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# Fourierinformationsir

### Maths

The last time I wrote about this, I’d found a formula that you can solve algorithmically to find the minimum number of shuffles it takes to resurrect an even numbered deck of cards. I’d also found two solution series – for decks of size 2N , and 2N -2. Since then, i’ve made some exciting progress!
Continue reading “Card Shuffling – oh, how the turn tables”

When we left off, we had a formula to solve that would give the length of a single loop:

P*2N mod (C+1) = P

P is the position of a card in the loop, N is the number of shuffles, and C is the number of cards. Essentially, you take the starting position, and keep doubling it – if it goes above the number of cards in the deck, take the remainder after dividing it by (C+1). Once it hits it’s starting position again, the loop has ended.
Continue reading “Some more card shuffling”

Yes, that’s right – i’m jumping on the Vsauce bandwagon. Well, not entirely – this is actually a different method for answering this question, which I was given in an assignment, and I thought I’d share.

If you haven’t seen the Vsauce video, the problem goes something like this. You take a sphere, and push a cylinder through it; the cylinder carves out a chunk of the sphere.

NOTE: I did not make this explanation. I based it on the explanation from the book “Journey through Genius” by William Dunham, and changed a few parts – I liked how much clearer this book’s explanation was than the other, more formal proofs I have seen.

## What is the Basel problem?

$\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...$

My old high school has a program where they invite primary school kids to the school to do science and maths activities – and this year, they asked me to help run one of these, the Mini Mathematicians class. This problem actually came out of a class we did there, on card shuffling.
Continue reading “Card shuffling – it’s actually really mathematical”

I want to have a look at a very famous (and bizarre) result, the infinite sum

$\sum_{n=1}^\infty n = 1+2+3+4+5+... = -\frac{1}{12}$

At first, it doesn’t seem to make any sense. If you add all these numbers up, the result keeps getting larger, so how does it approach a finite number – and a negative one for that matter?

In this post, I want to have a look at a really neat part of calculus, Taylor series. It’s essentially a process you can use to approximate some graph, e.g. ex, using a polynomial – the idea is, the higher order the polynomial, the more accurate the approximation gets.

## Let’s dive in! (This gets a bit tricky)

How does it work? First, some terminology – the first order approximation is a linear approximation (so ax+b), second order is quadratic (ax2 + bx + c), and so on. So let’s start with a first order approximation of, say, ex.