Plugging Matrices into Functions

[Joshua2]

Are you saying my usage was correct?

[R.D. Silverman]

Originally Posted by m_f_h

Robert, he said: the trace of (A^T * A). Let B=A^T, then
tr(A^T A) = tr(BA) = b_ji a_ij = a_ij a_ij = sum( a_ij^2 ; i=1..n, j=1..m)
(Both indices are repeated.)

But he did not write a_ij a_ij. He wrote:

"(a sub ij for every ij)^2".

[m_f_h]

Originally Posted by R.D. Silverman

But he did not write a_ij a_ij. He wrote:

"(a sub ij for every ij)^2".

Yes, that's not nice. But just for the records and to try a "positive feedback" pedagogical approach, I wanted to ascertain that

tr(A^T A) = a_ij a_ij = \sum_{i,j} a_ij^2 = sum of squares of all matrix elements of A

are all correct.
Btw, that's the square of the Euclid (L²) norm of A (when M_{m,n} is seen as IR^{m x n}.
(Not to be confused with the norm /induced/ on M_{mn} by the L² norms on R^m and R^n, which is max{ | Ax | ; |x|=1 }.)

PS: and depending on the error-tolerance of the interpreter, a_ij^2 could be understood as shorthand for a_ij a_ij , especially if i,j don't appear as free indices on the l.h.s.

[m_f_h]

Originally Posted by davieddy

What is the nicest way of explaining why e is special?

I'll start the ball rolling with d(e^x)/dx = e^x

IMHO "e" is not that special, it just happens to be (the limit) exp(1),
where exp(x) = sum x^n/n!.

Everything else comes from the properties of this power series, and also everything which is abusively written as e^(...) (like e^{i phi}, e^A, ...)
has really nothing to do with 2.718281828, but refers to that power series.
IMHO it makes absolutely no sense at all to write 2.71828^{i pi} or 2.71828^A.

[wblipp]

Originally Posted by m_f_h

Everything else comes from the properties of this power series

lim (1+1/n)^n = e

Is there a simple way to deduce this from the properties of the power series?

[m_f_h]

Originally Posted by wblipp

lim (1+1/n)^n = e

Is there a simple way to deduce this from the properties of the power series?

Although this is not difficult to deduce(*), I agree that it is a less trivial property and makes 2.71828... indeed a little more "special".

--

(*) If I made no error in stupidly calculating the difference between partial sums S(n) = sum( k=0..n, 1/k! ) and (1+1/n)^n, this should be equal to
sum( k=1..n, 1/k! - binomial(n,k)/n^k)
= sum( k=1..n, (1 - n!/(n-k)!/n^k )/k! )
which is easily seen to go to zero: The first term is zero,
and the sum of the other terms is O() of (actually less than)
1/n (1/2 + 1/3! + 1/4! +...) = 0.71828/n
(I hope the gerbils won't introduce subtile errors in the above argumentation and calculation which initially at least was of course absolutely correct...)

[__HRB__]

Did someone mention a matrix?

http://www.youtube.com/watch?v=pbpnsc6-HAs