# LATEX rules!

Ah, then! We do have LATEX support in wordpress… although I’ve taken nearly a year to realise this 😀

So, then we can have $E=mc^2$, $\hat{H}|\psi\rangle = i\hbar\frac{d}{dt}|\psi\rangle$ and what not, all written in these blogs…

# π day

π = 3.14159 26535 89793 23846 26433 …

3.14 = March 14 happens to be pi-day every year. Curiously enough, it also happens to be Einstein’s birthday!

http://www.angio.net/pi/bigpi.cgi gives you a search engine to search for digits of π. For example, a search for 15081947 gives

“The string 15081947 occurs at position 604,341 counting from the first digit after the decimal point. The 3. is not counted.

The string and surrounding digits:

99538905839657483550 15081947 01003364946075415680″

If you are quite crazy about π, you might want to pick up this geeky T-shirt (I created something very similar on photoshop, yet to print it!) from thinkgeek.

Indeed an enigmatic constant!

# The Similarity Transform Song

(to the tune of “If You’re Happy and You Know It“)

Note: This song describes what to do for a similarity transformation to find the eigenvalues of a matrix A. A similarity transform changes the matrix but does not alter its eigenvalues. Ideally, if you could find a similar matrix that was diagonal, you could pick the eigenvalues right off the diagonal. A similarity transform is of this form: B = T^{-1} A T, where T is the similarity transform matrix, and B is the matrix that is similar to A.

If your matrix has distinct eigenvalues,
It is very much apparent what to use:
Choose nonsingular for T,
and get a diagonal B,
If your matrix has distinct eigenvalues.

If you have a real symmetric matrix A,
Choose orthogonal for T: it’s the best way!
Because, after all
B is real diagonal
If you have a real symmetric matrix A.

If it’s complex and Hermitian, never fear!
The solution to your problem is quite clear:
Use a unitary T,
to get a real diagonal B
If it’s complex and Hermitian, never fear!

If your matrix A is normal as can be,
Take the following advice from me:
B’s diagonal (how scary!)
If your matrix A is normal as can be.

If A is chosen arbitrarily,
and you find a T that’s unitary,
Then B has the form of Schur
(that’s upper triangular)
If A is chosen arbitrarily.

If A is once again arbitrary,
Then you choose “nonsingular” for T,
B is almost diagonal,
(that’s Jordan form, y’all!)
If A is once again arbitrary.

“The Similarity Transform Song” Copyright (c) 2000 Rebecca Hartman-Baker.(from http://www.cse.uiuc.edu/~rjhartma/eigensong.html)

Found this while surfing. Was really funny and interesting.