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How to make this float precision without using print

I'm working on assignment, it's about numerical method regarding to trapezoidal rule
def trapezoidalRule(F,a,b,n):
h = float(b-a)/n
f_sum = 0
for i in range(1, n, 1):
x = a + i * h
f_sum = f_sum + f(x)
return h * (0.5 * f(a) + f_sum + 0.5 * f(b))
def f(x):
return x**3
a = 2
b = 10
n = 512
print('%.16f' %trapezoidalRule(f, a, b, n))
And the output is
2496.0058593750000000
My question is, how do i get a precission like that.. without using print('%.16f' %trapezoidalRule(f, a, b, n)). I want to append the result to the list, with exact value like that..
I already tried to google it, but i found nothing related to this problem, can somebody tell me the solution if i want to it ?

Change your return statement in trapezoidalRule to be formatted with 16 points of precision, do note that this is going to cause it to become a string as if you cast it back to float you'll lose the trailing 0's.
def trapezoidalRule(F,a,b,n):
h = float(b-a)/n
f_sum = 0
for i in range(1, n, 1):
x = a + i * h
f_sum = f_sum + f(x)
return format((h * (0.5 * f(a) + f_sum + 0.5 * f(b))), '.16f')
def f(x):
return x**3
a = 2
b = 10
n = 512
See the return line in trapezoidalRule so now if I print the exact output of trapezoidalRule like so: print(trapezoidalRule(f, a, b, n)) with no formatting I get:
2496.0058593750000000

To increase precision try using decimal module
import decimal
def trapezoidalRule(F,a,b,n):
h = decimal.Decimal(float(b-a)/n)
f_sum = 0
for i in range(1, n, 1):
x = a + i * h
f_sum = f_sum + f(x)
return h * (decimal.Decimal(0.5) * f(a) + f_sum + decimal.Decimal(0.5) * f(b))
def f(x):
return decimal.Decimal(x**3)

Algorithm for integer solutions of a circle?

I am trying to search for integer solutions to the equation:
y^2 + x^2 = 2n^2
If I search this in wolfram alpha, they are all found almost immediately even for very large n. When I implemented a brute force approach it was very slow:
def psearch(n, count):
for x in range(0, n):
for y in range(0, n):
if x*x + y*y == 2*n**2:
print(x,y)
count += 1
return count
So I assume there is a much faster way to get all of the integer solutions to the equation above. How can I do this in python so that it will have much lower runtime?
Note: I have seen this question however it is about finding lattice points within a circle not the integer solutions to the equation of the circle. Also I am interested in finding the specific solutions not just the number of solutions.
Edit: I am still looking for something an order of magnitude faster. Here is an example: n=5 should have 12 integer solutions to find what those should be search this equation on Wolfram alpha.
Edit 2: #victor zen gave a phenomenal answer to the problem. Can anyone think of a way to optimize his solution further?

In your algorithm, you're searching for all possible y values. This is unnecessary. The trick here is to realize that
y^2 + x^2 = 2n^2
directly implies that
y^2 = 2n^2-x^2
so that means you only have to check that 2n^2-x^2 is a perfect square. You can do that by
y2 = 2*n*n - x*x
#check for perfect square
y = math.sqrt(y2)
if int(y + 0.5) ** 2 == y2:
#We have a perfect square.
Also, in your algorithm, you are only checking x values up to n. This is incorrect. Since y^2 will always be positive or zero, we can determine the highest x value we need to check by setting y^2 to its lowest value (i.e 0). Consequentially, we need to check all integer x values satisfying
x^2 <= 2n^2
which reduces to
abs(x) <= sqrt(2)*n.
Combine this with the optimization of only checking the top quadrant, and you have an optimized psearch of
def psearch(n):
count = 0
top = math.ceil(math.sqrt(2*n*n))
for x in range(1, top):
y2 = 2*n*n - x*x
#check for perfect square
y = math.sqrt(y2)
if int(y + 0.5) ** 2 == y2:
count+=4
return count

