How to find the value of a variable in MATLAB that exists on both sides of th eequal sign?

23 vues (au cours des 30 derniers jours)
Does anybody know how we can find the value of a variable in MATLAB that exists on both sides of equal sign, e.g. n in 2n+q=n+r (n is the only variable)?
In this case it is so simple; n=r-q. But what if the equation is very complex and we are not able to transfer the variable to one side?

Réponse acceptée

Star Strider
Star Strider le 16 Mai 2014
If you have the Symbolic Math Toolbox:
syms n q r
Eq = 2*n+q == n+r
n = solve(Eq, n)
produces:
n =
r - q
To solve them numerically for n, make the entire equation equal to zero and use the fzero function:
r = 3;
q = 5;
f = @(n) [2*n+q - (n+r)];
n = fzero(f, 1)
produces:
n =
-2
  4 commentaires
Kourosh
Kourosh le 19 Mai 2014
Dear Star,
I have one specific problem in my case. What you mentioned is correct. But in my case with the funntion
su=1; b=0.5; l=1.571; p=0.16; t=atan(2*pi()/p); Landa=-tan((pi()/2)-t);
pn=p/l; B=asin(pn);
q=0.49*pn^-0.55; r=0.76*pn^-0.55;
Nmax=4.266*cos (B)*((b/l)/0.318)*su*l^2; Tmax=0.191*((b/l)/0.318)*su*l^3;
f = @(Nn) [((Nn/Nmax)^(q-1))/(1-(Nn/Nmax)^q)^((r-1)/r)-(r*Landa*Nmax/(q*Tmax))];
Nn = fzero(f, 0)
where I know the result is definitely between 0 and 10, the result by Matlab is NaN.
I am pretty sure the equation is correct. Is there any other way to solve it?
Cheers, Cyrus
Star Strider
Star Strider le 19 Mai 2014
I don’t know what the equation is, but plotting it:
Nn = linspace(-10,10);
figure(1)
plot(Nn, real(f(Nn)))
grid
shows that the real part doesn’t have a zero crossing anywhere in the region. (It has a small region in the Nn domain where it generates complex values.)

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Plus de réponses (1)

Brian B
Brian B le 16 Mai 2014
If you have the Symbolic Toolbox, you can use:
>> syms n q r
>> solve('2*n+q=n+r','n')
ans =
r - q

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