 Abdiel Mays

2022-11-08

I want to know why the equation ${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$ gives an approximate square.
Background
I was just playing around with functions and I wanted to see if $y=|\mathrm{sin}\left(\frac{\pi x}{2}\right)|$ (radians) would give a semicircle for the interval [0,2] as the distance of (1,0) is the same from (0,0), (2,0) and (1,1), all of which will lie on the curve. The equation of a unit semicircle with its centre at (1,0) is $y=\sqrt{2x-{x}^{2}}$
I know that the curves of both the equations don't resemble each other much but I still thought of approximating the sine function using this because I thought that it could still be combined with another approximation to make a better approximation. Anyway, I did it and for , the value of sinϕ can to be approximately $\frac{2}{\pi }\sqrt{\pi x-{x}^{2}}$. It looked like a semi-ellipse and so I verified it to find that it was a semi-ellipse. I thought of using this to derive the equation for an ellipse with it's centre at the origin and the value of $a$ and b being $\frac{\pi }{2}$ and 1 respectively.
The equation came out to be : ${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$
Finally, I thought of playing with this equation and changed the exponent of x. I observed that as I increased the power, keeping it even, the figure got closer and closer to a square.
${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$ gave a good approximation of a square. For the exponent of x being some power of 10 greater than 1012, a part of the curve began to disappear. cenjene9gw

First, let's determine the possible values for x.
Since ${y}^{2}$ is non-negative, we have:
$1\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}\phantom{\rule{thickmathspace}{0ex}}\ge \phantom{\rule{thickmathspace}{0ex}}0$
${x}^{{10}^{12}}\phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thickmathspace}{0ex}}\frac{{\pi }^{2}}{4}$
$-{\left(\frac{{\pi }^{2}}{4}\right)}^{{10}^{-12}}\phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thickmathspace}{0ex}}{\left(\frac{{\pi }^{2}}{4}\right)}^{{10}^{-12}}$
$-1.0000000000009031654105793\dots \phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thickmathspace}{0ex}}1.0000000000009031654105793\dots$
For the decimal approximation used above, see this WolframAlpha computation.
Note that for $x=±{\left(\frac{{\pi }^{2}}{4}\right)}^{{10}^{-12}}\stackrel{\text{def}}{=}\phantom{\rule{thickmathspace}{0ex}}±\beta ,$ we have ${y}^{2}=0,$ and hence y=0.
When $x=±\phantom{\rule{thinmathspace}{0ex}}0.999999,$ we find that $\phantom{\rule{thickmathspace}{0ex}}{y}^{2}\approx 1\phantom{\rule{thickmathspace}{0ex}}–\phantom{\rule{thickmathspace}{0ex}}{10}^{-434,000}\phantom{\rule{thickmathspace}{0ex}}$ and $\phantom{\rule{thickmathspace}{0ex}}y\approx ±\left(1\phantom{\rule{thickmathspace}{0ex}}–\phantom{\rule{thickmathspace}{0ex}}{10}^{-217,000}\right)$. The table below shows the result of several similar calculations.
$\begin{array}{|ccc|}\hline x& {y}^{2}& y\\ & & \\ 0& 1& ±\phantom{\rule{thinmathspace}{0ex}}1\\ ±\phantom{\rule{thinmathspace}{0ex}}0.9& 1-{10}^{-45,700,000,000}& ±\left(1-{10}^{-22,900,000,000}\right)\\ ±\left(1-{10}^{-6}\right)\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}±\phantom{\rule{thinmathspace}{0ex}}0.999999& 1-{10}^{-434,000}& ±\left(1-{10}^{-217,000}\right)\\ ±\left(1-{10}^{-10}\right)\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}±\phantom{\rule{thinmathspace}{0ex}}0.9999999999& 1\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}2.5×{10}^{-44}& ±\left(1\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}1.2×{10}^{-22}\right)\\ ±\left(1-{10}^{-12}\right)& 0.8509\dots & ±\phantom{\rule{thinmathspace}{0ex}}0.9224\dots \\ ±\left(1-{10}^{-15}\right)& 0.5951\dots & ±\phantom{\rule{thinmathspace}{0ex}}0.7714\dots \\ ±\phantom{\rule{thinmathspace}{0ex}}1& 0.5947\dots & ±\phantom{\rule{thinmathspace}{0ex}}0.7711\dots \\ ±\phantom{\rule{thinmathspace}{0ex}}1.000000000000903& 0.000165\dots & ±\phantom{\rule{thinmathspace}{0ex}}0.012860\dots \\ ±\phantom{\rule{thinmathspace}{0ex}}\beta & 0& 0\\ \hline\end{array}$
Thus, using the fact that ${y}^{2}$ is a decreasing function of $|x|$ for $-\beta it follows that the points (x,y) on the graph form two nearly horizonal arcs and two nearly vertical arcs. The upper arc is concave down, has endpoints $\left(-\beta ,0\right)$ and $\left(\beta ,0\right),$, reaches a maximum height above the x-axis at the point (0,1), and visually it will look like a horizontal segment for $-\beta \approx -1 along with a pair of vertical segments, one at $x=1\approx \beta$ and the other at $x=-1\approx -\beta .$. The lower arc is the reflection of the upper arc about the x-axis.
Visually, the upper arc will look like the upper horizontal and two vertical sides of a rectangle whose vertices are (−1,0) and (−1,1) and (1,1) and (1,0). Visually, the lower arc will look like the lower horizontal and two vertical sides of a rectangle whose vertices are (−1,−1) and (−1,0) and (1,0) and (1,−1). Together, these two arcs will visually look like the four sides of a square whose vertices are (−1,−1) and (−1,1) and (1,1) and (1,−1).

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