Why the Highest Derivative Needs Its Own Bound
A Taylor polynomial approximates a function near a center \(a\). The Lagrange remainder formula expresses the error at \(x\) using a value of the next derivative somewhere between \(a\) and \(x\). To turn that expression into a numerical error bound, we need an upper bound for the absolute value of that derivative throughout the interval.
The important phrase is throughout the interval. The largest derivative value might occur at an endpoint, at an interior critical point, or be easiest to control using a global inequality. Checking only the derivative at \(x\), for example, is not enough unless we have established that its absolute value is largest there.
With this \(M\), the Lagrange remainder formula gives $$ |R_n(x)|\leq \frac{M|x-a|^{n+1}}{(n+1)!}. $$ Our task here is to find \(M\). The derivative to study is always \(f^{(n+1)}\), one order higher than the Taylor polynomial degree.
First Check the Whole Interval
For a fixed Taylor approximation, define \(h(t)=f^{(n+1)}(t)\). The quantity to bound is \(|h(t)|\), not necessarily \(h(t)\) itself. A large negative value of \(h\) can have a large absolute value, so finding the largest signed value alone may miss the required bound.
Continuity on a closed interval guarantees that \(|h|\) attains a maximum somewhere on that interval. This ensures that an exact maximum exists. It does not, by itself, tell us where the maximum occurs or how to calculate it. Useful approaches include checking monotonicity, examining endpoints and critical points, and applying a simple inequality that holds throughout the interval.
Using Monotonicity
If \(|h(t)|\) is increasing throughout the interval, its maximum is at the right endpoint; if it is decreasing, its maximum is at the left endpoint. To use this approach, establish the monotonicity on the entire interval. Sometimes \(h\) has a fixed sign, so \(|h|\) is either \(h\) or \(-h\), and its monotonicity follows from the derivative of \(h\).
Worked Example: A Decreasing Absolute Derivative
Find a valid \(M\) for the Taylor remainder of degree \(2\) for \(f(t)=1/(1+t)\), centered at \(a=0\), when evaluated at \(x=1\). Here \(n=2\), so we need the third derivative on \([0,1]\).
Differentiate in order: $$ f(t)=(1+t)^{-1},\qquad f'(t)=-(1+t)^{-2}, $$ $$ f''(t)=2(1+t)^{-3},\qquad f'''(t)=-6(1+t)^{-4}. $$ Thus the quantity to bound is $$ |f'''(t)|=\frac{6}{(1+t)^4}. $$ Its derivative on \([0,1]\) is $$ \frac{d}{dt}\left(\frac{6}{(1+t)^4}\right) =-\frac{24}{(1+t)^5}<0. $$ Therefore \(|f'''(t)|\) decreases across the interval. Its largest value occurs at \(t=0\), and $$ M=|f'''(0)|=\frac{6}{1^4}=6. $$ The result is a valid bound because the derivative of \(|f'''|\) is negative throughout \([0,1]\). As a check, the Taylor remainder bound at \(x=1\) is \(6(1-0)^3/3!=1\). The derivative used to obtain \(M\) is indeed the third derivative, as required for degree \(2\).
In this example, the largest absolute value is at the center \(a\), not at the evaluation point \(x\). Monotonicity establishes this rather than assuming it.
Worked Example: A Derivative That Increases on the Interval
Find the exact maximum of \(|f^{(4)}(t)|\) for \(f(t)=\sin t\) on \([0,\pi/3]\). This is the derivative needed for a Taylor polynomial of degree \(3\).
Differentiate: $$ f'(t)=\cos t,\qquad f''(t)=-\sin t,\qquad f'''(t)=-\cos t,\qquad f^{(4)}(t)=\sin t. $$ On \([0,\pi/3]\), \(\sin t\geq 0\), so \(|f^{(4)}(t)|=\sin t\). Its derivative is \(\cos t\), and \(\cos t>0\) throughout this interval. Hence \(\sin t\) increases from the left endpoint to the right endpoint. Its maximum is $$ M=\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}. $$ Checking the endpoints gives \(\sin 0=0\) and \(\sin(\pi/3)=\sqrt{3}/2\), consistent with the monotonicity argument. Therefore \(M=\sqrt{3}/2\) is the exact maximum on the interval.
Using Endpoints and Interior Critical Points
Monotonicity is not always enough to identify the maximum directly. If \(h\) is differentiable and \(|h|\) has a positive maximum at an interior point \(t_*\), then \(h(t_*)\neq 0\). Near that point, \(h\) keeps its sign, so \(|h|\) is differentiable there. The condition for an interior maximum then implies \(h'(t_*)=0\). Consequently, to find the exact maximum of \(|h|\), it is enough to compare its values at the endpoints and at interior points where \(h'=0\). If \(h\) is identically zero, the maximum is simply zero.
