## Pi and the Golden Ratio

Two of my favorite numbers are pi:

$\pi = 3.14159...$

and the golden ratio:

$\displaystyle{ \Phi = \frac{\sqrt{5} + 1}{2} } = 1.6180339...$

They’re related:

$\pi = \frac{5}{\Phi} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \Phi}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}}} \cdots$

Greg Egan and I came up with this formula last weekend. It’s probably not new, and it certainly wouldn’t surprise experts, but it’s still fun coming up with a formula like this. Let me explain how we did it.

History has a fractal texture. It’s not exactly self-similar, but the closer you look at any incident, the more fine-grained detail you see. The simplified stories we learn about the history of math and physics in school are like blurry pictures of the Mandelbrot set. You can see the overall shape, but the really exciting stuff is hidden.

François Viète is a French mathematician who doesn’t show up in those simplified stories. He studied law at Poitiers, graduating in 1559. He began his career as an attorney at a quite high level, with cases involving the widow of King Francis I of France and also Mary, Queen of Scots. But his true interest was always mathematics. A friend said he could think about a single question for up to three days, his elbow on the desk, feeding himself without changing position.

Nonetheless, he was highly successful in law. By 1590 he was working for King Henry IV. The king admired his mathematical talents, and Viète soon confirmed his worth by cracking a Spanish cipher, thus allowing the French to read all the Spanish communications they were able to obtain.

In 1591, François Viète came out with an important book, introducing what is called the new algebra: a symbolic method for dealing with polynomial equations. This deserves to be much better known; it was very familiar to Descartes and others, and it was an important precursor to our modern notation and methods. For example, he emphasized care with the use of variables, and advocated denoting known quantities by consonants and unknown quantities by vowels. (Later people switched to using letters near the beginning of the alphabet for known quantities and letters near the end like $x,y,z$ for unknowns.)

In 1593 he came out with another book, Variorum De Rebus Mathematicis Responsorum, Liber VIII. Among other things, it includes a formula for pi. In modernized notation, it looks like this:

$\displaystyle{ \frac2\pi = \frac{\sqrt 2}2 \cdot \frac{\sqrt{2+\sqrt 2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt 2}}}{2} \cdots}$

This is remarkable! First of all, it looks cool. Second, it’s the earliest known example of an infinite product in mathematics. Third, it’s the earliest known formula for the exact value of pi. In fact, it seems to be the earliest formula representing a number as the result of an infinite process rather than of a finite calculation! So, Viète’s formula has been called the beginning of analysis. In his article “The life of pi”, Jonathan Borwein went even further and called Viète’s formula “the dawn of modern mathematics”.

How did Viète come up with his formula? I haven’t read his book, but the idea seems fairly clear. The area of the unit circle is pi. So, you can approximate pi better and better by computing the area of a square inscribed in this circle, and then an octagon, and then a 16-gon, and so on:

If you compute these areas in a clever way, you get this series of numbers:

$\begin{array}{ccl} A_4 &=& 2 \\ \\ A_8 &=& 2 \cdot \frac{2}{\sqrt{2}} \\ \\ A_{16} &=& 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2 + \sqrt{2}}} \\ \\ A_{32} &=& 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2 + \sqrt{2}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \end{array}$

and so on, where $A_n$ is the area of a regular n-gon inscribed in the unit circle. So, it was only a small step for Viète (though an infinite leap for mankind) to conclude that

$\displaystyle{ \pi = 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2 + \sqrt{2}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \cdots }$

or, if square roots in a denominator make you uncomfortable:

$\displaystyle{ \frac2\pi = \frac{\sqrt 2}2 \cdot \frac{\sqrt{2+\sqrt 2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt 2}}}{2} \cdots}$

The basic idea here would not have surprised Archimedes, who rigorously proved that

$223/71 < \pi < 22/7$

by approximating the circumference of a circle using a regular 96-gon. Since $96 = 2^5 \times 3$, you can draw a regular 96-gon with ruler and compass by taking an equilateral triangle and bisecting its edges to get a hexagon, bisecting the edges of that to get a 12-gon, and so on up to 96. In a more modern way of thinking, you can figure out everything you need to know by starting with the angle $\pi/3$ and using half-angle formulas 4 times to work out the sine or cosine of $\pi/96$. And indeed, before Viète came along, Ludolph van Ceulen had computed pi to 35 digits using a regular polygon with $2^{62}$ sides! So Viète’s daring new idea was to give an exact formula for pi that involved an infinite process.

