Tangent Series

Approach:

The sum of the tangent series is simply $\tan x$. Substitute $x = \frac{\pi}{4}$ into the series and evaluate the tangent function directly.

Steps:

1. Identify the series: $\sum_{n=1}^{\infty} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}$ is the tangent series, and its sum is $\tan x$.
2. Set the parameter: $x = \frac{\pi}{4}$.
3. First check that it lies inside the interval of convergence: $\frac{\pi}{4} < \frac{\pi}{2}$, so the series converges and may be evaluated.
4. Compute the sum: $S = \tan\left(\frac{\pi}{4}\right) = 1$.

Answer:

The sum is $1$.

Approach:

Substitute $n = 1, 2, 3$ into the general term. The Bernoulli numbers $B_2, B_4, B_6$ are $\frac{1}{6}, -\frac{1}{30}, \frac{1}{42}$ respectively, and each coefficient can then be computed.

Steps:

1. $n = 1$: $B_2 = \frac{1}{6}$, so the coefficient is $\frac{1}{6} \cdot (-4) \cdot (1 - 4) \div 2! = \frac{1}{6} \cdot \frac{12}{2} = 1$.
2. $n = 2$: $B_4 = -\frac{1}{30}$, so the coefficient is $-\frac{1}{30} \cdot 16 \cdot (1 - 16) \div 4! = -\frac{1}{30} \cdot \frac{240}{24} = \frac{1}{3}$.
3. $n = 3$: $B_6 = \frac{1}{42}$, so the coefficient is $\frac{1}{42} \cdot (-64) \cdot (1 - 64) \div 6! = \frac{1}{42} \cdot \frac{4032}{720} = \frac{2}{15}$.
4. Put the powers back: the $n$-th term carries $x^{2n-1}$, so the three terms are $x$, $x^3$ and $x^5$.

Answer:

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$; the first three terms are $x$, $\frac{x^3}{3}$ and $\frac{2x^5}{15}$.

Approach:

Approximating a function value by the first few terms of its series is the most direct application of a power series. The error comes from the truncated higher-order terms, so it can be estimated by comparing against the more accurate value of $\tan 0.5$.

Steps:

1. Substitute $x = 0.5$ into the first two terms: $\tan 0.5 \approx 0.5 + \frac{0.5^3}{3} = 0.5 + \frac{0.125}{3}$.
2. This gives $0.5 + 0.0416667 = 0.5416667$.
3. Compare with the exact value $\tan 0.5 \approx 0.5463025$: the absolute error is about $0.5463025 - 0.5416667 = 0.0046358$, roughly $0.0046$.
4. The relative error is about $\frac{0.0046358}{0.5463025} \approx 0.85\%$, so the two-term approximation is already accurate to about one part in a thousand; adding the third term $\frac{2}{15}x^5 = \frac{2}{15} \times 0.03125 = 0.0041667$ raises the approximation to $0.5458333$ and brings the error below $0.0005$.

Answer:

The approximation is $0.5417$, with an absolute error of about $0.0046$ (relative error about $0.85\%$); the error comes mainly from the truncated $x^5$ and higher-order terms.

Approach:

The radius of convergence of a power series is determined by the nearest singularity. Find the singularities of $\tan x$, then examine what happens at the endpoints.

Steps:

1. $\tan x = \frac{\sin x}{\cos x}$ is undefined wherever $\cos x = 0$, that is, at $x = \frac{\pi}{2} + k\pi$.
2. The two singularities closest to the origin are $x = \pm\frac{\pi}{2}$, so the radius of convergence is $R = \frac{\pi}{2}$ and the interval of convergence is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
3. Endpoints: at $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$ the function $\tan x$ itself has no value, so the sum of the series cannot be taken there; moreover $\tan x$ tends to $+\infty$ from the left of $\frac{\pi}{2}$ and to $-\infty$ from the right, so the singularity is not removable and the power series does not converge at those points.
4. Both endpoints are therefore open, giving the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ rather than a closed one.

Answer:

The interval of convergence is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: the radius is set by the nearest singularities of $\tan x$, namely $x = \pm\frac{\pi}{2}$. At both endpoints $\tan x$ is undefined with a non-removable singularity, so the series does not converge there and the interval is open at both ends.