Re: Solving this in mathematica?
- To: mathgroup at smc.vnet.net
- Subject: [mg128077] Re: Solving this in mathematica?
- From: Dana DeLouis <dana01 at me.com>
- Date: Fri, 14 Sep 2012 00:22:15 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
- Delivered-to: mathgroup-newsend@smc.vnet.net
> B[h_, r_] := Exp[(-alpha1[h]) (r) - psi[h]]
>
> Exp[Cbar - beta] Sum[(Pi^x) B[x, r], {x, 1, 1000}]
Hi. I don't have a solution, but perhaps some random suggestions:
> alpha1[n_] := alpha1[n] = alpha1[n - 1] + alpha2[n - 1]
I might break the problem down by rewriting some functions as follows:
For example:
equ = {
alpha1[n]==alpha1[n-1]+alpha2[n-1],
alpha2[n]==k *alpha2[n-1],
alpha1[0]==0,
alpha2[0]==1};
RSolve[equ,{alpha1[n],alpha2[n]},n]
{
alpha1[n] -> (-1+k^n)/(-1+k),
alpha2[n] -> k^n
}
I don't believe you gave a starting value for psi[0], so we're stuck at this point.
Another observation:
> Exp[Cbar - beta] Sum[(Pi^x) B[x, r], {x, 1, 1000}]
If we just look at this: (B for the function B[x,r])
Sum[(Pi^x) B, {x, 1, 1000}]
With a specific value for the upper limit (ie 1000), you are solving: (I'll use 10)
Sum[Pi^x*B, {x, 1, 10}]
B*Pi + B*Pi^2 + B*Pi^3 + B*Pi^4 + B*Pi^5 + B*Pi^6 +
B*Pi^7 + B*Pi^8 + B*Pi^9 + B*Pi^10
Sometimes (but not always) it's better to give a symbolic upper limit, then go back and change it:
Sum[B*Pi^x, {x, 1, ul}] /. ul -> 1000
(Pi*(Pi^1000 - 1)*B) / (Pi - 1)
However, that equation numerically is:
N[%]
2.071503628841560*10^ 497 B
This is a very large number. I would assume your B[x,r] would have to be very small.
Your variable inputs are all machine precision, so I wonder how accurate the solution would be. ??
Again, just some random thoughts. Good luck.
= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =
On Wednesday, September 12, 2012 3:03:51 AM UTC-4, Leon wrote:
> I have following code in mathematica:
>
>
>
> ------------------------------------------------------------
>
> rbar = 0.006236
>
> rt = r_bar
>
> k = 0.95
>
> sigmar = 0.002
>
> betazr = -0.00014
>
> sigmaz = 0.4
>
> pi = 0.99
>
> chi = 0.05
>
> Cbar = -3.7
>
>
>
> alpha1[n_] := alpha1[n] = alpha1[n - 1] + alpha2[n - 1]
>
> alpha2[n_] := alpha2[n] = k (alpha2[n - 1])
>
> sigma1sq[n_] :=
>
> sigma1sq[n] = sigma2sq[n - 1] + 2 sigma12[n - 1] + sigmaz^2
>
> sigma12[n_] :=
>
> sigma12[n] = k (sigma12[n - 1]) + k (sigma2sq[n - 1]) + betazr
>
> sigma2sq[n_] := sigma2sq[n] = (k^2) (sigma2sq[n - 1]) + sigmar^2
>
> phi1[n_] := phi1[n] = phi1[n - 1] + phi2[n - 1] + (0.5) (sigmaz^2)
>
> phi2[n_] := phi2[n] = k (phi2[n - 1]) + (1 - k) (rbar)
>
> psi[n_] := psi[n] = phi1[n] - (0.5) (sigma1sq[n])
>
>
>
> alpha1[0] = 0
>
> alpha2[0] = 1
>
> sigma1sq[0] = 0
>
> sigma12[0] = 0
>
> sigma2sq[0] = 0
>
> phi1[0] = 0
>
> phi2[0] = 0
>
>
>
> B[h_, r_] := Exp[(-alpha1[h]) (r) - psi[h]]
>
> Exp[Cbar - beta] Sum[(Pi^x) B[x, r], {x, 1, 1000}]
>
>
>
> ----------------------------------------------------------------------
>
>
>
> and I am wondering if it is possible to solve the last line such that I have "r" as a function of "beta", satisfying
>
>
>
> Exp[Cbar - beta] Sum[(Pi^x) B[x, r], {x, 1, 1000}] = 1
>
>
>
> Because ultimately, I would need to integrate a function J[r] over "beta", so if I don't have "r" as a function of "beta", I don't know how to do the integration of J[r].
>
>
>
> Thank you!! L.