From Remainder Formulas to Useful Bounds
The Peano remainder describes how an error behaves as \(x\) approaches the center \(a\), while the Lagrange form gives a formula for the error at a particular target. Taylor bounds turn that formula into a practical estimate: they can show that a Taylor polynomial lies above or below the function, and they can give a numerical interval containing the function value.
The key information is often not just the size of the next derivative, but also its sign. Recall Taylor’s Theorem, Lagrange Form: if \(f\) has derivatives through order \(n+1\) on an open interval containing \(a\) and \(x\), then, for \(x\ne a\),
for some \(c\) strictly between \(a\) and \(x\). The Taylor Remainder Bound, established earlier in this course, also says that if \(|f^{(n+1)}(t)|\leq M\) throughout that segment, then
These facts are useful only when their hypotheses hold throughout the segment joining the center to the target. Knowing a derivative’s sign or bound at the center alone is not enough to control the remainder at a different point.
One-Sided Taylor Bounds
A derivative sign determines the sign of the Lagrange remainder, except that an odd power changes sign when the target is to the left of the center. The next theorem records both directions in one statement.
Proof. By Taylor’s Theorem, Lagrange Form, there is a point \(c\) strictly between \(a\) and \(x\) such that $$ f(x)-T_{n,f,a}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$ Multiplying both sides by \((x-a)^{n+1}\) gives $$ \bigl(f(x)-T_{n,f,a}(x)\bigr)(x-a)^{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{2n+2}\geq0. $$ Here the derivative is nonnegative by hypothesis, the factorial is positive, and the even power is positive because \(x\ne a\). If \(x>a\), then \((x-a)^{n+1}>0\), so the remainder is nonnegative. If \(x<a\), the sign of \((x-a)^{n+1}\) is positive for even \(n+1\) and negative for odd \(n+1\), giving the stated alternatives. Replacing \(f^{(n+1)}\geq0\) with \(f^{(n+1)}\leq0\) reverses the sign of the remainder and hence reverses each inequality. \(\square\)
For targets to the right of the center, the rule is especially simple: a nonnegative next derivative makes the Taylor polynomial a lower bound, and a nonpositive next derivative makes it an upper bound. To the left, check the parity of the first omitted power as well as the derivative sign.
Worked Example: A Lower Bound for the Exponential
Let \(f(x)=e^x\), expand at \(a=0\), and take \(x=1/2\). The degree-two Taylor polynomial is $$ T_{2,f,0}(x)=1+x+\frac{x^2}{2}, \qquad T_{2,f,0}\left(\frac12\right)=1+\frac12+\frac18=\frac{13}{8}. $$ The third derivative is \(f'''(t)=e^t>0\) on \([0,1/2]\). Since the target is to the right of the center, the one-sided Taylor bound gives $$ e^{1/2}\geq\frac{13}{8}. $$ For a finite upper estimate as well, note that \(e^t\leq e^{1/2}\) on this segment. The Taylor Remainder Bound with \(M=e^{1/2}\) gives $$ 0\leq e^{1/2}-\frac{13}{8} \leq \frac{e^{1/2}(1/2)^3}{3!} =\frac{e^{1/2}}{48}. $$ Thus the polynomial gives a lower bound, and the derivative bound gives a certified upper bound on its error.
Combining a Sign with an Error Bound
An absolute error bound alone places \(f(x)\) in a symmetric interval centered at the Taylor value. If the remainder’s sign is known too, only one side of that interval is needed. The following result makes this combination explicit for targets to the right of the center.
Proof. In the first case, the Taylor Bound from a Derivative Sign gives \(f(x)-T_{n,f,a}(x)\geq0\). The Lagrange remainder formula gives a point \(c\in(a,x)\) such that $$ f(x)-T_{n,f,a}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$ The derivative is at most \(M\), and \(x-a>0\), so $$ 0\leq f(x)-T_{n,f,a}(x) \leq\frac{M(x-a)^{n+1}}{(n+1)!}. $$ Adding \(T_{n,f,a}(x)\) throughout yields the first enclosure. In the second case, apply the same argument to the nonnegative quantity \(T_{n,f,a}(x)-f(x)\): its value is the negative of the displayed remainder, and it lies between zero and \(M(x-a)^{n+1}/(n+1)!\). Rearranging gives the second enclosure. \(\square\)
The enclosure is one-sided because the derivative’s sign already determines which side of the Taylor value contains the function. The size bound limits how far away the function can be on that side. This can be sharper than using the absolute remainder bound to include values on both sides, even when that absolute bound has the same numerical error term.
