AP Calculus BC

When a Taylor Series Does Not Equal Its Function

Q: How do I bound the Lagrange error?

Find a number M with \left|f^{(n+1)}(t)\right| \le M for every t between a and x, then the error is at most \frac{M|x-a|^{n+1}}{(n+1)!}. For sine and cosine, M = 1 always works.

Q: Do the standard Maclaurin series equal their functions?

Yes. For e^x, \sin x and \cos x the remainder tends to zero for every real x, because the factorial in the denominator eventually beats any power. That is why those expansions are safe to use everywhere.

Having every derivative is not enough for a function to equal its own Taylor series. What is needed is that the remainder tends to zero, and there is a famous function with every derivative zero at the origin whose Maclaurin series is identically zero while the function is not.

Taylor's Theorem with Lagrange Remainder

f(x) = P_n(x) + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}

Taylor's Theorem writes a function as its degree n Taylor polynomial plus a remainder term, with the remainder controlled by the size of the next derivative somewhere between a and x.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The remainder tends to zero as the degree grows
$\lim_{n \to \infty} R_n(x) = 0$
Taylor's Theorem is exact for every finite degree, remainder included, and says nothing at all about the infinite series. Passing to the limit is a separate step, and it is legitimate exactly when the remainder collapses. Having every derivative does not deliver that.
Drop it and the theorem fails
$f(x) = e^{-1/x^{2}} \ \ (x \ne 0), \qquad f(0) = 0$
The function e to the minus one over x squared, at the origin
This function is infinitely differentiable everywhere, and every one of its derivatives at the origin is zero. So its Maclaurin series is $0 + 0x + 0x^2 + \cdots$ , the zero function, which converges everywhere. The check compares the two at $x = 1$ : the function is $e^{-1} \approx 0.368$ and the series is 0. They agree at exactly one point, the origin, and nowhere else.
2
f has n+1 derivatives on the interval between a and x
$f^{(n+1)} \text{ exists on the interval}$
The remainder formula names the $(n+1)$ st derivative at an unknown interior point, so that derivative has to exist across the whole interval for the formula to mean anything. This is the hypothesis that limits how far you can push the degree for a function with limited smoothness. It fails only for functions that run out of derivatives, which no single elementary counterexample illustrates better than the definition does, so none is displayed.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Fix $x$ and treat the remainder as an unknown constant $K$ defined by $f(x) = P_n(x) + K(x-a)^{n+1}$ .
Build the auxiliary function $g(t) = f(t) - P_n(t) - K(t-a)^{n+1}$ , which vanishes at $t = a$ along with its first n derivatives, by construction of the Taylor polynomial.
It also vanishes at $t = x$ , by the choice of K. Repeated application of Rolle's Theorem then produces a point $\xi$ between a and x where the $(n+1)$ st derivative of $g$ is zero.
Unpacking that condition gives $K = \frac{f^{(n+1)}(\xi)}{(n+1)!}$ , which is the Lagrange remainder. Note that the whole result rests on Rolle, which is why differentiability across the interval is required rather than just at the centre.

What it does not say

Every infinitely differentiable function equals its Taylor series.

False, and the counterexample above is the standard one. Functions that do equal their Taylor series on an interval are called analytic, and analytic is strictly stronger than infinitely differentiable.

The Taylor series converging means it converges to f.

Two different questions. The Maclaurin series above converges everywhere, to the zero function, and equals f at one point only. Convergence of the series and equality with the function have to be established separately.

The unknown point in the remainder can be found.

It cannot, in general, and it is not supposed to be. The remainder is used by BOUNDING the derivative on the interval, which is why exam questions ask for an error bound rather than an error.

Frequently asked questions

How do I bound the Lagrange error?

Find a number M with $\left|f^{(n+1)}(t)\right| \le M$ for every t between a and x, then the error is at most $\frac{M|x-a|^{n+1}}{(n+1)!}$ . For sine and cosine, $M = 1$ always works.

Do the standard Maclaurin series equal their functions?

Yes. For $e^x$ , $\sin x$ and $\cos x$ the remainder tends to zero for every real x, because the factorial in the denominator eventually beats any power. That is why those expansions are safe to use everywhere.