Choosing a Bound Is Part of the Approximation
A Taylor polynomial gives a computable approximation, but the polynomial alone does not show how accurate that approximation is. To certify an answer, you need an error bound that applies to the particular function, center, interval, and polynomial in the problem.
Two useful choices are the Lagrange remainder bound, which uses a bound on a derivative, and the alternating-series remainder bound, which uses the sizes and signs of the terms in a convergent alternating series. Neither is automatically available just because a Taylor polynomial has been written down. The conditions must be checked before the bound is used.
The Two Error-Bound Strategies
The Lagrange form of the Taylor remainder says that if \(f\) has continuous derivatives through order \(n+1\) on the interval between \(a\) and \(x\), then for some \(\xi\) between \(a\) and \(x\), $$ f(x)-P_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. $$ Consequently, if \(\lvert f^{(n+1)}(t)\rvert\le M\) throughout that interval, then $$ \lvert f(x)-P_n(x)\rvert\le \frac{M\lvert x-a\rvert^{n+1}}{(n+1)!}. $$ The derivative bound must hold everywhere on the interval, not just at the expansion center or at the target.
The alternating-series bound applies when the function is represented at the target by a convergent alternating series whose term magnitudes decrease to zero. If the first omitted term has magnitude \(b\), the absolute error after the retained terms is at most \(b\). To use this rule, verify that the series converges at the target, that its signs alternate, and that its term magnitudes decrease from the omitted term onward.
These methods answer the same practical question but use different information. A derivative bound can work without an alternating series, while an alternating-series estimate can be much sharper when its conditions hold. A valid strategy is the one whose hypotheses can be verified and whose bound is small enough for the requested tolerance.
Why the Alternating-Series Bound Works
Here is a short justification of the error estimate. Suppose the omitted tail has alternating terms with magnitudes \(b_m,b_{m+1},\ldots\), where \(b_m\ge b_{m+1}\ge\cdots\ge0\) and \(b_k\to0\). First suppose the first omitted term is positive. Grouping consecutive terms gives $$ b_m-b_{m+1}+b_{m+2}-b_{m+3}+\cdots. $$ Each pair \(b_{m+2j}-b_{m+2j+1}\) is nonnegative, so the tail is nonnegative. Also, after its first term, the remainder can be grouped as $$ -b_{m+1}+b_{m+2}-b_{m+3}+\cdots, $$ which is nonpositive by the same pairing argument. Thus the full tail lies between \(0\) and \(b_m\). The argument for a negative first omitted term follows by changing signs. Taking limits of the finite partial sums is justified by the assumed convergence. Therefore the tail has the sign of the first omitted term and magnitude at most that term.
A Practical Decision Process
For a mixed set of approximation tasks, use a consistent sequence of checks rather than choosing a method by habit.
Record the function, center, evaluation point or interval, and required error.
Be clear about its degree and which terms it includes, especially if some Taylor coefficients are zero.
For Lagrange, establish a derivative bound on the whole interval. For an alternating series, check convergence, alternating signs, and decreasing magnitudes.
A bound that is valid but too large does not certify the requested accuracy. Increase the degree or use another valid bound.
Give the exact polynomial and an inequality showing that its error bound is less than the tolerance.
Worked Example: Choosing an Alternating Bound for Sine
Worked Example: Certifying a Sine Approximation
Approximate \(\sin(0.5)\) with error less than \(10^{-6}\). The Maclaurin series is $$ \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots. $$ At \(x=0.5\), the term magnitudes decrease: the ratio of the magnitude of the term of degree \(k+2\) to that of degree \(k\) is $$ \frac{(0.5)^{k+2}/(k+2)!}{(0.5)^k/k!} =\frac{0.25}{(k+1)(k+2)}<1 $$ for the relevant nonnegative \(k\). The series converges, alternates in sign, and has decreasing magnitudes, so its alternating-series remainder bound applies.
Retain terms through degree \(7\). The exact polynomial value is $$ P_7(0.5)=0.5-\frac{0.125}{6}+\frac{0.03125}{120}-\frac{0.0078125}{5040}. $$ The terms are \(0.0208333333333\ldots\), \(0.0002604166667\ldots\), and \(0.0000015500992\ldots\), respectively. Therefore $$ P_7(0.5)=0.479425533234127\ldots. $$ The first omitted term has magnitude $$ \frac{(0.5)^9}{9!} =\frac{0.001953125}{362880} <5.384\times10^{-9}. $$ This is less than \(10^{-6}\). The complete certificate, stated against the exact polynomial rather than a rounded decimal, is $$ 0<\sin(0.5)-P_7(0.5)<\frac{(0.5)^9}{9!}<5.384\times10^{-9}<10^{-6}. $$ The error is positive because the first omitted term is positive. Thus the degree-7 polynomial meets the tolerance.
Worked Example: Using a Lagrange Bound for the Exponential
Worked Example: Certifying an Exponential Approximation
Approximate \(e^{0.4}\) with error less than \(2\times10^{-5}\), using a Maclaurin polynomial. Since every derivative of \(e^x\) is \(e^x\), for \(0\le t\le0.4\) we have \(\lvert f^{(n+1)}(t)\rvert=e^t\le e<3\). The familiar estimate \(e<3\) follows, for example, from the series for \(e\) and the geometric bound on its factorial terms. Thus \(M=3\) is a valid derivative bound on the full interval.