Now let’s see in detail how Viète’s formula works. Since there’s no need to start with a square, we might as well start with a regular n-gon inscribed in the circle and repeatedly bisect its sides, getting better and better approximations to pi. If we start with a pentagon, we’ll get a formula for pi that involves the golden ratio!

We have

$\displaystyle{ \pi = \lim_{k \to \infty} A_k }$

so we can also compute pi by starting with a regular n-gon and repeatedly doubling the number of vertices:

$\displaystyle{ \pi = \lim_{k \to \infty} A_{2^k n} }$

The key trick is to write $A_{2^k}{n}$ as a ‘telescoping product’:

$A_{2^k n} = A_n \cdot \frac{A_{2n}}{A_n} \cdot \frac{A_{4n}}{A_{2n}} \cdot \frac{A_{8n}}{A_{4n}}$

Thus, taking the limit as $k \to \infty$ we get

$\displaystyle{ \pi = A_n \cdot \frac{A_{2n}}{A_n} \cdot \frac{A_{4n}}{A_{2n}} \cdot \frac{A_{8n}}{A_{4n}} \cdots }$

where we start with the area of the n-gon and keep ‘correcting’ it to get the area of the 2n-gon, the 4n-gon, the 8n-gon and so on.

There’s a simple formula for the area of a regular n-gon inscribed in a circle. You can chop it into $2 n$ right triangles, each of which has base $\sin(\pi/n)$ and height $\cos(\pi/n)$, and thus area $n \sin(\pi/n) \cos(\pi/n)$:

Thus,

$A_n = n \sin(\pi/n) \cos(\pi/n) = \displaystyle{\frac{n}{2} \sin(2 \pi / n)}$

This lets us understand how the area changes when we double the number of vertices:

$\displaystyle{ \frac{A_{n}}{A_{2n}} = \frac{\frac{n}{2} \sin(2 \pi / n)}{n \sin(\pi / n)} = \frac{n \sin( \pi / n) \cos(\pi/n)}{n \sin(\pi / n)} = \cos(\pi/n) }$

This is nice and simple, but we really need a recursive formula for this quantity. Let’s define

$\displaystyle{ R_n = 2\frac{A_{n}}{A_{2n}} = 2 \cos(\pi/n) }$

Why the factor of 2? It simplifies our calculations slightly. We can express $R_{2n}$ in terms of $R_n$ using the half-angle formula for the cosine:

$\displaystyle{ R_{2n} = 2 \cos(\pi/2n) = 2\sqrt{\frac{1 + \cos(\pi/n)}{2}} = \sqrt{2 + R_n} }$

Now we’re ready for some fun! We have

$\begin{array}{ccl} \pi &=& \displaystyle{ A_n \cdot \frac{A_{2n}}{A_n} \cdot \frac{A_{4n}}{A_{2n}} \cdot \frac{A_{8n}}{A_{4n}} \cdots } \\ \\ & = &\displaystyle{ A_n \cdot \frac{2}{R_n} \cdot \frac{2}{R_{2n}} \cdot \frac{2}{R_{4n}} \cdots } \end{array}$

so using our recursive formula $R_{2n} = \sqrt{2 + R_n}$, which holds for any $n$, we get

$\pi = \displaystyle{ A_n \cdot \frac{2}{R_n} \cdot \frac{2}{\sqrt{2 + R_n}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + R_n}}} \cdots }$

I think this deserves to be called the generalized Viète formula. And indeed, if we start with a square, we get

$A_4 = \displaystyle{\frac{4}{2} \sin(2 \pi / 4)} = 2$

and

$R_4 = 2 \cos(\pi/4) = \sqrt{2}$

giving Viète’s formula:

$\pi = \displaystyle{ 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2 + \sqrt{2}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \cdots }$

as desired!

But what if we start with a pentagon? For this it helps to remember a beautiful but slightly obscure trig fact:

$\cos(\pi / 5) = \Phi/2$

and a slightly less beautiful one:

$\displaystyle{ \sin(2\pi / 5) = \frac{1}{2} \sqrt{2 + \Phi} }$

It’s easy to prove these, and I’ll show you how later. For now, note that they imply

$A_5 = \displaystyle{\frac{5}{2} \sin(2 \pi / 5)} = \frac{5}{4} \sqrt{2 + \Phi}$

and

$R_5 = 2 \cos(\pi/5) = \Phi$

Thus, the formula

$\pi = \displaystyle{ A_5 \cdot \frac{2}{R_5} \cdot \frac{2}{\sqrt{2 + R_5}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + R_5}}} \cdots }$

gives us

$\pi = \displaystyle{ \frac{5}{4} \sqrt{2 + \Phi} \cdot \frac{2}{\Phi} \cdot \frac{2}{\sqrt{2 + \Phi}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \Phi}}} \cdots }$

or, cleaning it up a bit, the formula we want:

$\pi = \frac{5}{\Phi} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \Phi}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}}} \cdots$

Voilà!

