Comparing Lagrange And Alternating-Series Bounds

Why Compare Two Error Bounds?

A Taylor polynomial gives an approximation to a function, but an error bound answers a separate question: how far from the true value could that approximation be? When more than one theorem applies, the bounds need not be equally sharp. Comparing them helps distinguish what each theorem guarantees and what information it uses.

For sine and cosine, there are two useful approaches. Lagrange’s remainder bound controls a derivative throughout the interval between the center and the point of approximation. The alternating-series remainder bound instead uses the sizes and signs of the terms in a convergent alternating series. Both give valid bounds when their respective hypotheses hold, but they need not give the same number.

Definition: If \(P_n(x)\) is a Taylor polynomial of degree \(n\) for \(f\) centered at \(0\), its remainder at \(x\) is \(R_n(x)=f(x)-P_n(x)\). An error bound is a number \(B\) for which \(|R_n(x)|\le B\).

What Each Bound Requires

For Lagrange’s bound, suppose \(f\) has a continuous derivative of order \(n+1\) on the interval between \(0\) and \(x\). If \(|f^{(n+1)}(t)|\le M\) throughout that interval, then

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

The value \(M\) must be a valid upper bound on the whole interval, not merely the derivative’s value at the endpoint. A larger-than-necessary \(M\) still gives a correct estimate, but it may make the result less sharp.

The alternating-series remainder bound applies when the function is represented at the point by a convergent alternating series whose term magnitudes decrease to zero. After keeping a finite number of terms, the error is at most the magnitude of the first omitted term. For sine and cosine at the values used below, their power series converge, alternate in sign for positive \(x\), and have decreasing term magnitudes.

Key distinction: Lagrange’s bound uses a bound on a derivative over an interval. The alternating-series bound uses the first omitted term of a suitable series. A smaller bound is more informative, but neither theorem can be used unless its own conditions hold.

Worked Example: Sine at \(x=0.5\)

Worked Example: Comparing bounds for a sine approximation

Approximate \(\sin(0.5)\) using the degree-five Taylor polynomial centered at \(0\). The polynomial is

P_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}.

First calculate the approximation:

P_5(0.5) =0.5-\frac{0.5^3}{6}+\frac{0.5^5}{120} =0.5-\frac{0.125}{6}+\frac{0.03125}{120} =0.5-0.020833333\ldots+0.000260416\ldots =0.479427083\ldots.

For the Lagrange bound with \(n=5\), \(f^{(6)}(t)=-\sin t\), so \(|f^{(6)}(t)|\le 1\). Thus \(M=1\) is valid on \([0,0.5]\), and

|R_5(0.5)| \le \frac{1(0.5)^6}{6!} =\frac{1/64}{720} =\frac{1}{46080} \approx 0.0000217014.

For the alternating-series bound, the sine series continues with \(-x^7/7!\). Its magnitude at \(x=0.5\) is

\frac{(0.5)^7}{7!} =\frac{1/128}{5040} =\frac{1}{645120} \approx 0.0000015501.

The sine terms decrease in magnitude at this input: each successive ratio is \(x^2/((2k+2)(2k+3))\), which is less than \(1\) when \(x=0.5\). The alternating-series bound is therefore valid. Comparing the two estimates, \(1/46080=14/645120\), so this alternating-series bound is 14 times smaller.

Check: The two bounds apply to the same polynomial and the same input, and both are positive upper bounds for \(|\sin(0.5)-P_5(0.5)|\). The comparison establishes that one estimate is sharper; it does not say that the actual error equals either bound.

Worked Example: Cosine at \(x=0.5\)

Worked Example: Comparing bounds when a coefficient is zero

Use the degree-four Taylor polynomial for cosine:

P_4(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}.

At \(x=0.5\), this gives

P_4(0.5) =1-\frac{0.25}{2}+\frac{0.0625}{24} =1-0.125+0.002604166\ldots =0.877604166\ldots.

For the degree-four Lagrange bound, \(f^{(5)}(t)=-\sin t\), and \(|f^{(5)}(t)|\le1\). Therefore

|R_4(0.5)| \le \frac{1(0.5)^5}{5!} =\frac{1/32}{120} =\frac{1}{3840} \approx 0.0002604167.

The next nonzero term in the cosine series is \(x^6/6!\). The intervening degree-five coefficient is zero, so the alternating series after \(P_4\) starts with this positive term. Its magnitude is

\frac{(0.5)^6}{6!} =\frac{1/64}{720} =\frac{1}{46080} \approx 0.0000217014.

The cosine-series terms decrease in magnitude at \(x=0.5\), so the alternating-series bound applies. Since \(1/3840=12/46080\), this bound is 12 times smaller than the degree-four Lagrange bound.

Check: The degree-four polynomial has no \(x^5\) term, but the degree-four Lagrange formula still uses a fifth derivative and a factor \(|x|^5/5!\). The alternating-series estimate starts from the next nonzero series term, which has degree six. The different powers come from the different information each estimate uses.

