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On Approximating Pi

So, you want to approximate $\pi$ from first principles (ie, no fancy formulas that you don’t know how to prove). One of the most straightforward ways to do this is through the method that Archimedes used thousands of years ago. He did this by looking at the perimeter of regular polygons, which he could compute. The more sides that the polygon had, the closer the perimeter was to $\pi$. In modern notation, his method amounts to the formula

$$ \lim_{x \rightarrow 0} \frac{\sin(\pi x)}{x} = \pi = \lim_{x \rightarrow 0} \frac{\tan(\pi x)}{x} $$

Since we can compute things like $\sin(\pi/2^n)$ without actually knowing what $\pi$ is, we can use this to find an approximation for $\pi$. But just knowing this formula is not good enough. For instance, how do we know how close we are to $\pi$? If we use $x=2^{-100}$ to do this, then exactly how close we are? If we want to get $\pi$ to, say, ten digits, then what power of two do we need in the denominator?

These questions can be answered by going through the proofs and keeping track of how these limits actually work. Through this, we can get a refined look at exactly how to approximate $\pi$. (And gain an appreciation for $\epsilon-\delta$ in limits.)

Our Fundamental Identity

Firstly, we need to show that for any $-\frac{\pi}{2} < x < \frac{\pi}{2}$ we have

$$ |x\cos(x)| \leq |\sin(x)| \leq |x| \leq |\tan(x)| \leq |x\sec(x)|$$

where we have equality only at $x=0$.

This is where the diagram in the picture comes into play. Note that we clearly have equality at $x=0$ and, by symmetry, we just have to check this for $x>0$. In the picture, we have a quarter-circle of radius 1, with a line from the origin to the circle that makes an angle of $\theta$ with the horizontal axis. We can make this to make a triangle as shown. The length of the horizontal leg of this triangle is $\cos(\theta)$ and the length of the vertical leg of this triangle is $\sin(\theta)$. The blue arc in the picture is clearly larger than the vertical leg of the triangle and, since we are in radians, this arc has length $\theta$. This gives us that $\sin(\theta)<\theta$. On the other hand, the pink highlighted arc has radius $\cos(\theta)$ which means that its length is $\theta\cos(\theta)$. Since this arc is less than the vertical leg of the triangle, we get the leftmost inequality. Finally, through similar reasoning with the larger triangle and arc, we get the final two inequalities.

The Limit Lemma

Next we will look at our first limit. We want to show that

$$\lim_{x\rightarrow 0}\cos(x) = 1 = \lim_{x\rightarrow 0} \sec(x) $$

This means working with $\epsilon$s and $\delta$s, and it will be important to keep track of this information, as it tells us how to actually do the approximation.

Let $\epsilon >0$. We have the basic trig identity: $1-\cos(x) = 2\sin^2(x/2)$. This means that, using our above inequality, that $|\cos(x)-1| <x^2/2$. Therefore if $0<|x|<\sqrt{2\epsilon}$, then $|\cos(x)-1|<\epsilon$. This proves the first limit.

We will use this to prove the second limit. We have $\sec(x)-1 = (1-\cos(x))/\cos(x)$. Suppose that $0<|x|<\sqrt{2\epsilon’}$. Using what we just did, we have $1-\cos(x) < \epsilon’$ and $1-\epsilon’ < \cos(x)$. Combining these gives $\sec(x)-1 < \frac{\epsilon’}{1-\epsilon’}$. Choose $\epsilon’=\epsilon/(1+\epsilon)$. Plugging this into $\frac{\epsilon’}{1-\epsilon’}$ simplifies to $\epsilon$. So, if $0<|x|< \frac{\epsilon}{1+\epsilon}$ then $|1-\sec(x)|<\epsilon$. This proves the second limit.