There’s a lot more to say, but let me just explain the slightly obscure trigonometry facts we needed. To derive these, I find it nice to remember that a regular pentagon, and the pentagram inside it, contain lots of similar triangles:

Using the fact that all these triangles are similar, it’s easy to show that for any one, the ratio of the long side to the short side is $\Phi$ to 1, since

$\displaystyle{\Phi = 1 + \frac{1}{\Phi} }$

Another important fact is that the pentagram trisects the interior angle of the regular pentagon, breaking the interior angle of $108^\circ = 3\pi/5$ into 3 angles of $36^\circ = \pi/5$:

Again this is easy and fun to show.

Combining these facts, we can prove that

$\displaystyle{ \cos(2\pi/5) = \frac{1}{2\Phi} }$

and

$\displaystyle{ \cos(\pi/5) = \frac{\Phi}{2} }$

To prove the first equation, chop one of those golden triangles into two right triangles and do things you learned in high school. To prove the second, do the same things to one of the short squat isosceles triangles:

Starting from these equations and using $\cos^2 \theta + \sin^2 \theta = 1$, we can show

$\displaystyle{ \sin(2\pi/5) = \frac{1}{2}\sqrt{2 + \Phi}}$

and, just for completeness (we don’t need it here):

$\displaystyle{ \sin(\pi/5) = \frac{1}{2}\sqrt{3 - \Phi}}$

These require some mildly annoying calculations, where it helps to use the identity

$\displaystyle{\frac{1}{\Phi^2} = 2 - \Phi }$

Okay, that’s all for now! But if you want more fun, try a couple of puzzles:

Puzzle 1. We’ve gotten formulas for pi starting from a square or a regular pentagon. What formula do you get starting from an equilateral triangle?

Puzzle 2. Using the generalized Viète formula, prove Euler’s formula

$\displaystyle{ \frac{\sin x}{x} = \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \cdots }$

Conversely, use Euler’s formula to prove the generalized Viète formula.

So, one might say that the real point of Viète’s formula, and its generalized version, is not any special property of pi, but Euler’s formula.

### 45 Responses to Pi and the Golden Ratio

1. […] Two of my favorite numbers are pi: and the golden ratio: They’re related: \$latex pi =… – Read full story at Hacker News […]

• Jon says:

Pi is equal to ( 9 + 3*(5^(1/2)) )/5

• Todd Trimble says:

No it’s not. It’s close though: overestimates by about .00005.

2. Paulo Santos says:

Congratulations!, Very interesting presentation of ideas, these is one of most favourites too, other is Sequence Limits of Euler and Fibonacci Numbers and the golden section in nature, art, geometry, architecture, music and even for calculating pi!

3. Cool !

4. John Cowan says:

$\pi$ was sometimes called “Ludolph’s number” for a while.

5. Alan White says:

The Golden Ratio is often used as the numerical value for the aspect ratio (Y Axis Height/ X Axis length) when graphing functions. Doesn’t work for the cycloid and it’s variants. The aspect ratio for the cycloid, a ‘wheel function’ must be set to Y = 1 / X = pi. Failing to do so locates the inflection point of Roberval’s ‘companion curve to the cycloid’ away from the center of the bounding rectangle.

6. jessemckeown says:

that first $\cos(x)$ in puzzle 2 is bothering me; it has zeros that the other side doesn’t.

• Greg Egan says:

I’m pretty sure that first $\cos(x)$ in puzzle 2 is a typo, and not actually part of Euler’s formula.

• John Baez says:

Thanks for catching that mistake! Fixed.

7. Mar Kac’s book “Statistical Independence in Probability, Analysis and Number Theory” starts out with Viete’s formula (essentially puzzle 2) and then uses Rademacher functions to ultimately derive the central limit theorem. It is a ripping read. Kac is also famous for his “Can you hearth shape of a drum paper?” (Answer is “no”)

8. Serge says:

¿Vòila? Voilà !

