it says ratio of averages on an interval equals ratio of slopes at some point), now gives us many ways of expressing the remainder $\rho(0)$.
By choosing $\phi$ appropriately, we get corresponding forms of the remainder. FOr example, choosing $\phi$ to be the identity function $\phi(s)=s$, this amounts to directly applying the MVT to $\rho$ and thus tells us the remainder is $\rho(0)=\frac{f^{(n+1)}(\lambda)(x-\lambda)^n}{n!}x$. If instead we choose $\phi(s)=(x-s)^{n+1}$, then we get $\rho(0)=\frac{f^{(n+1)}(\lambda)}{(n+1)!}x^{n+1}$, the Lagrange form of the remainder.

Alternatively, we can use the FTC (assuming $\rho'$ is Riemann-integrable, or equivalently $f^{(n+1)}$ is RIemann integrable) to say $\rho(0)=\rho(0)-\rho(x)=-\int_0^x\rho'(s)\,ds=\int_0^x\frac{f^{(n+1)}(s)}{n!}(x-s)^n\,ds$, which gives us the integral form of the remainder.
So, the point is that we have a function $\rho$ which at each $s$ gives the remainder $\rho(s)$ when $f$ is Taylor expanded about $s$. The goal is to express $\rho(0)$ in more manageable terms. This is of course a highly non-unique procedure; we can express the same quantity $\rho(0)$ in many different-looking ways by using different techniques (MVT vs Cauchy MVT with respect to a particular $\phi$, vs FTC).
