Inverse-Tangent Maclaurin Polynomials from Known Series

From a Geometric Series to Inverse Tangent

The inverse tangent function has a Maclaurin series with alternating signs and only odd powers. That structure makes it possible to build its Maclaurin polynomials directly and, on a useful interval, estimate how far each polynomial is from the function.

The series is closely connected to the geometric series for \(1/(1+u)\). A finite geometric identity gives a way to justify the inverse-tangent expansion and its error, including at the endpoints. Those details matter: a power series’ endpoint behavior must be checked separately from its behavior inside its radius of convergence.

Constructing the Maclaurin Polynomials

A Maclaurin polynomial of degree \(n\) keeps the terms of degree at most \(n\) in a function’s Maclaurin series. For the inverse tangent, the nonzero powers are \(1,3,5,\ldots\); every even-power coefficient is zero. Thus, for an integer \(n\ge 1\),

The largest power in the polynomial is the largest odd integer not exceeding \(n\). Consequently, consecutive degrees sometimes give the same polynomial: \(P_2=P_1\), \(P_4=P_3\), and so on. For degree zero, \(P_0(x)=0\).

Why the Series Represents \(\arctan(x)\)

For \(|t|<1\), the geometric series with ratio \(-t^2\) gives \(1/(1+t^2)=1-t^2+t^4-t^6+\cdots\). To justify both the resulting inverse-tangent series and its endpoints, start with the finite identity below. It holds for every real \(t\), since \(1+t^2\ne0\):

The derivative of \(\arctan(t)\) is \(1/(1+t^2)\), and \(\arctan(0)=0\). Integrate the finite identity from \(0\) to \(x\). Integrating the finite sum requires only the power rule, and yields

This is an exact finite formula: the integral is the remainder after the first \(N+1\) terms. When \(|x|\le1\), its absolute value satisfies

Here the inequality uses \(1+t^2\ge1\), and the last equality follows by integrating the power. Since \(|x|^{2N+3}/(2N+3)\) tends to zero as \(N\) increases for every \(|x|\le1\), the remainder tends to zero. Therefore the finite sums converge to \(\arctan(x)\) throughout the closed interval, endpoints included.

Convergence at the Endpoints

At \(x=1\), the series becomes \(1-\frac13+\frac15-\frac17+\cdots\). Its term magnitudes \(1/(2k+1)\) decrease to zero, so the alternating series test guarantees convergence. The finite-integral argument above also shows that its sum is \(\arctan(1)=\pi/4\).

At \(x=-1\), each odd power changes sign, so the series is \(-1+\frac13-\frac15+\frac17-\cdots\). Its terms also alternate, with magnitudes decreasing to zero. Its sum is \(\arctan(-1)=-\pi/4\). In both cases the series is not absolutely convergent: the absolute-value series \(\sum_{k=0}^{\infty}1/(2k+1)\) diverges, since \(1/(2k+1)\ge1/(2k+2)\) and the harmonic series diverges. Thus the endpoint convergence is conditional.

Using the First Omitted Term

For \(0<|x|\le1\), the magnitudes of consecutive terms decrease. In fact, the ratio of the \((k+1)\)st term magnitude to the \(k\)th is

The ratio calculation is stated only for \(0<|x|\le1\), where its denominator is nonzero. At \(x=0\), the series and every partial sum are zero, so the error is zero. The alternating-series remainder rule now applies for \(0<|x|\le1\): the error is no larger in magnitude than the first omitted term and has the same sign as that term. The finite-integral estimate gives the same magnitude bound.

A Useful Endpoint Approximation

At \(x=1\), a polynomial gives a familiar approximation to \(\pi/4\). For example, \(P_7(1)\) keeps the four terms through degree seven:

The first omitted term is \(1/9\), which is positive. Hence \(\pi/4\) is greater than \(76/105\), and the difference is at most \(1/9\). This bound is valid but not especially tight; including more terms improves the approximation and decreases the first-omitted-term bound.

Common Pitfall: Forgetting the Conditions

A ratio test calculation for term magnitudes can be mishandled at \(x=0\). For example, directly substituting zero into the ratio \[ \frac{|x|^{2k+3}/(2k+3)}{|x|^{2k+1}/(2k+1)} \] produces \(0/0\), which is undefined. It is incorrect to claim that the ratio formula applies at every \(x\) with \(|x|\le1\). The correction is to use the ratio only when \(0<|x|\le1\), as above, and handle \(x=0\) separately: every series term is zero, every partial sum is zero, and the error is zero. Similarly, the sign-of-the-first-omitted-term statement applies when \(x\ne0\); zero error has no positive or negative sign.

Another common error is to substitute \(x=1\) into a formula valid only for \(|x|<1\) and assume convergence follows automatically. Endpoint behavior must be checked. Here the alternating series test proves convergence at each endpoint, while the finite-integral argument confirms that the endpoint sums equal the corresponding inverse-tangent values.

Why These Polynomials Are Useful

The inverse-tangent Maclaurin polynomials connect a familiar function to a geometric-series pattern, while the remainder estimate turns a polynomial approximation into a controlled numerical estimate. The interval restriction is essential: these polynomials can approximate \(\arctan(x)\) through this series for \(|x|\le1\), but the series does not converge when \(|x|>1\). For inputs in the interval, the sign and size of the first omitted term provide a practical check on an approximation’s accuracy.