A Zero Coefficient Can Also Change the Lagrange Comparison

The cosine example has an additional useful feature. Since its degree-five Taylor coefficient at zero is also zero, \(P_4\) is exactly the same polynomial as the degree-five Taylor polynomial \(P_5\). Applying Lagrange’s theorem with \(n=5\), rather than \(n=4\), gives a different valid estimate for this same polynomial.

Now \(f^{(6)}(t)=-\cos t\), so \(|f^{(6)}(t)|\le1\) on \([0,0.5]\). Hence

|R_5(0.5)| \le \frac{1(0.5)^6}{6!} =\frac{1}{46080}.

This equals the alternating-series bound found above. There is no contradiction: the degree-four and degree-five Taylor polynomials are identical, but the Lagrange theorem is being applied with different values of \(n\), which changes the derivative and power in its bound. When comparing bounds, identify both the polynomial and the order used to obtain each estimate.

Worked Example: Sine at \(x=1\)

Worked Example: Comparing bounds at a larger input

Use the degree-three Taylor polynomial for sine, \(P_3(x)=x-x^3/3!\). At \(x=1\),

P_3(1)=1-\frac{1}{6}=\frac{5}{6}.

For the Lagrange bound, \(f^{(4)}(t)=\sin t\), and \(|f^{(4)}(t)|\le1\) on \([0,1]\). Thus

|R_3(1)| \le \frac{1(1)^4}{4!} =\frac{1}{24} \approx 0.0416667.

The first omitted sine term is \(x^5/5!\), with magnitude

\frac{1^5}{5!} =\frac{1}{120} \approx 0.0083333.

At \(x=1\), successive sine-series term magnitudes decrease: the ratio from the term of degree \(2k+1\) to the next is \(1/((2k+2)(2k+3))<1\). The alternating-series bound applies, and \(1/24=5/120\). It is five times smaller than the Lagrange bound.

Check: The approximation is \(5/6\), and both methods bound the absolute error relative to \(\sin(1)\). The comparison concerns the guaranteed upper bounds, not a claim that the error has a particular value or sign.

Why the Bounds Differ

Lagrange’s theorem gives a remainder involving a derivative evaluated somewhere between the center and the target point. To turn that statement into a numerical estimate, one bounds the derivative over the full interval. This can discard useful information about the pattern of later Taylor terms. The estimate also depends on which degree \(n\) is used in the theorem.

The alternating-series bound uses more specific information: the function’s series has alternating signs and decreasing term magnitudes. Under those conditions, cancellation among the omitted terms ensures that the remainder’s magnitude is no greater than the first omitted term. The bound can be much smaller than a derivative estimate, but it is unavailable if the relevant series does not alternate with decreasing terms.

One way to understand the alternating-series conclusion is to group consecutive terms in the tail. When the terms alternate and decrease, pairs of consecutive terms have a predictable sign and magnitude; successive partial sums move toward the series value. The limiting sum therefore lies between consecutive partial sums, whose separation is the magnitude of the next term. That is why the first omitted term bounds the remainder.

Practical rule: Verify the hypotheses of each method separately, calculate each valid bound for the same polynomial and input, then compare the numerical results. If a Taylor coefficient is zero, check whether the same polynomial can also be treated as a higher-degree Taylor polynomial before deciding which Lagrange estimate to report.

Common Pitfall: Comparing Unlike Remainders

A tempting but incorrect comparison is to use the degree-five Lagrange bound for the sine polynomial in the first example and then treat the degree-six Lagrange bound as if it were a different approximation. The calculations are

P_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}, \qquad P_6(x)=P_5(x)+0x^6=P_5(x).

Thus they approximate with the same polynomial. The degree-six Lagrange bound is available because the sixth Taylor coefficient is zero, but its hypotheses and derivative order must still be checked. It gives \(0.5^7/7!\), not the degree-five bound \(0.5^6/6!\).

Another error is to assume the alternating-series estimate applies just because a Taylor series contains alternating signs. Decreasing term magnitudes and convergence are also required. Conversely, the Lagrange estimate does not require an alternating series, but it does require a suitable derivative and a valid bound for it throughout the interval. Keep these conditions attached to the method they justify.

Check Your Understanding

For each question, identify which remainder estimate is being used and state the condition that makes it valid.

For the degree-five sine polynomial at \(x=0.5\), calculate the first omitted term’s magnitude and compare it with the degree-five Lagrange bound.
Why does the degree-four cosine polynomial at \(x=0.5\) have an alternating-series bound involving a sixth power?
What must \(M\) bound in the Lagrange remainder formula?
What two properties of the series terms are needed to use the alternating-series remainder bound?
How can the same cosine polynomial be viewed as both a degree-four and a degree-five Taylor polynomial?