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Remainders and Error Estimates · Tutorial 865 of 998

Comparing Error Bounds With Actual Errors

Compute actual Taylor truncation errors and compare them with rigorous bounds to distinguish guaranteed accuracy from observed accuracy.

Advanced 10 min read

What You'll Learn

  • Define actual truncation error and distinguish it from an error bound.
  • Compare Taylor errors with Lagrange remainder estimates.
  • Apply alternating-series bounds when the terms satisfy the required conditions.
  • Compute and check actual errors for sine, cosine, and logarithm approximations.
  • Interpret the gap between a guaranteed maximum error and the observed error.

Actual Error and Guaranteed Error

A Taylor polynomial gives an approximation to a function value, while a remainder estimate gives a guarantee about how far that approximation can be from the true value. Those quantities answer different questions. The actual error is the difference at the particular input; an error bound is a number that the actual error is proved not to exceed.

Comparing the two helps explain what an estimate does and does not tell us. A bound may be close to the actual error, or it may be much larger. Either way, it is useful if it is valid: it certifies a maximum possible error without requiring the exact value of that error to be known in advance.

Definition: If \(P_n(x)\) is a Taylor polynomial used to approximate \(f(x)\), the signed truncation error is \(R_n(x)=f(x)-P_n(x)\). The actual absolute error is \(|R_n(x)|\). An error bound \(B\) is a proved inequality \(|R_n(x)|\le B\).

The sign records whether the polynomial is below or above the function value. The absolute error measures the size of the discrepancy without regard to direction. In comparing an actual error with a bound, compare \(|R_n(x)|\) with the nonnegative bound, not the signed error itself.

Two Kinds of Guarantees

For a Taylor polynomial of degree \(n\) centered at \(a\), the Lagrange remainder theorem says that, if \(f\) has a continuous derivative of order \(n+1\) on the interval between \(a\) and \(x\), then for some \(\xi\) between those points,

$$ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. $$

Consequently, if \(|f^{(n+1)}(t)|\le M\) throughout that interval, then

$$ |R_n(x)|\le \frac{M|x-a|^{n+1}}{(n+1)!}. $$

This is a guarantee based on a derivative bound over the whole interval. It does not generally give the actual error, because the intermediate point \(\xi\) is not usually known.

A different guarantee applies when the approximation comes from an alternating series. Suppose its terms alternate in sign, their absolute values decrease to zero, and the series converges. After truncating the series, the absolute value of the remaining tail is at most the absolute value of the first omitted term. This estimate applies only when those conditions hold; it cannot be used just because a polynomial has been truncated.

Keep the quantities distinct: The actual error depends on the function value at the selected input. The Lagrange bound depends on a derivative bound over an interval. The alternating-series bound depends on the first omitted term and the behavior of the series terms.

Why the Alternating Bound Works

Consider an alternating tail whose term magnitudes satisfy \(a_1\ge a_2\ge a_3\ge\cdots\ge0\) and tend to zero. Its terms have signs \(+,-,+,-,\ldots\), or the opposite signs. For the first sign pattern, its even partial sums after the first term can be grouped as

$$ (a_1-a_2)+(a_3-a_4)+\cdots, $$

so they are nonnegative. Its odd partial sums can be written as \(a_1-(a_2-a_3)-\cdots\), and are at most \(a_1\), since each parenthesized difference is nonnegative. Because the series converges, the tail is the limit of these partial sums. It therefore lies between \(0\) and \(a_1\), so its absolute value is at most \(a_1\). Reversing every sign gives the same absolute-value conclusion for the opposite sign pattern.

This explains both the size guarantee and the sign information: the remainder has the sign of the first omitted term, and its magnitude cannot exceed that term. Neither statement says that the error equals the first omitted term.

Worked Comparison: Sine at One

Worked Example: Comparing Two Bounds for \(\sin(1)\)

Use the degree-three Maclaurin polynomial for sine to approximate \(\sin(1)\), then compare the actual error with both estimates.

The polynomial is \(P_3(x)=x-\frac{x^3}{3!}\). At \(x=1\),

$$ P_3(1)=1-\frac{1}{6}=\frac{5}{6}=0.8333333333\ldots $$

The function value is \(\sin(1)\approx0.8414709848\). Thus the signed error and its magnitude are

$$ R_3(1)=0.8414709848-0.8333333333 \approx0.0081376515, \qquad |R_3(1)|\approx0.0081376515. $$

For the Lagrange estimate, the fourth derivative is \(f^{(4)}(x)=\sin x\). On \([0,1]\), \(|\sin x|\le1\), so we may take \(M=1\). The degree is \(3\), the center is \(0\), and \(x=1\):

$$ |R_3(1)| \le\frac{1\cdot|1-0|^4}{4!} =\frac{1}{24} \approx0.0416666667. $$

The sine series alternates with decreasing term magnitudes at \(x=1\). The first nonzero term omitted after \(x-\frac{x^3}{3!}\) is \(\frac{x^5}{5!}\). Therefore the alternating-series estimate is

$$ |R_3(1)|\le\frac{1}{5!}=\frac{1}{120} \approx0.0083333333. $$

Check: The actual error is positive, as expected because the first omitted term is positive. Numerically, \(0.0081376515<0.0083333333<0.0416666667\). The approximation and both bounds are consistent; the alternating-series bound is much tighter here.