Worked Example: Bracketing a Sine Value with Two Taylor Polynomials
Take \(f(x)=\sin x\), \(a=0\), and \(x=1/2\). The degree-two Taylor polynomial is \(T_{2,f,0}(x)=x\). Since \(f'''(t)=-\cos t<0\) on \([0,1/2]\), the derivative-sign theorem gives $$ \sin\left(\frac12\right)\leq T_{2,f,0}\left(\frac12\right)=\frac12. $$ For a lower bound, use the degree-three Taylor polynomial $$ T_{3,f,0}(x)=x-\frac{x^3}{6}, \qquad T_{3,f,0}\left(\frac12\right)=\frac12-\frac{1}{48}=\frac{23}{48}. $$ The fourth derivative is \(f^{(4)}(t)=\sin t\geq0\) on \([0,1/2]\). The target is to the right of the center, so the degree-three Taylor polynomial is a lower bound. Together, these estimates give $$ \frac{23}{48}\leq\sin\left(\frac12\right)\leq\frac12. $$ The two inequalities come from different Taylor degrees and different derivative signs; neither requires knowing the exact value of the sine.
Worked Example: A Logarithm Enclosed by Adjacent Polynomials
Worked Example: Bounding \(\ln(3/2)\)
Let \(f(x)=\ln(1+x)\), with center \(a=0\) and target \(x=1/2\). The degree-two and degree-three Taylor polynomials are $$ T_{2,f,0}(x)=x-\frac{x^2}{2}, \qquad T_{3,f,0}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}. $$ At the target, their values are $$ T_{2,f,0}\left(\frac12\right)=\frac12-\frac18=\frac38, \qquad T_{3,f,0}\left(\frac12\right)=\frac38+\frac1{24}=\frac5{12}. $$ The third derivative is \(f'''(t)=2/(1+t)^3>0\) on \([0,1/2]\), so the degree-two polynomial is a lower bound. The fourth derivative is \(f^{(4)}(t)=-6/(1+t)^4<0\) there, so the degree-three polynomial is an upper bound. Consequently, $$ \frac38\leq\ln\left(\frac32\right)\leq\frac5{12}. $$ The interval is obtained from derivative signs alone. A uniform bound on either derivative could further quantify the corresponding remainder, but is not needed to establish these two sides.
Choosing and Interpreting a Taylor Bound
A reliable estimate starts by identifying the entire segment between \(a\) and \(x\), then checking the relevant derivative there. A derivative that changes sign on that segment may not provide a one-sided bound, even if its value at the center has the desired sign. Likewise, a numerical error estimate requires a bound \(M\) valid throughout the segment, not merely at one point.
The parity of \(n+1\) matters when \(x<a\). For example, if \(f^{(n+1)}\geq0\), the sign of the remainder is the sign of \((x-a)^{n+1}\). Ignoring that power can reverse the claimed inequality. The safest approach is to write the Lagrange remainder and determine the signs of both factors.
Finally, distinguish a one-sided Taylor estimate from a two-sided error estimate. Knowing \(T_{n,f,a}(x)\leq f(x)\) does not by itself give a useful upper bound on \(f(x)\); a bound on the derivative supplies a maximum possible remainder. Conversely, an absolute error bound does not say whether the Taylor polynomial is above or below the function. When both derivative sign and size are available, the one-sided enclosure combines the two kinds of information.
Check Your Understanding
Use the Lagrange remainder and the derivative bounds to check your understanding.
- If \(x>a\) and \(f^{(n+1)}\geq0\) between \(a\) and \(x\), which direction does the Taylor inequality take?
- Why can the same derivative sign lead to a different inequality when \(x<a\)?
- What additional information, beyond the sign of the next derivative, is needed to give a numerical error bound?
- For the sine example, which derivative signs establish the lower and upper bounds?
- Why must a derivative bound be valid on the whole segment from the center to the target?