1. Where do people get their inspirations for $\pi$ formulas?
  2. Where do they begin with these ideas?

Equations such as$$\dfrac 2\pi=1-5\left(\dfrac 12\right)^3+9\left(\dfrac {1\times3}{2\times4}\right)^3-13\left(\dfrac {1\times3\times5}{2\times4\times6}\right)^3+\&\text{c}.\tag{1}$$$$\dfrac {2\sqrt2}{\sqrt{\pi}\Gamma^2\left(\frac 34\right)}=1+9\left(\dfrac 14\right)^4+17\left(\dfrac {1\times5}{4\times8}\right)^4+25\left(\dfrac {1\times5\times9}{4\times8\times12}\right)^4+\&\text{c}.\tag{2}$$$$\dfrac \pi4=\sum\limits_{k=1}^\infty\dfrac {(-1)^{k+1}}{2k-1}=1-\dfrac 13+\dfrac 15-\&\text{c}.\tag{3}$$ Have always confused me as to where Mathematicians always get their inspirations or ideas for these kinds of identities.

The first one was found by G. Bauer in $1859$ (something I still want to know how to prove. I've found this recently asked question still open for proofs), the second was found by Ramanujan. And has a relation with Hypergeometrical series.

I'm wondering whether people see $\pi$ in other formulas, such as$$\sum\limits_{k=1}^{\infty}\dfrac 1{k^2}=\dfrac {\pi^2}6\implies\pi=\sqrt{\sum\limits_{k=1}^\infty\dfrac 6{k^2}}\tag{4}$$ And isolate $\pi$, or if something new comes up and they investigate it?

For example, I'm wondering if it's possible to manipulate the expansion of $\ln m$

$$\ln m=2\left\{\dfrac {m-1}{m+1}+\dfrac 13\left(\dfrac {m-1}{m+1}\right)^3+\dfrac 15\left(\dfrac {m-1}{m+1}\right)^5+\&\text{c}.\right\}\tag{5}$$

To get a $\pi$ formula. Or the series$$\sum\limits_{k=1}^{\infty}\dfrac 1{k^p}=\dfrac {\pi^p}n\tag{6}$$ Which converges faster and faster as $p$ gets larger and larger.

  • $\begingroup$ $(3)$ comes from the arctangent, and from it, you can derive yet another not mentioned pi formula, but the other formulas you have are not as simply derived. $(3)$ is also derived in the mentioned link. $\endgroup$ Feb 20, 2017 at 21:11
  • $\begingroup$ Hopefully though, the first step is to find some geometric argument. $\endgroup$ Feb 20, 2017 at 21:12
  • $\begingroup$ Most of equations like that are simply special cases of trogonometric or related function infinite series that result in some expression that contains $\pi$. Unless you look at Ramanujan's or Chudnovsky formulas. Those are purely genius work of Ramanujan who had extremely good number intuition. No one really knows how he discovered his formula. Then there is also BPP formula that can give you nth dogit of pi. This one was discovered by accident. $\endgroup$
    – KKZiomek
    Feb 20, 2017 at 21:12
  • $\begingroup$ $\pi$ is used in a lot of mathematical concepts. For example, the ancients found out that we could use it to express the area of a circle. As our math gets better, we found more ways to manipulate these concepts. We now know a lot of ways to find the area of a circle without using $\pi$. Ergo, we can find $\pi$. Because $\pi$ is transcendental, we can't express it with a closed formula. But there's many other ways. $\endgroup$
    – Kaynex
    Feb 20, 2017 at 21:18
  • 1
    $\begingroup$ The series expansion for $\ln\left(\frac{1+z}{1-z}\right)$ can be used in conjuction with Euler's continued fraction formula to obtain a continued fraction for $\pi$. $\endgroup$
    – dxdydz
    Feb 20, 2017 at 23:09

4 Answers 4


Here are three more gems presented in chronological order.

Viète and the first infinite product in the history of mathematics (1593)

\begin{align*} \frac{2}{\pi}=\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots\tag{1} \end{align*}

Here we understand (1) as infinite product \begin{align*} \frac{2}{\pi}=u_1u_2u_3\cdots=\lim_{n\rightarrow\infty}(u_1u_2\cdot u_n) \end{align*} where \begin{align*} u_1=\sqrt{\frac{1}{2}}\qquad\text{and}\qquad u_n=\sqrt{\frac{1}{2}(1+u_{n-1})}\qquad \text{for }n>1 \end{align*}

In 1593 Viète expressed the number $\pi$ as an infinite product, which is the first infinite product in the history of mathematics. His proof was done in the spirit of Archimedes ideas: $\pi$ is the limit of the areas $A_{2n}$ of the regular $2^n$-gons inscribed in the circle of radius one.

