Also, using solveset we can get the solution pretty fast :
```
solveset(omega_nf - 942.5 , J_u)
{0.00235331614197391}
```
Regards,
Shekhar
On Thursday, 6 June 2019 02:50:27 UTC+5:30, Aaron Meurer wrote:
>
> What version of SymPy are you using? For me in 1.4, solve(omega_nf -
> 942.5, J_u, dict=True) returns [{J_u: 0.00235331614197392}]
>
> In general, solve() only returns closed-form solutions, so if it
> doesn't return a solution, it may just mean that it couldn't find one
> in closed-form. If you know that you want a numeric solution, you may
> be better off starting with nsolve.
>
> I believe nsolve() gives the wrong answer because you passed
> verify=False. Without it, it gives an error that it couldn't find the
> root. The default Newton's method solver has a hard time with this
> equation, but you can use a different one. For instance,
> nsolve(omega_nf - 942.5, [0, 1], solver='bisect') immediately returns
> the solution 0.00235331614197389. See the docstring of nsolve.
>
> Aaron Meurer
>
> On Wed, Jun 5, 2019 at 7:20 AM pull_over93 <[email protected]
> <javascript:>> wrote:
> >
> > Probably I'm not using sympy correctly, here I have this equation :
> >
> > omega_nf = sqrt(2)*sqrt(87791997.5351708 -
> 12563210.5479217*sqrt(-0.00144380926150678*J_u + 48.4180817181289*(J_u +
> 0.00027375075)**2 - 3.58991000729848e-7)/(J_u + 0.00027375075) +
> 2392.05862861605/(J_u + 0.00027375075))/2
> > and I want to solve this equation :
> > 942.5 = omega_nf
> >
> > I have tried the classic sympy.solve()
> >
> > eq_solution = sym.solve(omega_nf - 942.5 , J_u, dict = True)
> >
> >
> > But I do not have any output.
> > Than I've tried using sympy.nsolve()
> > eq_solution = sym.nsolve(omega_nf - 942.5, J_u, 0.0023, verify=False)
> >
> > But I don't get the right answer. (Right answer: J_u = 0.0023)
> > There's a smarter way to use sympy ?
> >
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