To summarize the important bits of what we just did, since $ \frac{\epsilon}{1+\epsilon} < \epsilon$, it follows that if $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}} $ then

$$ |1-\cos(x)|< \epsilon \;\;\;\;\&\;\;\;\;\; |1-\sec(x)| < \epsilon$$

The Main Limit

We’re now done with lemmas and can prove the main result, which is: For $0<|x|<\frac{\pi}{2}$ we have that

$$ \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x}$$

Moreover, we have the limits

$$ \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 = \lim_{x\rightarrow 0}\frac{\tan(x)}{x}$$

Dividing our original inequality through by $x$ gives

$$ \cos(x) < \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x} < \sec(x)$$

This holds for all $x$ in our range because we are able to use the evenness of these functions to get rid of the absolute value. We then immediately have the first statement of this, and we just have to do the limit.

For the limits, we will keep track of our $\epsilon$s and $\delta$s. Obviously, using the divided expression, the Squeeze Lemma will give our result, but we want a more refined result, with errors in place. Take the first two inequalities, and rearrange to get

$$ 0 < 1-\frac{\sin(x)}{x} < 1-\cos(x)$$

By subtracting through by $1$ and multiplying by $-1$. Similarly for the latter two inequalities we have:

$$ 0 < \frac{\tan(x)}{x} -1 < \sec(x) -1$$

If we then take $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}}$, we find, using our previous results, that

$$ 0 < 1- \frac{\sin(x)}{x} < \epsilon$$

$$ 0 < \frac{\tan(x)}{x} - 1 < \epsilon$$

This proves the result, and we know how the error works.

Approximating Pi

We will now use this to approximate $\pi$. In our final result, replace $x$ with $\pi x$, multiply our equations through by $\pi$ and replace $\epsilon$ with $\epsilon/\pi$. This results in the statement: If $0<|x|<\frac{1}{\pi} \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ then

$$ 0 < \pi- \frac{\sin(\pi x)}{x} < \epsilon$$

$$ 0 < \frac{\tan(\pi x)}{x} - \pi < \epsilon$$

This means that the sine inequality gives a lower bound for $\pi$ with error $\epsilon$ and the tangent inequality gives an upper bound for $\pi$ with error $\epsilon$. Explicitly: Whenever $0<|x|<\frac{1}{\pi} \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ we have

$$\pi\in \left( \frac{\sin(\pi x)}{x}, \frac{\tan(\pi x)}{x} \right) \subseteq (\pi-\epsilon , \pi +\epsilon)$$

Follow this recipe and you’ll be able to approximate $\pi$ to your heart’s content.

Example

We can compute trig functions of angles $\pi/(3\cdot 2^n)$ by hand. If we want to compute $\pi$ to within $k$ digits, then the question we need to answer is: What does $n$ have to be?

Using $\epsilon = 1/10^{k+1}$, $x=3^{-1}2^{-n}$, and the trivial bound $3<\pi$, in our main result, we find that:

$$ \frac{1}{2\cdot 3^n} < \frac{1}{3} \sqrt{\frac{2\cdot 10^{-k-1}}{3+10^{-k-1}}}$$

which simplifies to

$$n> \frac{k+1}{2}\log_2(10) + \frac{1}{2}\log_2(3/2) $$

$$n > 1.66097k +1.95345$$

This gives a linear relationship between the number of digits that we get and how many powers of two we have to take.So, if, say, we want to get $k=3$ digits, then we need $n>6.9$, which means that we need $x=\frac{1}{384}$ will do. Archimedes did it for $x=\frac{1}{96}$, which means that $n=5$, which means that we would have to have $k<1.83$, or we can be sure his result is accurate with error $\epsilon = 0.01$. This isn’t too tight of a bound, unfortunately, but it’s what we can prove using elementary trigonometry.

The record is $k=22,459,157,718,361$ digits. If we were to do this using this method, we would have to have $n=3.73\times 10^{13}$, which means $x\approx 3^{-1}\cdot 2^{-10^{13}}$, which would be really tough.

Using Calculus and Taylor Series, we can show that the error is actually much tighter. The question and challenge would then be to figure out if you can tighten this error using elementary methods.

#pi #math #maths #approximation #trig #archimedes #limits

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