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local-to-global-blog · 7 years

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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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icarolorran · 7 years

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Complex Numbers and the Stereographic projection

One of the most famous functions in the field of complex analysis is the following:

$$\begin{matrix} f: \mathbb{C}^2 \to \mathbb{C}\\\\ f(x,y) = \mathrm{e}^{i(x+iy)} = \cos{(x+iy)} + i\sin{(x+iy)} \end{matrix}$$

This function can simplify very complicated polynomial and differential equations, define closed forms for inverse trigonometric functions such as the arc tangent, and easen up the study of waves in several fields of physics. However, the most interesting property of this function is that it is defined everywhere in the complex plane with exception of the origin. When the argument is restricted only to real values, every pair $(x,y)$ on $\mathbb{R}^2$ can be brought to a point in the complex plane through it. In order to make the function injective, it is often convenient to restrict the domain of the function only to $\{(y,x)\in\mathbb{R}^2 | -\pi < y \leq \pi\}$ even though the particular choice for $y$ to range from $-\pi$ to $\pi$ is entirely conventional and can be taken elsewhere as long as the endpoints differ by $2\pi$; I will call this special region "$\mathcal{I}$"

Restricting our function to the aforementioned conditions allows us to rewrite it in the following manner:

$$ \begin{matrix} f: \mathcal{I} \to \mathbb{C} \backslash \{0\} \\\\ (x,y) \mapsto {e^{-x}}\left({\cos{y} + i\sin{y}}\right) \end{matrix} $$

Now let us look at what the image of the above function looks like for different values of $x$ and $y$. By keeping $y$ fixed while variating $x$ the function covers a straight line departing from zero (open) and extending to infinity while the angle it makes with the real axis is $y$:

If the opposite is done (now varying y) the image is a circle with center in the origin of radius $e^{-x}$. Notice that negative values of $x$ take place outisde the unitary circle whereas positive values are found inside of it:

Edit: In this image, read x<0 as the blue part and x>0 as the orange part.

There is something special at the mesh our function makes in the complex plane, it isn't only the mesh of an ordinary polar mapping. The fact that points inside the untary circle are assigned a positive $x$ while the outermost ones have a negative $x$ are analogous to a stereographic projection with the equator of the sphere being projected coinciding with the unitary circle in the plane.

In case you don't know what a stereographic projection is, it works in the following way: You place a sphere on top of a plane, a line coming from the north pole pierces the sphere in an assigned point and it eventually reaches the plane itself. With this construction, each point in the sphere can be associated with a point in the plane. However, our stereographic projection will have to differ slightly from the conventional one as our sphere (assumed to be unitary) will have half of its hemisphere on top of the complex plane so that the circle at the equator will coincide exactly with the unitary circle in the complex plane just like in the picture below.

Evidently, it seems that the complex plane can be mapped into a sphere through an stereographic projection. Our goal is to prove this hypothesis.

First, we need to define a space to work on that has three dimensions so that we can house our sphere and the complex plane inside of it. Let us then define a metric vector space, namely $\mathcal{A}$ that will be the usual euclidian space with the addition of product between complex numbers. In symbols:

$$\begin{matrix} \mathcal{A} = (\mathbb{E}^3, \circ) \\\\ \circ: \mathcal{A^2} \to \mathcal{A} \\\\ ((x,y,w),(a,b,w)) \mapsto (ax-by,xb+ya,w) \\\\ \end{matrix}$$

Notice that the notion of product is only defined between points in the same 'z' plane.

Also, the idea of distance is defined in the usual form:

$$|(x,y,z) - (a,b,w)| = \sqrt{(x-a)^2+(y-b)^2+(z-w)^2}$$

You can check by yourself that the complex plane, if regarded as a $\mathbb{R}^2$ vector space, is a subspace of $\mathcal{A}$ and it still holds its properties as a field. In fact, it corresponds to the set $\{(x,y,0)/x \land y \in \mathcal{A}\}$

