In an Old Notebook (Generating Random Sequences VI)

April 4, 2017

Looking for something else in old notebooks, I found a diagram with no other indication, but clearly a low-cost random generator.

So, why not test it?

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The 1 bit = 6 dB Rule of Thumb, Revisited.

March 28, 2017

Almost ten years ago I wrote an entry about the “1 bit = 6 dB” rule of thumb. This rule states that for each bit you add to a signal, you add 6 dB of signal to noise ratio.

The first derivation I gave then was focused on the noise, where the noise maximal amplitude was proportional to the amplitude represented by the last bit of the (encoded) signal. Let’s now derive it from the most significant bit of the signal to its least significant.

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Choosing the Right Pseudoinverse

January 17, 2017

On a number of previous occasions, I have used the pseudoinverse of a matrix solve systems of equations, and do other things such as channel mixing. However, the demonstration I gave before isn’t entirely correct. Let’s see now why it’s important to make the difference between a left and a right pseudoinverse.

otter

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Logarithms (Part I?)

January 3, 2017

The traditional—but certainly not the best—way to compute the value of the logarithm of some number x is to use a Taylor series, for example

\displaystyle \ln x = (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\cdots

but that expansion is only valid for 0<x\leqslant{2}, or so, because it is the Taylor expansion of \ln x "around 1", and the convergence radius of this particular expression isn't very large. Furthermore, it needs a great deal of terms before converging.

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π, π, Archimedes!

December 20, 2016

This week, another derivation for a famous formula: Archimedes’
formula for π.

arkimedes-konkani_vishwakosh

Some time in the 3rd century BC, Archimedes used the perimeter of a regular polygon, starting with an hexagon and repeatedly doubling the number of sides, to estimate the value for π. He arrived at the approximation

\displaystyle 3\frac{10}{71}=\frac{223}{71}<\pi<\frac{22}{7}=3\frac{10}{70}.

How he arrived to this result is a bit mysterious until we completely understand how he got that result. Let’s see together how he did it.

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Binet’s Formula

December 13, 2016

You will encounter a large number of formulas in your life, and quite many of them just seem to come out of the blue. That’s fortunately quite false, despite it being not always easy to retrace the formulas’ authors’ steps. One that appears as suspicious as it seems magic, is Binet’s Fibonacci Number formula:

\displaystyle F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}},

where

\displaystyle\phi=\frac{1+\sqrt{5}}{2}

is the golden number.

But it’s not quite out of nowhere, and, if you know how to solve recurrences using the characteristic equation, it’s in fact quite straightforward. Let’s see how.

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The infinite radicals of Ramanujan

November 29, 2016

What if I asked you to find the numerical value of

\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{\cdots}}}}}} ?

If you have difficulty figuring out what this thing is, don’t worry. You’re not the only one. This question is one the problems posed by Srinivasa Ramanujan, one of the most brilliant and, well, mysterious mathematicians of all time. In fact, it was head enough that Ramanujan had to give the answer a few months later.

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