Why the Interval Conditions Matter
A Taylor polynomial is built from derivatives at its center, but the Lagrange remainder theorem makes a claim about the error at a separate target point. The hypotheses therefore concern more than whether the derivatives at the center exist: they concern the behavior of the function across the interval connecting center and target.
This distinction matters when that interval contains a singularity, a point where a piecewise formula changes, or an endpoint at which a derivative does not exist. The theorem is not automatically available just because the Taylor polynomial can be written down.
The Lagrange Remainder and Its Hypotheses
Let \(n\) be a nonnegative integer, let \(a\) be the Taylor center, and let \(x\ne a\) be the target. The degree-\(n\) Taylor polynomial of \(f\) centered at \(a\) is
provided the derivatives in this expression exist. The remainder is the difference between the function value and this polynomial:
The following conditions are a useful sufficient version of the Lagrange remainder theorem. They are stated for the whole closed segment joining \(a\) and \(x\), regardless of which point is larger.
In particular, the theorem does not require \(f^{(n+1)}\) to exist at either endpoint. For \(n\ge 1\), the derivatives through order \(n\) must behave continuously up to the endpoints in this version. A familiar, stronger condition that ensures all the hypotheses is that \(f\) has continuous derivatives through order \(n+1\) on an open interval containing the segment.
All these conditions are local to the interval between center and target. A singularity somewhere else in the function’s domain does not prevent applying the theorem on a segment that avoids it. A singularity inside the segment, however, can prevent the theorem from applying.
Why the Theorem Works
The key idea is Rolle’s theorem: a continuous function that is differentiable in the interior and has equal values at the endpoints has a point in the interior where its derivative is zero. The Taylor polynomial matches the function’s derivatives at \(a\), so an auxiliary function can be arranged to have matching zeros at the center and target.
Set \(K=R_n(x)/(x-a)^{n+1}\), and define \(g(t)=f(t)-P_n(t)-K(t-a)^{n+1}\) on the segment.
At \(t=a\), the Taylor polynomial matches \(f\) and its derivatives through order \(n\), so \(g^{(k)}(a)=0\) for \(0\le k\le n\). At \(t=x\), the choice of \(K\) gives \(g(x)=0\).
The zero at \(a\), together with the zero at \(x\), allows Rolle’s theorem to be applied successively. It yields an interior point \(c\) where \(g^{(n+1)}(c)=0\).
Since \(P_n^{(n+1)}=0\) and the \((n+1)\)st derivative of \(K(t-a)^{n+1}\) is \((n+1)!K\), the equality \(g^{(n+1)}(c)=0\) gives \(f^{(n+1)}(c)=(n+1)!K\). Substituting the definition of \(K\) proves the formula.
Repeated applications of Rolle’s theorem require continuity on the closed segment and differentiability in its interior at each stage. The hypotheses above provide exactly that for the auxiliary function and its successive derivatives. The argument also explains why the theorem produces an interior point \(c\), not necessarily either endpoint.
Worked Example: A Derivative Missing at an Endpoint
Worked Example: A Derivative Missing at an Endpoint
Apply the degree-zero remainder theorem to \(f(t)=\sqrt{t}\) on \([0,1]\). Here \(a=0\), \(x=1\), and \(n=0\). The function is continuous on the closed interval and differentiable at every interior point \(0<t<1\), even though its derivative does not exist at \(t=0\).
The degree-zero Taylor polynomial is \(P_0(1)=f(0)=0\). Thus $$ R_0(1)=f(1)-P_0(1)=1-0=1. $$ For an interior point \(t\), \(f'(t)=1/(2\sqrt{t})\). The Lagrange formula requires \(1=f'(c)(1-0)\), so $$ 1=\frac{1}{2\sqrt{c}} \quad\Longrightarrow\quad 2\sqrt{c}=1 \quad\Longrightarrow\quad \sqrt{c}=\frac12 \quad\Longrightarrow\quad c=\frac14. $$ This point is in \((0,1)\). Checking it, $$ f'\left(\frac14\right)(1-0)=\frac{1}{2\sqrt{1/4}}=\frac{1}{2(1/2)}=1=R_0(1). $$ The example confirms that the theorem can apply without differentiability at the endpoint when \(n=0\).
Worked Example: A Piecewise Function That Meets the Conditions
Worked Example: A Piecewise Function That Meets the Conditions
Define \(f(t)=t^4\) for \(t\le 0\), and \(f(t)=2t^4\) for \(t>0\). Use \(a=0\), \(x=1\), and \(n=2\). At the joining point, \(f(0)=0\), \(f'(0)=0\), and \(f''(0)=0\): for example, the difference quotient for \(f'(0)\) is \(f(h)/h\), which equals \(h^3\) for \(h\le0\) and \(2h^3\) for \(h>0\), and tends to zero from both sides.
