atreyasha.github.io - Atreya Shankar

It's base 9+1!

Tue, 03 Sep 2019 04:20:00 +0000

As a disclaimer, this post can be considered a sort of mathematical rant. In daily life, it is a more or less agreed upon fact that we communicate natural numbers (eg. $1,2,3,…$) in what we call “base-10”. This is more or less ubiquitous around the world. If one thinks about this “base-10” phenomenon, it starts to seem weird and even paradoxial. Let me explain why.

Numerical symbols

Let’s think of our notion of numbers as symbolisms to represent quantities. For a moment, forget what numbers mean to you, and try to only associate numbers with quantities, as shown below:

$$ \underbrace{\bigotimes}_{1} \quad \underbrace{\bigotimes \bigotimes}_2 \quad \underbrace{\bigotimes \bigotimes \bigotimes}_3 \quad ... \nonumber $$

If we restrict ourselves to only associating numbers as symbols, we start to put our attention on the actual quantities that these numbers represent. As shown above, we can essentially go on with our representations. If you think deeply about it, the final unique symbol that we have for such quantities is 9.

Birth of 10

So what comes after 9? Well, I would argue 9+1. But we don’t have a (layman/ubiquitous) symbol for such a quantity. So we use (what we know as) 10.

But what is 10 really? Well 10 can be broken down into a 1 and 0 (reading from left to right). 10 decodes itself to 1 aggregated quantity and 0 basic quantities. What an aggregated quantitiy is; is open to our definition.

Let’s say we use base-2, then 2 is our aggregated quantity and 10 represents 2. If we use base-3, 10 in base-3 represents 3.

Base 9+1

So we can see that 10, rather than being an actual symbolic number (like 0 to 9), is actually a referential number. 10 requires a clearly defined base and it represents aggregated quantities of that base.

With that idea in mind, it makes no sense to me to say that we use base-10. 10 is not a symbolic number, it is a referential number. So in my opinion, we should not refer to our layman number system as base-10. A more correct representation would be base 9+1, since 9+1 best represents what 10 means to us in daily life. Either that, or come up with a widely accepted symbol which represents the quantity encoded by 9+1.

Derivation of the Cauchy-Riemann Equations

Mon, 02 Sep 2019 04:20:00 +0000

Recently, I had started exploring the concept of replacing the real-valued components of deep-learning networks with complex numbers instead. For this, the concept of holomorphic functions becomes important. It took me some time to find a complete and concise proof of complex-valued differentiability. Here I will summarize what holomorphic functions are, which will ultimately lead us to the Cauchy-Riemann Equations.

Firstly, in order for us to have a complete complex-differentiable function, we would need to show that a function $f(z)$ has a well-defined derivative $f’(z)$. We can define the derivate $f’(z)$ below:

\[ f’(z)=\lim_{\delta z \rightarrow 0} \frac{f(z+\delta z)-f(z)}{\delta z} \]

In order for $f’(z)$ to be well-defined, the value of $f’(z)$ must be the same regardless of how we approach the limit. In this case, $\delta z$ can approach $0$ in many different ways. To investigate the different approaches, we can first redefine our complex function in terms of its real and “imaginary” parts:

$$ \begin{eqnarray} &z \in \{x+iy~|~x,y \in \mathbb{R}, ~ i = \sqrt{-1}\} \\ \nonumber \\ &f(z)\in \{u(z)+iv(z)~|~ u,v \in \mathbb{R},~ i = \sqrt{-1}\} \\ \nonumber \\ &f(x,y) = u(x,y) + iv(x,y) \label{4} \end{eqnarray} $$

Approach 1: Via the real-axis

If we approach the limit via the real-axis, $\delta z \rightarrow 0$ will deform to $\delta x \rightarrow 0$.

$$ \begin{align} f'(x,y)&=\lim_{\delta x \rightarrow 0} \frac{f(x+\delta x,y)-f(x,y)}{\delta x}\\ &=\lim_{\delta x \rightarrow 0} \frac{u(x+\delta x,y)+iv(x+\delta x,y)-u(x,y)-iv(x,y)}{\delta x} \nonumber\\ &=\lim_{\delta x \rightarrow 0} \frac{u(x+\delta x,y)-u(x,y)}{\delta x} + \lim_{\delta x \rightarrow 0} \frac{iv(x+\delta x,y)-iv(x,y)}{\delta x} \nonumber \\ &=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} \nonumber \end{align} $$

Approach 2: Via the imaginary-axis

If we approach the limit via the imaginary-axis, $\delta z \rightarrow 0$ will deform to $\delta iy \rightarrow 0$.

$$ \begin{align} f'(x,y)&=\lim_{\delta iy \rightarrow 0} \frac{f(x,y+\delta y)-f(x,y)}{\delta iy}\\ &=\lim_{\delta iy \rightarrow 0} \frac{u(x,y+\delta y)+iv(x,y+\delta y)-u(x,y)-iv(x,y)}{\delta iy} \nonumber\\ &=\lim_{\delta iy \rightarrow 0} \frac{u(x,y+\delta y)-u(x,y)}{\delta iy} + \lim_{\delta iy \rightarrow 0} \frac{iv(x,y+\delta y)-iv(x,y)}{\delta iy} \nonumber \\ &=\frac{1}{i}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \nonumber \\ &=-i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \nonumber \\ \end{align} $$

Deductions

If the function $f$ is differentiable in complex space, both formulations of the differential (via the real and imaginary axes) must be the same. Therefore we apply the equality:

$$ \begin{align} \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} &= \frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y} \\ \nonumber \\ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},& \quad \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \quad \square \end{align} $$

And with that, we arrive at the Cauchy-Riemann Equations, which essentially define a natural constraint on the real and imaginary parts of complex functions in order for such a complex function to be complex-differentiable or holomorphic.