• John Baez says:

I knew something looked funny about that. I’ll fix it. Thanks!

• francisco says:

voilò? voilà!

• John Baez says:

Wow, amazing! This time it’s really fixed.

In my defense, to write “Voilà” here I actually type “Voil&agrave;”, and that makes it a bit easier to screw up.

9. Greg Egan says:

Curiously, there is another infinite product formula for $\frac{\sin(x)}{x}$ that is also attributed to Euler:

$\displaystyle{\frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \left(1-\frac{x^2}{\pi^2 n^2}\right)}$

and setting $x=\frac{\pi}{2}$ in this one leads to the Wallis formula for $\pi$ that JB discussed here:

• John Baez says:

Yes, this other infinite product formula

$\displaystyle{\frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \left(1-\frac{x^2}{\pi^2 n^2}\right)}$

is more famous: the young Euler used it to show

$\displaystyle{\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}}$

thus meeting Bernoulli’s challenge to the world to sum this series:

• John Baez, The Riemann zeta function, 21 December 2003.

Euler simply guessed this product formula by creating an ‘infinite polynomial’ with zeros at the same places as $\sin x / x$ (though I imagine he checked his guess in various ways). Later Weierstrass came up with a nice general theory of such product expansions:

• Wikipedia, Weierstrass factorization theorem.

But I’ve never seen anything written about

$\displaystyle{ \frac{\sin x}{x} = \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \cdots }$

except in the Wikipedia article on Vieète’s formula, so I’m a bit curious about its history.

• Man of Fold says:

It is interesting to note that Ramanujan was indirectly aware of Euler’s proof through Carr’s synopsis. I left a somewhat embarrassing question on MSE that puts Carr’s proof on display. https://math.stackexchange.com/questions/1885685/important-identities-that-can-be-obtained-by-manipulating-the-function-fracx
I guess it isn’t so surprising that Carr gives this proof but it’s fun to think about how much it would have delighted Ramanujan.

BTW, I first learned about Ramanujan from your lecture about the number 5, so thanks for that.

• John Baez says:

I’m glad to have introduced you to Ramanujan—he’s a really fascinating mathematician. My lecture about the number 5 mentioned this formula of his:

$\mathbf{ \frac{1} {1 + \frac{e^{-2\pi}} {1 + \frac{e^{-4\pi}} {1 + \frac{e^{-6\pi}} {1 + \frac{e^{-8\pi}} {1 + \frac{e^{-10\pi}}{._{._.}}} } } } } = \left(\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{1 + \sqrt{5}}{2}\right) e^{2 \pi/5} }$

That makes me very happy to have stumbled on this vaguely similar—but much less profound—formula:

$\pi = \frac{5}{\Phi} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \Phi}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \Phi}}}}} \cdots$

Thanks for telling me about Carr’s work!

• Todd Trimble says:

I’m a little late to this party. Neat! Another way of proving the identity in Puzzle 2 would be to use the Weierstrass infinite product formula for the sine and combine that with $\cos x = \frac{\sin(2x)}{2\sin x}$ to get the infinite product for $\cos(x)$. With that, the Puzzle 2 identity basically follows from the fact that any integer has a unique representation as a power of 2 times an odd number (how it follows could be left as a puzzle).

10. Friedmann and Hagen have got Wallis formula for pi from a variational computation of the spectrum of the hydrogen atom. Check out the proof here:

http://dx.doi.org/10.1063/1.4930800

• John Baez says:

I will need to go through that sometime. People keep telling me about it, but I haven’t checked out the calculation! By doing a calculation in a sufficiently complicated way you can relate it to almost anything, so the question is how elegantly they make the connection.

• Th only step I couldn’t immediately replicate was the expectation of the Hamiltonian and I have never got back to it – the rest is correct. I must do it but it is 40 years since I did that stuff.

11. francisco says:

I believe it should read
$\cos(\pi/5) = \frac{\Phi}{2}$
$\cos(2\pi/5) = \frac{1}{2\Phi}$
because $\Phi > 1$.

• John Baez says:

You’re right, I got them switched. Earlier, when I actually used them, I had ’em right.

Fixed.

