Page 7 of 7

Re: Silly questions

Posted: February 13th, 2020, 11:57 am
by Cuchulainn
Well, if we are going to be so literal, then since the original post did not want an answer but a vote I am going to vote yes!! This is maths 21st century style!
That's exactly what happened on LI... everyone voted on some beautiful formula by RAMANUJAN (as in the movie). 
Most people on LI were not interested in accuracy, but in beauty. It has become art, like physics 60 years ago. Even I have a t-shirt with E = mc^2 !
Hollywood has destroyed mathematics.

How can you differentiate wrt an integer?

n! is a special case of gammaish function.(z) when z is an integer, yes?

Re: Silly questions

Posted: February 13th, 2020, 12:20 pm
by Cuchulainn
Here's the formula! Who can explain it?

Image

Re: Silly questions

Posted: February 13th, 2020, 1:22 pm
by FaridMoussaoui
That's exactly what happened on LI... everyone voted on some beautiful formula by RAMUJAN (as in the movie).
Ramanujan

Re: Silly questions

Posted: February 13th, 2020, 1:50 pm
by Cuchulainn
I demand a recount! Anyway, it's not even 6am -- don't you people sleep?
Isaiah 57:20 

Re: Silly questions

Posted: February 13th, 2020, 1:53 pm
by Cuchulainn
That's exactly what happened on LI... everyone voted on some beautiful formula by RAMUJAN (as in the movie).
Ramanujan
Corrected! It's the first time that I wrote this name. Number theory..

Re: Silly questions

Posted: March 1st, 2020, 3:43 pm
by Cuchulainn
Does the solution of 

[$](1 - \tau)^2 \frac{\partial u}{\partial \tau} = \Delta u[$] (where [$]\Delta[$] is the Laplacian) with appropriate initial and boundary conditions converge as [$]\tau \rightarrow  1[$]? ([$]\tau[$] is false/pseudo time)

And what does it converge to?

// aka
[$]\bigtriangledown^2 \equiv \Delta[$]
nabla ...

Re: Silly questions

Posted: March 1st, 2020, 9:06 pm
by Alan
Pretty sure: in 1D, the transition density [$]p(x,t)[$], starting from an arbitrary density at t=0, converges to 0 for all real x, as [$]t \uparrow 1[$]. Loosely, all the mass went to [$]\pm \infty[$]. 

Proof: the Green function [$]G(x,T|x_0) \propto \frac{e^{-(x-x_0)^2/(2 v_T)}}{\sqrt{v_T}}[$], where [$]v_T = T/(1-T)[$].

Re: Silly questions

Posted: March 1st, 2020, 9:28 pm
by katastrofa
Does the solution of 

[$](1 - \tau)^2 \frac{\partial u}{\partial \tau} = \Delta u[$] (where [$]\Delta[$] is the Laplacian) with appropriate initial and boundary conditions converge as [$]\tau \rightarrow  1[$]? ([$]\tau[$] is false/pseudo time)

And what does it converge to?

// aka
[$]\bigtriangledown^2 \equiv \Delta[$]
nabla ...
For tau =0 you have a time-independent heart/wave equation. I'm not sure why it wouldn't converge?

Re: Silly questions

Posted: March 1st, 2020, 9:30 pm
by Alan
Well, if my solution is correct, the Green function cannot be continued beyond [$]\tau = 1[$], so there is an issue there.

Re: Silly questions

Posted: March 1st, 2020, 9:43 pm
by katastrofa
If you change variables to z = 1/(1-tau), you obtained a standard heat equation. If tau goes from 0 to 1, z goes from 0 to infty. The physical solution for u at infty is a uniform heat distribution, modulo boundary conditions.
Just some quick thoughts.

Re: Silly questions

Posted: March 1st, 2020, 10:36 pm
by Alan
Nice! Another proof that, on the whole line, the density -> 0 as tau -> 1.

Plus, shows the answer is the same (0) in D dimensions!

Re: Silly questions

Posted: March 2nd, 2020, 4:33 pm
by Cuchulainn
Elliptic PDE are more difficult to solve than parabolic PDE.