It is enough to search inside the first octant y>0, x<y (the four solutions (±n, ±n) are obvious and by symmetry a solution (x, y) yields 8 copies (±x, ±y), (±y, ±x)).
By monotonicity, for a given y there is at most one x. You can find it by following the circular arc incrementally, decreasing y then adjusting x. If you maintain the condition x²+y²≤2n² as tightly as possible, you get the code below which is optimized to use only elementary integer arithmetic (for efficiency, 2x is used instead of x).
x, y, d= 2 * n, 2 * n, 0
while y > 0:
y, d= y - 2, d - y + 1
if d < 0:
x, d= x + 2, d + x + 1
if d == 0:
print(x >> 1, '² + ', y >> 1, '² = 2.', n, '²', sep='')
Here are all solutions for n between 1 and 100:
7² + 1² = 2.5²
14² + 2² = 2.10²
17² + 7² = 2.13²
21² + 3² = 2.15²
23² + 7² = 2.17²
28² + 4² = 2.20²
31² + 17² = 2.25²
35² + 5² = 2.25²
34² + 14² = 2.26²
41² + 1² = 2.29²
42² + 6² = 2.30²
46² + 14² = 2.34²
49² + 7² = 2.35²
47² + 23² = 2.37²
51² + 21² = 2.39²
56² + 8² = 2.40²
49² + 31² = 2.41²
63² + 9² = 2.45²
62² + 34² = 2.50²
70² + 10² = 2.50²
69² + 21² = 2.51²
68² + 28² = 2.52²
73² + 17² = 2.53²
77² + 11² = 2.55²
82² + 2² = 2.58²
84² + 12² = 2.60²
71² + 49² = 2.61²
79² + 47² = 2.65²
85² + 35² = 2.65²
89² + 23² = 2.65²
91² + 13² = 2.65²
92² + 28² = 2.68²
98² + 14² = 2.70²
103² + 7² = 2.73²
94² + 46² = 2.74²
93² + 51² = 2.75²
105² + 15² = 2.75²
102² + 42² = 2.78²
112² + 16² = 2.80²
98² + 62² = 2.82²
97² + 71² = 2.85²
113² + 41² = 2.85²
115² + 35² = 2.85²
119² + 17² = 2.85²
123² + 3² = 2.87²
119² + 41² = 2.89²
126² + 18² = 2.90²
119² + 49² = 2.91²
133² + 19² = 2.95²
137² + 7² = 2.97²
124² + 68² = 2.100²
140² + 20² = 2.100²

You can optimize this algorithm maybe by considering only one quadrant only and the multiplying by 4.
import math
def psearch(n, count):
for x in range( 0 , 2*n + 1):
ysquare = 2*(n**2) - x * x
if (ysquare <0):
break
y = int(math.sqrt(ysquare))
if ysquare == y * y :
print(x,y)
count+=1
return count
print(psearch(13241324,0) * 4)
OUTPUT
(1269716, 18682964)
(1643084, 18653836)
(11027596, 15134644)
(12973876, 13503476)
(13241324, 13241324)
(13503476, 12973876)
(15134644, 11027596)
(18653836, 1643084)
(18682964, 1269716)
36

Remove mixed-variable terms in SymPy series expansion

Consider two functions of SymPy symbols e and i:
from sympy import Symbol, expand, Order
i = Symbol('i')
e = Symbol('e')
f = (i**3 + i**2 + i + 1)
g = (e**3 + e**2 + e + 1)
z = expand(f*g)
This will produce
z = e**3*i**3 + e**3*i**2 + e**3*i + e**3 + e**2*i**3 + e**2*i**2 + e**2*i + e**2 + e*i**3 + e*i**2 + e*i + e + i**3 + i**2 + i + 1
However, assume that e and i are both small and we can neglect both terms that are order three or higher. Using Sympy’s series tool or simply adding an O-notation Order class can handle this:
In : z = expand(f*g + Order(i**3) + Order(e**3))
Out: 1 + i + i**2 + e + e*i + e*i**2 + e**2 + e**2*i + e**2*i**2 + O(i**3) + O(e**3)
Looks great. However, I am still left with mixed terms e**2 * i**2. Individual variables in these terms are less than the desired cut-off so SymPy keeps them. However, mathematically small²·small² = small⁴. Likewise, e·i² = small·small² = small³.
At least for my purposes, I want these mixed terms dropped. Adding a mixed Order does not produce the desired result (it seems to ignore the first two orders).
In : expand(f*g + Order(i**3) + Order(e**3) + Order((i**2)*(e**2)))
Out: 1 + i + i**2 + i**3 + e + e*i + e*i**2 + e*i**3 + e**2 + e**2*i + e**3 + e**3*i + O(e**2*i**2, e, i)
Question: Does SymPy have an easy system to quickly remove the n-th order terms, as well as terms that are (e^a)·(i^b) where a+b > n?
Messy Solution: I have found a way to solve this, but it is messy and potentially not general.
z = expand(f*g + Order((e**2)*i) + Order(e*(i**2)))
zz = expand(z.removeO() + Order(e**3) + Order(i**3))
produces
zz = 1 + i + i**2 + e + e*i + e**2 + O(i**3) + O(e**3)
which is exactly what I want. So to specify my question: Is there a way to do this in one step that can be generalized to any n? Also, my solution loses the big-O notation that indicates mixed-terms were lost. This is not needed but would be nice.

