Background
I've been working for some time on attempting to solve the (notoriously painful) Time Difference of Arrival (TDoA) multi-lateration problem, in 3-dimensions and using 4 nodes. If you're unfamiliar with the problem, it is to determine the coordinates of some signal source (X,Y,Z), given the coordinates of n nodes, the time of arrival of the signal at each node, and the velocity of the signal v.
My solution is as follows:
For each node, we write (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 = (v(t_i - T)**2
Where (x_i, y_i, z_i) are the coordinates of the ith node, and T is the time of emission.
We have now 4 equations in 4 unknowns. Four nodes are obviously insufficient. We could try to solve this system directly, however that seems next to impossible given the highly nonlinear nature of the problem (and, indeed, I've tried many direct techniques... and failed). Instead, we simplify this to a linear problem by considering all i/j possibilities, subtracting equation i from equation j. We obtain (n(n-1))/2 =6 equations of the form:
2*(x_j - x_i)*X + 2*(y_j - y_i)*Y + 2*(z_j - z_i)*Z + 2 * v**2 * (t_i - t_j) = v**2 ( t_i**2 - t_j**2) + (x_j**2 + y_j**2 + z_j**2) - (x_i**2 + y_i**2 + z_i**2)
Which look like Xv_1 + Y_v2 + Z_v3 + T_v4 = b. We try now to apply standard linear least squares, where the solution is the matrix vector x in A^T Ax = A^T b. Unfortunately, if you were to try feeding this into any standard linear least squares algorithm, it'll choke up. So, what do we do now?
...
The time of arrival of the signal at node i is given (of course) by:
sqrt( (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 ) / v
This equation implies that the time of arrival, T, is 0. If we have that T = 0, we can drop the T column in matrix A and the problem is greatly simplified. Indeed, NumPy's linalg.lstsq() gives a surprisingly accurate & precise result.
...
So, what I do is normalize the input times by subtracting from each equation the earliest time. All I have to do then is determine the dt that I can add to each time such that the residual of summed squared error for the point found by linear least squares is minimized.
I define the error for some dt to be the squared difference between the arrival time for the point predicted by feeding the input times + dt to the least squares algorithm, minus the input time (normalized), summed over all 4 nodes.
for node, time in nodes, times:
error += ( (sqrt( (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 ) / v) - time) ** 2
My problem:
I was able to do this somewhat satisfactorily by using brute-force. I started at dt = 0, and moved by some step up to some maximum # of iterations OR until some minimum RSS error is reached, and that was the dt I added to the normalized times to obtain a solution. The resulting solutions were very accurate and precise, but quite slow.
In practice, I'd like to be able to solve this in real time, and therefore a far faster solution will be needed. I began with the assumption that the error function (that is, dt vs error as defined above) would be highly nonlinear-- offhand, this made sense to me.
Since I don't have an actual, mathematical function, I can automatically rule out methods that require differentiation (e.g. Newton-Raphson). The error function will always be positive, so I can rule out bisection, etc. Instead, I try a simple approximation search. Unfortunately, that failed miserably. I then tried Tabu search, followed by a genetic algorithm, and several others. They all failed horribly.
So, I decided to do some investigating. As it turns out the plot of the error function vs dt looks a bit like a square root, only shifted right depending upon the distance from the nodes that the signal source is:
Where dt is on horizontal axis, error on vertical axis
And, in hindsight, of course it does!. I defined the error function to involve square roots so, at least to me, this seems reasonable.
What to do?
So, my issue now is, how do I determine the dt corresponding to the minimum of the error function?
My first (very crude) attempt was to get some points on the error graph (as above), fit it using numpy.polyfit, then feed the results to numpy.root. That root corresponds to the dt. Unfortunately, this failed, too. I tried fitting with various degrees, and also with various points, up to a ridiculous number of points such that I may as well just use brute-force.
How can I determine the dt corresponding to the minimum of this error function?
Since we're dealing with high velocities (radio signals), it's important that the results be precise and accurate, as minor variances in dt can throw off the resulting point.
I'm sure that there's some infinitely simpler approach buried in what I'm doing here however, ignoring everything else, how do I find dt?
