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Remainders and Error Estimates · Tutorial 872 of 998

Common Mistakes In Taylor Error Estimates

Use the Lagrange remainder carefully, check its hypotheses, and distinguish a guaranteed error bound from the error itself.

Advanced 10 min read

What You'll Learn

  • Write the Taylor polynomial and Lagrange remainder with the correct derivative order and factorial.
  • Identify the full interval on which a derivative must be bounded.
  • Diagnose why a derivative value at the center alone may not give a valid bound.
  • Distinguish an actual approximation error from a guaranteed upper bound.
  • Check error estimates with exact formulas and numerical values.

Why Error Estimates Go Wrong

Taylor polynomials can approximate a function accurately, but an error estimate is useful only if it is applied correctly. A calculation may look plausible while using the wrong derivative, evaluating that derivative at only one point, or confusing a guaranteed maximum error with the error that actually occurs.

This tutorial focuses on those common mistakes. The central tool is the Lagrange form of the Taylor remainder. Its derivative order, factorial, and interval condition work together; changing or omitting one of them can invalidate a conclusion. A bound also says only that the error is no larger than a stated quantity. It does not usually tell us the exact error.

The Taylor Polynomial and Its Remainder

Let \(f\) have derivatives through order \(n\) at a center \(a\). Its degree-\(n\) Taylor polynomial about \(a\) is

$$ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$

The actual remainder at \(x\) is the signed difference \(R_n(x)=f(x)-P_n(x)\). The absolute error is \(|R_n(x)|\). These are related, but they are not interchangeable: the signed remainder can be positive or negative, while the absolute error is nonnegative.

Theorem: Suppose \(f\) has continuous derivatives through order \(n+1\) on the closed interval with endpoints \(a\) and \(x\). Then, for some point \(\xi\) between \(a\) and \(x\), $$ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. $$ Consequently, if \(\left|f^{(n+1)}(t)\right|\le M\) throughout that interval, then $$ |R_n(x)|\le \frac{M|x-a|^{n+1}}{(n+1)!}. $$

The condition is about a bound over the whole interval between the center and the evaluation point, not just about the derivative at \(a\). For \(x=a\), the remainder is zero directly. When \(x\ne a\), the theorem identifies a point \(\xi\) in the interval; it does not generally identify \(\xi\) explicitly.

Why the Lagrange Formula Works

A short proof explains where the derivative order and factorial come from. Let \(x\ne a\), and choose \(K\) so that \(K(x-a)^{n+1}=R_n(x)\). Define an auxiliary function by subtracting both the Taylor polynomial and a term that matches the remainder at \(x\):

$$ g(t)=f(t)-P_n(t)-K(t-a)^{n+1}. $$

At \(t=a\), the Taylor polynomial matches \(f\) and its derivatives through order \(n\). Thus \(g(a)=g'(a)=\cdots=g^{(n)}(a)=0\). By the choice of \(K\), we also have \(g(x)=0\). Repeated applications of Rolle’s theorem give a point \(\xi\) between \(a\) and \(x\) with \(g^{(n+1)}(\xi)=0\).

The \((n+1)\)st derivative of \(P_n\) is zero, since its degree is \(n\), and the \((n+1)\)st derivative of \(K(t-a)^{n+1}\) is \(K(n+1)!\). Therefore

$$ 0=g^{(n+1)}(\xi)=f^{(n+1)}(\xi)-K(n+1)!, \qquad K=\frac{f^{(n+1)}(\xi)}{(n+1)!}. $$

Substituting this value of \(K\) into \(R_n(x)=K(x-a)^{n+1}\) gives the stated remainder formula. Taking absolute values and using the bound \(M\) gives the error inequality. The factorial is therefore part of the exact formula, not an optional adjustment.

Worked Example: Using the Correct Derivative Order

Worked Example: Using the Correct Derivative Order

Approximate \(\sin(0.5)\) by its degree-2 Taylor polynomial about \(0\), and estimate the absolute error.

The needed derivatives at \(0\) are \(f(0)=0\), \(f'(0)=1\), and \(f''(0)=0\). Hence

$$ P_2(x)=0+1x+\frac{0}{2!}x^2=x, \qquad P_2(0.5)=0.5. $$

For a degree-2 polynomial, the Lagrange remainder uses the third derivative, not the second. Here \(f'''(t)=-\cos t\), so \(|f'''(t)|\le 1\) on \([0,0.5]\). Thus \(M=1\), and

$$ |R_2(0.5)|\le \frac{1(0.5)^3}{3!} =\frac{0.125}{6} =\frac{1}{48} \approx 0.0208333. $$

The actual error is \(\sin(0.5)-0.5\), which is negative because \(\sin(0.5)\approx0.4794255\). Its absolute value is approximately \(0.0205745\), which is less than \(1/48\). The estimate checks.

A wrong derivative order can produce a false bound. If one uses the second derivative instead, then \(|f''(t)|=|\sin t|\le \sin(0.5)\) on this interval. Inserting that value in the remainder formula as though it were the required derivative gives \(\sin(0.5)(0.5)^3/6\approx0.009988\), smaller than the actual error \(0.0205745\). The resulting claim fails. The degree \(n\) determines the required derivative order: it is always \(n+1\).

