Why the First Omitted Term Can Select a Degree
A Taylor polynomial is useful only if its error is controlled. When a Taylor series at the target has alternating terms whose magnitudes decrease to zero, the remainder after truncation is no larger than the first omitted term. This turns a tolerance requirement into a concrete degree-selection test.
The test is simple to state, but it is only valid when its hypotheses hold for the series tail at the evaluation point. In particular, one must check the signs and sizes of the omitted terms, and confirm that the series represents the function there. The degree selected this way is the smallest degree certified by this alternating-series bound; it need not be the smallest degree whose actual error happens to meet the tolerance.
The Selection Rule and Its Conditions
Suppose the Taylor polynomial through degree \(N\) is \(T_N(x)\), and write the evaluated series as a sum of terms \(a_k(x)\). If the first omitted nonzero term has index \(j\), the relevant quantity is its magnitude \(|a_j(x)|\). For a requested absolute error tolerance \(\varepsilon>0\), the test is:
The strict inequality matters when the goal is to guarantee error below the tolerance. Since the remainder has magnitude at most \(|a_j(x)|\), this strict test guarantees \(|f(x)-T_N(x)|<\varepsilon\). If the first omitted term equals the tolerance, the bound only guarantees error at most the tolerance.
Before using the test, check the following at the target:
- The series represents \(f(x)\), so its tail is the error \(f(x)-T_N(x)\).
- The omitted terms alternate in sign, with positive magnitudes \(b_k\).
- Those magnitudes are nonincreasing from the first omitted term onward.
- The magnitudes tend to zero, ensuring convergence of the alternating tail.
For a power series with zero coefficients, degree and term index can differ. For example, the cosine series has only even powers. Its degree-six Taylor polynomial includes terms through \(x^6\), and the next nonzero term is the \(x^8\) term. The first omitted nonzero term, not a missing zero-coefficient term, supplies the bound.
Why the Bound Holds
Here is the argument behind the selection rule. Consider an alternating tail whose first term has positive sign; changing the overall sign does not affect its magnitude. Write the tail as \(b_1-b_2+b_3-b_4+\cdots\), where \(b_1\ge b_2\ge\cdots\ge0\) and \(b_k\to0\). The alternating-series test ensures that the tail converges. Grouping consecutive terms gives the inequalities below.
Every parenthesis is nonnegative because the magnitudes are nonincreasing. Also, subtracting the tail from its first term leaves another alternating tail of the same form:
Thus \(0\le R\le b_1\). If the original tail begins with a negative term, multiply it by \(-1\) to obtain this case; its absolute value is unchanged. Applying the result with \(b_1\) equal to the first omitted magnitude proves the remainder bound.
Worked Examples
Worked Example: Selecting a Degree for \(\ln(1+x)\)
Approximate \(\ln(1.5)\) using its Taylor series at zero, with absolute error below \(0.001\). Here \(x=0.5\), and the series is
At \(x=0.5\), the magnitude of term \(k\) is \(b_k=0.5^k/k\). Its ratio to the preceding magnitude is
So the magnitudes decrease, and \(b_k\to0\). The alternating series represents \(\ln(1.5)\) at this interior point, so the remainder bound applies. For degree \(N\), the first omitted term has index \(N+1\), with magnitude \(0.5^{N+1}/(N+1)\).
Test \(N=6\):
Degree six does not meet the bound. Test the next degree, \(N=7\):
Thus degree seven is the least degree certified by the test. The polynomial and its value at the target are
The alternating-series remainder bound gives \(|\ln(1.5)-T_7(0.5)|\le1/2048\approx0.00048828<0.001\), as required.
Worked Example: An Endpoint Check for \(\arctan(1)\)
Choose the least Taylor polynomial degree certified to approximate \(\arctan(1)\) with error below \(0.1\). The series is
At the target \(x=1\), this becomes \(1-\frac13+\frac15-\frac17+\cdots\). The terms alternate, their magnitudes \(1/(2k+1)\) decrease to zero, and the alternating series converges. To check that its sum is indeed \(\arctan(1)\), use the finite geometric identity
Integrate from \(0\) to \(1\). The left side integrates to \(\arctan(1)\); the finite sum integrates to \(\sum_{k=0}^{m}(-1)^k/(2k+1)\). The absolute value of the remaining integral is at most
This verifies the series identity at the endpoint. A truncation through index \(m\) has degree \(2m+1\), and its first omitted magnitude is \(1/(2m+3)\). For \(m=3\), the degree is seven and the first omitted magnitude is \(1/9\approx0.11111>0.1\), so the bound is insufficient. For \(m=4\), the degree is nine and
Therefore degree nine is the least certified degree. Its value is
The remainder is at most \(1/11\), which is below \(0.1\). The endpoint check is essential: alternating signs alone would not establish that the series equals the target function or that the bound applies.
Worked Example: Accounting for Zero Coefficients in the Cosine Series
Find the least Taylor degree certified to approximate \(\cos(0.8)\) with error below \(0.0001\). Its Maclaurin series is
This series represents cosine at every real \(x\). At \(x=0.8\), the term magnitudes are \(b_k=0.8^{2k}/(2k)!\). For successive terms,
because the denominator is at least \(2\), while \(0.8^2=0.64\). The magnitudes therefore decrease to zero, and the alternating-series bound applies.
Truncating through \(k=2\) gives degree four. Its first omitted magnitude is the \(k=3\) term:
So degree four is insufficient. Truncating through \(k=3\) gives degree six, and the first omitted nonzero term is \(k=4\), of degree eight:
The least certified Taylor degree is six. The degree-six polynomial gives
Its error is at most \(b_4\approx0.00000416102\), which is below the requested tolerance. The next omitted power is \(x^8\), even though the polynomial is degree six, because the odd-power coefficients in the cosine series are zero.
A Common Mistake: Using the Bound Without Checking the Tail
A first omitted term is not automatically an error bound. For example, consider \(1-2+2-2+\cdots\). Its signs alternate, but the term magnitudes do not decrease to zero, and the series does not converge. The alternating-series remainder bound cannot be used. A small-looking selected term would not repair those failures.
There is a second, subtler mistake: treating the first omitted term as the exact error. In the logarithm example, the first omitted term after degree seven is \(1/2048\). The theorem guarantees that the error is no larger than this number; it does not say the error equals it. The alternating tail contains all later terms as well.
Finally, failure of the test at a shorter degree means only that this particular bound does not certify that degree. It does not prove that the actual error exceeds the tolerance. The selection rule identifies the least truncation certified by the first-omitted-term bound, not necessarily the least truncation that works in fact.
A Reliable Degree-Selection Routine
Identify the actual sequence of evaluated terms and the target function value it sums to.
Verify alternating signs, nonincreasing magnitudes, and convergence to zero from the proposed first omitted term onward.
Compare the first omitted magnitude with the tolerance using a strict inequality when error must be below it.
Check that the preceding truncation fails the same bound, while keeping in mind that this does not establish its actual error is too large.
Check Your Understanding
For each question, distinguish what the alternating-series bound guarantees from what it does not guarantee.
- What conditions must the omitted terms satisfy for the first-omitted-term bound to apply?
- If the first omitted term equals the tolerance, has the strict “below tolerance” requirement been certified?
- For the cosine series, why is the first omitted nonzero term after degree six the term of degree eight?
- At \(x=1\), what checks justify applying the alternating-series bound to the arctangent expansion?
- If a shorter truncation fails the first-omitted-term test, what can and cannot be concluded about its actual error?