My requirements:
Speed is of utmost importance
I have access only to pure Python and NumPy in the environment where this will be run
EDIT:
Here's my code. Admittedly, a bit messy. Here, I'm using the polyfit technique. It will "simulate" a source for you, and compare results:
from numpy import poly1d, linspace, set_printoptions, array, linalg, triu_indices, roots, polyfit
from dataclasses import dataclass
from random import randrange
import math
#dataclass
class Vertexer:
receivers: list
# Defaults
c = 299792
# Receivers:
# [x_1, y_1, z_1]
# [x_2, y_2, z_2]
# [x_3, y_3, z_3]
# Solved:
# [x, y, z]
def error(self, dt, times):
solved = self.linear([time + dt for time in times])
error = 0
for time, receiver in zip(times, self.receivers):
error += ((math.sqrt( (solved[0] - receiver[0])**2 +
(solved[1] - receiver[1])**2 +
(solved[2] - receiver[2])**2 ) / c ) - time)**2
return error
def linear(self, times):
X = array(self.receivers)
t = array(times)
x, y, z = X.T
i, j = triu_indices(len(x), 1)
A = 2 * (X[i] - X[j])
b = self.c**2 * (t[j]**2 - t[i]**2) + (X[i]**2).sum(1) - (X[j]**2).sum(1)
solved, residuals, rank, s = linalg.lstsq(A, b, rcond=None)
return(solved)
def find(self, times):
# Normalize times
times = [time - min(times) for time in times]
# Fit the error function
y = []
x = []
dt = 1E-10
for i in range(50000):
x.append(self.error(dt * i, times))
y.append(dt * i)
p = polyfit(array(x), array(y), 2)
r = roots(p)
return(self.linear([time + r for time in times]))
# SIMPLE CODE FOR SIMULATING A SIGNAL
# Pick nodes to be at random locations
x_1 = randrange(10); y_1 = randrange(10); z_1 = randrange(10)
x_2 = randrange(10); y_2 = randrange(10); z_2 = randrange(10)
x_3 = randrange(10); y_3 = randrange(10); z_3 = randrange(10)
x_4 = randrange(10); y_4 = randrange(10); z_4 = randrange(10)
# Pick source to be at random location
x = randrange(1000); y = randrange(1000); z = randrange(1000)
# Set velocity
c = 299792 # km/ns
# Generate simulated source
t_1 = math.sqrt( (x - x_1)**2 + (y - y_1)**2 + (z - z_1)**2 ) / c
t_2 = math.sqrt( (x - x_2)**2 + (y - y_2)**2 + (z - z_2)**2 ) / c
t_3 = math.sqrt( (x - x_3)**2 + (y - y_3)**2 + (z - z_3)**2 ) / c
t_4 = math.sqrt( (x - x_4)**2 + (y - y_4)**2 + (z - z_4)**2 ) / c
print('Actual:', x, y, z)
myVertexer = Vertexer([[x_1, y_1, z_1],[x_2, y_2, z_2],[x_3, y_3, z_3],[x_4, y_4, z_4]])
solution = myVertexer.find([t_1, t_2, t_3, t_4])
print(solution)
It seems like the Bancroft method applies to this problem? Here's a pure NumPy implementation.
# Implementation of the Bancroft method, following
# https://gssc.esa.int/navipedia/index.php/Bancroft_Method
M = np.diag([1, 1, 1, -1])
def lorentz_inner(v, w):
return np.sum(v * (w # M), axis=-1)
B = np.array(
[
[x_1, y_1, z_1, c * t_1],
[x_2, y_2, z_2, c * t_2],
[x_3, y_3, z_3, c * t_3],
[x_4, y_4, z_4, c * t_4],
]
)
one = np.ones(4)
a = 0.5 * lorentz_inner(B, B)
B_inv_one = np.linalg.solve(B, one)
B_inv_a = np.linalg.solve(B, a)
for Lambda in np.roots(
[
lorentz_inner(B_inv_one, B_inv_one),
2 * (lorentz_inner(B_inv_one, B_inv_a) - 1),
lorentz_inner(B_inv_a, B_inv_a),
]
):
x, y, z, c_t = M # np.linalg.solve(B, Lambda * one + a)
print("Candidate:", x, y, z, c_t / c)
My answer might have mistakes (glaring) as I had not heard the TDOA term before this afternoon. Please double check if the method is right.