12. domenico says:

I am thinking that it would be possible to write a pi calculus with a volume of a sphere, but the problem is with the polyhedrons that must be ever increasingly smaller (it is impossible to have more of five regular polyhedra, so that a generalization of Viete in three dimension is not possible).
I tried to generate a general irregular polyhedra using a Bravais lattice on the sphere (for increase the faces of the polyhedron reducing the primitive vectors), with the discrete translations replaced with the versors for the rotation along the edge of the spherical triangles, and I have used the Rodrigues rotation formula to obtain the points on the sphere (with a little program), so that there are an infinite number of tessellation using an arbitrary initial spherical triangle (that generate the primitive vectors of Bravais lattice) with a vertex in the zenit, an edge along a meridian, and the other edge with an arbitrary lengths and directions (to complete the scalene triangle): the edges of the triangles have different lengths in the primitive vectors, but have the same rotations in the primitive vectors; the volumes of the polyhedron are too complex to evaluate, but it seem that it is possible to obtain the tetrahedron, the octahedron and the cube using the Bravais lattice (so that could be possible for all the regular polyhedra).

• Ant says:

Perhaps archemedian solids would be of use in this investigation.

13. That generalized formula is so good. Puzzle-1 is simple enough but but nevertheless an excuse to write my first comment here in this blog;

$3 \frac{2}{\sqrt{2+\sqrt{3}}} \frac{2}{\sqrt{2+\sqrt{2+\sqrt{3}}}} ...$

I think there is a typo in the paragraph about the area of the regular n-gon, although it is correctly stated after the pentagon depiction. It should read n.sin()… instead of 2.sin()…

• John Baez says:

Hi, Akira. Thanks! I haven’t actually worked out the answer to Puzzle 1, but I’ll check your answer and see if it’s right. It’s certainly pretty.

Also thanks for catching that typo. I fixed it.

By the way, you used double dollar signs in your LaTeX; those don’t work here, alas. You can use \displaystyle{ } to make the text in an equation taller, and

div align=”center”< >/div<

around the equation to center it. With those tricks your equation becomes

$\displaystyle{ 3 \frac{2}{\sqrt{2+\sqrt{3}}} \frac{2}{\sqrt{2+\sqrt{2+\sqrt{3}}}} ... }$

or, even more gloriously:

gives

$\displaystyle{\pi = 3 \frac{2}{\sqrt{2+\sqrt{3}}} \frac{2}{\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdots }$

But these extra tricks are not very important!

14. John I read many of your Maths/physics papers and always take time for Azimuth in/email. I sent you a series of polyhedrons, one I described as a tesseract or hypercube, a 3D icosahedral/cuboid construct yielding eight tetrahedrons.

My working background is in building science, I’m a serious amateur mathematician with an emphasis on greenhouse and climate impacts, and number theory. its sometimes a bit awkward to explain but bricks, curvature and modularities are my areas of focused interest.

The PI : PHI golden section relationship in 3D solids is an example I find fascinating, and the maths reasoning is why I follow Azimuth to connect the environment with pure maths and physics.

A relationship between Phi and Pi; phi brick in ball, the brick in a sphere shows a simple relationship between Pi and Phi, using the values for the diagonal D that we have just found and the surface area S.
If we imagine the brick tightly packed into a sphere, the centre of the sphere will be half way along diagonal D.
So the radius of the sphere will be 1 and its surface area will be 4 Pi.

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/propsOfPhi.html#phiBrick

15. […] in order to learn something new about π, go read this lovely post by John […]

16. Eric Astor says:

The general formula IS lovely. I particularly like that you get the same result whether you start with an equilateral triangle (Puzzle 1) or a regular hexagon! (The triangle case just has one extra factor get cancelled out.)

I should try to figure out whether the formula for n always agrees with the formula for kn (which it should), and why the algebra works out; it seems like it must be a beautiful pattern.

• John Baez says:

I don’t think the formula for n will agree with that for kn unless k is a power of 2.

17. The formula that I use to compute the golden ratio – not sure if it’s commonly used as I just sort of dreamt it up – is:

$\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{ 2 - \cdots }}}}$

• John Baez says:

Does that really work? My favorite is

$\displaystyle{ \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}} \quad = \quad \frac{\sqrt{5} + 1}{2} }$

• Todd Trimble says:

Sure, it works. To check $x^2 = 2 + \sqrt{2 - x}$ for $x = \frac{1 + \sqrt{5}}{2}$, use $2 - x = \frac{3 - \sqrt{5}}{2} = \left(\frac{1 - \sqrt{5}}{2}\right)^2$. Only, the square root of $2 - x$ is taken to be the positive square root, so it’s $\sqrt{2 - x} = \frac{\sqrt{5} - 1}{2}$. Then $2 + \frac{\sqrt{5} - 1}{2} = \frac{3 + \sqrt{5}}{2} = x^2$, so it checks out. One possible reason for preferring Michael’s expression (haven’t checked this carefully; just a gut feeling) is that you get quicker convergence than with John’s preferred formula.

