A Tribute to an Ingenious Mind: Kishlaya Jaiswal

The solution for the first part of the question has been provided by my friend Kish below

The value of $\zeta= f(100)$ evaluates to be $102$ .

We all know that $\arcsin (n) = -i\cdot ln(i\cdot n \pm \sqrt{1-n^{2}}) \\= -i\cdot ln(i\cdot n \pm i\sqrt{m} \\= -i\cdot ln(i(n \pm \sqrt{m}))$

Now we use $e^{ix} = cos(x) + i\cdot sin(x) \\ \quad \text{and} \quad \\= ln(i\cdot x) = ln(x\cdot e^{i\cdot \frac{\pi}{2}}) \\= -i\cdot ( ln(x) + i\cdot \frac{\pi}{2}) \\= \frac{\pi}{2} - ln(x)$

So we get $\\ \rightarrow \frac{\pi}{2} - i\cdot ln(n \pm \sqrt{m})$

So we have $\arcsin (102) = \frac{\pi}{2} - i\cdot ln(102 + \sqrt{101\cdot 103})$ .

Therefore the product evaluates to $1666781.407$ and hence the final answer is $1666781$ .

Okay, so let's begin by defining $f(x)$ . Set $y = 0$ ; this gives $f(x) = x + f(0) = x + 2$ .

This gives us $\zeta = f(100) = 100 + 2 = 102$ . We want to find $sin^{-1}(102)$

Now comes the tricky part. I assume that the reader is familiar with Euler's formula, which is $e^{i\theta} = cos(\theta) + isin(\theta)$ ; from it we can derive the following:

$sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2}$

Let $e^{i*\theta} = y$ . We have: $y - \frac{1}{y} = 204i$ ; solving for y yields:

$y = \frac{204i \pm \sqrt{-41616 + 4}}{2} = \frac{204i \pm 2i\sqrt{10403}}{2}$

$y = 102i \pm i\sqrt{10403}$

Thus, $i\theta = ln(y) = ln (i) + ln (102 \pm \sqrt{10403})$ . I assume that the final answer you seek is the principal value of the expression; in other words, $ln (i) = \frac{i\pi}{2}$ .

Therefore, $\theta = \frac{\pi}{2} - i*ln(102 + \sqrt{10403})$ ; the product $\omega$ equals $\frac{102*10403*\pi}{2} \approx 1666781.4$ ; thus, the greatest integer smaller than $\omega$ is $1666781$ .