Question

For $k = 1, 2, ...,$ let $f_k$ be the number of times

$\sin(\frac{k \pi x}{2})$

attains its maximum value on the interval $x \in [0, 1]$. Compute

$\lim_{k \to \infty} \frac{f_k}{k}$.

Answer 1

1/4

Answer 2

To find the number of times $\sin(\frac{k \pi x}{2})$ attains its maximum value on the interval $x \in [0, 1]$, we first identify the maximum value of the sine function, which is 1.

The sine function attains its maximum value when its argument is of the form $\frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

So, we set the argument equal to this form:

$\frac{k \pi x}{2} = \frac{\pi}{2} + 2n\pi$

Divide by $\pi$:

$\frac{kx}{2} = \frac{1}{2} + 2n$

Multiply by 2:

$kx = 1 + 4n$

Solve for $x$:

$x = \frac{1 + 4n}{k}$

Now, we apply the given interval constraint $x \in [0, 1]$:

$0 \le \frac{1 + 4n}{k} \le 1$

Since $k$ is a positive integer ($k=1, 2, ...$), we can multiply the inequality by $k$ without changing its direction:

$0 \le 1 + 4n \le k$

We need to find the number of integer values of $n$ that satisfy this compound inequality.

From the left side of the inequality:

$1 + 4n \ge 0$

$4n \ge -1$

$n \ge -\frac{1}{4}$

Since $n$ must be an integer, the smallest possible value for $n$ is $0$.

From the right side of the inequality:

$1 + 4n \le k$

$4n \le k - 1$

$n \le \frac{k - 1}{4}$

Since $n$ must be an integer, the largest possible value for $n$ is $\lfloor \frac{k - 1}{4} \rfloor$.

So, the integer values of $n$ for which the maximum is attained are $0, 1, 2, \dots, \lfloor \frac{k - 1}{4} \rfloor$.

The number of such values, $f_k$, is given by:

$f_k = \left( \lfloor \frac{k - 1}{4} \rfloor \right) - 0 + 1 = \lfloor \frac{k - 1}{4} \rfloor + 1$

We need to compute the limit $\lim_{k \to \infty} \frac{f_k}{k}$.

$\lim_{k \to \infty} \frac{\lfloor \frac{k - 1}{4} \rfloor + 1}{k}$

We use the property of the floor function: for any real number $y$, $y-1 < \lfloor y \rfloor \le y$.

Let $y = \frac{k-1}{4}$. Then:

$\frac{k-1}{4} - 1 < \lfloor \frac{k-1}{4} \rfloor \le \frac{k-1}{4}$

Adding 1 to all parts of the inequality:

$\frac{k-1}{4} - 1 + 1 < \lfloor \frac{k-1}{4} \rfloor + 1 \le \frac{k-1}{4} + 1$

$\frac{k-1}{4} < f_k \le \frac{k-1+4}{4}$

$\frac{k-1}{4} < f_k \le \frac{k+3}{4}$

Now, divide all parts of the inequality by $k$ (since $k > 0$):

$\frac{k-1}{4k} < \frac{f_k}{k} \le \frac{k+3}{4k}$

Finally, take the limit as $k \to \infty$:

$\lim_{k \to \infty} \frac{k-1}{4k} = \lim_{k \to \infty} \frac{1 - \frac{1}{k}}{4} = \frac{1 - 0}{4} = \frac{1}{4}$

$\lim_{k \to \infty} \frac{k+3}{4k} = \lim_{k \to \infty} \frac{1 + \frac{3}{k}}{4} = \frac{1 + 0}{4} = \frac{1}{4}$

By the Squeeze Theorem, since both the lower and upper bounds approach $\frac{1}{4}$, the limit of $\frac{f_k}{k}$ must also be $\frac{1}{4}$.