How to write the codes for this gamma function?

I am trying to find the value for alpha and beta. the answer should be alpha=395.2359 and beta=2.0698

But i am not getting this answer.

I used this code.

syms a b
[a,b]=solve(a.*gamma((1/b) + 1) - 350.1 == 0, a.^2.*gamma((2/b) + 1) - 154056.7 == 0)

Can someone tell me where is the problem in my codes and what should be the right codes?

Thanks.

Answer 1

[John D'Errico]

Easy, peasy. Although since you are using gamma as a FUNCTION, it is a REALLY bad idea to use beta, another closely related special function as a variable.

For the changed problem, I now have:

First, eliminate a. Square the first equation, then divide one into the other. This is valid as long as we create no zero divides.

We get

gamma(2/beta + 1)/gamma(1/beta + 1)^2 = 154056.7/350.1^2

Solve for beta. But before we bother to try that, plot the function. Does it EVER cross zero? Certainly, does it cross zero near 2? (NO.)

bfun = @(bet) gamma(2./bet + 1)./gamma(1./bet + 1).^2 - 154056.7/350.1^2;
fplot(bfun,[1,3])
yline(0);

So now, we see a solution bet beta, roughly near 2.

format long g
bet = fzero(bfun,2)

bet = 
          2.06980749988105

Solving for alpha is now easy. (Again, alpha is ALSO a function in MATLAB.

alph = 350.1/gamma(1/bet + 1)

alph = 
           395.23590882898

Could you have used the symbolic toolbox? Trivially easy too. But since an analytical solution will not exist, you will use vpasolve. Since vpasolve is a numerical solver, you will do best if you provide initial estimates of the unknowns as multiple solutions can exist.

syms a b real positive
[a,b] = vpasolve(a.*gamma((1/b) + 1) - 350.1 == 0, a.^2.*gamma((2/b) + 1) - 154056.7 == 0,[a,b],[300,2])

a = 

395.23590882898036747077460275278

b = 

2.0698074998810479469634339822283

[MD. Rokibujjaman sovon]

hi,

not 154.056, it would be 154056.7

i made a mistake during the latex writing.

[John D'Errico]

With a valid value for the constant, now a solution exists. I show two solution methods, though I could also have used fsolve.

[MD. Rokibujjaman sovon]

why do you use [300, 2]? if i don't use that, isn't it possible? because i do some other calculation using 'solve', they results same without initial estimates.

[Walter Roberson]

As John wrote, "Since vpasolve is a numerical solver, you will do best if you provide initial estimates of the unknowns as multiple solutions can exist"

[MD Rokibujjaman Sabuj]

If I don’t want to use initial estimates, how can i solve this?

[Walter Roberson]

vpasolve does not demand initial estimates. But it might happen to find a different solution, possibly even one that involves complex values.

Also, in general some equations are nonlinear enough that vpasolve does a poor job of finding any solution at all. Sometimes it is necessary to provide a guess that is pretty close. A couple of weeks ago, one user had a system that vpasolve failed on unless the initial guess was within 0.1 on one of the values. Just "wanting" not to use initial values is not always enough to get a solution in practice.

[MD. Rokibujjaman sovon]

Thanks

Answer 2

[Alan Stevens]

Compare your coded value of 154056.7 with the equation's value of 154.056.

[MD. Rokibujjaman sovon]

i edited the question.