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Create a function that plots the $\infty$ symbol when plotted.

My function is $$|y|=|\sin x|$$ For $\{x: -3\le x \le 3\}$

Bonus points if your function is NOT mathematically equivalent to mine.

Double bonus points if the function is simpler than mine, or draws a more beautiful symbol.

Tobi Alafin
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  • https://en.wikipedia.org/wiki/Lemniscate – MJD Nov 29 '16 at 20:54
  • There is no such function. What you stated is not a function. It fails the vertical line test. The infinity symbol fails the vertical line test so no function graphs to it. – fleablood Nov 29 '16 at 21:51
  • Your relation extends infinitely in either direction and at the points where it crosses the axis is not circular but pinched. So it isn't an infinity sign. (It's more like a double helix....) – fleablood Nov 29 '16 at 21:55
  • I don't know what the vertical line test is. What about those functions that draw logos, breasts, etc. – Tobi Alafin Nov 29 '16 at 21:56
  • If we restrict the domain of $x$, can't we plot the $\infty$ sign? I don't have access to a tool to allow me plot this relation. I merely theorised that it would plot infinity. I noticed that $|y| = |x|$ gave a cross sign. And since $sin(x)$ gives a horizontal "S", I theorised |y| = |sin x| will give an infinity symbol. – Tobi Alafin Nov 29 '16 at 22:01
  • @fleablood, I've modified my answer, so that my relation draws the infinity sign. – Tobi Alafin Nov 29 '16 at 23:58
  • @TobiAlafin You can plot graphs at Wolfram Alpha like this. – dxiv Nov 30 '16 at 03:52
  • @fleablood It fails the vertical line test Any closed curve will fail the vertical line test. This includes the standard lemniscate most often identified as the infinity sign $(x^2+y^2)^2=2a^2(x^2-y^2)$. – dxiv Nov 30 '16 at 04:23
  • the lemniscate isn't a function. No closed curve is the Cartesian graph of a function. None of those logos, breasts etc. are functions either. – fleablood Nov 30 '16 at 04:27
  • |y| = |sin x| for 0 <= x <= 2 pi is not a function. It is a curve. It is a reflected sin wave but the the tips will be pinched into a right angle and not a rounded circular curve. – fleablood Nov 30 '16 at 04:32
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    Those that are complaining that it is not a function it is trival to convert those definitions to a function $ \mathbb{R} \to \mathbb{R} \times \mathbb{R} ) $ – Q the Platypus Nov 30 '16 at 04:39

2 Answers2

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Take any smooth, concave curve which passes through the origin at a finite, positive slope, then has a vertical tangent at its next root, for example $y = x \sqrt{1-x}$:

enter image description here

Then mirror it along both axes to complete an infinity-like sign e.g. $|y| = |x| \sqrt{1-|x|}$:

enter image description here

dxiv
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Viviani's Curve is a nice way to see the infinity in a 3 dimensional figure by intersecting a sphere of radius $2a$ with a cylinder of centered at $(a,0,0)$ of radius $a$ this intersection:

$$V(t)=\bigg < a(1+ \cos(t)), a\sin(t), 2a\sin\Big(\frac t2\Big)\bigg>$$

MJD
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OLE
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