• Scott Hotton says:

For fast convergence we can apply Newton’s
method to the minimal polynomial for the golden
ratio.

$x \mapsto N(x) = \frac{x^2+1}{2x-1}$

The intermediate convergents of the golden ratio
are ratios of consecutive Fibonacci numbers and
Newton’s method applied to a polynomial is a
rational function. A rational function of a
rational number is rational number. Not
surprisingly then, it turns out that Newton’s
method allows us to jump through the sequence of
Fibonacci numbers.

$N(F_{n+1}/F_n) = F_{2n+1}/F_{2n}$

This is related to the tendency of Newton’s method
to double the number of digits of precision with
each iteration. This example is easier to state
and prove though so its useful for introducing
students to Newton’s method.

18. wilhelmtell says:

According to Wikipedia, the first infinite series recorded was indeed an estimate of Pi, but it was of Archimedes, in the third century BC.

https://en.m.wikipedia.org/wiki/Series_(mathematics)#Development_of_infinite_series

19. When we make a logistic map indicating bifurcation leading to chaos, the number of nodes at each step together form a Fibonacci sequence and as this leads to the golden ratio phi… we can physically interpret this to the mean radius of the KAM tori representing the conditionally periodic process of the system under question whose phase space contain the tori. This is a possible interpretation of your formula Prof. John Baez.

20. helloparth says:

Thank you for this post, I feel like I learnt alot in a relatively short space of time. Never knew about Viete’s geometric formula, I associate Viete with the formulae relating roots of polynomials to coefficient.
While it’s true that the golden number is a special case, the way it enters into the formula i.e. cos(pi/5)=phi/2 is elegant. This relates a simple trig function of pi with phi, and indeed a simple trig function can be related to a simple shape like a regular pentagon (and vice versa). The golden number is a simple expression in terms of a radical [sqrt(5)] unlike pi; in this sense pi is the greater enigma. The open question is: what other elegant ways are there of proving this connection?

21. […] Voila~ kita sudah menemukan lagi kaitan antara dan . Masih banyak lagi cara untuk menuliskan ke dalam persamaan yang memuat atau sebaliknya, namun ada satu persamaan paling cantik yang menghubungkan dengan , yang mana rumus ini saya temukan di dalam blog Azimuth, […]

22. zeeni says:

Im not a mathematician but need help with this please, anybody?

I saw this video about how if we draw a spiral from a circle to a bigger circle in the ration of 1.68, then the LENGTH of that spiral is pi itself.

The maker of this video doesn’t have any contact. Can someone please confirm this??

And if yes, what is the unit of measuring the 3.14??

• Todd Trimble says:

It’s correct, and mildly amusing, but not terribly interesting to a mathematician. The YouTube comment by Linus Brendel from 2 years ago is exactly right. The narrator (who oddly pronounces it “Fibonicci” and not “Fibonacci”, and who obscurely refers to “cubes”) is adding up lengths of just six quarter-circle arcs to get from A to B. The length of the largest quarter circle arc is $\pi \phi/4$, and each successive arc has $1/\phi$ the length of the previous. So the claim is that $\pi = \pi \phi/4 \cdot (1 + 1/\phi + 1\phi^2 + 1/\phi^3 + 1/\phi^4 + 1/\phi^5)$. A good high school mathematics student, given some time, can prove that with a little algebra and knowing that $\phi^2 = \phi + 1$.

23. James Borger says:

What about for the 17-gon, using Gauss’s formula?

• John Baez says:

That case is left as an exercise for the reader, along with the 257-gon and 65537-gon.

According to Wikipedia:

Although it was known to Gauss by 1801 that the regular 65537-gon was constructible, the first explicit construction of a regular 65537-gon was given by Johann Gustav Hermes (1894). The construction is very complex; Hermes spent 10 years completing the 200-page manuscript.

As Greg noted (somewhere, years ago), the appearance of $\Phi$ is a bit of a red herring here. The ‘generalized Viète formula” which lets us compute pi starting from any n-gon is equivalent to Euler’s formula

$\displaystyle{ \frac{\sin x}{x} = \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \cdots }$

and this formula is probably the most elegant distillation of what’s going on.

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