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Is there a programmable method for calculating the exponent value of a power sum

Say I have an equation:
a^x + b^x + c^x = n
Since I know a, b, c and n, is there a way to solve for x?
I have been struggling with this problem for a while now, and I can't seem to find a solution online.
My current method is to iterate over X until the left side is "close enough" to n. The method is pretty slow and in an already computationally difficult algorithm.
Example:
3^x + 5^x + 7^x = 83
How do i go about solving for x. (2 in this case)
I tried the equation in WolframAlpha and it seems to know how to solve it, but any other program fails to do so.
I probably should also mention that X is not an integer (mostly in 0.01 to 0.05 range in my case).

You can use scipy library. You can install it using command pip install scipy
Then, this code will work:
from scipy.optimize import root
def eqn(x):
return 3**x + 5**x + 7**x - 83
myroot = root(eqn, 2)
print(myroot.x)
Here, root takes two arguments root(fun, x0) where fun is the function of the equation and x0 is an rough estimate of the root value. For example if you know that your root will fall in range of (0,1) then you can enter 0 as rough estimate.
Also make sure the equation entered in the code is such that R.H.S. is equal to 0.
In our case 3^x + 5^x + 7^x = 83 becomes 3^x + 5^x + 7^x - 83 = 0
Reference Documentation

If you want to stick to base Python, it is easy enough to implement Newton's method for this problem:
from math import log
def solve(a,b,c,n,guess,tol = 1e-12):
x = guess
for i in range(100):
x_new = x - (a**x + b**x + c**x - n)/(log(a)*a**x + log(b)*b**x + log(c)*c**x)
if abs(x-x_new) < tol: return x_new
x = x_new
return "Doesn't converge on a root"
Newton's method might fail to converge in some pathological cases, hence an escape valve for such cases. In practice it converges very rapidly.
For example:
>>> solve(3,5,7,83,1)
2.0
Despite all this, I think that Cute Panda's answer is superior. It is easy enough to do a straight-forward implementation of such numerical algorithms, one that works adequately in most cases, but naive implementations such as the one give above tend to be vulnerable to excessive round-off error as well as other problems. scipy uses highly optimized routines which are implemented in a much more robust way.

Writing code to integrate 1D Fresnel Diffraction in Python

my task is to integrate the following equation for 1d Fresnel diffraction, in red in:
The point is that you are fourier transforming the aperture to a pattern on the screen, but for now just focussing on integrating the horizontal strip in 1d (ignoring the height for now). yprime is thus ignored. You also have fixed z and k, and j is an imaginary number. I have written the following code for it:
import math
import numpy as np
import cmath
k=5
z=5
x=0
j=cmath.sqrt(-1)
func=math.exp((j*k/2*z)(x-xp)*(x-xp))
def X(xp1,xp2,function,N):
h=(xp2-xp1)/N
y=0.0
xp=xp1
for x in np.arange(1, N/2 +1): #summing odd order y terms
y+=4*f(xp)
xp+=2*h
xp=xp1+2*h
for x in np.arange(0, N/2): #summing even order y terms
y+=2*f(x)
xp+=2*h
integral= (h/3)*(y+f(xp1)+f(xp2))
return integral
print(simpson(0,5,func,10))
however, it is saying that xp is not defined. but I clearly have defined xp in the function.
Does anyone have an idea what could be wrong?
Thanks
EDIT: here is a neater version of my code. But it's still asking me to define xp..
import math
import cmath
lamda=0.2
k=(2*math.pi)/lamda
z=0.1
def expfunc(x, xp):
func = math.exp(((1j)*k/2*z)(x-(xp))*(x-(xp)))
return(func)
def X(xp1,xp2,x,f,N):
h=(xp2-xp1)/N
y=0.0
xp=xp1
for i in np.arange(1, N/2 +1): #summing odd order y terms
y+=4*f(xp)
xp+=2*h
xp=xp1+2*h
for i in np.arange(0, N/2): #summing even order y terms
y+=2*f(xp)
xp+=2*h
integral= (h/3)*(y+f(xp1)+f(xp2))
return integral
print(X(0,1,x,expfunc,10))

