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Aug
20
awarded  Autobiographer
Aug
17
comment Why does this “miracle method” for matrix inversion work?
@DavidZhang: The end result should be the same; what I'm saying is if you carry out the procedure assuming B is a matrix then I don't think anywhere during the procedure you will need a miracle in order to obtain the same result.
Aug
16
comment Why does this “miracle method” for matrix inversion work?
@DavidZhang: Did you try actually computing the series expansion of $B^{-1} I$? I haven't tried it but I suspect if you do, you would get what is on the right hand side without any miracles along the way.
Aug
16
answered Why do we use “congruent to” instead of equal to?
Aug
14
comment Can a logarithm have a function as a base?
@mattecapu: What if $f$ is a matrix? Do you still consider it wrong?
Aug
14
comment Can a logarithm have a function as a base?
The base doesn't have to be a number, does it? Like, considering that $f^2 \equiv f \circ f$, I wouldn't mind saying $\log_f f^n = n$ if $f$ is a function...
Aug
12
revised Is “the nth root of x” well-defined without further qualification?
added 12 characters in body
Aug
12
revised Is “the nth root of x” well-defined without further qualification?
added 7 characters in body
Aug
12
comment Is “the nth root of x” well-defined without further qualification?
@peterwhy: Whoops! Yeah, I can't think today... fixed, thanks.
Aug
12
comment Is “the nth root of x” well-defined without further qualification?
I feel like this answer just repeats my question...
Aug
12
revised Is “the nth root of x” well-defined without further qualification?
I can't math
Aug
12
comment Is “the nth root of x” well-defined without further qualification?
"The nth root" always refers to the principal nth root, which is exactly the issue that I was not talking about. But I think your second paragraph does address the question.
Aug
12
asked Is “the nth root of x” well-defined without further qualification?
Jul
30
comment Continuity Must Hold in an Entire Open Set?
Would be cool if you could show the second example too!
Jul
29
comment When I was teaching absolute function properties, I suddenly made this question …
I was confused what you meant by the "absolute" function. I think you mean the "absolute value" function...
Jul
26
comment Intuition for the Cauchy-Schwarz inequality
@IttayWeiss: OK, I see what you're saying, but while the proof you mentioned is trivial to justify, I think it's not at all trivial to to derive in the first place. How is a student supposed to know that he should go through the extra step of decomposing the vector with respect to an entire orthonormal basis merely to show that the weight of one vector in that orthonormal basis is the desired dot product?
Jul
26
comment Intuition for the Cauchy-Schwarz inequality
@DavidZ: Well if you define dot product that way then I guess my problem is that it's not obvious to me why the dot product is the sum of the componentwise products.
Jul
26
comment Intuition for the Cauchy-Schwarz inequality
@DavidZ: Nope, read my previous comments again. I literally said "the hard part is understanding why dot product has anything to do with projection in the first place". i.e., no, I don't already "know" that the dot product has to do anything whatsoever with the projection or with the angle. That fact is precisely what I'm calling out as non-obvious here.
Jul
26
comment Intuition for the Cauchy-Schwarz inequality
@DavidZ: That's not quite what I meant. I meant that (geometrically and intuitively) projection has to do with the (cosine of) the angle between the vectors. Why should the (cosine of) the angle have anything whatsoever to do with the dot product of the two vectors? It's not at all obvious to me that the two might have any significant relationship, yet they do.