One approximation, many target points
A Taylor polynomial is often used to approximate a function at a particular input. But in applications, the input may be uncertain, or the approximation may need to be reliable over a whole range. In that setting, a point-by-point error estimate is not enough: we want a single bound that covers every target point in the interval.
The Lagrange remainder provides a way to do this. The key is to control two quantities at once: how far any target point can be from the Taylor center, and how large the relevant derivative can be anywhere in the interval. Once both have uniform bounds, their combination gives one error bound for the entire interval.
The Lagrange remainder on an interval
Let \(I=[u,v]\) be a closed interval, and let \(c\in I\) be the Taylor center. Suppose \(f\) has continuous derivatives through order \(n+1\) on an open interval containing \(I\). Its degree-\(n\) Taylor polynomial about \(c\) is
The remainder at \(x\) is the difference between the function and this polynomial:
The Lagrange remainder theorem says that for each \(x\in I\), there is a point \(\xi\) between \(c\) and \(x\) such that
The point \(\xi\) may depend on \(x\). We do not need to find it. Instead, we bound the derivative at every possible \(\xi\) in \(I\). Since \(f^{(n+1)}\) is continuous and \(I\) is closed and bounded, its absolute value has a maximum there. Write
The greatest distance from \(c\) to any point of \(I\) is
Every point between the endpoints is no farther from \(c\) than the more distant endpoint. Therefore \(|x-c|\le d\) for all \(x\in I\). Together, these facts produce one bound that works across the interval.
Why the uniform bound works
The conclusion follows directly from the Lagrange remainder formula. For any target \(x\in I\), the associated point \(\xi\) lies between \(c\) and \(x\). Because both points are in \(I\), so is \(\xi\). Thus \(|f^{(n+1)}(\xi)|\le M\). Also, \(|x-c|\le d\). Taking absolute values in the remainder formula and applying both inequalities gives
This reasoning applies to every \(x\in I\), not just to one chosen target. That is the essential distinction between a pointwise estimate and a uniform one: the same numerical upper bound is valid everywhere in the stated interval.
Make sure the Taylor center lies in the interval being certified.
Set \(d\) equal to the larger distance from the center to either endpoint.
Find \(M\) so that the absolute value of the \((n+1)\)st derivative is at most \(M\) at every point in the interval.
Use \(M d^{n+1}/(n+1)!\) as the single error bound for all target points.
Worked Example: A sine approximation on a symmetric interval
Worked Example: A sine approximation on a symmetric interval
Use the degree-three Taylor polynomial for \(\sin x\) about \(0\) to bound the error for every \(x\in[-1/2,1/2]\). Decide whether the error is less than \(0.003\) throughout this interval.
The derivatives at \(0\) give
Thus the degree-three polynomial is
The fourth derivative of \(\sin x\) is \(\sin x\). Since \(|\sin x|\le 1\) for all real \(x\), we can take \(M=1\) on the stated interval. The center is \(c=0\), and the farthest endpoint is at distance \(d=1/2\). The uniform Lagrange bound is therefore
To compare this exact bound with \(0.003=3/1000\), cross-multiply positive denominators:
The inequality holds. The degree-three approximation therefore has error less than \(0.003\) at every point of the interval. The final statement is uniform because both the derivative bound and the maximum distance were chosen for the entire interval.
As a check on the polynomial, differentiating \(T_3(x)=x-x^3/6\) gives \(T_3'(x)=1-x^2/2\), \(T_3''(x)=-x\), and \(T_3'''(x)=-1\). At \(x=0\), these match the first three derivatives of \(\sin x\), as required for its degree-three Taylor polynomial.
Worked Example: An exponential approximation on a one-sided interval
Worked Example: An exponential approximation on a one-sided interval
Use the degree-two Taylor polynomial for \(e^x\) about \(0\) to give one error bound for every \(x\in[0,1/2]\). Then certify that the error is less than \(0.042\).
Since every derivative of \(e^x\) equals \(e^x\), the polynomial is
The third derivative is \(e^x\), which is increasing. Its maximum on \([0,1/2]\) is \(e^{1/2}\), so we may take \(M=e^{1/2}\). The farthest point from \(0\) is \(1/2\), giving \(d=1/2\). Hence
To obtain a simple numerical guarantee, note that \(e<3\), so \(e^{1/2}<\sqrt{3}<2\). Consequently,
Finally, \(1/24<0.042\), since \(1/24\approx0.04167\). The error is therefore less than \(0.042\) throughout the interval.
For a polynomial check, \(T_2(0)=1\), \(T_2'(x)=1+x\), and \(T_2''(x)=1\). At zero these agree with \(e^0\), \((e^x)'|_{x=0}\), and \((e^x)''|_{x=0}\), respectively. The derivative bound used in the estimate is also valid on the whole interval: \(e^x\le e^{1/2}\) there.
Worked Example: A rational function on an interval around zero
Worked Example: A rational function on an interval around zero
Find a uniform degree-two Taylor error bound for \(f(x)=1/(1-x)\) on \([-1/4,1/4]\), using the Taylor center \(0\).
The first derivatives are
At zero, the function and its first two derivatives are \(1\), \(1\), and \(2\). Thus
On \([-1/4,1/4]\), the smallest value of \(1-x\) is \(3/4\). Therefore the largest value of \(6/(1-x)^4\) on the interval is
The distance from the center to either endpoint is \(d=1/4\). The Lagrange bound becomes
As an algebraic check, subtract the polynomial directly:
On this interval, \(|x|^3\le1/64\) and \(1-x\ge3/4\), so the absolute difference is at most \((1/64)/(3/4)=1/48\). Since \(1/48<4/81\), this check is consistent with the uniform Lagrange bound. The Taylor polynomial also checks directly: \(T_2'(x)=1+2x\) and \(T_2''(x)=2\), matching the corresponding derivatives of \(f\) at zero.
A common mistake: using the derivative only at the center
In the exponential example, a tempting but invalid calculation is to use \(f'''(0)=1\) as the derivative bound \(M\). That would give
This does not provide a Lagrange bound on \([0,1/2]\). The remainder theorem involves \(f'''(\xi)\) for a point \(\xi\) that may be anywhere between \(0\) and the target \(x\), and \(e^\xi\) is greater than \(1\) when \(\xi>0\). The value at the center alone does not bound the derivative throughout the interval. The correction is to use a valid interval-wide bound, \(M=e^{1/2}\), giving \(e^{1/2}/48\).
Why the interval and the center both matter
A uniform bound can be conservative, but it is reliable: it certifies every input in the specified interval using the same estimate. The interval determines both the largest distance factor and the region over which the derivative must be bounded. Changing the interval can change either quantity, even when the Taylor polynomial stays the same.
In particular, if the center is not the midpoint of the interval, the two endpoint distances need not be equal. The correct distance is still the larger of the two, not an average and not the distance to a convenient endpoint. Similarly, the derivative bound must cover the full interval between the center and all target points. A bound valid only near the center cannot be used to certify targets farther away.
Check Your Understanding
For each question, identify which quantities must be controlled to make an error estimate valid throughout an interval.
- For a Taylor center \(c\) in \([u,v]\), how do you calculate the greatest distance from \(c\) to a point in the interval?
- Why does the Lagrange remainder theorem require a bound on the \((n+1)\)st derivative over an interval rather than just its value at the center?
- For a degree-four Taylor polynomial, which derivative appears in the Lagrange remainder bound?
- If the interval is \([-1/3,1/2]\) and the center is \(0\), what is the maximum distance factor \(d\)?
- What two interval-wide bounds combine to give the uniform error estimate?
