Pumpkin Pi


Fig. 1

Among many images of carved pumpkin, I like the one above (see Fig. 1) the most. It shows Leibniz’s formula for calculating the value of \pi. Namely,


To derive this formula, we begin with finding the derivative of \arctan{x}:

Let y = \arctan{x}, we have x=tan(y), and

\frac{d}{dx}x=\frac{d}{dx}\tan{y}=\frac{d}{dy}\tan{y}\frac{dy}{dx}=\sec^{2}{y}\frac{dy}{dx} = (1+tan^{2}{y})\frac{dy}{dx}

= (1+x^{2})\frac{dy}{dx}.

Since \frac{d}{dx}x=1,


It follows that by (1) and the Fundamental Theorem of Calculus,

\int\limits_{0}^{1}\frac{1}{1+x^2}dx = \arctan{x}\bigg|_{0}^{1}=\frac{\pi}{4}


\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad\quad\quad\quad(2)

From carrying out polynomial long division, we observe

\frac{1}{1+x^2} = 1 + \frac{-x^2}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + \frac{x^4}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + x^4  + \frac{-x^6}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +  \frac{x^8}{1+x^2}.

It seems that

\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}{(-1)^{k+1}x^{2k-2}} + \frac{(-1)^{n} x^{2n}}{1+x^2}\quad\quad\quad\quad\quad\quad(3)

Assuming (3) is true, we integrate it with respect to x from 0 to 1,

 \int\limits_{0}^{1}\frac{1}{1+x^2}dx=\int\limits_{0}^{1}\sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{1}\frac{(-1)^{n} x^{2n}}{1+x^2}dx

= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{1}x^{2k-2}dx +(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx

= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{1}+(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx .

As a result of integration,  (2) becomes

\frac{\pi}{4} = \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} + (-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx,


\frac{\pi}{4} - \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} =(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx.


|\frac{\pi}{4} -  \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|=|(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx | = \int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx < \int\limits_{0}^{1}x^{2n}dx

=\frac{x^{2n+1}}{2n+1}\bigg|_{0}^{1}= \frac{1}{2n+1}.

Moreover, \forall \epsilon > 0, we obtain n > \frac{1}{2}(\frac{1}{\epsilon}-1) through solving \frac{1}{2n+1} < \epsilon. It means that \forall \epsilon > 0, \exists n^*=\frac{1}{2}(\frac{1}{\epsilon}-1) such that  for all n > n^*, |\frac{\pi}{4} -  \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|<\epsilon, i.e.,

\lim\limits_{n\to\infty}{ \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}}=\frac{\pi}{4}.


\pi = 4 \sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}\quad\quad\quad(4)

The numerical value of \pi is therefore approximated according to (4) by the partial sum

 4 \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}=4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots+(-1)^{n+1}\frac{1}{2n-1})\quad\quad\quad\quad(5)

Its value converges to \pi as n increases.

However, (5) is by no means a practical way of finding the value of \pi, since its convergence is so slow that many terms must be summed up before a reasonably accurate result emerges (see Fig. 2)

Screen Shot 2017-08-10 at 9.33.13 PM.png

Fig. 2

I doubt Leibniz has ever used his own formula to obtain the value of \pi !

Let me leave you with an exercise: Prove (3)


1 thought on “Pumpkin Pi

  1. Pingback: Piece of Pi | Vroom

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