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Computing pi with tools from Calculus

Computing $\pi$

This note was originally written in the context of my fall Math 100 class at Brown University. It is also available as a pdf note.

While investigating Taylor series, we proved that
\frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} + \cdots
Let’s remind ourselves how. Begin with the geometric series
\frac{1}{1 + x^2} = 1 – x^2 + x^4 – x^6 + x^8 + \cdots = \sum_{n = 0}^\infty (-1)^n x^{2n}. \notag
(We showed that this has interval of convergence $\lvert x \rvert < 1$). Integrating this geometric series yields
\int_0^x \frac{1}{1 + t^2} dt = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag
Note that this has interval of convergence $-1 < x \leq 1$.

We also recognize this integral as
\int_0^x \frac{1}{1 + t^2} dt = \text{arctan}(x), \notag
one of the common integrals arising from trigonometric substitution. Putting these together, we find that
\text{arctan}(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag
As $x = 1$ is within the interval of convergence, we can substitute $x = 1$ into the series to find the representation
\text{arctan}(1) = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{1}{2n+1}. \notag
Since $\text{arctan}(1) = \frac{\pi}{4}$, this gives the representation for $\pi/4$ given in \eqref{eq:base}.

However, since $x=1$ was at the very edge of the interval of convergence, this series converges very, very slowly. For instance, using the first $50$ terms gives the approximation
\pi \approx 3.121594652591011. \notag
The expansion of $\pi$ is actually
\pi = 3.141592653589793238462\ldots \notag
So the first $50$ terms of \eqref{eq:base} gives two digits of accuracy. That’s not very good.

I think it is very natural to ask: can we do better? This series converges slowly — can we find one that converges more quickly?


As an aside: one might also ask if we can somehow speed up the convergence of the series we already have. It turns out that in many cases, you can! For example, we know in alternating series that the sum of the whole series is between any two consecutive partial sums. So what if you took the average of two consecutive partial sums? [Equivalently, what if you added only one half of the last term in a partial sum. Do you see why these are the same?]

The average of the partial sum of the first 49 terms and the partial sum of the first 50 terms is actually
3.141796672793031, \notag
which is correct to within $0.001$. That’s an improvement!

What if you do still more? More on this can be found in the last Section.

Estimating $\pi$ through a different series

We return to the question: can we find a series that gives us $\pi$, but which converges faster? Yes we can! And we don’t have to look too far — we can continue to rely on our expansion for $\text{arctan}(x)$.

We had been using that $\text{arctan}(1) = \frac{\pi}{4}$. But we also know that $\text{arctan}(1/\sqrt{3}) = \frac{\pi}{6}$. Since $1/\sqrt{3}$ is closer to the center of the power series than $1$, we should expect that the convergence is much better.

Recall that
\text{arctan}(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n + 1}}{2n + 1}. \notag
Then we have that
\text{arctan}\left(\frac{1}{\sqrt 3}\right) &= \frac{1}{\sqrt 3} – \frac{1}{3(\sqrt 3)^3} + \frac{1}{5(\sqrt 3)^5} + \cdots \notag \\
&= \frac{1}{\sqrt 3} \left(1 – \frac{1}{3 \cdot 3} + \frac{1}{5 \cdot 3^2} – \frac{1}{7 \cdot 3^3} + \cdots \right) \notag \\
&= \frac{1}{\sqrt 3} \sum_{n = 0}^\infty (-1)^n \frac{1}{(2n + 1) 3^n}. \notag
Therefore, we have the equality
\frac{\pi}{6} = \frac{1}{\sqrt 3} \sum_{n = 0}^\infty (-1)^n \frac{1}{(2n + 1) 3^n} \notag
or rather that
\pi = 2 \sqrt{3} \sum_{n = 0}^\infty (-1)^n \frac{1}{(2n + 1) 3^n}. \notag
From a computation perspective, this is far superior. For instance, based on our understanding of error from the alternating series test, using the first $10$ terms of this series will approximate $\pi$ to within
2 \sqrt 3 \frac{1}{23 \cdot 3^{11}} \approx \frac{1}{26680}. \notag

