One more derivative, bounded over the whole interval BC only

If $P_n$ is the $n$th-degree Taylor polynomial for $f$ centred at $a$, then for any $x$,

$$\left|f(x) - P_n(x)\right| \le \frac{M}{(n+1)!}\left|x-a\right|^{n+1},$$

where $M$ is any number with $\left|f^{(n+1)}(t)\right| \le M$ for all $t$ between $a$ and $x$. Four parts, each a separate opportunity:

1. The derivative order is $n+1$. One past the degree you used, exactly as the alternating bound uses one term past the ones you summed.
2. The factorial is $(n+1)!$, matching that order.
3. The power is $n+1$, matching both.
4. $M$ bounds the derivative on an interval, not at a point.

The first three move together, so a single off-by-one usually corrupts all three at once, which at least makes it visible. The fourth is independent and is the one that gets skipped.

The remainder theorem says there exists some $c$ between $a$ and $x$ with

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

You do not know which $c$. That is the entire reason a bound is needed rather than a value: to be safe for every possible $c$, $M$ has to be at least as large as $\left|f^{(n+1)}\right|$ everywhere it could land.

So evaluating the derivative at $a$, or at $x$, is not enough. If $f^{(n+1)}$ happens to be larger somewhere in between, the resulting number is not a bound at all and the answer is wrong in the unsafe direction.

Three practical ways to get $M$:

1. A universal bound. For $\sin$ and $\cos$ every derivative is bounded by $1$, so $M = 1$ always works.
2. Monotonicity. For $e^{x}$ on $[0, b]$ the derivative increases, so its maximum is at the right end and $M = e^{b}$.
3. Any convenient over-estimate. $M$ need not be the true maximum. A larger $M$ gives a looser but still valid bound, and $e^{1} < 3$ is often all a question needs.

Approximate $\cos 2$ using $P_2(x) = 1 - \frac{x^{2}}{2}$ centred at $0$, and bound the error.

1. Order: $n = 2$, so the remainder uses $f^{(3)}$.
2. $M$: $f^{(3)}(x) = \sin x$, and $\left|\sin x\right| \le 1$ everywhere, so $M = 1$.
3. Assemble: $\dfrac{1}{3!}\left|2 - 0\right|^{3} = \dfrac{8}{6} = \dfrac{4}{3} \approx 1.3333$.

The true error is $\left|\cos 2 - (-1)\right| \approx \left|-0.4161 + 1\right| = 0.5839$, comfortably inside the bound. As with the alternating bound, over-estimating is what a bound does, and the gap here is large because $M = 1$ is generous and because $\left|x - a\right| = 2$ is far from the centre.

That last point is worth isolating. The bound contains $\left|x-a\right|^{n+1}$, so it grows quickly with distance from the centre and shrinks quickly with degree. Both halves of "a Taylor polynomial is good near its centre" are visible in that one factor.

Two error bounds, one question. The frozen taxonomy pairs them because they are confused in both directions, and the distinction is not subtle once stated:

1. Does the series alternate at the point in question, with decreasing terms? Then the alternating bound applies and is far less work: it is a single term, already written.
2. Is it a Taylor approximation, with no useful alternation? Then Lagrange, and you will need a derivative bound.

Where both apply, either is correct. They give different numbers, since they are different over-estimates, and neither is wrong for being looser.

Two signals that settle it immediately. Being handed a bound on some $f^{(k)}$ means Lagrange, since the alternating bound has no derivative anywhere in it. Being asked about a numerical series rather than a function approximation means the alternating bound, since there is no $f$ to differentiate.