LATEX rules!

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

Check out

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! 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:
If your T is unitary
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

Found this while surfing. Was really funny and interesting.