0

I saw this $$\frac 65 \phi^2 \sim \pi$$ with $99.9985\%$ of accuracy. There are many more estimations like this between $\phi $ and $\pi$, I think?!. Now the question is about the origination of this(those) estimation. Is it a geometrical representation of this concept?
can someone get a clue?

Kamal Saleh
  • 6,497
Khosrotash
  • 24,922
  • 3
    Not just $\phi$ and $\pi$, but all kinds of unrelated irrational numbers have "close matches" when you allow powers and multiplication by rational numbers. Moreover all such operations can be represented geometrically, but if you do not have an exact equation, how is that representation meaningful to you? – David K Feb 09 '23 at 17:29
  • @DavidK: It was intersting to me because pi an phi are irrational, but say 99.9985 of acuracy. (truncation maybe used).Honestly I found it on web. I checked it by calculator. But I can't find a representation about it. – Khosrotash Feb 09 '23 at 17:36
  • 10
    Any irrational number can be approximated arbitrarily closely by rational numbers. Suppose the ratio $\phi/\pi$ is irrational. You can still set $m\phi \approx n\pi$ for integers $m$ and $n$ to any degree of accuracy you want. By chance, for some pairs of irrational numbers the numbers $m$ and $n$ will be small. I don't think the meaning is any deeper than that. – David K Feb 09 '23 at 17:41
  • 3
    It is very likely that this is just a coincidence. – Tuvasbien Feb 09 '23 at 17:42
  • 2
    @DavidK You don’t need to suppose $\phi/ \pi$ is irrational, it is irrational since $\pi$ is transcendental. – Vivaan Daga Feb 09 '23 at 17:48
  • 2
    @Shinrin-Yoku Yes, we know this ratio is irrational, but I'm making a more general argument that applies to any pair of irrational numbers, not just $\phi$ and $\pi.$ – David K Feb 09 '23 at 17:53
  • 3
    There are numerical coincidences all over the place in mathematics (there is even a short Wikipedia article on this topic). Expecting there to be something more than coincidence feels like numerology, to me. – Xander Henderson Feb 09 '23 at 17:53

4 Answers4

5

Maybe the golden triangle could allow for some geometric intuition.

You can use the cosine law on the golden triangle to show that $$ \frac{6}{5}\phi^2=\frac{3}{5(1-\cos(\frac{\pi}{5}))}\approx\pi. $$

As a side note, if you notice that $\cos(\frac{\pi}{5})=\frac{1+\sqrt{5}}{4}=\frac{1}{2}\phi$, we arrive at $$ \phi^2=\frac{1}{2-\phi}. $$

Scene
  • 1,566
4

There are a vast combination of real numbers $a,p,q$ such that $a\phi^p\pi^q\approx 1$. You have found the example $a=6/5~,~p=2~,~q=-1$. But there are others, like $a=111/250~,p=q=1/2$. So one might ask

I have seen that $\frac{111}{250}\sqrt{\phi}=\frac{1}{\sqrt{\pi}}$ to within less than a 1% error. Why???

Because.... numbers??

K.defaoite
  • 12,536
3

There are exact equivalences of $\pi^3$ using $\phi$ derived in this short paper. It states that $$\pi^3=\frac{125}{4}\frac{(3-\phi)^{3/2}}{\phi}\sum_{n=-\infty}^\infty\frac{1}{(1-5n)^3}$$and $$\pi^3=\frac{250}{\phi^3\sqrt{2+\phi}}\sum_{n=-\infty}^\infty\frac{1}{(1-10n)^3}$$ Take a look at this integral $$\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}dx=\frac{\pi}{2\sqrt{\phi}}$$From this answer. Here is another identity: $$\frac1{12}\int_0^{2\pi}\frac{x\,dx}{\phi-\cos^2 x}=\frac{\pi^2}6$$

Kamal Saleh
  • 6,497
-1

This can be interpreted by the fact that the area of an inscribed regular decagon is close to the area of the disk and that the side of the regular decagon $a$ and the radius of the circle are linked by the formula: $$R= a \phi.$$

mathcounterexamples.net
  • 70,018
  • 1
    I don't see how this answers the question. The number $6$ is nowhere in sight, whether we consider area or perimeter, inscribed or circumscribed. – mr_e_man Feb 09 '23 at 19:08
  • 1
    Given a regular decagon's circumradius $R$, the inradius is $r=R\cos(\pi/10)=\tfrac12R\sqrt{\varphi\sqrt5}$, and the edge length is $a=2R\sin(\pi/10)=R\varphi^{-1}$. The decagon's perimeter is $10a=10R\varphi^{-1}$, and its area is $20(\tfrac14ar)=\tfrac52R^2\sqrt{\varphi^{-1}\sqrt5}$. The inscribed circle's perimeter is $2\pi r=\pi R\sqrt{\varphi\sqrt5}$, and its area is $\pi r^2=\tfrac14\pi R^2\varphi\sqrt5$. The circumscribed circle's perimeter is $2\pi R$, and its area is $\pi R^2$. How do you get $\tfrac65\varphi^2$ from any of this? – mr_e_man Feb 09 '23 at 19:50
  • 1
    @amWhy No thanks. I'm happy, and also free by the way, to spend my time where I want to. – mathcounterexamples.net Feb 09 '23 at 20:27
  • @mr_e_man: Thank you for clarifying. – Khosrotash Feb 09 '23 at 21:10