In this note, we produce a proof of Taylor's Theorem. As in many proofs of Taylor's Theorem, we begin with a curious start and then follow our noses forward.
Is this a new proof? I think so. But I wouldn't bet a lot of money on it. It's certainly new to me.
Is this a groundbreaking proof? No, not at all. But it's cute, and I like it.1 1Though I must admit that it is not my favorite proof of Taylor's Theorem.
We begin with the following simple observation. Suppose that $f$ is two times continuously differentiable. Then for any $t \neq 0$, we see that \begin{equation} f'(t) - f'(0) = \frac{f'(t) - f'(0)}{t} t. \end{equation} Integrating each side from $0$ to $x$, we find that \begin{equation} f(x) - f(0) - f'(0) x = \int_0^x \frac{f'(t) - f'(0)}{t} t dt. \end{equation} To interpret the integral on the right in a different way, we will use the mean value theorem for integrals.
Mean Value Theorem for Integrals Suppose that $g$ and $h$ are continuous functions, and that $h$ doesn't change sign in $[0, x]$. Then there is a $c \in [0, x]$ such that \begin{equation} \int_0^x g(t) h(t) dt = g(c) \int_0^x h(t) dt. \end{equation}
For this application, let $g(t) = (f'(t) - f'(0))/t$ for $t \neq 0$, and $g(0) =f'{}'(0)$. The continuity of $g$ at $0$ is exactly the condition that $f'{}'(0)$exists. We also let $h(t) = t$.
For $x > 0$, it follows from the mean value theorem for integrals that there exists a $c \in [0, x]$ such that \begin{equation} \int_0^x \frac{f'(t) - f'(0)}{t} t dt = \frac{f'(c) - f'(0)}{c} \int_0^x t dt = \frac{f'(c) - f'(0)}{c} \frac{x^2}{2}. \end{equation} (Very similar reasoning applies for $x < 0$). Finally, by the mean value theorem (applied to $f'$), there exists a point $\xi \in (0, c)$ such that \begin{equation} f'{}'(\xi) = \frac{f'(c) - f'(0)}{c}. \end{equation} Putting this together, we have proved that there is a $\xi \in (0, x)$ such that \begin{equation} f(x) - f(0) - f'(0) x = f'{}'(\xi) \frac{x^2}{2}, \end{equation} which is one version of Taylor's Theorem with a linear approximating polynomial.
This approach generalizes. Suppose $f$ is a $(k+1)$ times continuously differentiable function, and begin with the trivial observation that \begin{equation} f^{(k)}(t) - f^{(k)}(0) = \frac{f^{(k)}(t) - f^{(k)}(0)}{t} t. \end{equation} Iteratively integrate $k$ times: first from $0$ to $t_1$, then from $0$ to $t_2$, and so on, with the $k$th interval being from $0$ to $t_k = x$.
Then the left hand side becomes \begin{equation} f(x) - \sum_{n = 0}^k f^{(n)}(0)\frac{x^n}{n!}, \end{equation} the difference between $f$ and its degree $k$ Taylor polynomial. The right hand side is \begin{equation}\label{eq:only}\underbrace{\int _0^{t_k = x} \cdots \int _0^{t _1}} _{k \text{ times}} \frac{f^{(k)}(t) - f^{(k)}(0)}{t} t \, dt \, dt _1 \cdots dt _{k-1}.\end{equation}
To handle this, we note the following variant of the mean value theorem for integrals.
Mean value theorem for iterated integrals Suppose that $g$ and $h$ are continuous functions, and that $h$ doesn't change sign in $[0, x]$. Then there is a $c \in [0, x]$ such that \begin{equation} \underbrace{\int_0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} g(t) h(t) dt =g(c) \underbrace{\int _0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} h(t) dt. \end{equation}
In fact, this can be proved in almost exactly the same way as in the single-integral version, so we do not repeat the proof.
With this theorem, there is a $c \in [0, x]$ such that we see that \eqref{eq:only} can be written as \begin{equation} \frac{f^{(k)}(c) - f^{(k)}(0)}{c} \underbrace{\int _0^{t _k = x} \cdots \int _0^{t _1}} _{k \; \text{times}} t \, dt \, dt _1 \cdots dt _{k-1}. \end{equation} By the mean value theorem, the factor in front of the integrals can be written as $f^{(k+1)}(\xi)$ for some $\xi \in (0, x)$. The integrals can be directly evaluated to be $x^{k+1}/(k+1)! $.
Thus overall, we find that \begin{equation} f(x) = \sum_{n = 0}^n f^{(n)}(0) \frac{x^n}{n!} + f^{(k+1)}(\xi) \frac{x^{k+1}}{(k+1)!} \end{equation} for some $\xi \in (0, x)$. Thus we have proved Taylor's Theorem (with Lagrange's error bound).
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