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bio website youtube.com/zonerarefatte
location San Diego, CA
age 25
visits member for 2 years, 7 months
seen yesterday

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


Dec
13
comment Finite differences and conservation law
I edited the question to avoid getting too much into the detail of my specific problem, I hope that increases the interest.
Dec
13
comment Finite differences and conservation law
I edited the question to avoid getting too much into the detail of my specific problem, I hope that rises the interest.
Dec
13
comment Finite differences and conservation law
It seems to me that here the stiff term will have trouble, because there is no implicit part in your scheme.
Dec
12
comment Finite differences and conservation law
I tried to impose conservation on the whole r.h.s, by discretising it in conserved form, i.e. as a first derivative, but that gives me an unstable thing (I see sawtooth modes entering the grid)
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 $\sum_{1}^{\infty}$!