Re: Solving this in mathematica?

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• Subject: [mg128077] Re: Solving this in mathematica?
• From: Dana DeLouis <dana01 at me.com>
• Date: Fri, 14 Sep 2012 00:22:15 -0400 (EDT)
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```> 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.

```

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