The expression (1) is a particular case of a more general formula, given two hundred years later by Euler: \begin{align*} \frac{\sin\theta}{\theta}&=\cos\frac{\theta}{2}\cos\frac{\theta}{4}\cos\frac{\theta}{8}\cdots\cos\frac{\theta}{2^n}\cdots =\lim_{n\rightarrow \infty}(v_1v_2\cdots v_n)\qquad \text{for }\theta\ne 0 \end{align*} where \begin{align*} v_1=\cos\frac{\theta}{2}\qquad\text{and}\qquad v_n=\cos\frac{\theta}{2^n}&=\sqrt{\frac{1}{2}\left(1+\cos\frac{\theta}{2^{n-1}}\right)}\\ &=\sqrt{\frac{1}{2}(1+v_{n-1})} \qquad\text{for}\qquad n>1 \end{align*} The proof of Euler's formula is simple: Applying repetitively the trigonometric identity $$\sin\theta=2\cos\frac{\theta}{2}\sin\frac{\theta}{2}$$ we obtain \begin{align*} \frac{\sin\theta}{\theta}&=\cos\frac{\theta}{2}\cdot\frac{\sin(\theta/2)}{\theta/2} =\cos\frac{\theta}{2}\cos\frac{\theta}{4}\cdot\frac{\sin(\theta/4)}{\theta/4} =\cdots\\ &=\cos\frac{\theta}{2}\cos\frac{\theta}{4}\cos\frac{\theta}{8}\cdot\cos\frac{\theta}{2^n}\frac{\sin(\theta/2^n)}{\theta/2^n} \end{align*} When $n$ tends to infinity, $x=\frac{\theta}{2^n}$ tends to $0$, thus $\frac{\sin x}{x}$ tends to $1$, which proves the claim.

Ramanujan, Elliptic Integrals and $\pi$ (1914)

\begin{align*} \frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{n=0}^\infty\frac{(4n!)}{4^{4n}(n!)^4}\left[1103+26390n\right]\left(\frac{1}{99^4}\right)^n\tag{2} \end{align*}

This incredible identity is one of $17$ identities of $\frac{1}{\pi}$ stated by Ramanjuan 1914 in the article Modular equations and approximations to $\pi$ without proof.

The mathematicians of the second half of the 19th century and the beginning 20th century developed methods for the efficient calculation of certain constants which appear in the theory of elliptic functions. Ramanjuan had a special mastery of such calculations.

A proof of (2) was not known until 1987 the Borwein brothers reconstituted a full proof, published in Pi and the AGM. The proof is complicated, requires a lot of work and it is not easy even to state the essential ideas. The main ingredients are generalized elliptic integrals, modular functions, theta functions and the AGM.


  • In section 5.8 An overview of the proof by J. Borwein and P. Borwein in the book The number $\pi$ by P. Eymard and J-P. Lafon present the authors on a few pages a summary of the essential ideas in form of propositions which are too technical to state them here isolated, without context.

  • The paper Ramanujan's Series for $\frac{1}{\pi}$: A Survey contains nice information around (2) and friends.

An interesting fact is that series of type (2) can be formidably used to approximate $\pi$ due to their fast convergence behaviour. With the methods the Borwein brothers used for their proof they and independently the brothers Chudnovsky discovered the identity \begin{align*} \frac{1}{\pi}=12\sum_{n=0}^\infty\frac{(-1)^n(6n)!(13591409+545140134n)}{(n!)^3(3n)!(640320^3)^{n+1/2}} \end{align*} which is even more efficient, since each term of the series provides $14$ additional digits of $\pi$.

Another astonishing series due to Ramanujan is \begin{align*} \frac{1}{\pi}=\sum_{n=0}^\infty\binom{2n}{n}^3\frac{42n+5}{2^{12n+4}} \end{align*} The denominator of the general term contains the number $16\cdot2^{12n}$. This means that one can calculate the digits in base $2$ from the $n$-th to the $2n$-th without having to calculate the first $n$ digits beforehand. This is the main theme of the third gem.

The formula of Bailey, Borwein and Plouffe (1997)

\begin{align*} \pi=\sum_{n=0}^\infty\frac{1}{16^n}\left(\frac{4}{8n+1}-\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)\tag{3} \end{align*}

This rather simple formula was of particular interest to the three mathematicians due to the factor $\frac{1}{16^n}$ in the general term, as a factor of an arithmetically very simple function of $n$.