Let’s check this.
2 \sqrt 3 \left(1 – \frac{1}{3\cdot 3} + \frac{1}{5 \cdot 3^2} + \cdots + \frac{1}{21 \cdot 3^{10}}\right) = 3.1415933045030813. \notag
Look at how close that approximation is, and we only used the first $10$ terms!
Roughly speaking, each additional 2.5 terms yields another digit of $\pi$. Using the first $100$ terms would give the first 48 digits of $\pi$.
Using the first million terms would give the first 47000 (or so) digits of $\pi$ — and this is definitely doable, even on a personal laptop. (On my laptop, it takes approximately 4 milliseconds to compute the first 20 digits of $\pi$ using this technique).

Even Better Series

I think it is very natural to ask again: can we find an even faster converging series? Perhaps we can choose better values to evaluate arctan at? This turns out to be a very useful line of thought, and it leads to some of the best-known methods for evaluating $\pi$. Through clever choices of values and identities involving arctangents, one can construct extremely quickly converging series for $\pi$. For more information on this line of thought, look up Machin-like formula.

Patterns in the Approximation of $\pi/4$


Looking back at the approximation of $\pi$ coming from the first $50$ terms of the series
1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \cdots
we found an approximation of $\pi$, which I’ll represent as $\widehat{\pi}$,
\pi \approx \widehat{\pi} = 3.121594652591011. \notag
Let’s look very carefully at how this compares to $\pi$, up to the first $10$ decimals. We color the incorrect digits in ${\color{orange}{orange}}$.
\pi &= 3.1415926535\ldots \notag \\
\widehat{\pi} &= 3.1{\color{orange}2}159{\color{orange}4}65{\color{orange}2}5 \notag
Notice that most of the digits are correct — in fact, only three (of the first ten) are incorrect! Isn’t that weird?

It happens to be that when one uses the first $10^N / 2$ terms (for any $N$) of the series \eqref{eq:series_pi4_base}, there will be a pattern of mostly correct digits with disjoint strings of incorrect digits in the middle. This is an unusual and surprising phenomenon.

The positions of the incorrect digits can be predicted. Although I won’t go into any detail here, the positions of the errors are closely related to something called Euler Numbers or, more deeply, to Boole Summation.

Playing with infinite series leads to all sorts of interesting patterns. There is a great history of mathematicians and physicists messing around with series and stumbling across really deep ideas.

Speeding up computation

Take an alternating series
\sum_{n = 0}^\infty (-1)^{n} a_n = a_0 – a_1 + a_2 – a_3 + \cdots \notag
If ${a_n}$ is a sequence of positive, decreasing terms with limit $0$, then the alternating series converges to some value $S$. And further, consecutive partial sums bound the value of $S$, in that
\sum_{n = 0}^{2K-1} (-1)^{n} a_n \leq S \leq \sum_{n = 1}^{2K} (-1)^{n} a_n. \notag
For example,
1 – \frac{1}{3} < \sum_{n = 0}^\infty \frac{(-1)^{n}}{2n+1} < 1 – \frac{1}{3} + \frac{1}{5}. \notag

Instead of approximating the value of the whole sum $S$ by the $K$th partial sum $\sum_{n \leq K} (-1)^n a_n$, it might seem reasonable to approximate $S$ by the average of the $(K-1)$st partial sum and the $K$th partial sum. Since we know $S$ is between the two, taking their average might be closer to the real result.

As mentioned above, the average of the partial sum consisting of the first $49$ terms of \eqref{eq:base} and the first $50$ terms of \eqref{eq:base} gives a much improved estimate of $\pi$ than using either the first $49$ or first $50$ terms on their own. (And indeed, it works much better than even the first $500$ terms on their own).

