Solving a two equation system yields strange results
- Subject: Solving a two equation system yields strange results
- From: Stavros Macrakis
- Date: Wed, 9 May 2012 17:48:59 -0400
As far as I know, 'solve' knows nothing about indexed sums, so Maxima will
treat your equation
exp(-1/b) / sum(i! * b^i, i, 0, k -1) = 0.02
roughly the way it treats:
exp(-1/b) / f(b,k) = 0.02
namely, not something that Maxima can do anything useful with....
However, you can work on this problem in the following simple way:
(%i1) eq1: b*(k - 1) = 0.1$
I will express eq2 as its left-hand-side minus its right-hand-side for
convenience later.
(%i2) eq2: exp(-1/b) * sum(1/(i! * b^i), i, 0, k -1) - 0.02$
(%i3) solve(eq1,b);
rat: replaced -0.1 by -1/10 = -0.1
(%o3) [b=1/(10*k-10)]
(%i4) subst(%o3,eq2);
(%o4) %e^(10-10*k)*(sum((10*k-10)^i/i!,i,0,k-1))-0.02
Now we have an expression in one integer variable, so we can just evaluate
it at the relevant integers....
(%i9) makelist(subst(j,k,%o4),j,1,10);
(%o9)
[-0.02,%e^(-10)*(sum(10^i/i!,i,0,1))-0.02,%e^(-20)*(sum(20^i/i!,i,0,2))-0.02,%e^(-30)*(sum(30^i/i!,i,0,3))-0.02,%e^(-40)*(sum(40^i/i!,i,0,4))-0.02,%e^(-50)*(sum(50^i/i!,i,0,5))-0.02,%e^(-60)*(sum(60^i/i!,i,0,6))-0.02,%e^(-70)*(sum(70^i/i!,i,0,7))-0.02,%e^(-80)*(sum(80^i/i!,i,0,8))-0.02,%e^(-90)*(sum(90^i/i!,i,0,9))-0.02]
Now expand the sums and calculate numerical results:
(%i10) %,sum,numer;
(%o10)
[-0.02,-.01950060077261267,-.01999954448504944,-0.0199999995338968,-.01999999999949795,-.01999999999999945,-0.02,-0.02,-0.02,-0.02]
It doesn't look as though this is ever near zero, so no solutions.
-s
On Wed, May 9, 2012 at 2:47 PM, Jason Filippou <jason.filippou at gmail.com>wrote:
> Pardon, the second equation (2) should read (as is implied by the
> output that I typed in):
>
> (%i7) exp(-1/b) * sum(1/(i! * b^i), i, 0, k -1) = 0.02;
> k - 1
> ====
> - 1/b \ 1
> (%o7) %e > ----- = 0.02
> / i
> ==== b i!
> i = 0
>
>
>
> Still no luck though.
>
> Jason
>
> On Wed, May 9, 2012 at 9:46 PM, Jason Filippou <jason.filippou at gmail.com>
> wrote:
> > Good afternoon.
> >
> > I've been using Maxima 5.21.1 in a Debian GNU / Linux 2.6.32-5-686
> > system to solve a particular two equation system that I have to. The
> > system is as follows:
> >
> > (1): b*(k - 1) = 0.1
> >
> > (2): exp(-1/b) / sum(i! * b^i, i, 0, k -1) = 0.02
> >
> > I've been using solve/2 for this, but I've been returned an empty
> > solution set. Namely:
> >
> > %i9) solve([b * (k - 1) = 0.1, exp(-1/b) / sum(i! * b^i, i, 0, k -1) =
> > 0.02], [b, k]);
> >
> > rat: replaced -0.1 by -1/10 = -0.1
> >
> > rat: replaced -0.02 by -1/50 = -0.02
> > (%o9) []
> >
> > Now, normally I would assume that this means that the system doesn't
> > have a solution, but after substituting beta with its equivalent from
> > the first equation, i.e 0.1 / (k - 1), I noticed that the evaluation
> > of the second equation halts after a couple of steps:
> >
> > (%i5) solve([exp(-(k - 1) / 0.1) * sum(1/(i! * (0.1 / (k-1))^i), i, 0,
> > k -1) = 0.02], [k]);
> >
> > rat: replaced -0.02 by -1/50 = -0.02
> >
> > rat: replaced 0.1 by 1/10 = 0.1
> >
> > rat: replaced 10.0 by 10/1 = 10.0
> > %i %pi
> > ------
> > 5
> > 1 - k %e
> > (%o5) [%e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 2 %i %pi
> > --------
> > 5
> > 1 - k %e
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 3 %i %pi
> > --------
> > 5
> > 1 - k %e
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 4 %i %pi
> > --------
> > 5
> > 1 - k %e
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 1 - k 1
> > %e = - -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 5 k - 4 %i %pi - 5
> > - ------------------
> > 5 1
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 5 k - 3 %i %pi - 5
> > - ------------------
> > 5 1
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 5 k - 2 %i %pi - 5
> > - ------------------
> > 5 1
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 5 k - %i %pi - 5
> > - ----------------
> > 5 1
> > %e = -----------------------------------,
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> > 1 - k 1
> > %e = -----------------------------------]
> > k - 1
> > ==== i i
> > 1/10 1/5 \ 10 (k - 1) 1/10
> > 2 5 ( > ------------)
> > / i!
> > ====
> > i = 0
> >
> > And then if I try to "solve" it any further I get no solutions:
> >
> > (%i6) solve(%, [k]);
> > (%o6) []
> >
> > What am I doing wrong here?
> >
> > Thanks,
> >
> > Jason
> >
> > --
> > Jason Filippou
> > Research Associate
> > NCSR Demokritos
> > NCSR Webpage: http://users.iit.demokritos.gr/~jfilip/
> > D.I.T Webpage: http://cgi.di.uoa.gr/~std06142/
> > LinkedIn! Profile:
> http://www.linkedin.com/profile/view?id=132927442&trk=tab_pro
> >
> >
> > --
> > Jason Filippou
> > Research Associate
> > NCSR Demokritos
> > NCSR Webpage: http://users.iit.demokritos.gr/~jfilip/
> > D.I.T Webpage: http://cgi.di.uoa.gr/~std06142/
> > LinkedIn! Profile:
> http://www.linkedin.com/profile/view?id=132927442&trk=tab_pro
>
>
>
> --
> Jason Filippou
> Research Associate
> NCSR Demokritos
> NCSR Webpage: http://users.iit.demokritos.gr/~jfilip/
> D.I.T Webpage: http://cgi.di.uoa.gr/~std06142/
> LinkedIn! Profile:
> http://www.linkedin.com/profile/view?id=132927442&trk=tab_pro
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