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Solving multivariate equation for a subset of variables

I am using sympy to solve some equations and I am running into a problem. I have this issue with many equations but I will illustrate with an example. I have an equation with multiple variables and I want to solve this equation in terms of all variables but one is excluded. For instance the equation 0 = 2^n*(2-a) - b + 1. Here there are three variables a, b and n. I want to get the values for a and b not in terms of n so the a and b may not contain n.
2^n*(2-a) - b + 1 = 0
# Since we don't want to solve in terms of n we know that (2 - a)
# has to be zero and -b + 1 has to be zero.
2 - a = 0
a = 2
-b + 1 = 0
b = 1
I want sympy to do this. Maybe I'm just not looking at the right documentation but I have found no way to do this. When I use solve and instruct it to solve for symbols a and b sympy returns to me a single solution where a is defined in terms of n and b. I assume this means I am free to choose b and n, However I don't want to fix n to a specific value I want n to still be a variable.
Code:
import sympy
n = sympy.var("n", integer = True)
a = sympy.var("a")
b = sympy.var("b")
f = 2**n*(2-a) - b + 1
solutions = sympy.solve(f, [a,b], dict = True)
# this will return: "[{a: 2**(-n)*(2**(n + 1) - b + 1)}]".
# A single solution where b and n are free variables.
# However this means I have to choose an n I don't want
# to that I want it to hold for any n.
I really hope someone can help me. I have been searching google for hours now...

Ok, here's what I came up with. This seems to solve the type of equations you're looking for. I've provided some tests as well. Of course, this code is rough and can be easily caused to fail, so i'd take it more as a starting point than a complete solution
import sympy
n = sympy.Symbol('n')
a = sympy.Symbol('a')
b = sympy.Symbol('b')
c = sympy.Symbol('c')
d = sympy.Symbol('d')
e = sympy.Symbol('e')
f = sympy.sympify(2**n*(2-a) - b + 1)
g = sympy.sympify(2**n*(2-a) -2**(n-1)*(c+5) - b + 1)
h = sympy.sympify(2**n*(2-a) -2**(n-1)*(e-1) +(c-3)*9**n - b + 1)
i = sympy.sympify(2**n*(2-a) -2**(n-1)*(e+4) +(c-3)*9**n - b + 1 + (d+2)*9**(n+2))
def rewrite(expr):
if expr.is_Add:
return sympy.Add(*[rewrite(f) for f in expr.args])
if expr.is_Mul:
return sympy.Mul(*[rewrite(f) for f in expr.args])
if expr.is_Pow:
if expr.args[0].is_Number:
if expr.args[1].is_Symbol:
return expr
elif expr.args[1].is_Add:
base = expr.args[0]
power = sympy.solve(expr.args[1])
sym = expr.args[1].free_symbols.pop()
return sympy.Mul(sympy.Pow(base,-power[0]), sympy.Pow(base,sym))
else:
return expr
else:
return expr
else:
return expr
def my_solve(expr):
if not expr.is_Add:
return None
consts_list = []
equations_list = []
for arg in expr.args:
if not sympy.Symbol('n') in arg.free_symbols:
consts_list.append(arg)
elif arg.is_Mul:
coeff_list = []
for nested_arg in arg.args:
if not sympy.Symbol('n') in nested_arg.free_symbols:
coeff_list.append(nested_arg)
equations_list.append(sympy.Mul(*coeff_list))
equations_list.append(sympy.Add(*consts_list))
results = {}
for eq in equations_list:
var_name = eq.free_symbols.pop()
val = sympy.solve(eq)[0]
results[var_name] = val
return results
print(my_solve(rewrite(f)))
print(my_solve(rewrite(g)))
print(my_solve(rewrite(h)))
print(my_solve(rewrite(i)))

Converting recursive algorithm to Iterative

I have done the Recursive function in Python that works:
def Rec(n):
if (n<=5):
return 2*n
elif (n>=6):
return Rec(n-6)+2*Rec(n-4)+4*Rec(n-2)
print (Rec(50))
But I can't think of an iterative one
I am sure I will need to use a loop and possibly have 4 variables to store the previous values, imitating a stack.

For your particular question, assuming you have an input n, the following code should calculate the function iteratively in python.
val = []
for i in range(6):
val.append(2*i)
for i in range(6,n+1):
val.append( val[i-6] + 2*val[i-4] + 4*val[i-2] )
print(val[n])

I get this answer:
$ python test.py
Rec(50) = 9142785252232708
Kist(50) = 9142785252232708
Using the code below. The idea is that your function needs a "window" of previous values - Kn-6, Kn-4, Kn-2 - and that window can be "slid" along as you compute new values.
So, for some value like "14", you would have a window of K8, K9, ... K13. Just compute using those values, then drop K8 since you'll never use it again, and append K14 so you can use it in computing K15..20.
def Rec(n):
if (n<=5):
return 2*n
elif (n>=6):
return Rec(n-6)+2*Rec(n-4)+4*Rec(n-2)
def Kist(n):
if n <= 5:
return 2 * n
KN = [2*n for n in range(6)]
for i in range(6, n+1):
kn = KN[-6] + 2 * KN[-4] + 4 * KN[-2]
KN.append(kn)
KN = KN[-6:]
return KN[-1]
print("Rec(50) =", Rec(50))
print("Kist(50) =", Kist(50))