You try to use the variable xp before you have defined it.
import math
import numpy as np
import cmath
k=5
z=5
x=0
j=cmath.sqrt(-1)
func=math.exp((j*k/2*z)(x-xp)*(x-xp)) #xp is not defined yet
You gave initial values for everything else except xp.

when you define func as you did
func=math.exp((j*k/2*z)(x-xp)*(x-xp))
you define a single value called func. What you probably want is something like that:
func = lambda x,xp : math.exp((j*k/2*z)(x-xp)*(x-xp))
and then change call of func to
y+=4*f(x, xp)
and
y+=2*f(x, xp)

I believe the issue is y+=4*f(xp) inside the first for loop of the function X.
At the very end you have print(X(0,1,x,expfunc,10)) where expfunc is acting as f in the bit of code y+=4*f(xp). The function expfunc takes two arguments, one of them defined as xp. Although the variable you pass in f is defined with the name xp, the function only sees that you have passed in the first argument x and not the argument xp.
Further, I do not see the variable x in print(X(0,1,x,expfunc,10)) defined anywhere.
Also, the second snipit of code is much different than the first. If the same questions apply then you should remove the first snipit altogether and/or rephrase your questions because from what I see in the second chuck the error you claim to be getting should not be raised.

Sympy series expansion with numerical integration

I want to make a series expansion for a function F(e,Eo) up to a certain power of e and integrate over the Eo variable numerically.
What I thought was using SymPy to make the power series in e, and then use MPMath for the numerical integration over Eo.
Below is an example code. I receive the message that it can not create mpf from the expression. I guess the problem has to do with the fact that with the series from SymPy has an O(e**5) term at the end, and later that I want the numerical integration to show a function of e instead of a number.
import sympy as sp
import numpy as np
from mpmath import *
e = sp.symbols('e')
Eo = sp.symbols('Eo')
expr = sp.sin(e-2*Eo).series(e, 0, 5)
F = lambda Eo : expr
I = quad(F, [0, 2*np.pi])
print(I)
It’s evident that I need a different approach, but I would still need the numerical integration for my actual code because it has much more complicated expressions that could not be integrated analytically.
Edit: I should have chosen a complex function of real variables for the example code, I am trying this (the expansion and integration) for functions such as:
expr = (cos(Eo) - e - I*sqrt(1 - e ** 2)*sin(Eo)) ** 2 * (cos(2*(Eo - e*sin(Eo))) + I*sin(2*(Eo - e*sin(Eo))))/(1 - e*cos(Eo)) ** 4

Here is an approach similar to Wrzlprmft's answer but with a different way of handling coefficients, via SymPy's polynomial module:
from sympy import sin, pi, symbols, Integral, Poly
def integrate_coeff(coeff):
partial_integral = coeff.integrate((Eo, 0, 2*pi))
return partial_integral.n() if partial_integral.has(Integral) else partial_integral
e,Eo = symbols("e Eo")
expr = sin(e-sin(2*Eo))
degree = 5
coeffs = Poly(expr.series(e, 0, degree).removeO(), e).all_coeffs()
new_coeffs = map(integrate_coeff, coeffs)
print((Poly(new_coeffs, e).as_expr() + e**degree).series(e, 0, degree))
The main code is three lines: (1) extract coefficients of e up to given degree; (2) integrate each, numerically if we must; (3) print the result, presenting it as a series rather than a polynomial (hence the trick with adding e**degree, to make SymPy aware that the series continues). Output:
-6.81273574401304e-108 + 4.80787886126883*e + 3.40636787200652e-108*e**2 - 0.801313143544804*e**3 - 2.12897992000408e-109*e**4 + O(e**5)

