Understanding the Taylor Series Expansion of tan(x)

The question asks about the interval of convergence for the Taylor series expansion of the function f(x) = \tan(x). The Taylor series of a function f(x) centered at a point a is given by the formula:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

When the series is centered at a=0, it is called a Maclaurin series, which is a special case of the Taylor series:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

The Taylor series expansion of \tan(x) centered at a=0 exists, but its coefficients involve the derivatives of \tan(x) evaluated at x=0. The first few terms are:

\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \dots

This is a power series. A power series converges for |x-a| < R, where R is the radius of convergence. The interval of convergence is typically (a-R, a+R), possibly including endpoints.

Determining the Interval of Convergence for tan(x) Taylor Series

For a Taylor series centered at a, the radius of convergence R is the distance from a to the nearest point where the function f(x) is not analytic. For real functions, this usually means the nearest singularity or point where the function is not defined or differentiable.

The function f(x) = \tan(x) is defined as \frac{\sin(x)}{\cos(x)}. It is not defined (has singularities) where the denominator \cos(x) is zero. The cosine function is zero at odd multiples of \frac{\pi}{2}.

The points where \cos(x) = 0 are x = \frac{\pi}{2}, -\frac{\pi}{2}, \frac{3\pi}{2}, -\frac{3\pi}{2}, \dots, which can be written as x = \frac{\pi}{2} + n\pi for any integer n.

The Taylor series for \tan(x) is typically centered at x=0. We need to find the distance from the center a=0 to the nearest singularities of \tan(x).

• Distance from 0 to \frac{\pi}{2} is |\frac{\pi}{2} - 0| = \frac{\pi}{2}.
• Distance from 0 to -\frac{\pi}{2} is |-\frac{\pi}{2} - 0| = \frac{\pi}{2}.
• Distance from 0 to \frac{3\pi}{2} is |\frac{3\pi}{2} - 0| = \frac{3\pi}{2}.
• Distance from 0 to -\frac{3\pi}{2} is |-\frac{3\pi}{2} - 0| = \frac{3\pi}{2}.

The smallest distance to a singularity from the center x=0 is \frac{\pi}{2}. This distance is the radius of convergence, R = \frac{\pi}{2}.

The interval of convergence for the Taylor series centered at a=0 is (a-R, a+R), which is (0 - \frac{\pi}{2}, 0 + \frac{\pi}{2}) = (-\frac{\pi}{2}, \frac{\pi}{2}).

This interval is expressed in radians. The options are given in degrees. We need to convert the interval from radians to degrees.

We know that \pi radians is equal to 180 degrees.

• \frac{\pi}{2} radians = \frac{180^\circ}{2} = 90^\circ.
• -\frac{\pi}{2} radians = -\frac{180^\circ}{2} = -90^\circ.

So, the interval of convergence (-\frac{\pi}{2}, \frac{\pi}{2}) in radians corresponds to (-90^\circ, 90^\circ) in degrees.

Looking at the options:

1. (-\pi, \pi) radians = (-180^\circ, 180^\circ) - Incorrect.
2. (0, 90) degrees - Incorrect.
3. (-90, 90) degrees - Correct. This matches (-\frac{\pi}{2}, \frac{\pi}{2}).
4. Infinite field - Incorrect. The series does not converge for all real numbers.

Therefore, the interval of the Taylor series expansion of \tan(x) centered at 0 is (-90^\circ, 90^\circ).

Revision Table: Key Concepts

Additional Information on Taylor Series and Convergence

The Taylor series expansion of \tan(x) centered at x=0 is a fascinating example because the function has poles (vertical asymptotes) at x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \dots. These poles are the singularities that limit the interval of convergence.

The radius of convergence R of a power series \sum c_n (x-a)^n can often be found using the Ratio Test or the Root Test. However, for Taylor series derived from a function f(x) where the function has singularities, the radius of convergence centered at a is precisely the distance to the nearest singularity in the complex plane. For real functions and real variables, we often only consider real singularities, but the concept extends to complex numbers.

In the case of \tan(x) centered at x=0, the nearest singularities are at x = \pm \frac{\pi}{2} on the real axis. The distance from 0 to either of these points is \frac{\pi}{2}. Hence, the radius of convergence is R = \frac{\pi}{2}.

The interval of convergence is (-\frac{\pi}{2}, \frac{\pi}{2}). At the endpoints x = \pm \frac{\pi}{2}, the function \tan(x) is undefined, so the series cannot converge to \tan(x) at these points. Therefore, the interval of convergence is open.

It's worth noting that if you were to center the Taylor series of \tan(x) at a different point, say a=1, the interval of convergence would be centered at 1. The radius of convergence would be the distance from 1 to the nearest singularity, which is |1 - \frac{\pi}{2}| (since \frac{\pi}{2} \approx 1.57 is closer to 1 than -\frac{\pi}{2} or \frac{3\pi}{2}). The interval would then be (1 - |\frac{\pi}{2}-1|, 1 + |\frac{\pi}{2}-1|).