Before we go on, let’s introduce a little notation. Let $S_K$ denote the partial sum of the terms up to $K$, i.e.
S_K = \sum_{n = 0}^K (-1)^{n} a_n. \notag
Then the idea is that instead of using $S_{K}$ to approximate the wholse sum $S$, we’ll use the average
\frac{S_{K-1} + S_{K}}{2} \approx S. \notag

Averaging once seems like a great idea. What if we average again? That is, what if instead of using the average of $S_{K-1}$ and $S_K$, we actually use the average of (the average of $S_{K-2}$ and $S_{K-1}$) and (the average of $S_{K_1}$ and $S_K$),
\frac{\frac{S_{K-2} + S_{K-1}}{2} + \frac{S_{K-1} + S_{K}}{2}}{2}.
As this is really annoying to write, let’s come up with some new notation. Write the average between a quantity $X$ and $Y$ as
[X, Y] = \frac{X + Y}{2}. \notag
Further, define the average of $[X, Y]$ and $[Y, Z]$ to be $[X, Y, Z]$,
[X, Y, Z] = \frac{[X, Y] + [Y, Z]}{2} = \frac{\frac{X + Y}{2} + \frac{Y + Z}{2}}{2}. \notag
So the long expression in \eqref{eq:avgavg} can be written as $[S_{K-2}, S_{K-1}, S_{K}]$.

With this notation in mind, let’s compute some numerics. Below, we give the actual value of $\pi$, the values of $S_{48}, S_{49}$, and $S_{50}$, pairwise averages, and the average-of-the-average, in the case of $1 – \frac{1}{3} + \frac{1}{5} + \cdots$.
\begin{equation} \notag
& \text{Value} & \text{Difference from } \pi \\ \hline
\pi & 3.141592653589793238462\ldots & \phantom{-}0 \\ \hline
4 \cdot S_{48} & 3.1207615795929895 & \phantom{-}0.020831073996803617 \\ \hline
4 \cdot S_{49} & 3.161998692995051 & -0.020406039405258092 \\ \hline
4 \cdot S_{50} & 3.121594652591011 & \phantom{-}0.01999800099878213 \\ \hline
4 \cdot [S_{48}, S_{49}] & 3.1413801362940204 & \phantom{-}0.0002125172957727628 \\ \hline
4 \cdot [S_{49}, S_{50}] & 3.1417966727930313 & -0.00020401920323820377 \\ \hline
4 \cdot [S_{48}, S_{49}, S_{50}] & 3.141588404543526 & \phantom{-}0.00000424904626727951 \\ \hline
So using the average of averages from the three sums $S_{48}, S_{49}$, and $S_{50}$ gives $\pi$ to within $4.2 \cdot 10^{-6}$, an incredible improvement compared to $S_{50}$ on its own.

There is something really odd going on here. We are not computing additional summands in the overall sum \eqref{eq:base}. We are merely combining some of our partial results together in a really simple way, repeatedly. Somehow, the sequence of partial sums contains more information about the limit $S$ than individual terms, and we are able to extract some of this information.

I think there is a very natural question. What if we didn’t stop now? What if we took averages-of-averages-of-averages, and averages-of-averages-of-averages-of-averages, and so on? Indeed, we might define the average
[X, Y, Z, W] = \frac{[X, Y, Z] + [Y, Z, W]}{2}, \notag
and so on for larger numbers of terms. In this case, it happens to be that
[S_{15}, S_{16}, \ldots, S_{50}] = 3.141592653589794,
which has the first 15 digits of $\pi$ correct!

By repeatedly averaging alternating sums of just the first $50$ reciprocals of odd integers, we can find $\pi$ up to 15 digits. I think that’s incredible — it seems both harder than it might have been (as this involves lots of averaging) and much easier than it might have been (as the only arithmetic input are the fractions $1/(2n+1)$ for $n$ up to $50$.

Although we leave the thread of ideas here, there are plenty of questions that I think are now asking themselves. I encourage you to ask them, and we may return to this (or related) topics in the future. I’ll see you in class.

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Towards an Expression for pi II

Continuing from this post

We start with $latex \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … \cos(\dfrac{\xi}{2^n})$. Recall the double angle identity for sin: $latex \sin 2 \theta = 2\sin \theta \cos \theta $. We will use this a lot.