On the positive side, \(f'(t)=8t^3\), \(f''(t)=24t^2\), and \(f'''(t)=48t\). The derivatives through order two are continuous on \([0,1]\), and the third derivative exists throughout \((0,1)\). The Taylor polynomial is \(P_2(1)=0\), so \(R_2(1)=f(1)-P_2(1)=2\).
The theorem predicts an interior \(c\) satisfying $$ 2=\frac{48c}{3!}(1-0)^3=8c, $$ so \(c=1/4\). Checking, $$ \frac{f'''(1/4)}{3!}(1-0)^3 =\frac{48(1/4)}{6} =\frac{12}{6} =2 =R_2(1). $$ The change in formula at zero causes no problem: the required derivatives are continuous there, and the highest derivative is required only on the open segment.
Worked Example: An Interior Point Where the Required Derivative Fails
Worked Example: An Interior Point Where the Required Derivative Fails
Consider \(f(t)=|t|^3\) on \([-1,1]\), with center \(a=-1\), target \(x=1\), and \(n=2\). This function has two continuous derivatives, but its third derivative does not exist at the interior point \(t=0\). Indeed, \(f'''(t)=-6\) for \(t<0\), while \(f'''(t)=6\) for \(t>0\).
At the center, \(f(-1)=1\), \(f'(-1)=-3\), and \(f''(-1)=6\). Therefore $$ P_2(1)=1+(-3)(1-(-1))+\frac{6}{2}(1-(-1))^2 =1-6+12=7. $$ Since \(f(1)=1\), the actual remainder is \(R_2(1)=1-7=-6\).
If the Lagrange formula held for some interior \(c\ne0\), its right-hand side would be either $$ \frac{-6}{6}(2)^3=-8 \quad\text{if }c<0, \qquad\text{or}\qquad \frac{6}{6}(2)^3=8 \quad\text{if }c>0. $$ Neither value equals \(-6\). There is no eligible \(c\), and the missing third derivative at the interior point is exactly the failed hypothesis. This example shows that the conclusion can fail when a required condition is absent.
Worked Example: A Singularity Inside the Segment
Worked Example: A Singularity Inside the Segment
Take \(f(t)=1/t\), \(a=-1\), \(x=1\), and \(n=0\). Although \(f\) is defined at both endpoints, it is undefined at \(t=0\), which lies between them. It is therefore not continuous on the closed segment, and the degree-zero remainder theorem cannot be applied.
The endpoint difference is $$ R_0(1)=f(1)-f(-1)=1-(-1)=2. $$ For any interior \(c\ne0\), \(f'(c)=-1/c^2\), so the proposed remainder expression would be $$ f'(c)(1-(-1))=-\frac{2}{c^2}<0. $$ It cannot equal the actual remainder \(2\). The computation confirms why the interior singularity cannot be ignored merely because the center and target values exist.
What to Check Before Using the Bound
The Lagrange remainder bound follows from the remainder formula when the \((n+1)\)st derivative is bounded in absolute value on the interior of the segment. If \(|f^{(n+1)}(t)|\le M\) there, then
Before using this estimate, first verify that the remainder theorem applies. Check that the function is continuous across the entire segment; that the derivatives needed to form the Taylor polynomial have the required continuity; and that the next derivative exists at every interior point. Only then should you establish a suitable bound for that derivative.
A common mistake is to check only the derivatives at the center. For example, \(1/t\) has derivatives of every order at \(a=-1\), so its Taylor polynomial there is well-defined. But those center derivatives do not repair the singularity at zero on the way to \(x=1\). The theorem’s hypotheses concern the interval, not just the center.
Check Your Understanding
For each question, consider the whole interval between the center and target, not only the center.
- For the degree-zero remainder theorem, which continuity and differentiability conditions are required on the segment?
- Does the degree-zero theorem require \(f'\) to exist at either endpoint? Explain.
- Why does \(f(t)=1/t\) prevent applying the theorem on the segment from \(-1\) to \(1\), even though both endpoint values exist?
- For \(f(t)=|t|^3\), \(a=-1\), \(x=1\), and \(n=2\), what is the actual remainder, and which required derivative fails inside the segment?
- Why can a piecewise formula still satisfy the theorem’s hypotheses at its joining point?
