Understanding the Taylor Series Error Bound

You have the first five terms of the Maclaurin series for e^x, and at x=1 your calculator shows 2.71667. The actual e is 2.71828. The difference is 0.00161. Is that close enough? For a physics lab, yes. For a financial model compounding every nanosecond, no. What you need is not the error itself but a bound on the error, a guarantee that the true value lies within a range you can state. That guarantee is the Taylor series error bound, and it can be computed for any truncated series.

When you cut off a power series after N terms, the difference between the true function f(x) and the partial sum S_N(x) is the remainder R_N(x). OpenStax 'Calculus Volume 2', Section 6.3, calls it the Taylor remainder. The formula is exact: R_N(x) = f(x) - S_N(x). You cannot compute it exactly without knowing f(x), but you can bound it. That bound is what you need for AP Calculus BC (Unit 10, Topic 10.12) and for any practical approximation where safety matters.

The Distinction Between Remainder and Error

The remainder is the real difference. The error bound is a number you can prove is larger than |R_N(x)|. A tight bound means your approximation is reliable with few terms. A loose bound may still be useful, it tells you the worst case. Never confuse the two. The calculator on many sites shows a numeric error, but that is just the difference between its high-precision value and the partial sum. It is not a bound. The bound is what you compute with the Lagrange or alternating series formulas.

For a Taylor polynomial of degree N centered at a, the Lagrange error bound is |R_N(x)| ≤ M |x-a|^(N+1) / (N+1)!, where M is the maximum value of |f^(N+1)(t)| on the closed interval between a and x. OpenStax Section 6.3 gives this as Taylor's theorem with remainder. The College Board (Topic 10.12) requires you to apply it.

Worked Example 1: Approximating e^x at x=0.5 with a 3rd-degree Maclaurin Polynomial

The Maclaurin series for e^x is Σ x^n/n!. The 3rd-degree polynomial is P_3(x) = 1 + x + x^2/2 + x^3/6. At x=0.5, P_3(0.5) = 1 + 0.5 + 0.125 + 0.0208333 = 1.6458333. The actual e^0.5 is about 1.648721. The remainder R_3(0.5) is approximately 0.002888. To bound it, use the 4th derivative f^(4)(x) = e^x. On [0, 0.5], e^x ≤ e^0.5 ≈ 1.64872. So M = 1.64872. Then |R_3(0.5)| ≤ 1.64872 * (0.5)^4 / 24 = 1.64872 * 0.0625 / 24 = 0.004295. The true error 0.002888 is less than 0.004295, as required.

Worked Example 2: Approximating ln(1+x) at x=0.3 with a 2nd-degree Taylor Polynomial Centered at 0

The Maclaurin series for ln(1+x) is x - x^2/2 + x^3/3 - ... The 2nd-degree polynomial is P_2(x) = x - x^2/2. At x=0.3, P_2(0.3) = 0.3 - 0.045 = 0.255. The actual ln(1.3) is about 0.262364. The remainder R_2(0.3) is about 0.007364. For the bound, the 3rd derivative f^(3)(x) = 2/(1+x)^3. On [0, 0.3], this is decreasing, so its maximum is at t=0: f^(3)(0) = 2. So M = 2. Then |R_2(0.3)| ≤ 2 * (0.3)^3 / 6 = 2 * 0.027 / 6 = 0.009. The true error 0.007364 is within that bound. Notice the bound is looser than the actual error because M is the maximum, not the value at some specific point.

When the series you truncate is alternating, terms alternate in sign, a simpler bound exists. The alternating series estimation theorem (OpenStax Section 5.5, College Board Topic 10.6) states that if the terms satisfy the alternating series test conditions (decrease in magnitude to zero), then |R_N| ≤ |a_(N+1)|, the absolute value of the first omitted term. No Lagrange-style derivative needed.

Worked Example 3: Approximating sin(1) with the First 3 Nonzero Terms of Its Maclaurin Series

The Maclaurin series for sin x is x - x^3/6 + x^5/120 - x^7/5040 + ... For sin(1), the first three nonzero terms (N=2, meaning up to x^5) give S_2(1) = 1 - 1/6 + 1/120 = 1 - 0.166667 + 0.008333 = 0.841666. The actual sin(1) is about 0.841471. The true error is about 0.000195. For a series like the alternating harmonic series for ln(2), the bound is also tight but the convergence is slow.

You set a tolerance, say 0.001, and need the smallest N such that the error bound is below that tolerance. For a non-alternating series, use the Lagrange bound. This often requires trial and error because the factorial grows fast. The ratio test from OpenStax Section 5.6 tells you the radius of convergence, but not the number of terms for accuracy, that is a separate step.

When the Series Is Not Alternating

For a series like the geometric series for x=0.5, the bound decreases as |x|/(1-x) = 1, so the bound does not shrink with N, the geometric series converges but slowly near the edge of its interval. For x=0.9, you need a much larger N for the same bound.

Many sites that evaluate power series display columns labeled 'Series Sum', 'Actual Value', 'Absolute Error', and 'Relative Error'. The absolute error is the difference between the calculator's high-precision evaluation and your partial sum. This is the actual remainder, not a bound. The typical failure: the calculator does not compute the bound at all, it shows only the numeric error. When you see an absolute error of 0.002, ask yourself: is that the true remainder or just the difference between two approximations? Only a bound from the Lagrange or alternating series theorem gives you a guarantee.

What the Calculator Cannot Do

A typical online power series tool (like the one on powerseriescalculator.com) evaluates preset series at a given x. It does not derive the Lagrange bound for an arbitrary function. It does not compute the interval of convergence, it only evaluates at one point. It does not tell you how many terms you need for a given tolerance. You must do that yourself using the formulas above. The tool is useful for checking your partial sum against a high-precision value, but it is not a substitute for error analysis. Apply the alternating series bound only to series that are alternating; the exponential series is not. OpenStax Section 6.1 and the College Board (Topic 10.11) require this check. Check your work against a simple case like e^0, where the bound should collapse to zero.

The single most practical thing to do next: pick a function and a tolerance, compute the Lagrange bound for N=3, N=4, N=5, and compare with the actual error. That loop, compute bound, check against actual, increase N, recompute, is how you gain confidence that you are applying the theorem correctly.