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.
The traditional—but certainly not the best—way to compute the value of the logarithm of some number is to use a Taylor series, for example
but that expansion is only valid for , or so, because it is the Taylor expansion of "around 1", and the convergence radius of this particular expression isn't very large. Furthermore, it needs a great deal of terms before converging.
This week, another derivation for a famous formula: Archimedes’
formula for π.
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
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.
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:
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.
What if I asked you to find the numerical value of
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.
Last week we examined the complexity of obtaining in the decomposition for some integer . This week, we’ll have a look at how we can use it to speed-up square root extraction. Recall, we’re interested in because
with , which allows us to get easy bounds on . Better, we also have that
and we know how to compute (somewhat efficiently! Let’s combine all this and see how it speeds up things.