Maclaurin Series vs Taylor Series

Most textbooks present the Maclaurin series as a separate thing, which makes it look like a different method. It is not. A Maclaurin series is a Taylor series centered at zero. That is the entire difference. The formula for a Taylor series is sum_{n=0}^{∞} f^{(n)}(a)/n! (x-a)^n. Set a=0 and you get sum_{n=0}^{∞} f^{(n)}(0)/n! x^n, which is the Maclaurin series. That is the full relationship between the two. The confusion comes from forgetting that the center matters. Pick a=0 and you have a Maclaurin series. Pick any other a and you have a Taylor series. The rest of the mechanics, finding derivatives, using the ratio test, bounding the error, are identical.

The Taylor series formula is the only one you need to remember. Write it once, then substitute the value of a you want. For a Maclaurin series, a=0. For a Taylor series centered at a=1, plug 1 into the same formula. The sigma notation is the same either way.

What changes is the pattern of the terms. The Maclaurin series for e^x is sum x^n/n!. The Taylor series for e^x centered at a=1 is sum e (x-1)^n/n!, because every derivative of e^x evaluated at 1 is e. The ratio test gives the same radius of convergence, infinity, for both, because the factorial growth rate in the denominator is faster than exponential. The factorial n! grows faster than any power of x, so the series converges for all real x.

The common mistake is to assume the Maclaurin series and the Taylor series are different tools for different functions. They are the same tool with a different center. If you can compute a Taylor series, you can compute a Maclaurin series. If you can compute a Maclaurin series, you can compute a Taylor series centered at any point a. The only extra step is evaluating the derivatives at a instead of at 0.

Not every function has a Maclaurin series. The function ln x has no Maclaurin series because ln 0 is undefined. You cannot take derivatives at zero, they do not exist. The same problem hits sqrt x. The derivative at zero blows up. These functions can still be represented by a Taylor series, but you must center them away from zero.

For ln x, the common choice is a=1. The Taylor series for ln x centered at a=1 is sum (-1)^{n+1} (x-1)^n/n, which converges for 0 < x ≤ 2. That series is derived from the Maclaurin series for ln(1+x) by substituting x-1. The Maclaurin series for ln(1+x) is sum (-1)^{n+1} x^n/n, with interval of convergence (-1,1]. The substitution shifts the interval to (0,2]. For sqrt x, the trick is to rewrite it as (1+(x-1))^{1/2} and use the binomial series, which converges for |x-1| < 1. The binomial series itself is a Maclaurin series for (1+x)^p, converging for |x| < 1.

If you try to force a Maclaurin series onto a function that has no derivatives at zero, every term is undefined. The series does not exist. The failure mode is not a divergent series, it is a series that cannot be written at all. Check the domain before you start. If the function is undefined at zero, pick a center where it is not.

Center Near Your X For Faster Convergence

The practical reason to choose one center over another is how fast the series converges for your x value. A Taylor series centered at a converges fastest near a. The terms shrink because the numerator is (x-a)^n, so the closer x is to a, the fewer terms you need for a given accuracy.

If you need to approximate ln 1.2, skip the Maclaurin series. The Maclaurin series for ln(1+x) works for x near 0, but you are evaluating at x=0.2, which is far from where the series is most accurate. You would need many terms to get three decimal places. A Taylor series centered at a=1 gives (x-1)=0.2, which is small. The terms drop off faster and the alternating series error bound is tighter. You get the same accuracy with fewer terms.

For a function like sin x, the Maclaurin series works well for any x because the factorial growth rate handles even large x, but the number of terms needed still rises with x. For x=10, the Maclaurin series requires about 20 terms for 3-digit accuracy. For x=1, three terms suffice. There is no rule that says you must use a Maclaurin series. If your x is far from zero, shift the center.

The failure case is picking a=0 because it is the default. The series still converges, but you may need an impractical number of terms. For a function like arctan x, the Maclaurin series sums (-1)^n x^{2n+1}/(2n+1) and converges for |x| ≤ 1. At x=1, it converges conditionally to π/4. To get 3-digit accuracy at x=1, you need 1000 terms. A Taylor series centered at a=0.5 would converge much faster for x=1. The ratio test limit L is lower when (x-a) is smaller.

