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Dec
25
revised A relationship between matrices, bernoulli polynomials, and binomial coefficients
added 1217 characters in body
Dec
25
comment A relationship between matrices, bernoulli polynomials, and binomial coefficients
@AndrewGibson: Many thanks for double checking my work! I really appreciate it :)
Dec
25
comment A relationship between matrices, bernoulli polynomials, and binomial coefficients
Oh, I hate to burst the magic: your matrix factorization is incorrect. If you carry out the matrix multiplication, you don't recover the correct matrix :(
Dec
25
answered A relationship between matrices, bernoulli polynomials, and binomial coefficients
Dec
25
comment A relationship between matrices, bernoulli polynomials, and binomial coefficients
This phenomena is unique to 4 dimensions, it fails in 5 dimensions (although, I openly confess, I haven't done intense linear algebra calculations in a while---so I may have committed an error!).
Dec
25
comment A relationship between matrices, bernoulli polynomials, and binomial coefficients
+1 for a great question!
Dec
25
suggested suggested edit on A relationship between matrices, bernoulli polynomials, and binomial coefficients
Dec
25
comment when does a separate-variable series solution exist for a PDE
A good reference on this is Methods of Theoretical Physics by Philip McCord Morse and Herman Feshbach.
Dec
24
suggested suggested edit on Interpreting a limit as a derivative
Dec
23
comment Division into $x(x-1)$
@Tomasz but $g=4$ doesn't cleanly divide $3(3-1)=6$ for $x=3$...
Dec
23
revised Division into $x(x-1)$
TeX-ed it up
Dec
23
comment Division into $x(x-1)$
Ah yes, well played, @Marvis!
Dec
23
comment Division into $x(x-1)$
<del>Also $g$ must be even...otherwise no such integer $x$ exists. So therefore $g\gt 2$, thanks to @Marvis' observation.</del> This is wrong, thanks to a simple counter-example $g=15$, $x=6$ (thanks @Marvis!).
Dec
23
suggested suggested edit on Division into $x(x-1)$
Dec
23
comment Homologies of the pairs are same but they are not homotopy equivalent as pairs.
But $D^{n}-\{0\}$ is homotopy equivalent to $S^{n-1}$! It's $D^{n}$ which is not homotopy equivalent to $S^{n-1}$...and, naturally, $D^{n}-\{0\}$ is not homotopy equivalent to $D^{n}$. I think this should clear up the matter...
Dec
10
comment Why is a singular matrix rare?
Hint: construct equivalence classes of matrices determined by their determinant. You'll see that singular matrices are comparatively small (i.e., represented by a single class) out of uncountably many equivalence classes (assuming you are working over $\mathbb{R}$ or $\mathbb{C}$)...
Dec
4
comment Advice about taking mathematical analysis class
+1 for "write the book" advice, a seldom noted nugget of wisdom :)
Dec
2
comment Vector spaces, subspace.
Yeah, let $A\in W$. Then $(-A)\in W$. But $A+(-A)=0\notin W$, contradiction, therefore etc. etc. etc.
Dec
2
comment Notation: What's $]a,b[$
This $]a,b[$ European notation (French, if I recall correctly) for the open interval $(a,b)$. It's sometimes still used in the literature, but I think parentheses carried the day.
Nov
30
answered Using Octave to solve systems of two non-linear ODEs