I want the numerical integration to show a function of e instead of a number.
This is fundamentally impossible.
Either your integration is analytical or numerical, and if it is numerical it will only handle and yield numbers for you (the words numerical and number are similar for a reason).
If you want to split the integration into numerical and analytical components, you have to do so yourself – or hope that SymPy automatically splits the integration as needed, which it unfortunately is not yet capable of.
This is how I would do it (details are commented in the code):
from sympy import sin, pi, symbols, Integral
from itertools import islice
e,Eo = symbols("e Eo")
expr = sin(e-sin(2*Eo))
# Create a generator yielding the first five summands of the series.
# This avoids the O(e**5) term.
series = islice(expr.series(e,0,None),5)
integral = 0
for power,summand in enumerate(series):
# Remove the e from the expression
Eo_part = summand/e**power
# … and ensure that it worked:
assert not Eo_part.has(e)
# Integrate the Eo part:
partial_integral = Eo_part.integrate((Eo,0,2*pi))
# If the integral cannot be evaluated analytically, …
if partial_integral.has(Integral):
# … replace it by the numerical estimate:
partial_integral = partial_integral.n()
# Re-attach the e component and add it to the sum:
integral += partial_integral*e**power
Note that I did not use NumPy or MPMath at all (though SymPy uses the latter under the hood for numerical estimates). Unless you really really know what you’re doing, mixing those two with SymPy is a bad idea as they are not even aware of SymPy expressions.

Avoiding hard coding a lot of coupled ODEs in python

First of, I'm sorry if the title is not entirely fitting, I had a hard time finding an appropriate one (which might have also effect my searching efficiency for already asked questions like this :/ ).
The problem is the following. While it is comparably easy to solve coupled ODE's in python with Scipy, I still have to write down my ODE in the form explicitly. For example for a coupled ODE of the form
d/dt(c_0)=a(c_0)+b(c_1) and d/dt(c_1)=c(c_0)
I would set up sth like:
import numpy as np
from scipy.integrate import ode
a=1
b=2
c=3
val=[]
def dC_dt(t, C):
return [a*C[0]+b*C[1],
c*C[0]]
c0, t0 = [1.0,0.0], 0
r = ode(dC_dt).set_integrator('zvode', method='bdf',with_jacobian=False)
r.set_initial_value(c0, t0)
t1 = 0.001
dt = 0.000005
while r.successful() and r.t < t1:
r.integrate(r.t+dt)
val.append(r.y)
However, now I have coupled ODE's of the rough form
d/dt(c_{m,n})=a(c_{m,n})+b(c_{m+1,n-1})+k(c_{m-1,n+1})
with c_{0,0}=1 and I have to include orders with m^2+n^2-mn smaller than a max value.
For a small max, what I did, is using a dictionary to use a notation with two indices and map it on a 1D list
dict_in={'0,0':0,'-1,0':2,...}
and then I entered the ODE for each order
def dC_dt(t,C):
return[a*C[dict_in['0,0']]+b*C[dict_in['1,-1']]...
Now I basically have to do that for some 100 coupled equations, which I ofc do not want to hard code, so I was trying to figure out a way, to realize the ODE's with a loop or sth. However I couldn't yet find a way around the fact of having two indices in my coefficients together with the condition of only including orders with m^2+n^2-mn smaller than a max value.
As I am running in some deadlines, I figured it is time to ask smarter people for help.
Thanks for reading my question!

I had a similar problem. If you fill you dictionary you can just redeclare the function more times inside the loop. This is a silly example of how it works:
dict_in={'0,0':0,'-1,0':2}
for elem in dict_in:
def dC_dt(t,C):
#return[a*C[dict_in['0,0']]+b*C[dict_in['1,-1']]
return dict_in[elem]
t, C = 0, 0
print(dC_dt(t,C))
#r = ode(dC_dt).set_integrator('zvode', method='bdf',with_jacobian=False)
If you need to use more functions together you can use anonymous functions and store them in memory. Another example:
functions_list = list()
for i in range(4):
f = lambda n = i: n
functions_list.append(f)
for j in range(4):
print(functions_list[j]())
You can use a list or a generator too. For example you can write down the value on a txt file and read that with the readline function each time.
As pointed in the comments below, if you use lamda functions you should pay attention to references. See also https://docs.python.org/3/faq/programming.html#why-do-lambdas-defined-in-a-loop-with-different-values-all-return-the-same-result

