How to draw $\sin(x^2)$?

Question

I wonder if it is possible to draw $\sin(x^2)$ by hand, I think it is impossible, but I'm not sure, it would be very useful if anyone can clarify if it is possible or not, and show what is the way to draw it if it is possible.

Thanks

Answer 1

I did this with Python.

import matplotlib.pyplot as plt import math def sinsq(x): return math.sin(x*x) x = [k/100 for k in range(1000)] y = [sinsq(k) for k in x] plt.plot(x, y) plt.show()

Answer 2

it is an even function

$f(x) = \sin x^2\\ f(-x) = f(x)\\ f(0) = 0$

$f(x) > 0$ while $\sqrt \pi <x \sqrt \pi$

$f(x) = 0 $ when $x = \pm \sqrt {n\pi}$

$f(x) = 1$ when $x = \pm \sqrt {\frac {2n+1\pi}{2}}$

$f(x) = -1$ when $x = \pm \sqrt {\frac {2n-1\pi}{2}}$

That give you plenty of points.

For most of the curve, it will resemble a standard sine wave, except the frequency increases as you move away from 0.

So something like:

Perhaps not incredibly precise, but it gives the idea.

Answer 3

It is an even function whose roots occur at $\sqrt{\pi}\approx 1+\frac{3}{4}$, $\sqrt{2\pi}\approx 2+\frac{1}{2}$,$\sqrt{3\pi}\approx 3$,$\sqrt{4\pi}\approx 3+\frac{1}{2}$, $\sqrt{5\pi}\approx 4$. The first stationary point is located at $\sqrt{\frac{\pi}{3}}\approx 1$, the behaviour in a right neighbourhood of the origin is the one of $x^2$ and it is an approximate sine function on the interval between two consecutive roots, so something like this:

Oh, no, the first stationary point is actually located at $\sqrt{\frac{\pi}{2}}$, so something like this: