Recognizing a Maclaurin Series
A Maclaurin series is a Taylor series centered at \(x=0\). Its coefficients are determined by the derivatives of the function at zero. Once you recognize a derivative pattern, you can often write the series directly, then check its interval of convergence.
The series is written in powers of \(x\), beginning with the constant term. For a quick recognition, look for three features: the derivative values at zero, any pattern of zero coefficients, and any alternating signs or factorials. After writing the series, determine where it converges; an endpoint must be checked separately from the interior.
Exponential and Trigonometric Patterns
The exponential function has a particularly simple derivative pattern. Every derivative of \(e^x\) is \(e^x\), so every derivative evaluated at zero is \(1\). Its Maclaurin series is therefore
For \(x\ne 0\), the ratio of the absolute values of consecutive terms is \(\lvert x\rvert/(n+1)\), which tends to zero as \(n\) increases. Thus the series converges absolutely for every real \(x\); at \(x=0\), it is simply \(1\).
Sine and cosine have repeating derivative cycles. For sine the cycle is \(\sin x,\cos x,-\sin x,-\cos x\), and for cosine it is \(\cos x,-\sin x,-\cos x,\sin x\). Evaluating at zero shows that sine has nonzero derivatives only at odd orders, while cosine has nonzero derivatives only at even orders.
In each series, the ratio of successive nonzero term magnitudes tends to zero for every fixed real \(x\). Both series therefore converge for all real \(x\). Their missing powers are not accidental: they reflect the derivative values at zero, or equivalently the odd and even symmetry of sine and cosine.
Worked Example: Reading Coefficients from the Exponential Series
Find the coefficient of \(x^5\) in the Maclaurin series for \(e^x\), and state the interval of convergence.
Since the \(n\)th derivative of \(e^x\) is \(e^x\), we have \(f^{(5)}(0)=e^0=1\). The coefficient of \(x^5\) is therefore
For convergence, the \(n\)th term is \(a_n=x^n/n!\). When \(x\ne0\),
The ratio test gives convergence for every real \(x\). At \(x=0\), the series equals \(1\), so the interval of convergence is \((-\infty,\infty)\). As a check, the displayed series has \(x^5/5!=x^5/120\), matching the derivative calculation.
Why the Coefficients Have These Values
The coefficient rule is not a pattern-matching guess. For a Taylor polynomial centered at zero, the constant coefficient must equal \(f(0)\). Differentiating the polynomial once and evaluating at zero forces the linear coefficient to be \(f'(0)\); differentiating twice forces the quadratic coefficient to be \(f''(0)/2!\). The same reasoning continues at every order.
For example, in the exponential series the \(n\)th derivative at zero is \(1\), so \(c_n=1/n!\). For sine, the derivatives at zero alternate between \(1\), \(0\), \(-1\), and \(0\); that gives the odd powers with alternating signs and no even powers. This explains how derivatives identify the coefficients, while convergence tests address a separate question: for which \(x\) does the resulting infinite series converge?
Logarithm and Inverse-Tangent Series
Two other standard Maclaurin series are especially useful. Their coefficients have reciprocal integer denominators rather than factorial denominators:
The logarithm series converges when \(-1<x\leq 1\). The inverse-tangent series converges when \(-1\leq x\leq 1\). These intervals include different endpoints, so each endpoint must be tested rather than inferred from the behavior inside the interval.
Worked Example: Finding the Logarithm Series Interval
Determine the interval of convergence of the Maclaurin series for \(\ln(1+x)\).
For the \(n\)th term \(a_n=(-1)^{n+1}x^n/n\), the ratio of successive term magnitudes is
The ratio test gives absolute convergence when \(|x|<1\), or \(-1<x<1\). At \(x=1\), the series becomes
This alternating series converges: its positive term magnitudes decrease to zero. Thus \(x=1\) is included, and the series has value \(\ln 2\), in agreement with \(\ln(1+1)\).
At \(x=-1\), the series is
Its terms do approach zero, since their magnitudes are \(1/n\). Nevertheless, the series diverges: it is the negative harmonic series. Therefore \(x=-1\) is excluded, and the interval of convergence is \((-1,1]\).
As a check, both endpoints have been tested: the right endpoint converges and the left endpoint diverges. The resulting interval includes exactly the values supported by those tests and the ratio test.
Worked Example: Checking the Inverse-Tangent Endpoints
Determine the interval of convergence of the series for \(\arctan x\), including its endpoint behavior.
The general term is \(a_n=(-1)^nx^{2n+1}/(2n+1)\). For \(x\ne0\), compare successive term magnitudes:
Thus the series converges absolutely if \(|x|<1\) and diverges if \(|x|>1\). At \(x=1\), it becomes
The term magnitudes decrease to zero, so the alternating series converges. Its value is \(\arctan(1)=\pi/4\). At \(x=-1\), every odd power changes sign, giving
This also converges by the alternating-series test, with value \(\arctan(-1)=-\pi/4\). Hence both endpoints are included and the interval is \([-1,1]\). Substituting \(x=1\) and \(x=-1\) into the function confirms the stated endpoint sums.
A Reliable Recognition Process
When asked to recognize or construct a standard Maclaurin series, separate the coefficient work from the convergence work. First identify the function and its derivative or coefficient pattern. Then test the resulting series, paying special attention to its endpoints.
Check whether the derivatives repeat, alternate, or vanish at regular orders.
Use \(f^{(n)}(0)/n!\), or match the function to a standard expansion.
Use an appropriate convergence test to determine where the series converges for values strictly inside or outside its boundary.
Substitute each boundary value into the series and decide convergence separately.
Common Pitfall: Confusing Terms with a Series
A necessary condition for a series \(\sum a_n\) to converge is that \(a_n\) approaches zero. But that condition alone is not sufficient: terms approaching zero do not guarantee that their infinite sum converges. For instance, the harmonic series diverges even though its terms approach zero. Its negative also diverges for the same reason.
A common incorrect endpoint argument for the logarithm series is: “At \(x=-1\), the terms are \(-1/n\), so they do not approach zero; therefore the series diverges.” The conclusion is right, but the stated reason is false. In fact, \(-1/n\) does approach zero. The correct argument is that the series is the negative harmonic series, which diverges. Checking the actual series, rather than relying on a vague rule about endpoints, prevents this error.
Why These Series Matter
Recognizing standard series gives a compact description of a function near its center and, in these examples, specifies a range of inputs for which the infinite expansion converges. The coefficients also encode derivative information: factorial denominators signal the exponential or trigonometric patterns, while reciprocal denominators signal the logarithmic or inverse-tangent patterns. Knowing both the formula and its interval helps you avoid using a familiar-looking expansion outside the range where it converges.
Check Your Understanding
For each question, identify the coefficient pattern and pay attention to endpoint conditions where relevant.
- What is the coefficient of \(x^6\) in the Maclaurin series for \(e^x\)?
- Why are all even-power coefficients zero in the Maclaurin series for \(\sin x\)?
- Does the Maclaurin series for \(\ln(1+x)\) converge at \(x=-1\)? Give the correct reason.
- What is the interval of convergence of the Maclaurin series for \(\arctan x\)?
- Why does the ratio test alone not settle the endpoint behavior of the logarithm series?