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katastrofa
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Re: Proof of the risk-neutral assumption

December 23rd, 2020, 3:24 pm

Lie Trotter .

Just so happens that I am implementing  LT as we speak. It's first-order. or 2nd order, go for Strang-Marchuk operator Splitting. Nice thing is splitting {diffusion, convection}, {potential, kinetic energy} and not {x,y}.

Wow! I didnie know you physicists knew BCH et al.

https://www.asc.tuwien.ac.at/~ewa/semin ... ag_Exl.pdf
I'm making candied orange skin.

I know it all mostly from quantum spin systems theory (earlier applications in physics are related to Brownian motions). What's different about physics and other hard sciences is that the Nature actually does the job of performing efficiently under strong random fluctuations. You guys just crash and crash, whenever the heat goes up.

What's BCH? :-D
 
 
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katastrofa
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Re: Proof of the risk-neutral assumption

December 24th, 2020, 12:02 am

Thanks, platinum.

Cuchulainn, splitting methods are commonly used for solving Schroedinger equations of featured physical systems (e.g. Bose-Einstein condensate). They can be numerically cumbersome when the solution oscillates strongly (they're not precise enough), so more sophisticated methods are being developed.
 
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Cuchulainn
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Re: Proof of the risk-neutral assumption

December 24th, 2020, 10:24 am

Thanks, platinum.

Cuchulainn, splitting methods are commonly used for solving Schroedinger equations of featured physical systems (e.g. Bose-Einstein condensate). They can be numerically cumbersome when the solution oscillates strongly (they're not precise enough), so more sophisticated methods are being developed.
Pathological numerical problems? What causes the oscillations? Overshoot?

Splitting error is a fixed cost, the other error is truncation error for the separate operators.

[$]u(t) = e^{(A+B)t}[$] = [$]e^{At}e^{Bt}[$] if [$][A,B] = AB - BA = 0[$] for matrices [$]A[$] and [$]B[$]. [$]t[$] is a real number.

is it also if and only if (iff)?
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katastrofa
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Re: Proof of the risk-neutral assumption

December 24th, 2020, 11:53 pm

"Pathological numerical problems? What causes the oscillations? Overshoot?"

Electric field or multiparticle interactions. They sometimes can be removed by hand-waving, sorry coarse graining procedures.


It's iff, methinks. It's obvious that if A and B commute, higher orders of BCHD are zero. To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group. But that's only for small |t| and bounded operators. I think unboundedness can be covered by expanding the operator into Taylor series or a similar operation... It's been 10 years ago, anyone here remembers more?
 
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Cuchulainn
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Re: Proof of the risk-neutral assumption

January 5th, 2021, 3:23 pm

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group. 
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

[$]tr(e^{A+B}) \le tr(e^Ae^B)[$].

[$]tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii} [$]
QED
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Mars
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Re: Proof of the risk-neutral assumption

January 5th, 2021, 4:45 pm

"easier than a sledgehammer approach" Really? 
Second order term in time of exp(A+B)t exp(At) and exp(Bt) are quite easy to calculate using known Taylor expansion of exponentials, then second order polynomial multiplication (non commutative one) show that exponential equality implies that matrix A and B commute.
 
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katastrofa
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Re: Proof of the risk-neutral assumption

January 5th, 2021, 8:13 pm

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group. 
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

[$]tr(e^{A+B}) \le tr(e^Ae^B)[$].

[$]tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii} [$]
QED
The [$]tr(e^{A+B}) \le tr(e^Ae^B)[$] holds regardless of the commutativity. I have no idea what you QED-ed up there.
 
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Cuchulainn
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Re: Proof of the risk-neutral assumption

January 5th, 2021, 8:23 pm

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group. 
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

[$]tr(e^{A+B}) \le tr(e^Ae^B)[$].

[$]tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii} [$]
QED
The [$]tr(e^{A+B}) \le tr(e^Ae^B)[$] holds regardless of the commutativity. I have no idea what you QED-ed up there.
Read the 'equality' clause in 2. 
// A and B are Hemitian, which is very restrictive.
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Cuchulainn
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Re: Proof of the risk-neutral assumption

January 12th, 2021, 3:28 pm

"easier than a sledgehammer approach" Really? 
Second order term in time of exp(A+B)t exp(At) and exp(Bt) are quite easy to calculate using known Taylor expansion of exponentials, then second order polynomial multiplication (non commutative one) show that exponential equality implies that matrix A and B commute.
I agree.
But If we have two specific splitting operators A and B how can we see at a glance if AB = BA?
In general, Strang-Marchuk doesn't need commutativity, but it's nice for the 1st-order splitting schemes.
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gatarek
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Re: Proof of the risk-neutral assumption

March 1st, 2021, 6:21 pm

I used it on multiple occasions in physics, to give one of the more complex ones which curiously involved the name of one of the brightest WF stars:

"Continuous quantum jumps and infinite-dimensional stochastic equations"
Dariusz Gatarek
Systems Research Institute, Polish Academy of Sciences, 01-447 Warszuwa, Newelska 6, Poland
Nicolas Gisin
Groupe de Physique Appliquee, Universite de Geneve, 20 rue de I’Ecote de Medecine, 1211 Geneve 4,
Switzerland
Former stars