Symbolic toolbox solution
in [Matlab]
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From: Serdar GOKHAN on 9 Jul 2010 07:39 Hi All, How can we get analytical expressions of "kappa" and "gamma" from the below two equations by using Symbolic toolbox? Any kind of help is appreciated. Regards Serdar 1.st equation Ip0*gamma+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L) 2.nd equation -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma*z)*Is0*exp(-kappa*z)
From: Sean on 9 Jul 2010 09:44 "Serdar GOKHAN" <fs_gokhan(a)hotmail.com> wrote in message <i171oo$9av$1(a)fred.mathworks.com>... > Hi All, > > How can we get analytical expressions of "kappa" and "gamma" from the below two equations by using Symbolic toolbox? Any kind of help is appreciated. > Regards > Serdar > > 1.st equation > Ip0*gamma+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L) > > 2.nd equation > -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma*z)*Is0*exp(-kappa*z) > How are those "equations"? They're not equal to anything... Anyway, once you figure out what those are equal to: >>syms gamma_1 alpha_p Ip0 gb K L z alpha_s kappa %Note don't use 'gamma' it's a built-in function >>doc solve >>solve('eqn1','eqn2',var1,var2); %var1 = gamma1, var = kappa >>
From: Serdar GOKHAN on 9 Jul 2010 10:19 "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i17934$del$1(a)fred.mathworks.com>... > "Serdar GOKHAN" <fs_gokhan(a)hotmail.com> wrote in message <i171oo$9av$1(a)fred.mathworks.com>... > > Hi All, > > > > How can we get analytical expressions of "kappa" and "gamma" from the below two equations by using Symbolic toolbox? Any kind of help is appreciated. > > Regards > > Serdar > > > > 1.st equation > > Ip0*gamma+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L) > > > > 2.nd equation > > -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma*z)*Is0*exp(-kappa*z) > > > > How are those "equations"? They're not equal to anything... > > Anyway, once you figure out what those are equal to: > >>syms gamma_1 alpha_p Ip0 gb K L z alpha_s kappa %Note don't use 'gamma' it's a built-in function > > >>doc solve > >>solve('eqn1','eqn2',var1,var2); %var1 = gamma1, var = kappa > >> Thank you very much for your reply. All the equations are equal to zero. And I changed gamma to gamma1 Ip0*gamma1+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L)=0 -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma1*z)*Is0*exp(-kappa*z)=0 when I try to solve usig your way, i.e, solve('eqn1','eqn2',var1,var2); %var1 = gamma1, var = kappa it says, Unable to find closed form solution. Do you have alternative way? Thank you for your reply and assistance. Regards. Serdar
From: Walter Roberson on 9 Jul 2010 10:32 Serdar GOKHAN wrote: > Hi All, > > How can we get analytical expressions of "kappa" and "gamma" from the > below two equations by using Symbolic toolbox? Any kind of help is > appreciated. > Regards > Serdar > > 1.st equation > Ip0*gamma+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L) > > 2.nd equation > > -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma*z)*Is0*exp(-kappa*z) If those two formula are to be equated to zero and then the pair solved for gamma and kappa, then I do not believe there is an analytical formula for them. Isolating the two variables in terms of the others involves finding the root of something of the form c0 + c1 * x + c2 * exp(c3 + c4 * exp(x)) If the term inside the first exp() only involved x linearly and not exp(x), then the root could be determined analytically using the LambertW function, but I do not know of any equivalent function that can be used when there is an x + exp(exp(x)) form.
From: Sean on 9 Jul 2010 10:45
Walter Roberson <roberson(a)hushmail.com> wrote in message <1qGZn.10985$Hw.4661(a)newsfe10.iad>... > Serdar GOKHAN wrote: > > Hi All, > > > > How can we get analytical expressions of "kappa" and "gamma" from the > > below two equations by using Symbolic toolbox? Any kind of help is > > appreciated. > > Regards > > Serdar > > > > 1.st equation > > Ip0*gamma+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L) > > > > 2.nd equation > > > > -Is0*kappa*exp(-kappa*z)-alpha_s*Is0*exp(-kappa*z)+gb/K*Ip0*exp(gamma*z)*Is0*exp(-kappa*z) > > If those two formula are to be equated to zero and then the pair solved > for gamma and kappa, then I do not believe there is an analytical > formula for them. Isolating the two variables in terms of the others > involves finding the root of something of the form > > c0 + c1 * x + c2 * exp(c3 + c4 * exp(x)) > > If the term inside the first exp() only involved x linearly and not > exp(x), then the root could be determined analytically using the > LambertW function, but I do not know of any equivalent function that can > be used when there is an x + exp(exp(x)) form. In addition to this, if you run just the first equation and solve for both variables you'll see: >>vars = solve('Ip0*gamma1+alpha_p*Ip0+gb/K*Ip0*Is0*exp(kappa*L)=0',... 'gamma1','kappa'); >>vars.kappa ans = z %meaning that kappa is equivalent to z. I'm not sure if this is what you meant for and could be a typo in your equation. |