I could not find solution to your original problem of finding dt corresponding to the minimum error. My answer also deviates from the requirement that other than numpy no third party library had to be used (I used Sympy and largely used the code from here). However I am still posting this thinking that somebody someday might find it useful if all one is interested in ... is to find X,Y,Z of the source emitter. This method also does not take into account real-life situations where white noise or errors might be present or curvature of the earth and other complications.
Your initial test conditions are as below.
from random import randrange
import math
# SIMPLE CODE FOR SIMULATING A SIGNAL
# Pick nodes to be at random locations
x_1 = randrange(10); y_1 = randrange(10); z_1 = randrange(10)
x_2 = randrange(10); y_2 = randrange(10); z_2 = randrange(10)
x_3 = randrange(10); y_3 = randrange(10); z_3 = randrange(10)
x_4 = randrange(10); y_4 = randrange(10); z_4 = randrange(10)
# Pick source to be at random location
x = randrange(1000); y = randrange(1000); z = randrange(1000)
# Set velocity
c = 299792 # km/ns
# Generate simulated source
t_1 = math.sqrt( (x - x_1)**2 + (y - y_1)**2 + (z - z_1)**2 ) / c
t_2 = math.sqrt( (x - x_2)**2 + (y - y_2)**2 + (z - z_2)**2 ) / c
t_3 = math.sqrt( (x - x_3)**2 + (y - y_3)**2 + (z - z_3)**2 ) / c
t_4 = math.sqrt( (x - x_4)**2 + (y - y_4)**2 + (z - z_4)**2 ) / c
print('Actual:', x, y, z)
My solution is as below.
import sympy as sym
X,Y,Z = sym.symbols('X,Y,Z', real=True)
f = sym.Eq((x_1 - X)**2 +(y_1 - Y)**2 + (z_1 - Z)**2 , (c*t_1)**2)
g = sym.Eq((x_2 - X)**2 +(y_2 - Y)**2 + (z_2 - Z)**2 , (c*t_2)**2)
h = sym.Eq((x_3 - X)**2 +(y_3 - Y)**2 + (z_3 - Z)**2 , (c*t_3)**2)
i = sym.Eq((x_4 - X)**2 +(y_4 - Y)**2 + (z_4 - Z)**2 , (c*t_4)**2)
print("Solved coordinates are ", sym.solve([f,g,h,i],X,Y,Z))
print statement from your initial condition gave.
Actual: 111 553 110
and the solution that almost instantly came out was
Solved coordinates are [(111.000000000000, 553.000000000000, 110.000000000000)]
Sorry again if something is totally amiss.
I'm trying to solve an integral with sympy. But it gives me a wrong solution. Why?
import sympy
from sympy import Integral, exp, oo
x, y = sympy.symbols("x y", real=True)
b, u, l, t = sympy.symbols("b u l t ", real=True, positive=True)
Fortet = Integral(exp(-l * t) * (sympy.sqrt(2 * sympy.pi * t)) ** (-1) * exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, oo))
Fortet.doit()
Result (wrong):
Piecewise((-(-b/2 + y)*sqrt(2*l +
u**2)*(-sqrt(pi)*sinh(sqrt(2)*sqrt(b)*sqrt(l +
u**2/2)*sqrt(polar_lift(1 + y**2/(b*polar_lift(b -
2*y))))*sqrt(polar_lift(b - 2*y))) +
sqrt(pi)*cosh(sqrt(2)*sqrt(b)*sqrt(l + u**2/2)*sqrt(polar_lift(1 +
y**2/(b*polar_lift(b - 2*y))))*sqrt(polar_lift(b - 2*y))))*exp(b*u -
u*y)/(sqrt(pi)*(b - 2*y)*(l + u**2/2)), Abs(arg(1 +
y**2/(b*polar_lift(b - 2*y))) + arg(b - 2*y)) <= pi/2),
(Integral(sqrt(2)*exp(-l*t)*exp(-(b - t*u -
y)**2/(2*t))/(2*sqrt(pi)*sqrt(t)), (t, 0, oo)), True))
Expected (correct) solution:
Solution = (exp((-u)*(b - y)) * exp(sympy.sqrt(u**2 + 2*l)*(b-y)))/(sympy.sqrt(2*l + u**2)) #RIGHT solution
Both results are in fact the same. The first one is probably slightly more correct. You tend see these polar_lift functions whenever SymPy tries to do something like square rooting something when it does not know the signs of the things inside (after integrating)
A polar_lift does not appear below, but this basic Gaussian example shows that SymPy tries to be as general as possible:
from sympy import *
x = Symbol("x", real=True)
y = Symbol("y", real=True)
s = Symbol("s", real=True) # , positive=True
gaussian = exp(-((x-y)**2)/(2*(s**2)))
nfactor = simplify(integrate(gaussian, (x,-oo,oo)))
print(nfactor)
You need s to be declared as positive: s = Symbol("s", real=True, positive=True). A similar thing happens with these kinds of polar_lift(b - 2*y) functions in your example. It also happens with the question I reference below.
