## Thursday, March 30, 2017

### Spring is here

It is amazing how a couple of days of sunshine and warmer weather erased most of the evidence that we had a major winter storm 2 weeks ago.

## Tuesday, March 14, 2017

### Pi Day 2017

Happy Pi day (and a snowy one in the Northeastern US)! One of the early childhood mathematical memories I had was learning that π can be approximated by 22/7 and accurate up to 2 decimal places, a fact that was known since antiquity. In particular, Archimedes gave the first proof that 22/7 is strictly larger than π. The fifth century Chinese mathematician and astronomer Zu Chongzhi gave the approximation 355/113 that is accurate up to 6 decimal places. Today, sophisticated algorithms using Ramanujan's formula and its variants can compute trillions of digits of π.

## Thursday, March 9, 2017

### Binomial coefficients

When studying binomial coefficients and how they form Pascal's Triangle, we learn that they are the coefficients of $(x+y)^n$, i.e.
$$(x+y)^n = \sum_{k=0}^n \left(\begin{array}{c}n\\k\end{array}\right)x^ky^{n-k}$$
Replacing $y$ with $-y$ we get:
$$(x-y)^n = \sum_{k=0}^n -1^k\left(\begin{array}{c}n\\k\end{array}\right)x^{n-k}y^{k}$$

For example, the coefficients of $(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ are 1,4,6,4,1 which form the 5th row of Pascal's triangle.

What happens if the coefficients of the polynomial are repeated binomial coefficients such as 1,1,4,4,6,6,4,4,1,1? What is $$\sum_{k=0}^{2n+1} \left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^ky^{n-k}?$$

It is easy to show that this is equal to $y(x^2+y^2)^n + x(x^2+y^2)^n = (x+y)(x^2+y^2)^n$.

What about repeating coefficients with sign changes such as 1,1,-4,-4,6,6,-4,-4,1,1?
The same argument shows that

$$\sum_{k=0}^{2n+1} -1^{\lfloor k/2\rfloor}\left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^{n-k}y^{k}$$

is equal to $y(x^2-y^2)^n + x(x^2-y^2)^n = (x+y)(x^2-y^2)^n$ which can be further simplified as $(x+y)^{n+1}(x-y)^n$.

If we now change the sign of every other coefficient as well, i.e. 1,-1,-4,4,6,-6,-4,4,1, -1,
we get

$$\sum_{k=0}^{2n+1} -1^{\lfloor k/2\rfloor+n-k}\left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^{n-k}y^{k}$$

This can be shown to be equal to $x(x^2-y^2)^n -y(x^2-y^2)^n = (x-y)(x^2-y^2)^n$ which is equal to $(x-y)^{n+1}(x+y)^n$.

We can repeat the coefficients more than 2 times as well and show that

$$\sum_{k=0}^{m(n+1)-1} \left(\begin{array}{c}n\\\lfloor k/m\rfloor\end{array}\right)x^ky^{n-k}$$

is equal to $\sum_{t=0}^{t=m-1}x^ty^{m-1-t}(x^m+y^m)^n$.