Now, we can draw an sphere in this space with a parametrization that is slightly different from the usual spherical coordinates:

$$S(\theta,y) = (-\sin(2\theta)\cos(y),-\sin(2\theta)\sin(y),-\cos(2\theta))$$

With $\theta \in (0,\frac{\pi}{2})$ and $y \in (-\pi,\pi]$

This is indeed a sphere because the distance (as defined previously) from any point in this surface to the origin is given as follows:

$$|S(\theta,y)|^2 = (\sin(2\theta)\cos(y))^2 + (\sin(2\theta)\sin(y))^2 + (\cos(2\theta)))^2 = 1$$

The reason for the sphere to be parametrized this way is to make the meridians have their zero and range coinciding with the arguments of the complex numbers in $z = 0$ and to make the latitudinal coordinate be described by the angle a point makes with the center of the sphere at the north pole. With this parametrization in mind we can now proceed to construct the projection onto the complex plane.

For a fixed meridian, choose a point on it and call it $P$. Use the vectors coming from the north pole and the point $P$ to create a plane, namely $G$. Obviously, the north pole, the origin and $P$ are coplanar in this same plane. At this plane, construct a line passing through the north pole (call it $N$) and $P$ simultaneously and denote by $Q$ the point at which this line pierces the complex plane. From the fact that $Q$ is collinear with $P$ and $N$, $Q$ is also coplanar with $G$. As the angle between the north pole and any point on the complex plane is a right angle, which follows from the fact that $\mathcal{A}$ is still an euclidian space and that the line where $N$ is located is orthogonal to the comple x plane, then the triangle with vertices at the north pole, $Q$ and the origin form a rectangle triangle. From the parametrization of the sphere, the angle $P\hat{N}O$, where $O$ denotes the origin, is $\theta$ so that $Q\hat{N}O$ is also the same. As the segment $\overline{ON}$ is unitary, the segment $\overline{OQ}$ is then $\tan{\theta}$.

The geometry is basically done, we now just have to find a way to parametrize our findings. We can use the fact that the plane $G$ is a linear combination of the north pole and the point $P$. More specifically, as we only kept the meridian fixed, the coordintate $\theta$ does not matter. Another way to see this is to notice that any point $P$ can be written as a linear combination of another point in the equator with the same meridian $S(\frac{\pi}{4},y)$ and $N$:

$$S(\theta, y) = -\sin(2\theta)(\cos{y},\sin{y},0) -\cos(2\theta)(0,0,1) $$

Hence, any point on $G$ can be written in the form:

$$G(u,v) = u(0,0,1) + vS(\frac{\pi}{4},y)$$

From the fact that $Q$ is at $G$, that it is in the complex plane, so that it has $z=0$, and that it has a lenght equal to $\tan{\theta}$ we can find a pair of reals $u$ and $v$ that correspond to $Q$ in $G$. In fact, they are $u = 0$ and $v = \tan{\theta}$ so that $Q$ can be mapped as follows:

$$Q(\theta,y) = (\tan{\theta})(\cos(y),\sin(y))$$

Although $Q$ maps points from the sphere to points in the complex plane it is not exactly what we wanted, we need to reparametrize $Q$ such as to make an exponential appear. The easiest way is to associate an real $x$ to an $\theta$ as:

$$\tan{\theta} = \exp{-x}$$

We only have to show that the rule bringing $x$ to $\theta$ is one-to-one.

As $\theta$ is open at $\frac{\pi}{2}$ we can write our rule for all $x$ as

$$\begin{matrix} \theta:x \mapsto \arctan{(\exp(-x))} \end{matrix}$$

We have:

$$\begin{matrix} \arctan{(\exp(-x_1))} = \arctan{(\exp(-x_2))} \\\\ \exp(-x_1) = \exp(-x_2) \\\\ x_1 = x_2 \end{matrix}$$

Meaning that the mapping is one-to-one. Thus $Q$ can be reparametrized to:

$$Q(x,y) = {e^{-x}}(\cos(y),\sin(y))$$

Which, bringing back to the standard notation for the complex plane, is exactly the exponential function we defined earlier.This means that the exponential function on $\mathcal{I}$ can be seen as the stereographic projection of points in a sphere to the complex plane except at the North and South poles, which correspond to infinity and zero.

#math #complex numbers #physics

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