Multiply our expression by $latex \sin(\dfrac{\xi}{2^n})$. Then we have

$latex \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … \cos(\dfrac{\xi}{2^n})\sin(\dfrac{\xi}{2^n})$

Using the double angle identity, we can reduce this:

$latex = \dfrac{1}{2} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … \cos(\dfrac{\xi}{2^{n-1}})sin(\dfrac{\xi}{2^{n-1}}) =$
$latex = \dfrac{1}{4} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … \cos(\dfrac{\xi}{2^{n-2}})\sin(\dfrac{\xi}{2^{n-2}}) =$
$latex …$
$latex = \dfrac{1}{2^{n-1}}\cos(\xi / 2)\sin(\xi / 2) = \dfrac{1}{2^n}\sin(\xi)$

So we can rewrite this as

$latex \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … \cos(\dfrac{\xi}{2^n}) = \dfrac{\sin \xi}{2^n \sin( \dfrac{\xi}{2^n} )}$ for $latex \xi \not = k \pi$

Because we know that $latex lim_{x \to \infty} \dfrac{\sin x}{x} = 1$, we see that $lim_{n \to \infty} \dfrac{\xi / 2^n}{\sin(\xi / 2^n)} = 1$. So we see that

$latex \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) … = \dfrac{\xi}{\xi}$
$latex \xi = \dfrac{\sin(\xi)}{\cos(\dfrac{\xi}{2})\cos(\dfrac{\xi}{4})…}$

Now we set $latex \xi := \pi /2$. Also recalling that $latex \cos(\xi / 2 ) = \sqrt{ 1/2 + 1/2 \cos \xi}$. What do we get?

$latex \dfrac{\pi}{2} = \dfrac{1}{\sqrt{1/2} \sqrt{ 1/2 + 1/2 \sqrt{1/2} } \sqrt{1/2 + 1/2 \sqrt{ 1/2 + 1/2 \sqrt{1/2} …}}}$

This is pretty cool. It’s called Vieta’s Formula for $latex \dfrac{\pi}{2}$. It’s also one of the oldest infinite products.

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Towards an Expression for Pi

I have stumbled across something beautiful! I haven’t the time to write of it now, but I can allude to it without fear. Eventually, I will reproduce a very fascinating formula for $latex \pi$.

But first:

Consider the following expression:

$latex \cos \dfrac{\xi}{2} \cos \dfrac{\xi}{4} \cos \dfrac{\xi}{8} … \cos \dfrac{\xi}{2^n}$

It can be simplified into a very simple quotient of $latex sin$ in terms of $latex \xi$.

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An even later pi day post

In my post dedicated to pi day, I happened to refer to a musical interpretation of pi. This video (while still viewable from the link I gave) has been forced off of YouTube due to a copyright claim. The video includes an interpretation by Michael Blake, a funny and avid YouTube artist. The copyright claim comes from Lars Erickson – he apparently says that he created a musical creation of pi first (and… I guess therefore no others are allowed…). In other words, it seems very peculiar.

I like Vi Hart’s treatment of the copyright claim. For completeness, here is Blake’s response.

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A late pi day post

As this blog started after March 14th, it hasn’t paid the proper amount of attention to $latex \pi $. I only bring this up because I have just been introduced to Christopher Poole’s intense dedication to $latex \pi $. It turns out that Christopher has set up a $latex \pi $-phone, i.e. a phone number that you can call if you want to hear $latex pi $. It will literally read out the digits of $latex \pi $ to you. I’ve only heard the first 20 or so digits, but perhaps the more adventurous reader will find out more. The number is 253 243-2504. Call it if you are ever in need of some $latex \pi $.

Of course, I can’t leave off on just that – I should at least mention two other great $latex \pi $-day attractions (late as they are). Firstly, any unfamiliar with the $latex \tau $ movement should read up on it or check out Vi Hart’s pleasant video. I also think it’s far more natural to relate the radius to the circumference rather than the diameter to the circumference (but it would mean that area becomes not as pleasant as $latex \pi r^2 $).

Finally, there is a great musical interpretation and celebration of $latex \pi $. What if you did a round (or fugue) based on interpreting each digit of $latex \pi$ as a different musical note? Well, now you can find out!

Until $latex \tau $ day!

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