145 reputation
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bio website youtube.com/zonerarefatte
location San Diego, CA
age 25
visits member for 2 years, 1 month
seen Jul 27 at 5:42

Knowledge and understanding are quite different. Only understanding can lead to being, whereas knowledge is but a passing presence in it. (G.Gurdjieff)


Jul
9
comment Probability of a trajectory in Markov processes
@Did Well, I won't go through what has been said. But, simply, please - do not give anyone lessons about what math.SE is the site for. Or about what is a loss or time or not. I didn't find reading this q&a a loss of time, and there will surely be other people thinking like me. Reading your unexplainably arrogant comment, that was indeed a loss of time and a huge let-down.
Jul
8
comment Space-dependent diffusivity and finite-differences
Found the mistake, in the explicit part I used $u_{j+1}-u_j$ instead of $[u_{j+1}-u_{j-1}]/2$!
Jul
8
comment Space-dependent diffusivity and finite-differences
Usually, to test it in the simplest case, I use $D(x)=x$ or $D(x)=1-x$, to see how it behaves going to both boundaries.
Jun
5
comment Multiplying a vector and a matrix's rows
Could you please explicit that?
Oct
25
comment Vectorial derivative
@Nick Could you help me trying to explicit the calculation in components of with the index notation? Because I cannot see it, sorry.
Oct
25
comment Vectorial derivative
Why should I do that, mate?
Feb
21
comment Orthogonal Part Operator
$(\boldsymbol{p}\cdot\nabla\boldsymbol{p})_\alpha=p_\beta\partial_\beta p_\alpha$
Feb
21
comment Orthogonal Part Operator
No, I simply meant the application of $P$ to those terms: $P(\boldsymbol{p}\cdot\nabla\boldsymbol{p})$
Feb
21
comment Orthogonal Part Operator
Yes, I need to be clearer. $\boldsymbol{p}$ is a column vector of $\mathbb R^3$, function of $\boldsymbol{x}$, respect to which the differentiation in $\nabla$ is carried. The $P$ operator is exactly the matrix $P_{\alpha\beta}=\delta_{\alpha\beta}-p_\alpha p_\beta$
Jun
22
comment Closed form with of a series Mathematica
Yes, but what about the closed form if $x$ can be every value, so not necessarily $0<x<x_0$?
Jun
22
comment Closed form with of a series Mathematica
Ups, thank you so much!
Jun
22
comment Closed form with of a series Mathematica
Added input code and yes $A=x_0$, $B=h$!
Jun
22
comment Closed form with of a series Mathematica
Good idea! I didn't know about that!
Jun
18
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
yep, sorry. So it diverges. Thanks!
Jun
18
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
So, what is wrong in this proof of convergence? $$ S=\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_{0}|}=\frac{1}{x_{0}}\left(\sum_{k<x/‌​x_{0}}\frac{1}{x/x_{0}-k}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0}}\right)=\frac{1}{x_{‌​0}}\left(\sum_{-k'<x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0‌​}}\right)=\frac{1}{x_{0}}\left(\sum_{k'>-x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x‌​_{0}}\frac{1}{k-x/x_{0}}\right)\leq\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{‌​x/x_{0}+k}+\frac{1}{k-x/x_{0}}\right) =\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{k^{2}-x^{2}/x_{0}^{2}}\right) $$
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
This is my friend's proof of convergence: $$S=\sum_{k=-\infty}^{+\infty} \frac{1}{|x-kx_0|} = \frac{1}{x_0}\left(\sum_{k<x/x_0} \frac{1}{x/x_0-k}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right) = \frac{1}{x_0}\left(\sum_{-k'<x/x_0} \frac{1}{x/x_0+k'}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right)= \frac{1}{x_0}\left(\sum_{k'>-x/x_0} \frac{1}{x/x_0+k'}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right)\leq \frac{1}{x_0}\sum_{k>x/x_0}\left(\frac{1}{x/x_0+k}+\frac{1}{k-x/x_0}\right)= \frac{1}{x_0} \sum_{k>x/x_0} \left( \frac{1}{k^2-x^2/x_0^2} \right)$$
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
@Cocopuffs in this case the series is $\sum_{-\infty}^{\infty}$, not $\sum_{1}^{\infty}$!
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
Perfect, I arrive there. Then, how can I avoid the separation of the sum due to the study of the absolute value?
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
Perfectly captured. And thank you about this. So, the sum must diverge as one could aspect due to the similarity with the harmonic series? Actually, I'm absolutely with you on the error in my prof's proof, that I consider wrong.