how can i find values of R

A=[472.3; 861.8; 1675; 1586; 1365; 1076 ;855];
r=2;
T=915.6345*10^3;
k=A.*0.58;
((R^4-r^4)/R)=((2*T*k)/pi)

These are the our values then how can i find R values?(5th line is our equation)

[James Tursa]

Are you trying to find an R vector where each element of R solves the equation for each element of A?

[Ozan Öz]

yes i am

Answer 1

[John D'Errico]

A=[472.3; 861.8; 1675; 1586; 1365; 1076 ;855];
r=2;
T=915.6345*10^3;
k=A.*0.58;

While the correct answer is exactly as James has posed it, to use roots after multiplying by R, there are always simple alternatives. If we re-write the relation as

R = nthroot(R*2*T*k/pi + r^4,4);

then we can just iterate that line until the vector converges for all R. Thus...

R = ones(size(A));
dr = inf;
tol = 1e-14;
while dr > tol
  R0 = R;
  R = nthroot(R0*2*T.*k/pi + r^4,4);
  dr = norm(R-R0);
end

R
R =
     542.52
     662.95
     827.34
     812.41
     772.78
     713.86
      661.2

So just a classic fixed point iteration. Did it work? Of course. As long as I was careful in how I chose my iteration scheme. And that means you need to understand fixed point iteration, to know when it will be convergent.

((R.^4-r^4)./R) - ((2*T*k)/pi)
ans =
            0
            0
            0
  -1.1921e-07
  -1.1921e-07
            0
            0

Roots is a better choice. But then you will want to extract the positive root, distinguishing it from the complex ones and the negative root. And you will still want to loop. Could you have used fzero? Of course. But there too you will need a loop, solving for a new value for R for each value of A.

Answer 2

[James Tursa]

Multiply by R to get a polynomial in R, then use the roots() function.

[John D'Errico]

With only the caveat that since A is a vector, it is apparently needed to solve the problem 7 separate times. A loop will of course suffice.

[James Tursa]

Yes. I guess I needed to append my comment above with this answer.

[John D'Errico]

I still upvoted your answer, since it is how I would solve this problem, were it mine to solve.

Answer 3

[Walter Roberson]

a = sym([472.3; 861.8; 1675; 1586; 1365; 1076 ;855], 'r');
syms A R
r = sym(2, 'r');
T = sym(915.6345*10^3, 'r');
k = A.*sym(0.58, 'r');
sol = solve(((R^4-r^4)/R)==((2*T*k)/sym(pi)), R, 'MaxDegree', 4, 'ReturnConditions', true);
solA = simplify(subs(sol.R, A, a(:).'));

There are four solutions for each A value, which are listed down the columns of solA; each column corresponds to a different A value.

By the way: in each case, two of the values for R are complex, one is negative, and one is positive.

An example solution is

(2^(5/6)*3^(1/3)*((6^(2/3)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(2/3) - 1920000*pi^(4/3))^(3/4) - (1920000*pi^(4/3)*(6^(2/3)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(2/3) - 1920000*pi^(4/3))^(1/2) + 9029643160428*2^(1/2)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(1/2) - 6^(2/3)*(6^(2/3)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(2/3) - 1920000*pi^(4/3))^(1/2)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(2/3))^(1/2)))/(120*pi^(1/3)*(6^(2/3)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(2/3) - 1920000*pi^(4/3))^(1/4)*(3^(1/2)*(65536000000000000*pi^4 + 106865153207769915068040884980079778922926695307)^(1/2) + 566211497254612223230161)^(1/6))

Considering you used floating point values for your coefficients, the worth of these solutions is in question.