There is a separate factorial issue. If the \(3!\) is simply omitted here, the computed quantity is \(1(0.5)^3=0.125\), rather than \(1/48\). That larger number still happens to be an upper bound in this example, but it is not the Lagrange estimate and is less informative. Omitting the factorial does not automatically make a bound false; it does mean the stated formula has not been applied correctly.

Worked Example: A Derivative Bound Must Cover the Interval

Worked Example: A Derivative Bound Must Cover the Interval

Use a degree-1 Taylor polynomial for \(f(x)=1/(1-x)\), centered at \(a=0\), to approximate \(f(0.5)\). Compare a center-only calculation with a valid Lagrange bound.

At the center, \(f(0)=1\) and \(f'(x)=1/(1-x)^2\), so \(f'(0)=1\). Therefore \(P_1(x)=1+x\), and \(P_1(0.5)=1.5\). The exact value is \(f(0.5)=2\), so the actual absolute error is \(|2-1.5|=0.5\).

The required derivative is the second derivative: \(f''(x)=2/(1-x)^3\). A tempting but invalid calculation is to use only its center value \(f''(0)=2\) as \(M\). That gives

$$ \frac{2(0.5)^2}{2!}=0.25, $$

which is less than the actual error \(0.5\). The problem is that \(2\) is not an upper bound for \(|f''|\) throughout \([0,0.5]\).

On this interval, \(f''(x)=2/(1-x)^3\) is increasing, so its maximum is \(f''(0.5)=2/(0.5)^3=16\). We can therefore take \(M=16\). The valid estimate is

$$ |R_1(0.5)|\le \frac{16(0.5)^2}{2!} =\frac{16(0.25)}{2} =2. $$

The actual error \(0.5\) is no greater than \(2\), as required. The bound is not tight, but it is valid. In general, if a derivative reaches its largest absolute value away from the center, evaluating it only at the center can underestimate the error.

Worked Example: A Bound Is Not the Actual Error

Worked Example: A Bound Is Not the Actual Error

Use the degree-2 Taylor polynomial for \(e^x\) about \(0\) to approximate \(e^{0.5}\). Find a valid error bound and compare it with the actual error.

Since \(f(0)=f'(0)=f''(0)=1\), the polynomial is

$$ P_2(x)=1+x+\frac{x^2}{2}, \qquad P_2(0.5)=1+0.5+\frac{0.25}{2}=1.625=\frac{13}{8}. $$

The third derivative is \(f'''(t)=e^t\). On \([0,0.5]\), it is increasing, so \(M=e^{0.5}\). The Lagrange estimate is

$$ |R_2(0.5)|\le \frac{e^{0.5}(0.5)^3}{3!} =\frac{e^{0.5}}{48} \approx 0.03435. $$

Using \(e^{0.5}\approx1.6487213\), the actual absolute error is

$$ \left|e^{0.5}-\frac{13}{8}\right| \approx |1.6487213-1.625| \approx 0.0237213. $$

This is less than \(0.03435\), so the bound checks. It is not equal to the bound. The theorem guarantees that the error is at most the stated quantity; it does not claim that the error reaches it. Treating the bound as an exact error would overstate the error in this example.

Four Checks Before Trusting an Estimate

When an error estimate seems surprising, check its structure before trusting its numerical value:

  • Match the degree to the derivative: a degree-\(n\) Taylor polynomial requires a bound on the \((n+1)\)st derivative.
  • Keep the factorial: the denominator is \((n+1)!\), paired with the power \(|x-a|^{n+1}\).
  • Bound over the interval: the chosen \(M\) must satisfy \(|f^{(n+1)}(t)|\le M\) at every point between \(a\) and \(x\).
  • Read the inequality accurately: it gives \(|R_n(x)|\le\) a quantity, not an equation for the actual error.

These checks do not replace verifying the theorem’s differentiability conditions. The function must have the required derivatives throughout the interval, and the proposed derivative bound must genuinely hold there. A numerical value at one point, even at the center, is not by itself evidence of a bound over an interval.

Why the Distinction Matters

An invalid estimate can make an approximation appear more reliable than it is. The center-only calculation for \(1/(1-x)\), for instance, claimed an error no greater than \(0.25\) when the actual error was \(0.5\). Conversely, a valid but loose estimate can be much larger than the actual error. That does not make it false; it means only that the guarantee is conservative.

Keeping the formula’s parts distinct makes diagnosis straightforward: the polynomial degree fixes the derivative order, the theorem supplies the factorial, and the interval determines where the derivative must be bounded. The actual error can then be compared with the resulting bound without mistaking one for the other.

Check Your Understanding

For each question, explain the relevant condition as well as the calculation.

  1. For a degree-4 Taylor polynomial, which derivative order appears in the Lagrange remainder, and which factorial belongs in its denominator?
  2. Why is the value \(\left|f^{(n+1)}(a)\right|\) alone generally insufficient to use as \(M\)?
  3. A student uses a number \(M\) that bounds the derivative only at the evaluation point. What must be checked before that number can serve as a Lagrange bound?
  4. If an error estimate gives \(|R_n(x)|\le 0.01\), does it follow that the actual absolute error is \(0.01\)? Explain.
  5. In the sine example, why did using the second derivative give a quantity smaller than the actual error?

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