0
$\begingroup$

Why is the result not Indeterminate?

Solve[x == a Cos[ω t + ϕ] /. {x -> 0, t -> 0, ϕ -> π/2}, a]

when the result should be

0/Cos[π/2]

Power::infy: Infinite expression 1/0 encountered.

Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered.

Out:= Indeterminate

$\endgroup$
7
  • $\begingroup$ Quit the kernel and try again. I cannot reproduce your problem. Evaluating x == a Cos[ω t + φ] /. {x -> 0, t -> 0, φ -> π/2} simply gives True, and so you're effectively solving Solve[True, a], which has no solution: {{}}. $\endgroup$
    – Roman
    Jul 1, 2019 at 17:03
  • $\begingroup$ I already did that but it's still doing the same. My result is has no solution {{}} too, but Why? if the solution is indeterminate. $\endgroup$
    – Andres
    Jul 1, 2019 at 17:06
  • $\begingroup$ What is your $Version? $\endgroup$
    – Roman
    Jul 1, 2019 at 17:07
  • $\begingroup$ my version is the 12.0 $\endgroup$
    – Andres
    Jul 1, 2019 at 17:11
  • 1
    $\begingroup$ Again: you have the equation 0==0, which is True, and you're solving it for a, which gives no solution because there is no information about a in your equation. Why do you expect Indeterminate? $\endgroup$
    – Roman
    Jul 1, 2019 at 17:13

2 Answers 2

2
$\begingroup$

When I run your code, I get:

Solve[x == a Cos[ω t + ϕ] /. {x -> 0, t -> 0, ϕ -> π/2}, a]
{{}}

As @MichaelE2 points out, this means a "full-dimensional solution" (as documented in the help file for Solve). The reason this occurs is because the equation contains

a Cos[ω t + ϕ] /. {x -> 0, t -> 0, ϕ -> π/2}
0

which evaluates to zero. So you are effectively solving

Solve[0 == a*0, a]
{{}}

which similarly returns a full dimensional solution set. In this case, since a is the variable being solved for, this may be interpreted as "true for all a".

$\endgroup$
2
  • 1
    $\begingroup$ Note: {} indicates an empty solution set, while {{}} indicates a full-dimensional component. $\endgroup$
    – Michael E2
    Jul 1, 2019 at 17:28
  • $\begingroup$ If we solve to variable A : then 0 / cos 90 is equal a indeterminate, only that was my question. But I saw other result and this alarmed me. $\endgroup$
    – Andres
    Jul 1, 2019 at 17:52
1
$\begingroup$

ContourPlot shows the possible solutions:

ContourPlot[0 == a Cos[y] , {y, 0, Pi}, {a, -1, 1}, FrameLabel -> {y, a}]

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.