approximating pi part 2

_posts/2024-05-09-approximating-pi-pt-2.md

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+---
+layout: post
+title:  "Approximating π, part 2"
+tags: math calculus
+image: https://rhinopotamus.github.io/images/tfd-pi.gif
+katex: True
+---
+
+[Last time we met](2024-05-08-approximating-pi), we found a Taylor series for the arctangent function, and we used the fact that π = 4 arctan(1) to find approximations to π. Unfortunately, this representation had really poor convergence behavior, because 1 is way the hell out at the edge of the interval of convergence for this series. I'm loath to abandon a fun trick, though; is there a way we can iterate on this idea?
+
+## Approximating π better
+
+When I start thinking about π, I start thinking about radians. For instance, I know that $\tan\left(\frac\pi4\right)= 1$. Therefore, $\pi = 4\arctan(1)$, but that fact wasn't very helpful. 
+
+Another fact I know about the unit circle is that $\tan\left(\dfrac\pi6\right) = \dfrac{1}{\sqrt{3}}$. Therefore, $\pi = 6 \arctan\left(\dfrac{1}{\sqrt{3}}\right)$. I think this might yield a better approximation: I know off the top of my head[^1] that $\frac{1}{\sqrt{3}}$ is like 0.6, so it's substantially in the interior of the interval of convergence. Let's see how this works out; note that for algebra reasons I wish to write $\frac{1}{\sqrt{3}}$ as $3^{-1/2}$.
+
+$$
+\begin{align*}
+\pi = 6\arctan\left(3^{-1/2}\right) &= 6 \cdot \displaystyle\sum_{k=0}^\infty (-1)^k\cdot \dfrac{1}{2k+1} \cdot \left(3^{-1/2}\right)^{2k+1} \\
+&= 6 \cdot \displaystyle\sum_{k=0}^\infty (-1)^k \cdot \dfrac{1}{2k+1} \cdot 3^{-k-1/2} \\
+&= 6 \cdot \displaystyle\sum_{k=0}^\infty (-1)^k \cdot \dfrac{1}{2k+1} \cdot \underbrace{\dfrac{1}{3^k}}_{!!!}\cdot\dfrac{1}{\sqrt{3}}
+\end{align*}
+$$
+
+I'm already lots happier with this representation, because look at that $3^k$ in the denominator; very beautiful, very powerful. This is going to converge *way* faster. Let's also factor out that constant at the end:
+
+$$
+\pi = 6\arctan(3^{-1/2}) = 2\sqrt{3} \cdot \displaystyle\sum_{k=0}^\infty (-1)^k \cdot \dfrac{1}{2k+1} \cdot \dfrac{1}{3^k}.
+$$
+
+### The alternating series error bound
+
+Happily, this is still an alternating series, so we can still use the alternating series error bound $|S - S_n| < a_{n+1}$ to easily find reasonably tight[^2] bounds on the truncation error. In this case, we find that:
+
+$$|\pi - S_n| < \dfrac{2\sqrt{3}}{3^{n+1}\cdot(2n+3)}$$
+
+I could try to set this equal to whatever I want my error bound to be and solve for $n$, but unfortunately anything that looks like $b^n\cdot n$ [isn't going to have nice solutions](https://en.wikipedia.org/wiki/Lambert_W_function). Therefore, more fun approaches are to spreadsheet this to death and/or to look at a graph of this function: 
+
+| k           | 0        | 1        | 2        | 3        | 4        | 5        |
+|-------------|----------|----------|----------|----------|----------|----------|
+| error bound | 3.464102 | 0.384900 | 0.076980 | 0.018329 | 0.004752 | 0.001296 |
+
+Here's a graph:
+
+![A graph of f(n)=\frac{2\sqrt{3}}{3^{n+1}*(2n+3)}](/images/pi-sqrt-3-linear.png)
+
+As you can see, those outputs get real small real quick; a graph on a logarithmic scale may be a better choice:
+
+![A graph of f(n)=\frac{2\sqrt{3}}{3^{n+1}*(2n+3)} on a logarithmic scale](/images/pi-sqrt-3-log.png)
+
+Hell yeah, these error bounds kick ass. I'm already down to $10^{-6}$ by the tenth term! Let us therefore compute a few approximations of π; here's the first ten, rounded to the sixth decimal place:
+
+| k   | 0        | 1        | 2        | 3        | 4        | 5        | 6        | 7        | 8        | 9        |
+|-----|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|
+| π ≈ | 3.464102 | 3.079201 | 3.156181 | 3.137853 | 3.142605 | 3.141309 | 3.141674 | 3.141569 | 3.141600 | 3.141591 |
+
+This is much better, and much more fun, than the approximations we were getting before.
+
+### However...
+
+We got into this whole mess by trying to figure out a way to approximate the transcendental number π, and we've done a pretty nice job. There's a sneaky circular problem here, though: When we wrote
+
+$$
+\pi = 6\arctan(3^{-1/2}) = 2\sqrt{3} \cdot \displaystyle\sum_{k=0}^\infty (-1)^k \cdot \dfrac{1}{2k+1} \cdot \dfrac{1}{3^k}
+$$
+
+as a way to approximate the transcendental number π, we used the irrational number $\sqrt{3}$. 
+
+## How do you know what $\sqrt{3}$ is?
+
+![the oh-you dog meme with pi and the square root of 3 looking at each other](/images/pi-sqrt-3-oh-you.png)
+
+[^1]: Because that is who I am as a person.
+
+[^2]: In practice, the bounds you get from this technique tend to be off by a bit more than a factor of 2. Do you see why?