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These are questions asked by Annonymouse

I have a system of ODEs with parameters, p[i], and variables, x[i].

f := [

associated with the innitial conditions:
[x[1](0) = p[2], x[2](0) = p[3]].

I am interested in sets of parameters where the solution x[1](t) is the same; if [p[1],p[2],p[3]] is associated with a solution x(p,t), and [ph[1],ph[2],ph[3]] is asociated with the solution x(ph,t); then x[1](p,t)=x[1](ph,t) for all t if and only if

[ph[1] = ph[1],
ph[2] = p[2],
ph[3] = -p[1]*p[2]^2+p[2]^2*ph[1]+p[3]]

i.e. ph[1] takes any real value, ph[2] takes the same values as p[2] and ph[3] takes a value determined by the original parameter vector and ph[1].

In a previous question it was demonstrated that x[2](ph,t)/x[2](p,t) rapidly converge on p[1] as t increases for a specific parameter vector that was given in the question (see graph below)


This raises the question does this limit generally hold?

I have struggled to do this in maple and I am suspicious of the answer i have got
limit (x[2](ph,t)/x[2](p,t),t=infinity)=+/- infinity

My question is
+ when does a finite limit exist?
+ what is the finite limit?


I have a system of ODEs with parameters, p[i], and variables, x[i].

f := [

associated with the innitial conditions:
[x[1](0) = p[2], x[2](0) = p[3]].

I have sets of parameters that i am interested in

[p[1] = 1, p[2] = 2, p[3] = 3]
[p[1] = 2, p[2] = 2, p[3] = 7]
[p[1] = 3, p[2] = 2, p[3] = 11]
[p[1] = 4, p[2] = 2, p[3] = 15]
[p[1] = 5, p[2] = 2, p[3] = 19]
[p[1] = 6, p[2] = 2, p[3] = 23]
and so on.

The solutions for these equations appear to be anisotropic scalings of each other.


/xh[1]\ =/1 0\ /x[1]\      
\xh[2]/   \0 q/ \x[2]/

where x is the variable for one parameter vector and xh is the variable for another; and q is a nuimber I am interesting in working out (and think may be p[1]).

how do you work out q?

[this is in fact a complicated way of asking when I numerically integrate an ODE on maple how do i get access to the sequence of numbers in the result]


I have an ODE with 3 parameters
diff(B[1](t), t) = piecewise(t < 1000, kaC*(R-B[1](t))-k[d1]*B[1](t), 1000 < t, -k[d1]*B[1](t))

I'd like to make a graph that shows how its solution vary as Kac and Kd vary. This could have an axis of the form:

            | /k[d]
  B(t)  |/_ __

which could contain a surface composed of the solutions as k[d] varies. Then a series of surfaces could be put together on the same axis to show what happens as kaC varies.

Some typical values are:

kaC = 6*10^(-2),
k[d1] = 7*10^(-3),
R = 1

I'd like to graph everything  in two orders of magnitude of these values for KaC and k[d1].

Currently I think the key obstacle is making a surface of solutions to the ODE; as once I can do that I think making a sequence of them on the same axis should be quite simple with Display




I'm trying to better understand the Black and Scholes model; which is a scalar function on (positive reals)^5.
a maplesoft worksheet defines it as


I am trying to understand the parameter vectors (r ,T,K,S[0],sigma) that give the same values of BS_Price - and particularly whether these form curves, closed curves, surfaces or similar.

Right now, I am not sure how to procede.

I've just put together  a procedure that evaluates BS at points in R^5 - and i think i can move forward by using the curry or rcurry functions to get a 5d tensor of the values of BS, that i can start to look for patterns within.

BS_Price := proc (InterestRate, StockPrice, StrikePrice, Duration, Volatility) evalf(subs([r = InterestRate, S[0] = StockPrice, K = StrikePrice, T = Duration, sigma = Volatility], exp(-r*T)*(-(1/2)*erf((1/4)*sqrt(2)*(sigma^2*T-2*r*T+2*ln(K)-2*ln(S[0]))/(sigma*sqrt(T)))+1/2))) end proc

could anyone give me advice on doing this?

I have a hard to understand quotient of multivariate polynomials- my intuition is that the denominator nearly divides the numerator - and it could be rewritten as:

remainder+(much simpler numerator)/denominator

as far as I can see the functions quo and rem aren't designed for this - but I'm certain that people in the maple community must have overcome this kind of problem before

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