Deriving The Series For Ln X At A=1

The Taylor series formula is sum_{n=0}^{∞} f^{(n)}(a)/n! (x-a)^n. For a worked example, take f(x)=ln x with a=1.

Plug into the formula: term n=0 is f(1)=0. For n≥1, the term is [(-1)^{n-1} (n-1)! / n! ] (x-1)^n = (-1)^{n-1} (x-1)^n / n. The series is sum_{n=1}^{∞} (-1)^{n-1} (x-1)^n / n. That matches the series obtained by substituting into the Maclaurin series for ln(1+x), but derived directly from the Taylor series formula.

Interval Of Convergence And Error Bounds

The interval of convergence is found with the ratio test: lim |a_{n+1}/a_n| = |x-1|, which converges for |x-1| < 1, giving radius 1. The test is inconclusive at the endpoints x-1 = ±1. Check x=2: the series becomes sum (-1)^{n-1} 1/n, converges conditionally. Check x=0: the series becomes sum (-1)^{n-1} (-1)^n / n = sum (-1)^{2n-1}/n = sum -1/n, diverges. The interval is (0,2]. This is consistent with the Maclaurin series for ln(1+x), which converges on (-1,1] after substitution.

The Lagrange remainder for a Taylor polynomial of degree N centered at a=1 is R_N(x)=f^{(N+1)}(c)/(N+1)! (x-1)^{N+1} for some c between x and 1. To bound the error on [1, 1.5], you need the maximum of |f^{(N+1)}| on that interval. For ln x, the (N+1)th derivative is N! / x^{N+1}, which is largest at the left endpoint x=1. The Lagrange remainder bound is then |R_N| ≤ N! / (N+1)! (0.5)^{N+1} = (0.5)^{N+1} / (N+1). For N=5, that bound is (0.5)^6 / 6 ≈ 0.0026, which guarantees 2-digit accuracy. The actual error is usually smaller because the Lagrange remainder uses a worst-case M.

The alternating series error bound applies here because the series for ln x at a=1 alternates and terms decrease.At x=1.5, (0.5)^6/6 ≈ 0.0026, matching the Lagrange bound in this case. The alternating bound is easier to compute but only works for alternating series.

Use The Right Error Bound For Your Series

The Lagrange remainder and the alternating series error bound are the two tools for bounding the error of a truncated series. The Lagrange remainder works for any function with a continuous (N+1)th derivative. The alternating series error bound works only for alternating series whose terms decrease in magnitude and approach zero. Both give a number of terms needed for a given accuracy.

For a Maclaurin series of sin x, the alternating series error bound says the error after N terms is at most the magnitude of the next term. For x=0.5, the Maclaurin polynomial of degree 3 (using x - x^3/6) has error ≤ |x^5/120| = (0.5)^5/120 ≈ 0.00026. That is 3-digit accuracy. The same bound for x=2 gives error ≤ 32/120 ≈ 0.267, which is useless. You need more terms: degree 5 gives error ≤ 2^7/5040 ≈ 0.018, still not 3-digit. Degree 7 gives error ≤ 2^9/362880 ≈ 0.0014, good for 2 digits. The number of terms rises fast with x.

The radius of convergence tells you where the series converges at all, not how many terms you need. For sin x, the radius is infinite, so the series always converges, but the number of terms for 3-digit accuracy at x=10 is about 20. The factorial growth rate ensures eventual convergence, but the Taylor remainder error bound tells you when.

For a series centered at a non-zero a, the same logic applies. The Taylor remainder bound uses M, the maximum of the (N+1)th derivative on the interval between a and x. That M can be much smaller than the value at 0 if you center near x. For ln x near 1, the derivatives are powers of 1/x, which are smallest near x=1. That is why a Taylor series centered at a=1 converges faster for x near 1 than the Maclaurin series for ln(1+x) does for x near 0.

The common failure is using the Lagrange remainder without finding a true M. If you guess M, the bound is meaningless. The correct method is to find the maximum of |f^{(N+1)}| on the closed interval between a and x, then plug it in. For ln x on [1, 1.5], the maximum of the (N+1)th derivative is at the left endpoint c=1, giving M = N! / 1^{N+1} = N!.