In my example I add a false time derivative to give

[$]\frac{\partial u}{\partial t} = \Delta u[$] 

When [$] t \rightarrow \infty[$] the solution converges to the time-dependent solution. But [$]\infty[$] is big so I transform [$]\tau = t/(1+t)[$] and discretise up to (almost) [$]\tau = 1[$] (works well for other bespoke cases) The empty quarter [$]\tau \gt 1[$] leads to solution explosion, i.e. solution does not exist. This is easy to check by looking at the exact solution of [$]\frac {du}{dt} + u = 0[$] or more generally Picard-Lindelof iteration.

kats almost recovered (typo?) my heat equation for [$]t = \tau/(1 - \tau)[$]. The mapping [$]t = 1/(1 - \tau)[$] is not monotonic.

There's not much out there on this False Transient MOL but I can imagine that it would be more efficient than SSOR and that bunch.

.. that's the background ..

Re: Silly questions

Posted: March 17th, 2020, 9:45 am
by Cuchulainn
Alan,
I have revisited the Anchor pde; I have several remarks and questions. 
Anchor-Style PDE (See Anchor article)
2020-3-16


https://onlinelibrary.wiley.com/doi/epd ... wilm.10366
 
(A)  For large asymptotic t values the solution of the parabolic  PDE(x,y,t) converges to the solution of the corresponding elliptic PDE(x,y) (Friedman 1992).
(B)  The MOL solution leads to  system of ODEs [$]\frac{\partial u}{\partial t} + Au = F[$]. If [$]A[$] is an M-matrix then the solution for large [$]t[$] is the solution of [$]Au = F [$](Varga 1962).
(C)  The Saul’yev ADE method is accurate and very fast. For example, in the case T = 1, run-time is 0.15 seconds on a normal laptop. The MOL works well also but in general 60 times slower than ADE. I haven’t checked, but I suspect that other FD methods would also be slower, e.g. ADE B&C is [2.3] times slower than Saul’yev  ADE.
(D)  I have no idea how to measure the rate of convergence of the results in (A) and (B), although in case (B) there might be a glimpse of hope if we knew what the smallest eigenvalues of the matrix A are (slow transients).
(E)   For Anchor ADE, the boundary conditions in both x and y have little impact on accuracy.
(F)   At the time, why did the authors not use a similarity reduction R = S/A (admittedly, it remains a trick and breaks down in this case)?
(G) For T = 10, K = 140 (NX = NY = 300) MOL C = 28.7023, Time = 13734 seconds (Alan gets Call = 28.629 but no Time given). Table 2 of Anchor article. For ADE, we got
NX, NY, NT; Call, Timing (seconds)
300,300,2000;28.311,27
300,300,3000;28.442,?
300,300,5000;28.522,70
300,300,6000;28.58,?
300,300,7000;28.6003,97
300,300,8000;28.6155,114
300,300,9000;28.6275,125
300,300,10000;28.637,140 (rounding errors start to creep in?)
NX = NY = 300, NT = 9000 seems to be optimal (ADE is more accurate and 110 times faster than MOL in this case).
 
(H) [$]\tau = t/(1+t)[$] turns a ODE into a stiff ODE ("no free lunch in maths").
 
 
 Main Questions:
 
(Q1) Which value of T is infinity in the article (T = 100, T = 1000 etc.)?
(Q2) In section 10.2 I assume NDSolve is used for large t.  I assume it was used to generate prices in Table 4? Also, T = 1000 is essentially infinity?
(Q3) Instead of NDSolve, what about Gauss-Seidel/SSOR etc.  on the elliptic PDE (see remark (A))? It would probably 1) faster and 2) less rough-and-ready than choosing a random T? It might be interesting to try it out? (Someone mentioned that Mathematica not have elliptic solvers).
(Q4) How can we reproduce/find call and put prices in Table 2 for T = 100? Why are the values as they are, e.g. T = 100, K = 100, why is put = 0.1073?