The Lagrange Remainder as an Intermediate Derivative
In “Taylor Remainder Terms,” the signed remainder was defined by subtracting a Taylor polynomial from the function. Taylor’s Theorem, Lagrange Form, established earlier, gives a more specific description of that difference: for a target \(x\ne a\), the remainder is determined by the next derivative at some point between \(a\) and \(x\). This tutorial focuses on how to read and use that representation. The intermediate point need not be known explicitly, but its location and the sign of the derivative can still give precise information about the error.
Let \(T_{n,f,a}\) be the degree-\(n\) Taylor polynomial of \(f\) at \(a\), and let \(R_{n,f,a}(x)=f(x)-T_{n,f,a}(x)\). Under the hypotheses of Taylor’s Theorem, its Lagrange form says that when \(x\ne a\), there is a \(c\) strictly between \(a\) and \(x\) such that
The derivative is evaluated at \(c\), not automatically at \(a\) or \(x\). The point \(c\) may depend on the target \(x\) and on the degree \(n\); the theorem does not assert that it is unique. The value of the formula is that \(c\) lies in a known interval, so information about \(f^{(n+1)}\) on that interval constrains the remainder.
This normalization is useful because it separates two sources of the signed error. The factor \((x-a)^{n+1}\) records the direction and distance from the center, while \(f^{(n+1)}(c)\) records the behavior of the next derivative between the center and target. In particular, when \(x<a\), the displacement factor can be negative if \(n+1\) is odd. Ignoring that factor can lead to an incorrect conclusion about the sign of the error.
Derivative Bounds and the Sign of the Error
A direct consequence of the Lagrange form is that bounds on the next derivative bound the normalized remainder. This is an exact enclosure: it keeps the signed derivative information rather than replacing it immediately with an absolute-value estimate.
Proof. Taylor’s Theorem, Lagrange Form, supplies a point \(c\) strictly between \(a\) and \(x\) such that \(Q_{n,f,a}(x)=f^{(n+1)}(c)\). The assumed derivative bounds apply at \(c\), so $$ m\leq Q_{n,f,a}(x)\leq M. $$ The displayed identity for \(R_{n,f,a}(x)\) follows by rearranging the definition of \(Q_{n,f,a}(x)\). This proves both claims. \(\square\)
If the next derivative is nonnegative throughout the segment, the normalized remainder is nonnegative. The signed remainder itself has the sign of \((x-a)^{n+1}\), unless it is zero. Thus, for \(x>a\), a nonnegative next derivative gives a nonnegative remainder. For \(x<a\), the same conclusion holds when \(n+1\) is even, while the remainder is nonpositive when \(n+1\) is odd. Reversing the derivative sign reverses these conclusions. This reasoning uses the signed remainder, not just its absolute value.
Worked Example: A Degree-Two Approximation to the Exponential
Let \(f(x)=e^x\), with center \(a=0\), degree \(n=2\), and target \(x=1\). The derivatives through order two at zero are all \(1\), so $$ T_{2,f,0}(x)=1+x+\frac{x^2}{2}, \qquad T_{2,f,0}(1)=1+1+\frac12=\frac52. $$ The signed remainder is therefore $$ R_{2,f,0}(1)=e-\frac52. $$ Since \(f^{(3)}(t)=e^t>0\), the Lagrange form gives a point \(c\in(0,1)\) such that $$ e-\frac52=\frac{e^c}{3!}(1-0)^3=\frac{e^c}{6}. $$ The remainder is positive, as the derivative sign predicts. The formula identifies an intermediate derivative value but does not say that the remainder equals the degree-three Taylor term \(1/6\), which uses \(f^{(3)}(0)\) instead.
Worked Example: A Logarithm with a Negative Remainder
Take \(f(x)=\ln x\), center \(a=1\), degree \(n=1\), and target \(x=3/2\). Since \(f(1)=0\) and \(f'(1)=1\), the degree-one Taylor polynomial is \(T_{1,f,1}(x)=x-1\). Hence $$ R_{1,f,1}\left(\frac32\right) = \ln\left(\frac32\right)-\frac12. $$ The next derivative is \(f''(t)=-1/t^2<0\) for \(t>0\). The Lagrange form gives \(c\in(1,3/2)\) with $$ \ln\left(\frac32\right)-\frac12 = \frac{f''(c)}{2!}\left(\frac12\right)^2 = -\frac{1}{8c^2}<0. $$ Thus the signed error is negative, in agreement with the sign of the second derivative and the positive displacement squared. This establishes the sign without requiring a decimal approximation to the logarithm.
The derivative enclosure also gives inequalities for the remainder, but multiplying by a negative displacement power reverses their order. For example, if \(x<a\) and \(n+1\) is odd, then \((x-a)^{n+1}/(n+1)!<0\). Multiplying \(m\leq Q_{n,f,a}(x)\leq M\) by this negative number yields an upper bound from \(m\) and a lower bound from \(M\). Keeping the normalized quantity separate until the final multiplication is a reliable way to avoid reversing the inequality incorrectly.
Monotonicity Gives an Endpoint Comparison
Sometimes the next derivative is not merely bounded; it is monotone. Then the intermediate derivative value can be compared directly with its values at the two endpoints of the segment. This gives a useful estimate that retains local information about the derivative rather than using only a single uniform bound.