They were able to extract from this an algorithm (named BBP) to calculate the $d$-th digit of the expansion of $\pi$ individually in base $16$, and this even for very large $d$, without needing to know or calculate the digits preceding $d$. For example, the $10^{12}$th digit of $\pi$ in base $2$ is $1$.

The sum of the RHS of (3) is of the form \begin{align*} S=4S_1-2S_4-S_5-S6 \end{align*} with \begin{align*} S_k&=\sum_{n=0}^\infty\frac{1}{16^n(8n+k)}\\ &=(\sqrt{2})^k\sum_{n=0}^\infty\left[\frac{x^{8n+k}}{8n+k}\right]_{x=0}^{x=1/\sqrt{2}}\\ &=(\sqrt{2})^k\sum_{n=0}^\infty\int_0^{1/\sqrt{2}}x^{8n+k-1}\,dx\\ &=(\sqrt{2})^k\int_{0}^{1/\sqrt{2}}x^{k-1}\sum_{k=0}^\infty x^{8n}\,dx\\ &=(\sqrt{2})^k\int_0^{1/\sqrt{2}}\frac{x^{k-1}}{1-x^8}\,dx \end{align*} where $k=1,4,5,6$ .

Hence \begin{align*} S=\int_{0}^{1/\sqrt{2}}\frac{8x^5+4\sqrt{2}x^4+8x^3-4\sqrt{2}}{x^8-1}\,dx \end{align*}

We obtain \begin{align*} 8x^5+4\sqrt{2}x^4+8x^3-4\sqrt{2}=8(x^2+1)(x^2+\sqrt{2}x+1)\left(x-\frac{1}{\sqrt{2}}\right) \end{align*} and \begin{align*} x^8-1=(x^2+1)(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)(x^2-1) \end{align*} Setting $\sqrt{2}x=t$ we get \begin{align*} S&=8\int_0^{1/\sqrt{2}}\frac{x-\frac{1}{\sqrt{2}}}{(x^2-\sqrt{2}x+1)(x^2-1)}\,dx\\ &=16\int_0^{1}\frac{t-1}{(t^2-2t+2)(t^2-2)}\,dt\\ &=-2\int_0^1\frac{2t-2}{t^2-2t+2}\,dt+4\int_0^1\frac{dt}{1+(t-1)^2}+2\int_0^1\frac{-2t}{2-t^2}\,dt\\ &=[-2\log(t^2-2t+2)+4\arctan(t-1)+2\log(2-t^2)]_{t=0}^{t=1}\\ &=\pi \end{align*}

and (3) follows.

Note: These three gems are from The number $\pi$ by P. Eymard and J-P. Lafon which contains many more fascinating identities of $\pi$.

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    $\begingroup$ $(2)$ and this is why we should provide proofs when we discover things, or else it may take a couple centuries for someone to fully understand our work. $\endgroup$ Mar 5, 2017 at 22:53
  • $\begingroup$ A Simply Beautiful (art) answer! Well done! +1 $\endgroup$
    – Crescendo
    Mar 5, 2017 at 23:03
  • 1
    $\begingroup$ @Crescendo: Thanks a lot! It was really a challenge to decide which of the many fascinating identities should be presented. :-) $\endgroup$
    – epi163sqrt
    Mar 5, 2017 at 23:11
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    $\begingroup$ @SimplyBeautifulArt : Lack of many proofs in Ramanujan's work is not because he did not want to provide proofs, or not because he did not have proofs (like case of Fermat), but mainly because of lack of resources like paper and time. Hardy remarked that there was no point asking him proofs of each and every identity because he was generating 10-20 identities everyday. If Ramanujan devoted his short life in providing proofs, he would have created much less amount of mathematics. He did however give rough sketches for proofs (including those for the series like $(2)$). Cont'd... $\endgroup$
    – Paramanand Singh
    May 19, 2017 at 6:31
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    $\begingroup$ @SimplyBeautifulArt : my only lament is that he did not provide any insight or rough sketch about how he found his modular equations. Apart from this he did try to give details of his method in his Notebooks or papers. Maybe he did not want to show all those very long tedious calculations involved in the process. In cases where Ramanujan gave proofs they are much better and easier to understand compared to what modern mathematicians did with his work. $\endgroup$
    – Paramanand Singh
    May 19, 2017 at 6:34

