Why Expand About an Arbitrary Center?
A Taylor series centered at zero describes a function near zero. But an approximation may be most useful near a different point: perhaps the value of interest is close to \(a=1\), \(a=2\), or another real number. To build such a series for \(\arctan x\), write \(x=a+h\), where \(h\) measures the displacement from the center.
The derivative of \(\arctan x\) is especially suitable for this construction:
We will expand this derivative in powers of \(h\), then integrate the expansion and use the value \(\arctan a\) to set the constant term. The complex zeros of \(1+x^2\), at \(i\) and \(-i\), determine how far the resulting Taylor series can extend.
Expand the Derivative Around the Center
Set \(h=x-a\). The rational derivative can be separated into two terms:
For a nonzero complex number \(b\), the geometric series identity gives \(\frac{1}{b+h}=\sum_{n=0}^{\infty}\frac{(-1)^n h^n}{b^{n+1}}\) when \(|h|<|b|\). Apply it with \(b=a-i\) and \(b=a+i\). Since \(a\) is real, both distances are \(\sqrt{a^2+1}\). Thus both geometric expansions converge when \[ |h|<R,\qquad R=\sqrt{a^2+1}. \]
Subtracting the two series gives the derivative expansion. Its coefficients are real, even though the calculation uses complex numbers:
The two powers inside the parentheses are complex conjugates. Their difference is purely imaginary, so division by \(2i\) gives a real coefficient. More explicitly, if \(a+i=R e^{i\theta}\), with \(0<\theta<\pi\), then \(a-i=R e^{-i\theta}\), and
Integrate to Obtain the Arctangent Series
A power series can be integrated term by term inside its radius of convergence. In particular, on any closed interval \(|h|\le r<R\), the derivative series converges uniformly, so its integral from \(0\) to \(h\) is the series of the integrated terms. Since \(\arctan(a+0)=\arctan a\), integration yields:
Equivalently, using \(\theta=\arg(a+i)\), the coefficient of \(h^{n+1}\) is \(\frac{(-1)^n\sin((n+1)\theta)}{(n+1)R^{n+1}}\). The constant term \(\arctan a\) is essential: integrating a derivative series determines an antiderivative only up to a constant, and the value at the center fixes that constant.
Why the Radius Is Exactly This Large
The singularities of the complex rational function \(1/(1+z^2)\) occur at \(z=i\) and \(z=-i\). From a real center \(a\), both are the same distance away:
The two geometric expansions therefore provide convergence inside the disk of that radius. This is also the exact radius, not merely a lower bound. If the Taylor series for \(\arctan\) had a larger radius, its termwise derivative would also converge on a larger disk and define an analytic function there. Within the original disk, that derivative equals \(1/(1+z^2)\). It cannot extend analytically through either pole: near \(z=i\) or \(z=-i\), the rational function is unbounded. Hence the derivative series, and therefore the arctangent Taylor series, cannot have a radius larger than the distance to those poles.
This reasoning concerns the radius in the complex plane. Along the real axis, it guarantees convergence whenever \(|x-a|<\sqrt{a^2+1}\). The distance to a complex singularity, rather than the distance to a real point where \(\arctan x\) fails to exist, controls the radius.
Worked Examples
Worked Example: The Maclaurin Series
Take \(a=0\), so \(h=x\). The derivative has the geometric expansion \(\frac{1}{1+x^2}=\sum_{k=0}^{\infty}(-1)^k x^{2k}\) for \(|x|<1\). Integrating each term from \(0\) to \(x\), and using \(\arctan 0=0\), gives:
Check the result by differentiating the series inside its radius: \[ \frac{d}{dx}\left(x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots\right) =1-x^2+x^4-x^6+\cdots =\frac{1}{1+x^2}. \] The series also has value \(0\) at \(x=0\), so it has the correct derivative and center value. The singularities \(i\) and \(-i\) are each distance \(1\) from zero, confirming the radius \(1\).