I have no idea why, but N(polar_lift(x)) for any float or int x gives x again; yet, SymPy does not simplify nicely with symbolic x. Turns out if you keep on simplifying, you get nicer and nicer looking answers. I couldn't find anything about polar_lift related to pure math so I don't know what it actually does.
Remember for the simple example above how it gave a piece-wise? Same thing here. So we just take the first piece since the second piece is an un-evaluated integral.
In the code below, I use this question to remove the piece-wise function and then I simplify twice. And finally, I manually remove the polar_lift.
import sympy as sp
x, y = sp.symbols("x y", real=True)
b, u, l, t = sp.symbols("b u l t ", real=True, positive=True)
Fortet = sp.integrate(sp.exp(-l * t) * (sp.sqrt(2 * sp.pi * t)) ** (-1) *
sp.exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, sp.oo), conds='none')
incorrect = Fortet.simplify().simplify()
correct = eval(str(incorrect).replace("polar_lift", ""))
correct = correct.factor()
print(correct)
The result is:
exp(b*u)*exp(-u*y)*exp(-sqrt(2*l + u**2)*sqrt(b**2 - 2*b*y + y**2))/sqrt(2*l + u**2)
That is close enough to your expression. I couldn't make SymPy simplify the sqrt(b**2 - 2*b*y + y**2) to Abs(b-y) no matter how hard I tried.
Note that either SymPy is still wrong or you are wrong since the powers in the numerator are opposite. So I checked on the Desmos for a numeric answer (top one is yours):
I'm attempting to implement Vincenty's inverse problem as described on wiki HERE
The problem is that lambda is simply not converging. The value stays the same if I try to iterate over the sequence of formulas, and I'm really not sure why. Perhaps I've just stared myself blind on an obvious problem.
It should be noted that I'm new to Python and still learning the language, so I'm not sure if it's misuse of the language that might cause the problem, or if I do have some mistakes in some of the calculations that I perform. I just can't seem to find any mistakes in the formulas.
Basically, I've written in the code in as close of a format as I could to the wiki article, and the result is this:
import math
# Length of radius at equator of the ellipsoid
a = 6378137.0
# Flattening of the ellipsoid
f = 1/298.257223563
# Length of radius at the poles of the ellipsoid
b = (1 - f) * a
# Latitude points
la1, la2 = 10, 60
# Longitude points
lo1, lo2 = 5, 150
# For the inverse problem, we calculate U1, U2 and L.
# We set the initial value of lamb = L
u1 = math.atan( (1 - f) * math.tan(la1) )
u2 = math.atan( (1 - f) * math.tan(la2) )
L = (lo2 - lo1) * 0.0174532925
lamb = L
while True:
sinArc = math.sqrt( math.pow(math.cos(u2) * math.sin(lamb),2) + math.pow(math.cos(u1) * math.sin(u2) - math.sin(u1) * math.cos(u2) * math.cos(lamb),2) )
cosArc = math.sin(u1) * math.sin(u2) + math.cos(u1) * math.cos(u2) * math.cos(lamb)
arc = math.atan2(sinArc, cosArc)
sinAzimuth = ( math.cos(u1) * math.cos(u2) * math.sin(lamb) ) // ( sinArc )
cosAzimuthSqr = 1 - math.pow(sinAzimuth, 2)
cosProduct = cosArc - ((2 * math.sin(u1) * math.sin(u2) ) // (cosAzimuthSqr))
C = (f//16) * cosAzimuthSqr * (4 + f * (4 - 3 * cosAzimuthSqr))
lamb = L + (1 - C) * f * sinAzimuth * ( arc + C * sinArc * ( cosProduct + C * cosArc * (-1 + 2 * math.pow(cosProduct, 2))))
print(lamb)
As mentioned the problem is that the value "lamb" (lambda) will not become smaller. I've even tried to compare my code to other implementations, but they looked just about the same.