Additional investigation

One limitation of our proof is that we assumed that our only two approaches can be from the real or imaginary axes. But what about approaches from arbitrary diagonals or combinations of both axes?

To investigate such a possibility, we have to revisit our original definition in \eqref{4} and utilize what is called a complete differential. We also assume an arbitrary signal for $x$ and $y$ in $z$.

$$ \begin{eqnarray} &f(x,y) = u(x,y) + iv(x,y) \\ \nonumber \\ &\delta u = \frac{\partial u}{\partial x} \delta x + \frac{\partial u}{\partial y}\delta y \\ \nonumber \\ &\delta v = \frac{\partial v}{\partial x} \delta x + \frac{\partial v}{\partial y}\delta y \\ \nonumber \\ \end{eqnarray} $$

With arbitrary proportions of $u$ and $v$, we can assume an arbitrary approach of $\delta z \rightarrow 0$.

$$ \begin{align} \delta f = \delta u + i \delta v &= \frac{\partial u}{\partial x} \delta x + \frac{\partial u}{\partial y}\delta y + i \frac{\partial v}{\partial x} \delta x + i \frac{\partial v}{\partial y}\delta y \\ &= \Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big)\delta x + \Big(\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}\Big) \delta y \nonumber \\ \text{Using CR-equations}:&= \Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big)\delta x + \Big(-\frac{\partial v}{\partial x} + i \frac{\partial u}{\partial x}\Big) \delta y \nonumber \\ &= \Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big)\delta x + \Big(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\Big) i\delta y \nonumber \\ &= \Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big)\cdot(\delta x+i\delta y) \nonumber \\ &= \Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big)\cdot \delta z \nonumber \end{align} $$

Now, with the help of the CR-equations, we can find the derivative of $f$ with respect to an arbitrary $z$.

$$ \begin{align} \lim_{\delta z \rightarrow 0} \frac{\delta f}{\delta z} &= \lim_{\delta z \rightarrow 0} \frac{\Big(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \Big) \cdot \delta z }{\delta z} \label{13} \\ \nonumber\\ &= \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \nonumber \\ \text{Using CR-Equations}:&=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y} \quad \blacksquare \nonumber \end{align} $$

With \eqref{13}, we can show that as long as the Cauchy-Riemann Equations hold for a complex function, it will still be differentiable regardless of the complex path we take towards the specific limit. This would imply that investigating a complex function’s differentiability via the real and imaginary axes individually is sufficient in proving complex differentiability everywhere (or through all paths).

Alternative Formulation of the Arithmetic Mean

Sun, 01 Sep 2019 04:20:00 +0000

In the field of statistics, central tendency is a critical concept. Central tendency is basically a value or number around which quantitative data, or sets of data, tend to cluster. Central tendency can also be thought of as a number, or a set of numbers, that conveniently and accurately describes a set of data. There are many measures of central tendency. One simple and commonly used measure is the arithmetic mean. For a set of data (numbers) $a_1,a_2…a_n$, the arithmetic mean $(\overline{a})$ is given by the following formula:

\[ \begin{equation} \overline{a} =\frac{a_1+a_2+…+a_n}{n}=\frac{\sum_{i=1}^{n}a_i}{n} \end{equation} \]

A standard formulation of the arithmetic mean can result from expanding this formula. Here, the arithmetic mean can be defined as a weighted sum of the individual data instances, where each data instance receives an equal weight $(\frac{1}{n})$:

\[ \begin{equation} \overline{a} =\frac{a_1+a_2+…+a_n}{n}=\frac{1}{n}\cdot{a_1}+\frac{1}{n}\cdot{a_2}+…+\frac{1}{n}\cdot{a_n} \end{equation} \]

However, I believe that this formulation hides the actual concept of central tendency. For this reason, I want to present a reformulation of the arithmetic mean; one that clearly reflects the concept of central tendency. To start off, assume there exists some arbitrary sample of numerical data with natural number indices from $1$ to $n$:

Now, assume a central tendency value or an arithmetic mean for this sample exists, and it, being the central value, exists such that the absolute sum of the positive deviations and the sum of the negative deviations from it must be exactly the same. To put this more mathematically, the arbitrary data set will need some re-arrangement. The data that is greater than or equal to the arithmetic mean (assuming it exists) must be separated from the data that is less than the arithmetic mean in the following manner:

$$ \begin{align*} &a_{1},a_{2},a_{3}...a_s\geq \overline{a} \\ &a_{s+1},a_{s+2},a_{s+3}...a_n < \overline{a} \end{align*} $$

Now for the mathematical statement regarding the arithmetic mean:

$$ \begin{eqnarray} &\sum_{i=1}^{s}|a_i-\overline{a}|-\sum_{i=s+1}^{n}|a_i-\overline{a}| = 0 \\ \nonumber \\ &(a_1-\overline{a})+(a_2-\overline{a})+(a_3-\overline{a})+...+(a_s-\overline{a}) \\ &-(\overline{a}-a_{s+1})-(\overline{a}-a_{s+2})-(\overline{a}-a_{s+3})-...-(\overline{a}-a_n)=0 \nonumber \end{eqnarray} $$

Re-arranging the terms:

$$ \begin{align} &a_1+a_2+...+a_n-n\cdot\overline{a}=0 \\\nonumber\\ &n\cdot\overline{a}=a_1+a_2+...+a_n \\\nonumber\\ &\overline{a} = \frac{a_1+a_2+...+a_n}{n} \quad \square \end{align} $$