As you have a dual limit, you must specify both infinitesimal variables (e and i) in all Order objects, even if they don’t appear in the first argument.
The reason for this is that Order(expr) only automatically chooses those symbols as infinitesimal that actually appear in the expr and thus, e.g., O(e) is only for the limit e→0.
Now, Order objects with different limits don’t mix well, e.g.:
O(e*i)+O(e) == O(e*i) != O(e)+O(e*i) == O(e) # True
This leads to a mess where results depend on the order of addition, which is a good indicator that this is something to avoid.
This can be avoided by explicitly specifying the infinitesimal symbols (as addition arguments of Order), e.g.:
O(e*i)+O(e,e,i) == O(e,e,i)+O(e*i) == O(e,e,i) # True
I haven’t found a way to avoid going through all combinations of e and i manually, but this can be done by a simple iteration:
orders = sum( Order(e**a*i**(n-a),e,i) for a in range(n+1) )
expand(f*g+orders)
# 1 + i + i**2 + e + e*i + e**2 + O(e**2*i, e, i) + O(e*i**2, e, i) + O(i**3, e, i) + O(e**3, e, i)

Without using Order you might try something simple like this:
>>> eq = expand(f*g) # as you defined
>>> def total_degree(e):
... x = Dummy()
... free = e.free_symbols
... if not free: return S.Zero
... for f in free:
... e = e.subs(f, x)
... return degree(e)
>>> eq.replace(lambda x: total_degree(x) > 2, lambda x: S.Zero)
e**2 + e*i + e + i**2 + i + 1

There is a way about it using Poly. I have made a function that keeps the O(...) term and another that does not (faster).
from sympy import Symbol, expand, Order, Poly
i = Symbol('i')
e = Symbol('e')
f = (i**3 + i**2 + i + 1)
g = (e**3 + e**2 + e + 1)
z = expand(f*g)
def neglect(expr, order=3):
z = Poly(expr)
# extract all terms and keep the lower order ones
d = z.as_dict()
d = {t: c for t,c in d.items() if sum(t) < order}
# Build resulting polynomial
return Poly(d, z.gens).as_expr()
def neglectO(expr, order=3):
# This one keeps O terms
z = Poly(expr)
# extract terms of higher "order"
d = z.as_dict()
large = {t: c for t,c in d.items() if sum(t) >= order}
for t in large: # Add each O(large monomial) to the expression
expr += Order(Poly({t:1},z.gens).as_expr(), *z.gens)
return expr
print(neglect(z))
print(neglectO(z))
This code prints the following:
e**2 + e*i + e + i**2 + i + 1
1 + i + i**2 + e + e*i + e**2 + O(e**2*i, e, i) + O(e*i**2, e, i) + O(i**3, e, i) + O(e**3, e, i)

How to do "inverse" substitution in sympy?

I need to find and substitute subexpression with a symbol, doing an "inverse" substitution of sorts.
Here is direct substitution example:
(simplify and collect added to make the resulting expression have the form that I need to work with)
In [1]: from sympy.abc import a, b, x, y, z
...: expr = (1 + b) * z + (1 + b) * y
...: z_expr = a / (1 + b) + x
...: subs_expr = expr.subs(z, z_expr).simplify().collect(1+b)
...: print(expr)
...: print(z_expr)
...: print(subs_expr)
y*(b + 1) + z*(b + 1)
a/(b + 1) + x
a + (b + 1)*(x + y)
Now I want to go back, and subs does not do anything:
In [2]: orig_expr = subs_expr.subs(z_expr, z)
...: print(orig_expr)
a + (b + 1)*(x + y)
How can I get back to y*(b + 1) + z*(b + 1)?

The substitution attempt fails because subs_expr does not actually contain z_expr in its expression tree. "Substitute for an expression that isn't there" is not really a well-defined goal. A well-defined goal would be "eliminate a using the relation z = z_expr". That can be done as follows:
var('orig_expr')
orig_expr = solve([orig_expr - subs_expr, z - z_expr], [orig_expr, a])[orig_expr]
Now orig_expr is equal to b*y + b*z + y + z

Fetch certain parts of sympy solution

I have a huge symbolic sympy expression on the form
expression = factor * (f1*a + f2*b + f3*c + f4*d + f5*e)
where all of the factors a through e all consists of several terms. I.e:
a = exp(2x) + exp(3x) + sin(Ix).
I want to create en array on the form
array = factor * [a,b,c,d,e]
But dont see a cleaver way to do this. I´ve tried to use the factor function, but it only gives me the expression on the form of "expression" above.
Until now i have used
print(expression)
and then done some brute force copy paste of the factors a through e. Since I am going to get expressions with way more terms than in this example, I want to do this without the copy paste routine. Any ideas?

Here's a simple example you can extrapolate for more terms
import sympy as sp
x = sp.var('x')
f1, f2 = sp.symbols('f1:3')
factor = sp.symbols('factor')
a = x**2 + sp.sin(x) + sp.exp(sp.I * x)
b = sp.log(x)/(x+1)**2
# example expression:
expression = (factor * (f1 * a + f2 * b)).expand()
print(expression)
# collect coefficients of f1 and f2
coeffs = sp.collect(expression.expand(),[f1,f2], evaluate=False)
print(coeffs)
# show the coefficients w/o the factor factor
[(coeffs[f]/factor).simplify() for f in (f1,f2)]
f1*factor*x**2 + f1*factor*exp(I*x) + f1*factor*sin(x) + f2*factor*log(x)/(x**2 + 2*x + 1)
{f2: factor*log(x)/(x**2 + 2*x + 1), f1: factor*x**2 + factor*exp(I*x) + factor*sin(x)}
[x**2 + exp(I*x) + sin(x), log(x)/(x**2 + 2*x + 1)]