What am I doing wrong here? :-)
Thank you all!
First, you should convert you latitudes in radians too (you already do this for your longitudes):
u1 = math.atan( (1 - f) * math.tan(math.radians(la1)) )
u2 = math.atan( (1 - f) * math.tan(math.radians(la2)) )
L = math.radians((lo2 - lo1)) # better than * 0.0174532925
Once you do this and get rid of // (int divisions) and replace them by / (float divisions), lambda stops repeating the same value through your iterations and starts following this path (based on your example coordinates):
2.5325205864224847
2.5325167509030906
2.532516759118641
2.532516759101044
2.5325167591010813
2.5325167591010813
2.5325167591010813
As you seem to expect a convergence precision of 10^(−12), it seems to make the point.
You can now exit the loop (lambda having converged) and keep going until you compute the desired geodesic distance s.
Note: you can test your final value s here.
Even if it is correctly implemented, Vincenty's algorithm will fail to
converge for some points. (This problem was noted by Vincenty.)
I give an algorithm which is guaranteed to
converge in Algorithms for geodesics; there's a python
implementation available here. Finally, you can find more
information on the problem at the Wikipedia page,
Geodesics on an ellipsoid. (The talk page has examples
of pairs of points for which Vincenty, as implemented by the NGS,
fails to converge.)
I'm trying to symbolically solve a polynomial with complex numbers and their conjugates with SymPy. I think I've come a long way, but solve does not give me any solutions although the polynomial is solveable.
from sympy import *
# set up symbols
a, b = symbols("a b", real=True)
t = a+I*b
T = functions.conjugate(t)
# set up polynomial
a1=0.005+I*0.0009
a2=0.9+I*-0.9
a3=0.4+I*0.5
a4=8+I*-80
a5=284+I*-1.5
a6=27100+I*-11500
poly=t**2 * T * a1 + t * T * a2 + t**2 * a3 + T * a4 + t * a5 + a6
# Trying to solve symbolically...
solve([re(poly), im(poly)], a, b)
# Output: []
# Solving numerically works, but only finds one solution...
nsolve((re(poly), im(poly)), (a, b), (0, 0))
# Output: matrix(
# [['-137.962596090596'],
# ['52.6296963395752']])
# verify with two solutions obtained in Maxima
poly.subs({a:-137.9625935162095, b:52.6296992481203}).n()
# Output: 0.000540354631040322 + 0.00054727003909351*I
poly.subs({a:-332.6474382554614+I*-185.9848818313149, b:258.0065640091016+I*-272.3344263478699}).n()
# Output: -6.55448222470652e-12 - 1.41238056784605e-12*I
Any ideas?
One problem could be that your coefficients contain floating point numbers. This often does not work with symbolic software.
Using f=10000*simplify(re(poly)) and g=10000*simplify(im(poly)) and editing the results gives polynomials with integer coefficients. The CAS (Magma in my case) then can produce a triangular representation of the ideal of f and g, which is given as the polynomials
a - 2483798807340123709959247/13545514719183259347425004828125*b^4
+ 66732206412048596386372283/541820588767330373897000193125*b^3
- 3849759933277117021785191063/86691294202772859823520030900*b^2
+ 9245906471290310401430681453/1733825884055457196470400618*b
- 31414499425567273751868164900/866912942027728598235200309,
b^5 - 189465979625/206648369*b^4
+ 330827538698125/826593476*b^3
- 17645868534640625/206648369*b^2
+ 1724106750659765625/206648369*b
- 52548859891484375000/206648369
which tells us that 5 solutions exist. The numerical solutions for the second polynomial are
174.10461010682254746847015187264557067610513554291323564643772
+ 63.402741884833821878468926640811609033267039765104756747285816*i
174.104610106822547468470151872645570676105135542913235646437738
- 63.402741884833821878468926640811609033267039765104756747285804*i
258.006564009101655109715962546854008929462784282347754971379392
+ 272.334426347869856080204881056671278679761260094680345276069337*i
258.006564009101655109715962546854008929462784282347754971379382
- 272.334426347869856080204881056671278679761260094680345276069359*i
52.62969633957523864698147679803879873180829265956656342643376
resulting in exactly one real solution. The numerical result of sympy was correct and complete.
