Pumpkin Pi

pi.jpg

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,

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

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,

\frac{d}{dx}\arctan{x}=\frac{1}{1+x^2}\quad\quad\quad(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}

i.e.,

\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,

or,

\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.

Therefore,

|\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}.

Thus

\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)

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1 thought on “Pumpkin Pi

  1. Pingback: Piece of Pi | Vroom

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