Many such formulas come from the generalized binomial expansion theorem or geometric series and a bit of interpretation of the definition of $\pi$. One such example is the Leibniz formula for pi, which comes by noting that

$$\int_0^x\frac1{1+t^2}\ dt=\arctan(x)$$

From here, it follows that

$$\frac\pi4=\arctan(1)=\int_0^x\frac1{1+t^2}\ dt=\int_0^x\sum_{n=0}^\infty(-1)^nt^{2n}\ dt=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$

By applying an Euler transform to this, we get another representation of pi:


You could take the geometric meaning of pi as area (integral) of a circle to deduce that

$$\frac\pi4=\int_0^1\sqrt{1-x^2}\ dx=\int_0^1\sum_{n=0}^\infty\binom{1/2}n(-1)^nx^{2n}\ dx=\sum_{n=0}^\infty\binom{1/2}n\frac{(-1)^n}{2n+1}$$

You noted that


This is a special case of the Riemann zeta function, which yields another form after an Euler transform:

$$\frac{\pi^2}6=2\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^2}$$

which converges much more rapidly.

Other places pi may show up, relating especially to logarithms:


Which is famously known as Euler's formula.

Beyond this, I think the formulas get less and less intuitive and more like a race for the best formula to apply.

  • $\begingroup$ Nice answer, however, I'm still having trouble simplifying$$\sum\limits_{i=0}^{\infty}\dfrac {1}{2^{i+1}}\sum\limits_{j=0}^{\infty}\binom{i}{j}\dfrac {(-1)^j}{2j+1}=\sum\limits_{n=0}^{\infty}\dfrac {n!}{(2n+1)!!}$$Any help? $\endgroup$
    – Crescendo
    Mar 5, 2017 at 2:01
  • $\begingroup$ @Crescendo Did you check the link? math.stackexchange.com/questions/2149340/… $\endgroup$ Mar 5, 2017 at 2:07
  • $\begingroup$ Yes, but I'm still not sure how $$\sum_{k=0}^0\binom0k\frac{(-1)^k}{2k+1}=\frac11=2^0\frac{0!}{1!!}$$ $$\sum_{k=0}^1\binom1k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)=2^1\frac{1!}{3!!}$$ $$\sum_{k=0}^2\binom2k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)=2^2\frac{2!}{5!!}$$ $$\sum_{k=0}^3\binom3k\frac{(-1)^k}{2k+1}=\left[\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)\right]-\left[\left(\frac13-\frac15\right)-\left(\frac15-\frac17\right)\right]=2^3\frac{3!}{7!!}$$Helps you simplify the nested summations... $\endgroup$
    – Crescendo
    Mar 5, 2017 at 3:26
  • $\begingroup$ Well, notice the pattern, where we repeatedly subtract more and more terms. You could think of it like this:$$a_{k,1}=\frac1{2k+1}-\frac1{2k+3}$$ then the second sum is just $a_{0,1}$. Then define $a_{k,n+1}=a_{k,n}-a_{k+1,n}$ and show that $a_{0,n}$ is equal to our sums, finding the closed form by induction. $\endgroup$ Mar 5, 2017 at 12:00

One method, no doubt, is due to reasoning involving the classic definition of $pi$ (the ratio of the diameter to the circumference). For example the ratio of the diameter of a regular polygon to perimeter as the number of sides goes to infinity gives $\pi$. Starting with the square and doubling the number of sides of the polygon yields the sequence $$2\sqrt2$$ $$4\sqrt{2-\sqrt2}$$ $$8\sqrt{2-\sqrt{2+\sqrt2}}$$ $$16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt2}}}$$ $$\dots$$

You could derive something similar for $\pi^2$ or $\pi^3$ using the areas of regular polygons, surface areas/volumes of convex regular polyhedrons, etc.

Just speculating: one might get inspiration from physical phenomenon such as angular velocity vs linear velocity for an object traveling in a circle, angular momentum, torque, etc.


It's a little of both, really. Often what happens is that they find an equation for something, and see an opportunity to get $\pi$ out of it. My favorite example is the formula $\pi = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots$. This comes from the Taylor series expansion of $\arctan{x}$, which is $\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$. At some point, someone noticed that $\arctan{x}$ often outputs multiples of $\pi$; in particular, $\arctan{1} = \frac{\pi}{4}$. Plugging in $x = 1$ to the Taylor series and then multiplying both sides by $4$ gives the formula.

I don't know of any situation where someone really set out for a formula for $\pi$; it's usually a mathematician working on something else who happens across a new formula. But it's rarely as direct as $\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$; it's usually that they have to plug in particular values to make $\pi$ happen.


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