For degree \(5\), the Lagrange estimate is $$ \lvert e^{0.4}-P_5(0.4)\rvert \le\frac{3(0.4)^6}{6!} =\frac{3(0.004096)}{720} =\frac{0.012288}{720} =0.0000170666\ldots <2\times10^{-5}. $$ The polynomial value is $$ P_5(0.4)=1+0.4+\frac{0.16}{2}+\frac{0.064}{6}+\frac{0.0256}{24}+\frac{0.01024}{120} =1.4918186666\ldots. $$ Each contribution after the first two is \(0.08\), \(0.0106666666\ldots\), \(0.0010666666\ldots\), and \(0.0000853333\ldots\), giving the stated sum. The exact-polynomial certificate is $$ \left|e^{0.4}-P_5(0.4)\right| \le\frac{3(0.4)^6}{6!} <2\times10^{-5}. $$ This verifies the requested tolerance. With this derivative bound, degree \(4\) would give \(3(0.4)^5/5!=0.000256\), which is larger than the tolerance; that degree is not certified by this particular estimate.
Worked Example: Checking an Alternating Logarithm Series
Worked Example: Certifying a Logarithm Approximation
Approximate \(\ln(1.2)\) with error less than \(10^{-5}\). The series $$ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$ converges for \(x=0.2\). Its terms alternate, and their magnitudes \(b_k=(0.2)^k/k\) decrease because $$ \frac{b_{k+1}}{b_k} =0.2\frac{k}{k+1}<1. $$ So the alternating-series bound applies.
Retaining terms through degree \(6\) gives $$ P_6(0.2)=0.2-\frac{0.04}{2}+\frac{0.008}{3}-\frac{0.0016}{4}+\frac{0.00032}{5}-\frac{0.000064}{6}. $$ Evaluating in order gives \(0.18\), then \(0.1826666666\ldots\), then \(0.1822666666\ldots\), then \(0.1823306666\ldots\), and finally \(P_6(0.2)=0.18232\). The first omitted term has magnitude $$ \frac{(0.2)^7}{7}=\frac{0.0000128}{7}=0.0000018285714\ldots<10^{-5}. $$ Its sign is positive, so the remainder is positive. The certificate is $$ 0<\ln(1.2)-P_6(0.2)<\frac{(0.2)^7}{7}<10^{-5}. $$ Thus this degree-6 approximation meets the tolerance.
A Common Degree-Selection Mistake
A frequent error is to skip a nonzero term when identifying the first omitted term. For example, the degree-5 Taylor polynomial for \(\sin(0.5)\) is $$ P_5(0.5)=0.5-\frac{(0.5)^3}{3!}+\frac{(0.5)^5}{5!}. $$ One might incorrectly claim that its alternating-series error is at most \((0.5)^9/9!\), by looking at the degree-7 polynomial used in the earlier example. But the degree-5 polynomial omits the degree-7 term first. Its valid alternating-series bound is $$ \left|\sin(0.5)-P_5(0.5)\right| \le\frac{(0.5)^7}{7!} =\frac{0.0078125}{5040} =0.0000015500992\ldots, $$ which is greater than \(10^{-6}\). The claimed degree-5 certificate therefore fails. To meet this tolerance using the alternating bound, retain the degree-7 term and bound the error by the degree-9 term. Always identify the first omitted nonzero term of the polynomial actually used.
How to Present a Complete Certificate
A clear certificate should make the reasoning auditable. State the target and tolerance, define the polynomial being evaluated, and name the bound used. Then verify the hypotheses: continuity and a derivative maximum or bound for Lagrange; convergence, alternating signs, and decreasing magnitudes for the alternating-series rule. Finally, display the numerical inequality that compares the valid error bound with the tolerance.
Keep the exact polynomial distinct from its decimal display. A rounded polynomial value is useful for reporting an approximation, but the error theorem bounds the difference between the function value and the exact polynomial value. If a rounded output itself needs to be certified, the rounding error must also be included in the total error. In the sine example, for instance, the rigorous statement compares \(\sin(0.5)\) with \(P_7(0.5)\), not with a shortened decimal that might shift the difference.
The two methods may both apply, and one may provide a tighter bound. It is legitimate to compare them, but only after checking both sets of conditions. A failed or oversized estimate is not proof that the approximation is inaccurate; it means only that this bound does not certify the desired tolerance. Try a higher degree, sharpen a justified derivative bound, or use another method whose conditions hold.
Check Your Understanding
For each question, identify what must be checked before claiming an error estimate.
- For a Lagrange remainder estimate, where must the bound on the \((n+1)\)st derivative hold?
- What three properties of an alternating series are needed to use the first-omitted-term bound?
- For the degree-5 Maclaurin polynomial of \(\sin x\), what is the first omitted nonzero term?
- If a valid error bound is larger than the requested tolerance, what can and cannot be concluded?
- Why should a certificate compare the function value with the exact Taylor polynomial rather than an unqualified rounded decimal?