A complicated answer from 'solve' function

Here is a simple M code:

syms Omega t delta C M
tn=solve(Omega*sin(Omega*t-delta) + C*cos(Omega*t - delta)*1/(2*M), t)

Then, the answer from MATLAB Symbolic Toolbox is

(delta + log((- C^2 - 4*M^2*Omega^2)^(1/2)/(C + M*Omega*2i))*1i)/Omega
(delta + log(-(- C^2 - 4*M^2*Omega^2)^(1/2)/(C + M*Omega*2i))*1i)/Omega

Meanwhile, when I solve the equation in the M code by hand yields this:

1/Omega * atan(-C/2/M/Omega) + delta/Omega

I can't draw a straight line between my handwritten answer and the answer given by MATLAB. Would you help me to understand the answer from MATLAB?

In Kwon Park

Answer 1

[John D'Errico]

Very often, the answer is that the symbolic toolbox does not see what seems obvious to you.

Yes, it seems clear that by dividing by cos(Omega*t-delta), that you can collect those two terms into a tangent.

tan(Omega*t-delta) = -C/(2*M)/Omega

It requires that cos(Omega*t-delta) not be zero to be valid, or else you have a problem. Yes, we see that when the cos term in question is zero, then the tangent function goes through a singularity. Of course, given that, it is now possible to see how you solved the problem, and the symbolic toolbox has no issue, if I take that step for it:

tn=solve(tan(Omega*t-delta) + C/(2*M)/Omega,t)
tn =
(delta - atan(C/(2*M*Omega)))/Omega

But that is seen from a bird's eye view. Instead, what did the symbolic toolbox do? It appears it chose to avoid creating a singularity, but to use a different approach. One alternative approach might be to use the relation

cos(u) = +/- sqrt(1-sin(u)^2)

So, if u=Omega*t-delta and v=sin(u), then your problem reduces to something like

Omega*v = +/- C*sqrt(1-v^2)/(2*M)

Solve for v by squaring, and we would get a solution, although we would need to resolve any spurious roots created by that squaring. To me, the solution that you post seems vaguely quadratic, so suggestive of the second approach I took. Interestingly, when I try the same solve as you wrote, MATLAB finds no solution at all. So I'm not totally confident that you did truly get that solution. (In R2017b)

syms Omega t delta C M
tn=solve(Omega*sin(Omega*t-delta) + C*cos(Omega*t - delta)*1/(2*M), t)
tn =
Empty sym: 0-by-1

Computers don't always see the obvious, at least, not what we think of as obvious.

[IN KWON PARK]

Thank for your explanation. It hits the point regarding what might have happened on the MATLAB side. One strange point is that I'm using same MATLAB version as yours and it turned up with the answer I posted above. Meanwhile, is there any way to give the bird's eye hint to the MATLAB 'solve' function, which would gently nudge it to the direction more human?

[John D'Errico]

I'm a bit surprised why you got a result, while MATLAB just refuses to generate that result for me.

Part of me wants to substitute that solution into your equation and see if it reduces to zero. But that log term seems like it would be a problem.

As I showed, if you do divide by cos, then convert to a tangent, it does give the same solution. Sometimes symbolic solvers need a broad "hint" like that to get a solution in a form you want to see.

Even if I write it as:

tn=solve(sin(Omega*t-delta)/cos(Omega*t-delta) + C/(2*M)/Omega,t)
tn =
Empty sym: 0-by-1

Solve fails to see your solution. But if I convert to a tangent, then solve is happy, as you would expect.

[IN KWON PARK]

Here is my MATLAB version info: MATLAB Version: 9.3.0.713579 (R2017b). Besides, if the solution is as obvious as what I posted, then probably adding a small tweak like the one you've suggested might offer an edge to the MATLAB 'solve()' function. However, the very starting point of the equation was not what I've posted, but a solution of a 2nd order non-homogeneous ordinary differential equation.

[Walter Roberson]

In R2017b with OS-X I get the two solutions that IN KWON PARK gets.

John, is it possible that you had a symbolic variable with assumptions sitting around in your workspace?

[John D'Errico]

No.

R2017b. OSX (10.13.2)

clear
syms Omega t delta C M
tn=solve(Omega*sin(Omega*t-delta) + C*cos(Omega*t - delta)*1/(2*M), t)

tn =
Empty sym: 0-by-1

[Walter Roberson]

Odd -- also OS X, 2017b. Though I am running El Capitan rather than High Sierra

clear
syms Omega t delta C M
tn=solve(Omega*sin(Omega*t-delta) + C*cos(Omega*t - delta)*1/(2*M), t)
tn =
  (delta + log((- C^2 - 4*M^2*Omega^2)^(1/2)/(C + M*Omega*2i))*1i)/Omega
 (delta + log(-(- C^2 - 4*M^2*Omega^2)^(1/2)/(C + M*Omega*2i))*1i)/Omega

Note: if you did have an assumption it would have been saved with the symbolic engine, not at the MATLAB level. https://www.mathworks.com/help/symbolic/clear-assumptions-and-reset-the-symbolic-engine.html

reset(symengine)