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Feb
26
comment Elements of SO(n) is block-diagonalizable
Orthogonal matrices are normal, and hence can be diagonalized using unitary matrices. Maybe?????
Feb
10
comment Proving that Pythagorean triples are relatively prime if s and t are odd
I think the question might have been edited since I posted my answer. I was explaining how to show $\text{gcd}(\frac{s^2+t^2}2,\frac{s^2-t^2}2) = 1$.
Jan
24
awarded  Good Answer
Jan
20
revised Weak $L_1$ norm is different from $L_1$ norm on a probability space
j -> j/n, etc
Jan
20
comment Weak $L_1$ norm is different from $L_1$ norm on a probability space
I must have meant $[(j-1)/n,j/n)$.
Jan
5
comment What did Whitehead and Russell's “Principia Mathematica” achieve?
For an interesting (and perhaps contrarian) view, read the book by Morris Kline "Mathematics: The Loss of Certainty."
Jan
2
awarded  analysis
Dec
30
awarded  Yearling
Dec
21
comment single valued analytic branch of multivalued function
@User001 I am not taking the square root, I am taking a square root. Also, if $z$ is complex, then the square roots of $z^2$ are $\pm z$. The formula $\sqrt{z^2} = |z|$ is only valid if $z$ is real.
Dec
13
revised Absolute convergence of fourier series
Added hint
Dec
13
comment Properties of Hardy operator $T(u)(x)=\frac{1}{x}\int_0^x u(t)dt$
I don't see what your problem is.
Dec
13
comment Properties of Hardy operator $T(u)(x)=\frac{1}{x}\int_0^x u(t)dt$
@User1 You can choose any primitive you like, as long as you are consistent (use the same one throughout).
Dec
12
comment Particular solution
Look, I think you generally understand what is going on. But I have a lot going on in my life right now, so I cannot keep up this correspondence. Sorry.
Dec
11
comment Particular solution
If the $b_i$ are distinct, then yes, because $L\left(\sum_{i=1}^N \alpha_i e^{b_i x} \right) = \sum_{i=1}^N \gamma_i e^{b_i x}$ where the $\gamma_i$ are non-zero, and $e^{b_i x}$ are linearly independent.
Dec
11
comment Particular solution
Yes, that is definitely correct for constant coefficient ODE.
Dec
11
comment Particular solution
Yes. I think that is correct.
Dec
11
answered Particular solution
Dec
11
revised Absolute convergence of fourier series
Define f(t) = W(t) - W(1) t.
Dec
11
revised Absolute convergence of fourier series
Probability 1 -> non-zero probability
Dec
10
revised Absolute convergence of fourier series
Made Wiener process the main theme.