I'd like to implement Euler's method (the explicit and the implicit one)
(https://en.wikipedia.org/wiki/Euler_method) for the following model:
x(t)' = q(x_M -x(t))x(t)
x(0) = x_0
where q, x_M and x_0 are real numbers.
I know already the (theoretical) implementation of the method. But I couldn't figure out where I can insert / change the model.
Could anybody help?
EDIT: You were right. I didn't understand correctly the method. Now, after a few hours, I think that I really got it! With the explicit method, I'm pretty sure (nevertheless: could anybody please have a look at my code? )
With the implicit implementation, I'm not very sure if it's correct. Could please anyone have a look at the implementation of the implicit method and give me a feedback what's correct / not good?
def explizit_euler():
''' x(t)' = q(xM -x(t))x(t)
x(0) = x0'''
q = 2.
xM = 2
x0 = 0.5
T = 5
dt = 0.01
N = T / dt
x = x0
t = 0.
for i in range (0 , int(N)):
t = t + dt
x = x + dt * (q * (xM - x) * x)
print '%6.3f %6.3f' % (t, x)
def implizit_euler():
''' x(t)' = q(xM -x(t))x(t)
x(0) = x0'''
q = 2.
xM = 2
x0 = 0.5
T = 5
dt = 0.01
N = T / dt
x = x0
t = 0.
for i in range (0 , int(N)):
t = t + dt
x = (1.0 / (1.0 - q *(xM + x) * x))
print '%6.3f %6.3f' % (t, x)
Pre-emptive note: Although the general idea should be correct, I did all the algebra in place in the editor box so there might be mistakes there. Please, check it yourself before using for anything really important.
I'm not sure how you come to the "implicit" formula
x = (1.0 / (1.0 - q *(xM + x) * x))
but this is wrong and you can check it by comparing your "explicit" and "implicit" results: they should slightly diverge but with this formula they will diverge drastically.
To understand the implicit Euler method, you should first get the idea behind the explicit one. And the idea is really simple and is explained at the Derivation section in the wiki: since derivative y'(x) is a limit of (y(x+h) - y(x))/h, you can approximate y(x+h) as y(x) + h*y'(x) for small h, assuming our original differential equation is
y'(x) = F(x, y(x))
Note that the reason this is only an approximation rather than exact value is that even over small range [x, x+h] the derivative y'(x) changes slightly. It means that if you want to get a better approximation of y(x+h), you need a better approximation of "average" derivative y'(x) over the range [x, x+h]. Let's call that approximation just y'. One idea of such improvement is to find both y' and y(x+h) at the same time by saying that we want to find such y' and y(x+h) that y' would be actually y'(x+h) (i.e. the derivative at the end). This results in the following system of equations:
y'(x+h) = F(x+h, y(x+h))
y(x+h) = y(x) + h*y'(x+h)
which is equivalent to a single "implicit" equation:
y(x+h) - y(x) = h * F(x+h, y(x+h))
It is called "implicit" because here the target y(x+h) is also a part of F. And note that quite similar equation is mentioned in the Modifications and extensions section of the wiki article.
So now going to your case that equation becomes
x(t+dt) - x(t) = dt*q*(xM -x(t+dt))*x(t+dt)
or equivalently
dt*q*x(t+dt)^2 + (1 - dt*q*xM)*x(t+dt) - x(t) = 0
This is a quadratic equation with two solutions:
x(t+dt) = [(dt*q*xM - 1) ± sqrt((dt*q*xM - 1)^2 + 4*dt*q*x(t))]/(2*dt*q)
Obviously we want the solution that is "close" to the x(t) which is the + solution. So the code should be something like:
b = (q * xM * dt - 1)
x(t+h) = (b + (b ** 2 + 4 * q * x(t) * dt) ** 0.5) / 2 / q / dt
(editor note:) Applying the binomial complement, this formula has the numerically more stable form for small dt, where then b < 0,
x(t+h) = (2 * x(t)) / ((b ** 2 + 4 * q * x(t) * dt) ** 0.5 - b)