Calculus I · Building a Function From Its Derivatives

The honest part: how wrong is the approximation?

A Taylor polynomial is an approximation, and an approximation you can’t bound is a guess. So we account for the gap exactly. Write

$$f(x) = T_n(x) + R_n(x),$$

where $R_n(x)$ is the remainder — everything the polynomial missed. Taylor’s theorem gives it a stunningly clean form:

Taylor’s Theorem (Lagrange remainder). For some point $c$ between $a$ and $x$, $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$

Stare at that. It is exactly the next term of the series — the $(n+1)$-th — except the derivative is evaluated not at the center $a$ but at some unknown point $c$ in between. And now look at the smallest case, $n = 0$:

$$f(x) = f(a) + f'(c)(x - a).$$

That is the Mean Value Theorem. The $n=1$ case is $f(x) = f(a) + f'(a)(x-a) + \tfrac12 f''(c)(x-a)^2$ — the “second-order MVT” we previewed in Chapter 6. Taylor’s remainder is the Mean Value Theorem, escalated one derivative at a time. This is precisely why we proved the MVT — and its Cauchy cousin — before we got here: they are the engine that produces this error term.

The unknown $c$ might look like it ruins everything — how can a formula with a mystery point be useful? Because you don’t need to find $c$; you only need to bound $f^{(n+1)}$ over the interval. If $|f^{(n+1)}| \le M$ between $a$ and $x$, then

$$|R_n(x)| \le \frac{M\,|x-a|^{n+1}}{(n+1)!}.$$

Worked example — $e^{0.1}$ with a guarantee. Take $f = e^x$, $a = 0$, $n = 2$, so $T_2(0.1) = 1 + 0.1 + \tfrac{0.01}{2} = 1.105$. The remainder needs $f''' = e^x$, and on $[0, 0.1]$ we have $e^x \le e^{0.1} < 1.2$. So

$$|R_2(0.1)| \le \frac{1.2 \cdot (0.1)^3}{3!} = \frac{1.2 \times 0.001}{6} = 0.0002.$$

We have proven, without a calculator, that $1.105$ is within $0.0002$ of the true $e^{0.1}$. (It is: the true value is $1.10517$, off by $0.00017$.) That is the difference between Taylor as a suggestion and Taylor as mathematics: the remainder turns “approximately” into a number you can stand behind.

◆ Layer 2 — Proof of Taylor’s theorem with Lagrange remainder

Fix the target $x$. Choose the number $M$ that makes the equation exact: $f(x) = T_n(x) + M(x-a)^{n+1}$. We’ll show $M = \tfrac{f^{(n+1)}(c)}{(n+1)!}$ for some $c$. Define, as a function of a moving center $t$, $$g(t) = f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^k - M(x-t)^{n+1}.$$ Two endpoint checks: $g(x) = f(x) - f(x) - 0 = 0$ (only the $k=0$ term of the sum survives at $t=x$, giving $f(x)$), and $g(a) = 0$ by our choice of $M$. So $g(a) = g(x)$, and Rolle’s theorem (Chapter 6) gives a $c$ with $g'(c) = 0$. Now differentiate $g$. The sum telescopes — each term’s product-rule pieces cancel against the next term’s — leaving only $$\frac{d}{dt}\sum_{k=0}^n \frac{f^{(k)}(t)}{k!}(x-t)^k = \frac{f^{(n+1)}(t)}{n!}(x-t)^n.$$ So $g'(t) = -\tfrac{f^{(n+1)}(t)}{n!}(x-t)^n + M(n+1)(x-t)^n$. Setting $g'(c) = 0$ and canceling $(x-c)^n$: $$M(n+1) = \frac{f^{(n+1)}(c)}{n!} \qquad\Longrightarrow\qquad M = \frac{f^{(n+1)}(c)}{(n+1)!}. \qquad\blacksquare$$ One auxiliary function, one application of Rolle — the same tilt-and-flatten move that gave us the MVT itself, run at higher order.

◆ Layer 2 — Why $e^x$ actually equals its series everywhere

Having a series isn’t the same as the series being true. For $e^x$ it is, for every $x$, and the remainder proves it. With $a=0$, $|R_n(x)| = \tfrac{e^c\,|x|^{n+1}}{(n+1)!}$ for some $c$ between $0$ and $x$, so $|R_n(x)| \le \tfrac{e^{|x|}\,|x|^{n+1}}{(n+1)!}$. Fix $x$ and let $n\to\infty$: the factorial in the denominator overwhelms the power in the numerator (you saw in Chapter 1 how violently $n!$ outgrows any fixed base raised to $n$), so $R_n(x)\to 0$. The polynomials converge to $e^x$ itself. That is what licenses writing “$=$” instead of “$\approx$.”

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