Finding real roots of a cubic equation

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robin johansson
robin johansson 2016-12-19
评论： Massimo Zanetti 2016-12-20
I have the equation:
eqn = ((d^3*pi*((4251*d + 5951400)/(25*d))^(1/2))/(2*(d + 1400)))*(pi*(d^2)/4)-180 == 0
and want to find the real root/roots.
when i attempt to solve it with
x = vpasolve(eqn,d)
Matlab only return the imaginary part of the solution:
- 3.593452147062167996782934136112 - 1.3074862631155484137468498544529i
How do i find the real solution?

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回答（3 个）

Roger Stafford
Roger Stafford 2016-12-20
This is an equation that can be manipulated so that d is one of the roots of a ninth degree polynomial equation of which only one of its nine roots is real.
The original equation is:
((d^3*pi*((4251*d+5951400)/(25*d))^(1/2))/(2*(d+1400)))*(pi*(d^2)/4)==180
Since 4251*d+5951400 = 4251*(d+1400), the d+1400 partially cancels with the same quantity in the denominator and we get the equation
pi^2/40*d^5*(4251/(d*(d+1400)))^(1/2)==180
or
pi^2/40*d^5*4251^(1/2) == 180*(d*(d+1400))^(1/2)
Squaring both sides and transposing gives
4251/1600*pi^4*d^10-32400*d^2-45360000*d == 0
One factor d can be factored out since d = 0 is clearly not a solution of the original equation and that finally leaves the polynomial equation
4251/1600*pi^4*d^9-32400*d-45360000 == 0
The ‘roots’ function can be used for this and it shows that there is only one real root, namely
d = 3.82617917662798
Massimo Zanetti
Massimo Zanetti 2016-12-19
编辑：Massimo Zanetti 2016-12-19
To force the vpasolve finding only real solutions you cannot use assume. If you know the real solution is only one a workaround is to set search range to [-Inf,Inf]:
x = vpasolve(eqn,d,[-Inf,Inf])
x =
3.8261791766279723703687735908663
By the way, I suggest you to read documentation to get more insights: vpasolve
  2 个评论
robin johansson
robin johansson 2016-12-19
编辑：robin johansson 2016-12-19
Thank you for the answer! Just what i was looking for.
With this problem i did know that i only had 1 real solution. I've read the documentation and i still don't understand how i would go about if i didn't know how many real solutions i could find.
My idea of how to get around this would be
X = vpasolve(eqn,x,[-inf, inf],'random',true)
Y = vpasolve(eqn,x,[X, inf],'random',true)
Z = vpasolve(eqn,x,[-inf, X],'random',true)
etc.
Not sure if this works, could you suggest a smarter solution to finding all real values of a nonpolynomial equation?
Massimo Zanetti
Massimo Zanetti 2016-12-20
For non polynomial equations there is no general rule to find all solutions. Consider using solve function. The documentation is rich of useful examples that cover many potential applications.

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Walter Roberson
Walter Roberson 2016-12-19
One approach for polynomials is to use solve instead of vpasolve, to get all of the solutions, and then to use logical indexing to select the results whose imag()==0
  1 个评论
Walter Roberson
Walter Roberson 2016-12-20
sols = solve(eqn, x);
sols(imag(sols)~=0) = []; %remove the ones that have a non-zero imaginary component.
However, in R2016b (and perhaps some other releases), this code can fail due to a bug in the Symbolic Toolbox (I notified them of the bug a few weeks ago.) The work-around for the moment is
sols(imag(vpa(sols))~=0) = [];

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