Recent content by anuttarasammyak

Excellent. Let me follow you. T(x):=\sum_{n=1}^{\infty}\frac{\sin(n(\pi-x))}{n^3} for ##-\pi < x < \pi## Regarding this as Fourier series, ##n^{3}## in denominator shows that T(x) is tertial function of x from integration by parts in the calculation of components. We can write T(x) with...

I would try another transformation S=Re(-i\sum_{n=1}^\infty \frac{e^{\frac{\pi n i}{2}}}{n^3}) =\lim_{x\rightarrow 1} Re(-i (\frac{\pi i}{2})^3 \int dx \int dx \int dx \sum_{n=1}^\infty e^{\frac{\pi nx i}{2}}) with convention that all the integral constants are zero. S=\frac{\pi^3}{8}Re( -...

Then have you known osmotic behaviour of H3O+ ion, which could cease your question ?

So your question seems why only neutral atoms and molecules pass the membrane ? I say it in logic. I do not know the facts.

Do you mean any postive charged ions cannot pass membrane or hydrogen ion only ?

Though I am not good at chemistry, I think hydrogen ion, i.e. proton and proton-neutron core, pass the semipermeable membrane including ones in living organism. Why not ?

In a narrow sense of mathematics, you are right. But in physics almost all the numbers in calculation is approximate. I am afraid that thinking of integer 583 just is not practical.

I asked Wolfram as shown There seems no integers to satisfy the relation. In neighbor, (2,3,27) satisfies a^2+b^2+c^2=582 and (2,2,24),(6,8,22) satisfies a^2+b^2+c^2=584 where I excluded (0,10,22) which includes physically prohibited 0.

Thanks for pointing out the difficulty. From my last line \frac{8}{\pi^3}S=\frac{1}{2}\int_{0}^{\infty} p^2 dp \ (\sum_{n=1}^\infty \sin \frac{n\pi} {2}\sin \frac{pn\pi} {2} ) In here \sum_{n=1}^\infty \sin \frac{n\pi} {2}\sin \frac{pn\pi} {2}=\frac{1}{2}\sum_{n=1}^\infty [\cos...

I am afraid that in this way we would not succed to reduce the effort than solving the equation itself.

Inspired by answer of julian, let me show my sketch using Fourier transform and distribution.

The result of attempt 2 would be written as \frac{A}{s}+\frac{B}{s+c} where c=R_\rho^{-1}(C_\rho^{-1}+C_2^{-1}) You can get constants A and B by calculation. You find it sum of simple pole functions. You do not have to do this reduction in applying residue theorem. The result of attempt 2...

BLUE BOX should be R_\rho C_\rho +1/s which makes attempts 1 and 2 have same result.

Axes are x and y as I wrote x- y- directions of velocity in my post #2. In xy coordinate, ##P(x_p,0)##. The contact point on the slope ##C(x_C,y_C)## in xy axes coordinate is x_C=x_P+R\phi \cos \beta+x_{C0} y_C=-R\phi \sin \beta+y_{C0} where ##(x_{C0}, y_{C0})## is initial position of C when...

Does your "x" or "time derivative of x" have something to do with the Lagrangian of ? I do not find it in parameter of the Lagrangian.