Worked Example: Expand About \(a=1\)
Put \(h=x-1\). The denominator of the derivative is \(1+(1+h)^2=2+2h+h^2\). Write its reciprocal as \(\sum_{n=0}^{\infty}c_nh^n\). Multiplying by the denominator and matching coefficients with \(1\) gives \[ 2c_0=1,\qquad 2c_1+2c_0=0,\qquad 2c_2+2c_1+c_0=0,\qquad 2c_3+2c_2+c_1=0. \] Consequently, \[ c_0=\frac12,\qquad c_1=-\frac12,\qquad c_2=\frac14,\qquad c_3=0. \] Continuing one more step, \(2c_4+2c_3+c_2=0\), so \(c_4=-\frac18\). Integrating and using \(\arctan 1=\frac{\pi}{4}\) gives:
For a check, differentiating the displayed terms gives \(\frac12-\frac12h+\frac14h^2-\frac18h^4+\cdots\); the missing \(h^3\) coefficient is zero. These are exactly the coefficients of \(1/(2+2h+h^2)\) obtained by multiplication. At \(h=0\), the series gives \(\frac{\pi}{4}=\arctan 1\). The singularities are at \(x=\pm i\), each at distance \(\sqrt{2}\) from \(1\), so the radius is \(\sqrt{2}\).
Worked Example: Expand About \(a=2\)
Now let \(h=x-2\). The derivative denominator is \[ 1+(2+h)^2=5+4h+h^2. \] If its reciprocal is \(\sum_{n=0}^{\infty}c_nh^n\), coefficient matching gives \[ 5c_0=1,\qquad 5c_1+4c_0=0,\qquad 5c_2+4c_1+c_0=0,\qquad 5c_3+4c_2+c_1=0. \] Solving in order, \[ c_0=\frac15,\qquad c_1=-\frac{4}{25},\qquad c_2=-\frac{4(-4/25)+1/5}{5}=\frac{11}{125},\qquad c_3=-\frac{4(11/125)-4/25}{5}=-\frac{24}{625}. \] Integrate the derivative series and set the value at \(h=0\):
Differentiating the nonconstant terms gives \(\frac15-\frac{4h}{25}+\frac{11h^2}{125}-\frac{24h^3}{625}+\cdots\), which reproduces the coefficients of \(1/(5+4h+h^2)\). At \(h=0\), the expansion gives \(\arctan 2\), as required. The two complex singularities are each distance \(\sqrt{2^2+1^2}=\sqrt5\) from the center, so the radius is \(\sqrt5\).
A Common Geometric-Series Pitfall
The familiar expansion \[ \frac{1}{1+x^2}=1-x^2+x^4-\cdots \] is valid only when \(|x|<1\). It is centered at zero, and it does not directly give a Taylor expansion about an arbitrary \(a\). For example, substituting \(x=1+h\) into that series is not justified near \(h=0\): at the center \(x=1\), the required strict inequality \(|x|<1\) already fails.
A different direct geometric rewrite can also mislead. At \(a=1\), one may write \[ \frac{1}{2+2h+h^2} = \frac12\frac{1}{1+h+h^2/2}. \] The geometric expansion of the last factor requires \(|h+h^2/2|<1\). That condition describes where this particular geometric representation converges; it does not by itself identify the full disk of convergence of the Taylor series. The correct radius comes from the nearest singularities, at distance \(\sqrt2\), and the complex partial-fraction expansions establish convergence throughout the disk \(|h|<\sqrt2\).
Check Your Understanding
Use the derivative expansion and the distances to the complex singularities to answer these questions.
- For a real center \(a\), what is the distance from \(a\) to each of the singularities \(i\) and \(-i\)?
- Why must the constant term in the Taylor series about \(a\) be \(\arctan a\)?
- Write the first three nonconstant terms of the series about \(a=0\), and check them by differentiation.
- For an expansion about \(a=3\), what is the radius of convergence, and which complex points determine it?
- Why does substituting \(x=a+h\) into the Maclaurin series not automatically produce the Taylor series centered at \(a\)?
