Solving one equation with one unknown and get all possible solutions

Question

Hassan Alkomy 2018-10-25

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评论： Hassan Alkomy 2019-1-8

I have an equation and I need to get its solution. I think it has more than one solution, but using the command (solve) I can get only one solution.

Actually, it is expected to get real and complex solutions, but I am interested on the real solutions only.

How can I get this solution in Matlab.

the required unknown is (alphap) and my equation and the command that I have used is:

m=15; Ki=1.3908e+06; B=0.945e-1; db=0.79e-2; alphao=.2618;
x = solve(Pr == m*Ki*(B*db*(cos(alphao)/cos(alphap)-1))^(3/2)*sin(alphap),alphap)

The answer is:

x = 0.37336926931567958392238007768557i

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Dimitris Kalogiros 2018-10-26

what is the value of Pr ?

Hassan Alkomy 2019-1-4

Pr=10

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Answer 1

Stephan 2019-1-4

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编辑：Stephan 2019-1-4

在 MATLAB Online 中打开

Hi,

getting all possible soultions is a hard job, because you have an infinite bunch of real solutions:

You find them for example by using fzero:

m=15; 
Ki=1.3908e+06; 
B=0.945e-1;
db=0.79e-2; 
alphao=.2618;
Pr = 10;
format long
fun = @(alphap)Pr - m*Ki*(B*db*(cos(alphao)/cos(alphap)-1))^(3/2)*sin(alphap)
x1 = fzero(fun,0.5)

This code results in:

x1 =
   0.542034560066698

If you want more solutions just add or subtract integer multiples of 2*pi. Then you can construct as many real solutions as you want by yourself:

   x2 = fzero(fun,2*pi+x1)
   is_it_2_pi = (x2-x1)/(2*pi)

gives:

   x2 =
   6.825219867246284
is_it_2_pi =
     1

Best regards

Stephan

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Walter Roberson 2019-1-8

The way your problem is constructed, with the sin(alphap) and cos(alphap) you would expect the results to repeat exactly every 2 π radians, since sin(alphap + 2*pi) = sin(alphap) and cos(alphap + 2*pi) = cos(alphap) . In theory. In practice due to round-off error it does not hurt to use the +2*pi as the starting point and use fzero() to confirm the exact location to within numeric bounds.

Hassan Alkomy 2019-1-8