## May the power be with you

One reason I tell my students as to why they need to learn mathematics is that knowledge of mathematics gives a sense of power. It often gives a neat understanding of problems and aha insights into solutions. Here are a list of popular problems that students can handle with confidence provided they have a clear idea of how regression works.

(1) Was the coffee spiked with Barbiturates?

Forensic scientists regularly need to handle situation where they have to verify if some chemical is present in a given sample. The absorption spectrometry gives a set of linear equations connecting chemical concentration to absorbance. The best fit solution would yield the chemical concentrations in the sample.

(2) Does higher rain correlate with high temperatures?

Folklore says that harsher a summer is, better will be the monsoon. With data available online one may solve a least square fit for a model and figure out if the folklore makes sense.

(3) Best way to mix.

In a industrial set-up you want to mix several fluids and do it in energy efficient manner. A standard way is to compare the performance of different types of mixers and for each of they fit a curve to motor speed and energy used. The minimum of the curve gives an operating speed for the mixer for which energy usage is minimum.

(4) Do people living in big cities use less energy compared to people in towns?

This and similar other questions have been analyzed by Prof. Geoffrey West and his colleagues using best fit on data of (say) population density and electricity usage. You may check for yourself if city dwellers pay lesser electricity bills.

Often not only one is able to answer interesting questions using regression on data but also gain insights into efficient designs and behavior of complex systems. If you like to think about data and modeling do watch this TED talk. Mathematical skills are powerful tools, May the power be with you.

## Understanding Fourier Series

Comparing Functions and vectors.

$\vec{v}$ function f(x)

Finite dimensional Infinite dimensional

A vector can be written in the following different ways,
$\vec{V} = V_x \hat{x} + V_y \hat{y} + V_z \hat{z}$
$\hskip .5cm = (V \cdot \hat{x}) \hat{x} + (V \cdot \hat{y}) \hat{y} + (V \cdot \hat{z}) \hat{z}$
If the decomposition is along an orthogonal frame along the vectors $\vec{a},\, \vec{b}$ and $\vec{c}$ then the expression would be,
$\vec{V} = (\vec{V} \cdot \hat{a}) \hat{a} + (\vec{V} \cdot \hat{b}) \hat{b} + (\vec{V} \cdot \hat{c}) \hat{c}$
$\hskip .5cm = \frac{\vec{V} \cdot \vec{a}}{\vec{a}\cdot\vec{a}} \vec{a} + \frac{\vec{V} \cdot \vec{b}}{\vec{b}\cdot\vec{b}} \vec{b} + \frac{\vec{V} \cdot \vec{c}}{\vec{c}\cdot\vec{c}} \vec{c}$

In general the dot product of two $n-$dimensional vectors $\vec{V} = (V_1, V_2,...,V_n)$ and $\vec{W} = (W_1,W_2,...,W_n)$, can be written as,
$\vec{V} \cdot \vec{W} = \sum_{i=1}^n V_i W_i.$

It is useful to think of a real function $f(x)$ over an interval
$[a,b]$ as a vector with infinite components. Here the argument serves
as an index and the function value as the vector component. Analogous to vector dot product, the dot product between two functions $f$ and $g$ defined over the same interval can be written as,
$(f,g) = \int_a^b f(x) g(x) dx.$

Using this definition of the dot product, one can show that the following functions
are orthogonal to each-other (mutual dot products are zero) on the interval
$[0, 2\pi]$.
$f_1(x) = 1, \sin{x}, \sin{2x}, \sin{3x}, ...,\cos{x}, \cos{2x}, \cos{3x},...$

Thus in parallel with writing a vector in terms of it’s components, one can write any (finite, smooth and continuous on $[0, 2\pi]$ (I am not trying to be mathematically precise, the aim is to give an intuitive feel)) function in terms of
the above basis functions in the same manner,
$f(x) = \frac{(f(x),1)}{(1,1)} 1 + \frac{(f(x),\sin(x)}{(\sin(x), \sin(x))} \sin(x) + \frac{(f(x),\sin(2x)}{(\sin(2x), \sin(2x))} \sin(2x) + ...$
$\hskip.2cm + \frac{(f(x),\cos(x)}{(\cos(x), \cos(x))} \cos(x) + \frac{(f(x),\cos(2x)}{(\cos(2x), \cos(2x))} \cos(2x) +....$
Notice the similarity of the expression of a function in terms of it’s components and a vector in terms of it’s components. Hence decomposition of a function in its Fourier components is quite akin to decomposition of a vector in its Cartesian components.

## My clan

One amazing thing I find about fellow academics is the ease with which we connect.

Even when I meet them after decades, the shared knowledge goes beyond technical. There is an open-ness which allows us to trust each-other, joke about ourselves, gripe about students ( yes, most of the people I know do that while admitting that they themselves were the same) and discuss health, family and society.