Worked Comparison: Cosine at One

Worked Example: Comparing Two Bounds for \(\cos(1)\)

Use \(P_2(x)=1-\frac{x^2}{2!}\), the degree-two Maclaurin polynomial for cosine.

First evaluate the polynomial:

$$ P_2(1)=1-\frac{1}{2}=\frac{1}{2}=0.5000000000. $$

Since \(\cos(1)\approx0.5403023059\), the signed error is positive, and the actual absolute error is

$$ R_2(1)=0.5403023059-0.5000000000 \approx0.0403023059, \qquad |R_2(1)|\approx0.0403023059. $$

For the Lagrange bound, \(f^{(3)}(x)=\sin x\), so \(|f^{(3)}(x)|\le1\) on \([0,1]\). With degree \(2\),

$$ |R_2(1)| \le\frac{1\cdot|1-0|^3}{3!} =\frac{1}{6} \approx0.1666666667. $$

The cosine series alternates, and its term magnitudes decrease at \(x=1\). Its first omitted term is \(\frac{x^4}{4!}\), giving

$$ |R_2(1)|\le\frac{1}{4!} =\frac{1}{24} \approx0.0416666667. $$

Check: The actual error \(0.0403023059\) is positive and less than both bounds. It is quite close to the alternating-series bound, whereas the Lagrange bound is about four times as large. Both are valid guarantees, but they provide different levels of precision.

Worked Comparison: The Logarithm at Three Halves

Worked Example: Comparing Two Bounds for \(\ln(3/2)\)

For \(|x|<1\), the Maclaurin series for \(\ln(1+x)\) is \(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\). At \(x=\frac12\), its terms alternate and decrease in magnitude. Truncate after the cubic term:

$$ P_3\left(\frac12\right) =\frac12-\frac{(1/2)^2}{2}+\frac{(1/2)^3}{3} =\frac12-\frac18+\frac1{24} =\frac{12-3+1}{24} =\frac{10}{24} =\frac5{12} \approx0.4166666667. $$

The function value is \(\ln(3/2)\approx0.4054651081\). Thus

$$ R_3\left(\frac12\right) =0.4054651081-0.4166666667 \approx-0.0112015586, \qquad \left|R_3\left(\frac12\right)\right|\approx0.0112015586. $$

The negative sign agrees with the first omitted term, \(-\frac{x^4}{4}\). Its magnitude gives the alternating-series estimate:

$$ \left|R_3\left(\frac12\right)\right| \le\frac{(1/2)^4}{4} =\frac{1}{64} =0.015625. $$

For a Lagrange estimate, let \(f(x)=\ln(1+x)\). Its fourth derivative is \(f^{(4)}(x)=-\frac{6}{(1+x)^4}\). On \([0,\frac12]\), its absolute value is at most \(6\), so \(M=6\) is valid. Therefore,

$$ \left|R_3\left(\frac12\right)\right| \le\frac{6(1/2)^4}{4!} =\frac{6}{16\cdot24} =\frac{6}{384} =\frac{1}{64} =0.015625. $$

Check: The actual error is negative, and its magnitude \(0.0112015586\) is less than \(0.015625\). In this case the two bounds agree. That agreement is specific to this function, interval, and degree; it does not imply that the two methods always produce the same estimate.

Reading the Gap Correctly

The difference between a bound and the actual error is not a defect in the approximation. A bound answers a conservative question: what error size is guaranteed not to be exceeded under the stated conditions? It may use a maximum derivative that occurs somewhere in the interval but is not representative of the derivative value at the unknown point in the remainder formula. An alternating bound similarly uses the whole first omitted term, even though later terms partly cancel it.

A small actual error does not make a larger valid bound false. Conversely, a bound alone does not tell us the actual error, nor does it promise that the actual error is close to the bound. The examples illustrate all three possibilities: the bounds may be substantially larger than the observed error, a sharper bound may be close to it, and two distinct methods may happen to yield the same bound.

Interpretation: A bound is a certificate, not a prediction of the exact error. Treat \(|R_n(x)|\le B\) as a proven ceiling on the error, not as an equality or a claim that the error is approximately \(B\).

A Common Mistake

For the sine approximation above, a mistaken calculation might set the actual error equal to the first omitted term:

$$ |R_3(1)|\ \text{(incorrectly)}=\frac{1}{5!}=0.0083333333. $$

This confuses the alternating-series upper bound with the actual tail. The correct calculation subtracts the polynomial value from the function value:

$$ |R_3(1)|=|\sin(1)-5/6| \approx|0.8414709848-0.8333333333| \approx0.0081376515. $$

The first omitted term gives a valid upper bound, not generally the exact error. In fact, subsequent alternating terms reduce the magnitude of the tail in this example.

Check Your Understanding

Use the definitions and comparisons above to answer each question. Keep actual errors distinct from estimates.

  1. If \(P_n(x)=0.72\) and \(f(x)=0.70\), what are the signed error and actual absolute error?
  2. For the sine example, why is the alternating-series estimate smaller than the Lagrange estimate?
  3. What conditions on an alternating series allow the first omitted term to bound the absolute tail?
  4. If an actual error is less than a valid bound, does that mean the bound was inaccurate? Explain.
  5. For the logarithm example, what does the negative signed error tell you about the polynomial approximation?

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