Why does the graph of $y^2=1-\frac{4x^{10^{12}}}{\pi^2}$ look so much like a square?

Question

I want to know why the equation $y^2=1-\dfrac{4x^{10^{12}}}{\pi^2}$ gives an approximate square. (See the figure below.)

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Background

I was just playing around with functions and I wanted to see if $y=\left|\sin\bigg(\dfrac{\pi x}{2}\bigg)\right|$ (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 $\phi=x~\mathrm{radians}$, the value of $\sin\phi$ can to be approximately $\dfrac2\pi\sqrt{\pi x-x^2}$. 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 $\dfrac\pi2$ and $1$ respectively.

The equation came out to be : $y^2 = 1 - \dfrac{4x^2}{\pi^2}$

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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.

$y^2=1-\dfrac{4x^{10^{12}}}{\pi^2}$ gave a good approximation of a square. For the exponent of $x$ being some power of $10$ greater than $10^{12}$, a part of the curve began to disappear.

I want to know why this equation gives an approximate square.

Note : I would like to inform you that I have no experience with conic sections.

Thanks!

Answer 1

This is a rectangle, because for $x=0$ we get $|y|=1$, but for $y=0$ we obtain $$ x=\root{10^{12}}\of{\pi^2\over4}\approx 1.0000000000009031654. $$ For a square, you'd better replace ${4\over\pi^2}$ with $1$.

Answer 2

First, let's determine the possible values for $x.$ Thanks to @Intelligenti pauca for pointing out this oversight in my original answer, which caused significant qualitative errors in my original answer.

Since $y^2$ is non-negative, we have:

$$ 1 \; - \; \frac{4x^{{10}^{12}}}{{\pi}^2} \; \geq \; 0 $$

$$ x^{{10}^{12}} \; \leq \; \frac{{\pi}^2}{4} $$

$$ -\left(\frac{{\pi}^2}{4}\right)^{{10}^{-12}} \; \leq \; x \; \leq \; \left(\frac{{\pi}^2}{4}\right)^{{10}^{-12}} $$

$$ -1.0000000000009031654105793 \ldots \; \leq \; x \; \leq \; 1.0000000000009031654105793 \ldots $$

For the decimal approximation used above, see this WolframAlpha computation.

Note that for $x = \pm \left(\frac{{\pi}^2}{4}\right)^{{10}^{-12}} \stackrel{\text{def}}{=} \; \pm \beta,$ we have $y^2 = 0,$ and hence $y = 0.$

When $x = \pm \, 0.999999,$ we find that $\;y^2 \approx 1 \; – \; {10}^{-434,000}\;$ and $\;y \approx \pm \left(1 \; – \; {10}^{-217,000}\right)$. The table below shows the result of several similar calculations.

$$\begin{array}{|c|c|c|} \hline x & y^2 & y \\ \hline & & \\ \hline 0 & 1 & \pm \, 1 \\ \hline \pm \, 0.9 & 1 - {10}^{-45,700,000,000} & \pm \left(1 - {10}^{-22,900,000,000}\right) \\ \hline \pm \left(1 - {10}^{-6}\right) \; = \;\pm \, 0.999999 & 1 - {10}^{-434,000} & \pm \left(1 - {10}^{-217,000}\right) \\ \hline \pm \left(1 - {10}^{-10}\right) \; = \;\pm \, 0.9999999999 & 1 \; - \; 2.5\times{10}^{-44} & \pm \left(1 \; - \; 1.2\times{10}^{-22}\right) \\ \hline \pm\left(1 - {10}^{-12}\right) & 0.8509 \ldots & \pm \, 0.9224\ldots \\ \hline \pm \left(1 - {10}^{-15}\right) & 0.5951 \ldots & \pm \, 0.7714\ldots \\ \hline \pm \, 1 & 0.5947 \ldots & \pm \, 0.7711\ldots \\ \hline \pm \, 1.000000000000903 & 0.000165 \ldots & \pm \, 0.012860 \ldots \\ \hline \pm \, \beta & 0 & 0 \\ \hline \end{array}$$

Thus, using the fact that $y^2$ is a decreasing function of $|x|$ for $-\beta < x < \beta,$ 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 $(- \beta, 0)$ and $(\beta, 0),$ reaches a maximum height above the $x$-axis at the point $(0,1),$ and visually it will look like a horizontal segment for $-\beta \approx -1 < x < 1 \approx \beta$ 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 $(1,-1).$

Answer 3

HINT

$y=\pm 1$ is clearly a tendency around $x=0$ and the

$y=\log[(4/\pi)^2 x^{m}] $ tends to pass through $(x=1, x=-1)$ as $y\rightarrow 0$

Answer 4

This is related to what happens with the graphs of very high powers of $x,$ which in turn is related to exponential growth and decay.

Graph $y = x^2.$ Notice that the curve goes through $(0,0)$ at its low point, and goes through $(-1,1)$ on the left and $(1,1)$ on the right. And the graph has a tiny nearly level section very near the bottom.

Try $y = x^4.$ It's somewhat like $y=x^2$, but the sides are steeper at $(-1,1)$ and $(1,1)$ and the bottom is much flatter.

Try $y = x^{10}$. Steeper sides, flatter bottom than $x^4.$

As you try higher and higher powers of $x,$ you get a larger and larger "flat" part at the bottom of the curve. This part isn't really flat, it's just that for any number $x$ with $|x|<1,$ if you look at $x^n$ and increase the exponent $n$ you have a process of exponential decay where $x^n$ approaches zero. At some exponent the value of $x^n$ will be so small that you cannot see the difference between $x^n$ and zero on the graph.

For values of $x$ closer to $\pm 1$, $x^n$ decays slower and it takes a higher value of $n$ before $x^n$ gets close enough to zero to be indistinguishable from zero by your eye. But if you take really large values of $n$, such as $10^{12},$ the numbers near $\pm1$ for which $x^n$ is not visually indistinguishable from zero are so close to $\pm1$ that they are visually indistinguishable from $1$ and the graph looks like it has straight vertical sides there. In fact even at $n = 1000$ the graph looks pretty square at the bottom to me.

Now flip the graph over by taking $y = 1 - x^n$ for a very large value of $n.$ It still has that rectangular shape, but the flat level part is at $y = 1$ and the rest is below that, passing through $(-1,0)$ and $(1,0)$.

Now take $y = \sqrt{1 - x^n}.$ If $n$ is large enough this still looks rectangular, but the parts of the graph below the $x$ axis have disappeared because negative numbers do not have real square roots.

If you now square both sides, $y^2 = {1 - x^n},$ you get the same result above the $x$ axis, but since $(-y)^2 = y^2$ you get two symmetric values of $y$ for each value of $x,$ that is, the graph above the $x$ axis is mirrored below the $x$ axis, forming what looks like a square.

Multiplying $x^n$ by some positive constant $a$, as in $y^2 = {1 - ax^n},$ makes the graph wider or narrower in the $x$ direction. That is, you are graphing $y^2 = {1 - (a^{1/n}x)^n},$ so the graph is scaled by a factor of $a^{-1/n}$ in width. If $a$ is not too large (for example, $a = 4/\pi^2$) and $n$ is very large, $a^{-1/n}$ is extremely near $1$ (as other answers have pointed out).

For the exponent of $x$ being some power of $10$ greater than $10^{12}$, a part of the curve began to disappear.

I had a similar experience with extremely high powers of $x$, using the graphing calculator at Desmos.com. I suspect this is a limitation of the size of number that the calculator can deal with, or perhaps the horizontal step size (graph so steep that the software cannot increment $x$ slowly enough to plot a continuous curve).