A Taylor series for the function arctan
The integral If we invert y = arctan(x) to obtain x = tan y, then, by differentiating with respect to y, we find dx/dy = sec2 y = 1 + tan2 y = 1 + x2 . Thus we have (ignoring the constant of integration) y = arctan(x) = dx . 1 + x2 (1)

If we now differentiate y = arctan(x/a) with respect to x, where a is a constant, we have, by the chain rule, y = 1 a 1 a = 2 . 1 + (x/a)2 x + a2 (2)

Thus we obtain the indefinite integral 1 dx = arctan(x/a). x 2 + a2 a

(3)

The Taylor Series By expanding the integrand in (3) as a geometric series 1/(1 − r) = 1 + r + r 2 + . . ., |r| < 1, and then integrating, we can obtain a series to represent the function arctan(x/a). We use the dummy variable t for the integration on [0, x] and we first write x x

arctan(x/a) = a
0

dt 1 = t2 + a 2 a
0

dt 1 + (t/a)2

(4)

Substituting the geometric series with r = −(t/a)2 , we find 1 arctan(x/a) = a x ∞

(−t2 /a2 )n dt =
0 n=0

(−1)n x 2n + 1 a n=0

∞

2n+1

.

(5)

The radius of convergence of this series is the same as that of the original geometric series, namely R = 1, or, in terms of x, |x/a| < 1. The series is a convergent alternating series at the right-hand end point x = a; and it can be shown that sum equals the value of arctan(1) = π/4 (as we might hope). Thus we have the nice (but slowly converging) series for π given by π 1 1 1 = 1 − + − + .... (6) 4 3 5 7 By choosing partial sums of (5) we obtain a sequence of Taylor polynomial approximations to arctan. For example, the sum of three terms yields the Taylor polynomial of degree five given by: x 1 x 3 1 x 5 arctan(x/a) ≈ T (x) = − + . (7) a 3 a 5 a

1