Another proof of Taylor’s Theorem

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

We begin with the following simple observation. Suppose that $f$ is two times continuously differentiable. Then for any $t \neq 0$, we see that $$f'(t) – f'(0) = \frac{f'(t) – f'(0)}{t} t.$$ Integrating each side from $0$ to $x$, we find that $$f(x) – f(0) – f'(0) x = \int_0^x \frac{f'(t) – f'(0)}{t} t dt.$$ 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 $$\int_0^x g(t) h(t) dt = g(c) \int_0^x h(t) dt.$$

Suppose without loss of generality that $h(t)$ is nonnegative. Since $g$ is continuous on $[0, x]$, it attains its minimum $m$ and maximum $M$ on this interval. Thus $$m \int_0^x h(t) dt \leq \int_0^x g(t)h(t)dt \leq M \int_0^x h(t) dt.$$ Let $I = \int_0^x h(t) dt$. If $I = 0$ (or equivalently, if $h(t) \equiv 0$), then the theorem is trivially true, so suppose instead that $I \neq 0$. Then $$m \leq \frac{1}{I} \int_0^x g(t) h(t) dt \leq M.$$ By the intermediate value theorem, $g(t)$ attains every value between $m$ and $M$, and thus there exists some $c$ such that $$g(c) = \frac{1}{I} \int_0^x g(t) h(t) dt.$$ Rearranging proves the theorem.

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 $$\int_0^x \frac{f'(t) – f'(0)}{t} t dt = \frac{f'(c) – f'(0)}{c} \int_0^x t dt = \frac{f'(c) – f'(0)}{c} \frac{x^2}{2}.$$ (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 $$f'{}'(\xi) = \frac{f'(c) – f'(0)}{c}.$$ Putting this together, we have proved that there is a $\xi \in (0, x)$ such that $$f(x) – f(0) – f'(0) x = f'{}'(\xi) \frac{x^2}{2},$$ 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 $$f^{(k)}(t) – f^{(k)}(0) = \frac{f^{(k)}(t) – f^{(k)}(0)}{t} t.$$ 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 $$f(x) – \sum_{n = 0}^k f^{(n)}(0)\frac{x^n}{n!},$$ the difference between $f$ and its degree $k$ Taylor polynomial. The right hand side is
$$\label{eq:only}\underbrace{\int _0^{t_k = x} \cdots \int _0^{t _1}} _{k \text{ times}} \frac{f^{(k)}(t) – f^{(k)}(0)}{t} t \, dt \, dt _1 \cdots dt _{k-1}.$$

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 $$\underbrace{\int_0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} g(t) h(t) dt =g(c) \underbrace{\int _0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} h(t) dt.$$

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 $$\frac{f^{(k)}(c) – f^{(k)}(0)}{c} \underbrace{\int _0^{t _k = x} \cdots \int _0^{t _1}} _{k \; \text{times}} t \, dt \, dt _1 \cdots dt _{k-1}.$$ 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 $$f(x) = \sum_{n = 0}^n f^{(n)}(0) \frac{x^n}{n!} + f^{(k+1)}(\xi) \frac{x^{k+1}}{(k+1)!}$$ for some $\xi \in (0, x)$. Thus we have proved Taylor’s Theorem (with Lagrange’s error bound).

Footnotes

1. Though I must admit that it is not my favorite proof of Taylor’s Theorem.
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