Re: Difficult Set of Equations

On Dec 7, 9:43 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:
On Dec 7, 4:13 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:





On Dec 7, 1:07 pm, sonicb11 <williamp...@xxxxxxxxxxx> wrote:

On Dec 7, 4:05 pm, sonicb11 <williamp...@xxxxxxxxxxx> wrote:

On Dec 7, 3:23 pm, Maxime <lab_ma...@xxxxxxxxxxx> wrote:

To do that, you have to use a pc programm who is going to resolve it.
If you want, I can make it in my mind but to do that you have to
explain me what is "sum" and Wi. I'm a french man and I don't
understand it.

Wi are just some random variables W1, W2,..., Wn. The context isn't
really important here. sum(Wi) means W1+W2+...+Wn

Thanks.

Perhaps if someone could solve them and email it to me or post on this
board, I would be extremely appreciative. I don't have the software
avaiable at home or at work at the moment. Thanks a lot.

Maple 9.5 gets a lengthy solution in terms of roots of some 4th degree
polynomials. Let's use notation m instead of mu, c instead of a+b, and
call the four right-hand sides r1..r4. Here it comes:> eq1:=m*a/c=r1:
eq2:=(m^2+s^2)*a*(a+1)/c/(c+1)=r2:eq3:=(m^3+3*m*s^2)*mul((a+j)/(c+j),j=0..2-)=r3:
eq4:=(m^4+6*m^2*s^2+3*s^4)*mul((a+j)/(c+j),j=0..3)=r4:

m:=solve(eq1,m);
r1 c
m := ----
a
Maple automatically substitutes this into the other equations. Now
let's solve for s from eq(2):
solve(eq2,s): There are two roots, one >0 and one <0; let's select the>0 root:

s:=%[1];
2 2 1/2
((a + 1) c (-r1 c a - r1 c + r2 a c + r2 a))
s := -------------------------------------------------
(a + 1) a
Now these m and s are automatically put into equations 3 and 4 the
next time we call on them. So, lets solve equations 3 and 4 for a and
c:
eq34:=simplify({eq3,eq4}):
sol_ac:=solve(eq34,{a,c}):assign(sol_ac): <--- this give a and c their
"solution" values.
a;
2 2 2 2 4 2 3
RootOf((54 r4 r1 r2 - 126 r4 r1 r2 + 8 r4 r3 r1

4 3 2 3
+ 27 r1 r2 r4 - 6 r1 r4 r3 r2 - 18 r4 r3 r1 r2

2 3 5 2 4
+ 84 r4 r3 r2 r1 - 108 r1 r3 r4 r2 + 20 r4 r3 r1

2 2 3 4 3 2
- 3 r1 r2 r4 r3 - 9 r1 r2 r3 - 3 r1 r3 r2

2 2 3 5 3 4 2 2 2 2
- 18 r1 r3 r2 + 8 r1 r3 + 24 r1 r2 r3 - 6 r3 r2 r4

6 2 4 4 2 2 2 2 3
+ 72 r1 r4 ) _Z + (216 r1 r2 r3 - 144 r1 r3 r2

2 4 2 3 3
- 504 r4 r1 r2 + 16 r4 r3 r1 - 90 r4 r3 r1 r2

5 5 3 2 2
- 756 r1 r3 r4 r2 + 64 r1 r3 - 12 r1 r2 r4 r3

2 2 2 2 4 6 2
+ 162 r4 r1 r2 + 100 r4 r3 r1 + 360 r1 r4

2 2 3 4 4 3
- 18 r3 r2 r4 - 90 r1 r2 r3 + 216 r1 r2 r4

3 2 2 3 2 3
- 18 r1 r3 r2 + 504 r4 r3 r2 r1 - 6 r1 r4 r3 r2) _Z +

2 3 5 3
(8 r4 r3 r1 - 1836 r1 r3 r4 r2 - 108 r4 r3 r1 r2

2 4 2 2 2 4 3
+ 140 r4 r3 r1 + 108 r4 r1 r2 + 621 r1 r2 r4

2 4 2 2 2 2 3
- 630 r4 r1 r2 - 9 r1 r2 r4 r3 - 378 r1 r3 r2

6 2 5 3 4 2 2
+ 648 r1 r4 + 176 r1 r3 + 696 r1 r2 r3

3 4 2 3 3 2 2
- 333 r1 r2 r3 + 924 r4 r3 r2 r1 - 27 r1 r3 r2 ) _Z +

2 4 2 2 3 2 3
(60 r4 r3 r1 - 324 r1 r3 r2 + 504 r4 r3 r2 r1

4 3 2 4 3 4
+ 756 r1 r2 r4 - 252 r4 r1 r2 - 540 r1 r2 r3

6 2 4 2 2 5 3
+ 504 r1 r4 + 936 r1 r2 r3 + 192 r1 r3

5 3 4 5 3
- 1836 r1 r3 r4 r2) _Z - 324 r1 r2 r3 + 72 r1 r3

5 6 2 4 3
- 648 r1 r3 r4 r2 + 144 r1 r4 + 324 r1 r2 r4

4 2 2
+ 432 r1 r2 r3 )

It takes more than 100 pages to write out the solution value for 'c',
but there is an easier way: You now have 'a', and can try solving the
first three equations for m,s and c in terms of 'a', r1, r2, and r3.
Doing this gives c as the roots of the quadratic equation:

c;

RootOf(

2 3 3 2 3 2
(2 a r1 + 6 a r1 - 3 r1 r2 a + 4 r1 - 6 a r2 r1 + r3 a )

2
_Z + 2 r3 + (3 r3 a - 6 r2 r1 - 3 a r2 r1) _Z) * a

That is, c = a*(root of quadratic). The quadratic has no constant
term, so one of its roots is zero, and the other equals -(coeff of Z)/
(coeff of Z^2).

Sorry. That last statement is nonsense. I missed the constant term
2*r3 in the quadratic. Therefore, there will be two roots, for each
value of 'a'. Of course, there may be as many as 4 real roots for 'a',
so there could be lots of real solutions (or possibly none, if the
roots for 'a' or 'c' are all complex).

R.G. Vickson





R.G. Vickson- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Wow, thanks for your answer. Looks incredibly complicated. Looking for
something simpler. I am assuming a/(a+b) = k, say, is given. What
other conditions would I need, if any, in order to solve for the
parameters? Maybe I can only solve for a combination of the
parameters. For example, I can estimate b*k or a*k.
.

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