solving non-linear equations using scipy - scipy

Question

I'm trying to solve the following equation:
Where a list of A_e/A* values are given, and gamma=1.2, how should I solve this equation such that a list of M_e is returned corresponding to a list of A_e/A* values?
I thought about using scipy.optimize.newton, but it seems like this is not the right approach
def expr(x):
result = np.arange(1,1.25,step=0.004)-((1/x)*((2/(1.2+1))*(1+((1.2-1)/2)*x**2))**((1.2+1)/(2*1.2-1)))
return result.any()
scipy.optimize.newton(expr,1.1)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_4660/3619719442.py in <module>
----> 1 scipy.optimize.newton(expr,1.1,x1=1.2)
D:\Softwares\Anaconda\lib\site-packages\scipy\optimize\zeros.py in newton(func, x0, fprime, args, tol, maxiter, fprime2, x1, rtol, full_output, disp)
338 " Failed to converge after %d iterations, value is %s."
339 % (itr + 1, p1))
--> 340 raise RuntimeError(msg)
341 warnings.warn(msg, RuntimeWarning)
342 p = (p1 + p0) / 2.0
RuntimeError: Tolerance of 0.09999999999999987 reached. Failed to converge after 1 iterations, value is 1.2.
I used x to denote M_e, and I replaced Ae/A* with a list of values--np.arange(1,1.25,step=0.004) But, I guess the newton's method can only return 1 scalar value, but I defined the function with a sweeping list of Ae/A* values. How should I fix this?

Answer 1

newton works for scalar functions, and you are turning it into a vector function. Since you want the zero for different Ae values, include the Ae parameter in the function definition, then call newton several times (you can use the args keyword):
def expr(x, a):
result = a-((1/x)*((2/(1.2+1))*(1+((1.2-1)/2)*x**2))**((1.2+1)/(2*1.2-1)))
return result
[newton(expr, .01, args=(a,)) for a in np.arange(1,1.25,step=0.004)]
>>[0.9999999999999992, 0.9944379739232752,0.9889506795317498, 0.9835363421148852, ...