- If \(a<x\), then $$ f^{(n+1)}(a)\leq Q_{n,f,a}(x)\leq f^{(n+1)}(x). $$
- If \(x<a\), then $$ f^{(n+1)}(x)\leq Q_{n,f,a}(x)\leq f^{(n+1)}(a). $$
Proof. By Taylor’s Theorem, Lagrange Form, there is \(c\) strictly between \(a\) and \(x\) such that \(Q_{n,f,a}(x)=f^{(n+1)}(c)\). If \(a<x\), then \(a<c<x\). Since \(f^{(n+1)}\) is nondecreasing, $$ f^{(n+1)}(a)\leq f^{(n+1)}(c)\leq f^{(n+1)}(x), $$ which is the first assertion. If \(x<a\), then \(x<c<a\), so the same monotonicity gives $$ f^{(n+1)}(x)\leq f^{(n+1)}(c)\leq f^{(n+1)}(a). $$ Substituting \(Q_{n,f,a}(x)=f^{(n+1)}(c)\) proves the second assertion. \(\square\)
If \(f^{(n+1)}\) is nonincreasing instead, the endpoint order is reversed in each case. These comparisons constrain the normalized remainder, not automatically the signed remainder: to obtain inequalities for \(R_{n,f,a}(x)\), multiply by \((x-a)^{n+1}/(n+1)!\), checking whether that factor is positive or negative.
Worked Example: A Cubic Derivative Controls a Quartic Remainder
Let \(f(x)=x^4\), with center \(a=1\), degree \(n=2\), and target \(x=2\). We have \(f(1)=1\), \(f'(1)=4\), and \(f''(1)=12\), so $$ T_{2,f,1}(x)=1+4(x-1)+6(x-1)^2, \qquad T_{2,f,1}(2)=1+4+6=11. $$ Since \(f(2)=16\), the remainder is \(R_{2,f,1}(2)=5\), and the normalized remainder is $$ Q_{2,f,1}(2)=\frac{3!\cdot 5}{(2-1)^3}=30. $$ The third derivative is \(f^{(3)}(t)=24t\), which is nondecreasing on \([1,2]\). The endpoint comparison gives $$ 24=f^{(3)}(1)\leq Q_{2,f,1}(2)\leq f^{(3)}(2)=48, $$ and the calculated value \(30\) lies in this interval. In fact, the Lagrange form gives \(5=24c/6=4c\), so \(c=5/4\), which is strictly between \(1\) and \(2\).
Direction, Order, and Common Pitfalls
The parity of \(n+1\) matters only through the displacement factor; it does not change which derivative appears in the Lagrange form. If \(x>a\), then \((x-a)^{n+1}>0\) for every order, so the remainder has the sign of \(f^{(n+1)}(c)\). If \(x<a\), the displacement power is negative exactly when \(n+1\) is odd. A useful check is to identify the sign of the displacement power before drawing a conclusion about the signed error.
Worked Example: Approximating to the Left of the Center
Let \(f(x)=e^x\), center \(a=1\), degree \(n=0\), and target \(x=0\). The degree-zero Taylor polynomial is the constant \(T_{0,f,1}(x)=e\), so $$ R_{0,f,1}(0)=1-e. $$ The Lagrange form gives \(c\in(0,1)\) such that $$ 1-e=f'(c)(0-1)=-e^c<0. $$ Here \(f'(c)>0\), but the displacement \(0-1\) is negative. The negative remainder is therefore consistent with the formula; a positive next derivative alone does not guarantee a positive signed error when the target is to the left and the displacement power is odd.
A common misreading is to replace \(f^{(n+1)}(c)\) by \(f^{(n+1)}(a)\). The latter gives the next term in the Taylor polynomial, not generally the full remainder. Another is to assume the intermediate point is fixed as \(x\) changes. Taylor’s Theorem guarantees a suitable point for each target, but does not provide one common point for all targets. When a numerical error bound is needed, a uniform bound on the derivative over the whole segment is often simpler; when the derivative varies substantially, the endpoint comparison or another local estimate can be more informative.
The Lagrange form is valuable even when \(c\) cannot be calculated. Its location converts qualitative facts about the derivative—such as its sign, range, or monotonicity—into conclusions about the actual signed error. The normalized remainder makes this structure especially clear: it is an intermediate value of the next derivative, while the displacement factor restores the scale and direction of the error.
Check Your Understanding
Use the Lagrange form and the normalized remainder to answer these questions.
- For \(x\ne a\), write the definition of \(Q_{n,f,a}(x)\) and its value at an intermediate derivative point.
- If \(f^{(n+1)}\) is positive between \(a\) and \(x\), what determines whether the signed remainder is positive or negative?
- Why can multiplying a normalized-remainder inequality by \((x-a)^{n+1}\) require reversing the inequality?
- If \(f^{(n+1)}\) is nondecreasing and \(x<a\), which endpoint gives the lower bound for \(Q_{n,f,a}(x)\)?
- Does Taylor’s Theorem, Lagrange Form, assert that the intermediate point is unique or independent of the target \(x\)?