Speed up function evaluation for integration in scipy

I am trying to port code from Matlab to SciPy. Here is the simplified version of the code I have written so far: https://gist.github.com/atmo/01b6e007be9ef90e402c . However, Python version is considerably slower then Matlab. I've included profiling results in the gist and they show that almost 90% of time python spends evaluating function f. Is there any way to speed up its evaluation, except from rewriting it in C or Cython?

As I mentioned in the comments, you can get rid of about half the calls to quad (and consequently the complicated function f) if you take into account that the matrix is symmetric.
Further speed gains, still in pure python, are to be had by rewriting that complicated function. I did most of that in sympy.
Finally I tried to vectorize the call to quad using np.vectorize.
from scipy.integrate import quad
from scipy.special import jn as besselj
from scipy import exp, zeros, linspace
from scipy.linalg import norm
import numpy as np
def complicated_func(lmbd, a, n, k):
u,v,w = 5, 3, 2
x = a*lmbd
fac = exp(2*x)
comm = (2*w + x)
part1 = ((v**2 + 4*w*(w + 2*x) + 2*x*(x - 1))*fac**5
+ 2*u*fac**4
+ (-v**2 - 4*(w*(3*w + 4*x + 1) + x*(x-2)) + 1)*fac**3
+ (-8*(w + x) + 2)*fac**2
+ (2*comm*(comm + 1) - 1)*fac)
return part1/lmbd *besselj(n+1, lmbd) * besselj(k+1, lmbd)
def perform_quad(n, k, a):
return quad(complicated_func, 0, np.inf, args=(a,n,k))[0]
def improved_main():
sz = 20
amatrix = np.zeros((sz,sz))
ls = -np.linspace(1, 10, 20)/2
inds = np.tril_indices(sz)
myv3 = np.vectorize(perform_quad)
res = myv3(inds[0], inds[1], ls.reshape(-1,1))
results = np.empty(res.shape[0])
for rowind, row in enumerate(res):
amatrix[inds] = row
symm_matrix = amatrix + amatrix.T - np.diag(amatrix.diagonal())
results[rowind] = norm(symm_matrix)
return results
Timing results show me a speed increase of a factor 5 (you'll forgive me if I only ran it once, it takes long enough as it is):
In [11]: %timeit -n1 -r1 improved_main()
1 loops, best of 1: 6.92 s per loop
In [12]: %timeit -n1 -r1 main()
1 loops, best of 1: 35.9 s per loop
There was also a microgain to be had if you replaced v immediately by its square, because that's the only time it is used in that complicated function: as its square.
There's also an extreme amount of repetition in the calls to besselj, but I don't see how to avoid that, because quad will determine lmbd, so you can't easily precompute those values and then perform a lookup.
If you profile the improved_main, you'll see that the amount of calls to complicated_func has nearly decreased by a factor of 2 (the diagonal still needs to be computed). All the other speed gains can be attributed to np.vectorize and the improvements to complicated_func.
I don't have Matlab on my system, so I can't make any statements for its speed gain if you improve the complicated function there.

Your numpy version probably is comparable in to speed to older MATLAB runs. But new MATLAB versions do various forms of just-in-time compilation that speed up repeated calculations considerably.
My guess is that you can nibble away at the lambda and f code, and maybe cut their evaluation times in half. But the real killer is that you are calling f so many times.
For a start I'd try to precalculate things in f. For example define K1=K[1] and use K1 in the calculations. That will reduce the number of indexing calls. Are of the exponentials repeated? Maybe replace the lambda definition with a regular def, or combine it with f.