Let \(I\) be the closed interval from \(a\) to \(x\), and set \(h=f^{(n+1)}\).
Solve \(h'(t)=0\) for points inside \(I\), when this derivative is available.
Compute \(|h|\) at every interior candidate and at both endpoints.
The largest of these values is the exact maximum; any number at least this large is also a valid \(M\).
This critical-point method depends on differentiability of \(h\), meaning that the derivative \(f^{(n+2)}\) is available for the search. If that additional derivative is not available, a different argument, such as a known monotonicity result or a direct inequality, may still give a bound.
Worked Example: An Interior Maximum
Find the exact maximum of \(|f'''(t)|\) on \([0,2]\) for $$ f(t)=\frac{t^4}{12}-\frac{t^5}{60}. $$ This is the derivative needed for a Taylor polynomial of degree \(2\).
Differentiate step by step: $$ f'(t)=\frac{t^3}{3}-\frac{t^4}{12}, \qquad f''(t)=t^2-\frac{t^3}{3}, $$ $$ f'''(t)=2t-t^2. $$ On \([0,2]\), $$ f'''(t)=t(2-t)\geq 0, $$ so \(|f'''(t)|=2t-t^2\). Differentiate this function to find its interior critical points: $$ \frac{d}{dt}(2t-t^2)=2-2t. $$ Setting this equal to zero gives \(t=1\), which lies inside \([0,2]\). Evaluate at the endpoints and this critical point: $$ |f'''(0)|=0,\qquad |f'''(1)|=|2-1|=1,\qquad |f'''(2)|=|4-4|=0. $$ The exact maximum is therefore \(M=1\), attained at \(t=1\). As a check, differentiating \(f''(t)=t^2-t^3/3\) gives \(2t-t^2\), the third derivative used in the calculation. For the degree-\(2\) remainder from \(a=0\) to \(x=2\), this \(M\) gives the bound \(1\cdot 2^3/3!=4/3\).
Using a Simple Global Bound
An exact maximum is not always necessary. A convenient inequality can give a valid \(M\) immediately, even if it is not the smallest possible one. For instance, if the derivative is a sine or cosine, the standard inequality \(|\sin t|\leq 1\) and \(|\cos t|\leq 1\) holds for every real \(t\). Such a bound automatically holds on any interval. A global bound can be especially useful when finding critical points would require unnecessary work.
Worked Example: A Global Trigonometric Bound
Find a valid \(M\) for a Taylor polynomial of degree \(4\) for \(f(t)=\cos t\), on the interval \([0,2]\). We need the fifth derivative: $$ f'(t)=-\sin t,\qquad f''(t)=-\cos t,\qquad f'''(t)=\sin t, $$ $$ f^{(4)}(t)=\cos t,\qquad f^{(5)}(t)=-\sin t. $$ For every \(t\in[0,2]\), the global inequality \(|\sin t|\leq 1\) gives $$ |f^{(5)}(t)|=|\sin t|\leq 1. $$ Thus \(M=1\) is a valid bound. In fact, \(\pi/2\) lies in \([0,2]\), and \(|f^{(5)}(\pi/2)|=1\), so the bound is also the exact maximum. Differentiating \(f^{(4)}(t)=\cos t\) confirms that \(f^{(5)}(t)=-\sin t\).
A Common Mistake: Checking Only One Endpoint
Consider again \(f(t)=1/(1+t)\) on \([0,1]\). A tempting but incorrect step is to evaluate the third derivative only at the right endpoint and use that value as the maximum: $$ |f'''(1)|=\left|-\frac{6}{(1+1)^4}\right| =\frac{6}{16} =\frac{3}{8}. $$ This does not give a valid bound on the interval. At the other endpoint, $$ |f'''(0)|=\left|-\frac{6}{(1+0)^4}\right|=6, $$ which is greater than \(3/8\). The correction is to establish where the absolute derivative is largest. Since \(6/(1+t)^4\) decreases throughout \([0,1]\), its maximum is \(6\) at \(t=0\), not \(3/8\) at \(t=1\).
Check Your Understanding
For each question, identify the derivative whose absolute value must be bounded and explain how you would justify the bound on the full interval.
- For a Taylor polynomial of degree \(3\), which derivative must be bounded in a Lagrange remainder estimate?
- If \(|f^{(n+1)}(t)|\) is decreasing on the interval from \(a\) to \(x\), at which endpoint does its maximum occur?
- Why must an exact-maximum search include interior critical points as well as endpoints?
- What is the difference between an exact maximum and a valid upper bound \(M\)?
- Why is evaluating \(|f^{(n+1)}(x)|\) alone generally insufficient to establish a valid \(M\)?