Translating a function integrating another function from BASIC into Python

The book Calculus and Pizza by Clifford Pickover has a few code examples here and there, all written in some dialect of BASIC.
I wrote a Python version of the code example covering integration. His BASIC example goes like:
10 REM Integration
20 DEF FNY(X) = X*X*X
30 A = 0
40 B = 1
50 N = 10
55 R = 0
60 H = (B-A)/N
70 FOR X = A TO B - H/2 STEP H
80 R = R + FNY(X)
90 NEXT X
100 R = R * H
110 PRINT *INTEGRATION ESTIMATE*: R
I changed a few things here and there, allowing the user to specify the interval over which to take the integral, specify the function to be integrated as a lambda, and so forth. I knew right off the bat that the for loop wouldn't work as I have written it below. I'm just wondering if there's some direct or idiomatic translation of the BASIC for to a Python for.
def simpleintegration():
f = eval(input("specify the function as a lambda\n:%"))
a = int(input("take the integral from x = a = ...\n:%"))
b = int(input("to x = b = ...\n:%"))
n = 10
r = 0
h = (b-a)/n
for x in range(a,b-h/2,h):
r = r + f(x)
r = r * h
print(r)

Your translation isn't far off. The only difference between the for loop in other languages and Python's "loop-over-a-range" pattern is that the "stop" value is usually inclusive in other languages, but is exclusive in Python.
Thus, in most other languages, a loop including a and b looks like
for i = a to b step c
' Do stuff
next i
In Python, it would be
for i in range(a, b + 1, c):
# Do stuff

The formula is computing the Riemann sums using the values at the left end of the subdivision intervals. Thus the last used value for X should be B-H.
Due to floating point errors, stepping from A by H can give a last value that is off by some small amount, thus B-H is not a good bound (in the BASIC code) and B-H/2 is used to stop before X reaches B.
The Python code should work in the presented form for the same reasons, since the bound B-H/2 is unreachable, thus the range should stop with B-H or a value close by.
Using a slight modification you can actually compute the trapezoidal approximation, where you initialize with R=f(A)/2, step X from A+H to including B-H adding f(X) to R and then finish by adding f(B)/2 (which could already be done in the initialization). As before, the approximation of the integral is then R*H.

You can do as below, just changing iteration of 'i' in for loop.
def simpleintegration():
f = eval(input("specify the function as a lambda\n:%"))
a = int(input("take the integral from x = a = ...\n:%"))
b = int(input("to x = b = ...\n:%"))
n = 10
r = 0
h = (b-a)/n
for x = a to b-h/2 step h:
r = r + f(x)
r = r * h
print(r)

Leibniz formula for π - Is this any good? (Python)

I'm doing an exercise that asks for a function that approximates the value of pi using Leibniz' formula. These are the explanations on Wikipedia:
Logical thinking comes to me easily, but I wasn't given much of a formal education in maths, so I'm a bit lost as to what the leftmost symbols in the second one represent. I tried to make the code pi = ( (-1)**n / (2*n + 1) ) * 4, but that returned 1.9999990000005e-06 instead of 3.14159..., so I used an accumulator pattern instead (since the chapter of the guide that this was in mentions them as well) and it worked fine. However, I can't help thinking that it's somewhat contrived and there's probably a better way to do it, given Python's focus on simplicity and making programmes as short as possible. This is the full code:
def myPi(n):
denominator = 1
addto = 1
for i in range(n):
denominator = denominator + 2
addto = addto - (1/denominator)
denominator = denominator + 2
addto = addto + (1/denominator)
pi = addto * 4
return(pi)
print(myPi(1000000))
Does anyone know a better function?

The Leibniz formula translates directly into Python with no muss or fuss:
>>> steps = 1000000
>>> sum((-1.0)**n / (2.0*n+1.0) for n in reversed(range(steps))) * 4
3.1415916535897934

The capital sigma here is sigma notation. It is notation used to represent a summation in concise form.
So your sum is actually an infinite sum. The first term, for n=0, is:
(-1)**0/(2*0+1)
This is added to
(-1)**1/(2*1+1)
and then to
(-1)**2/(2*2+1)
and so on for ever. The summation is what is known mathematically as a convergent sum.
In Python you would write it like this:
def estimate_pi(terms):
result = 0.0
for n in range(terms):
result += (-1.0)**n/(2.0*n+1.0)
return 4*result
If you wanted to optimise a little, you can avoid the exponentiation.
def estimate_pi(terms):
result = 0.0
sign = 1.0
for n in range(terms):
result += sign/(2.0*n+1.0)
sign = -sign
return 4*result
....
>>> estimate_pi(100)
3.1315929035585537
>>> estimate_pi(1000)
3.140592653839794

Using pure Python you can do something like:
def term(n):
return ( (-1.)**n / (2.*n + 1.) )*4.
def pi(nterms):
return sum(map(term,range(nterms)))
and then calculate pi with the number of terms you need to reach a given precision:
pi(100)
# 3.13159290356
pi(1000)
# 3.14059265384

The following version uses Ramanujan's formula as outlined in this SO post - it uses a relation between pi and the "monster group", as discussed in this article.
import math
def Pi(x):
Pi = 0
Add = 0
for i in range(x):
Add =(math.factorial(4*i) * (1103 + 26390*i))/(((math.factorial(i))**4)*(396**(4*i)))
Pi = Pi + (((math.sqrt(8))/(9801))*Add)
Pi = 1/Pi
print(Pi)
Pi(100)

This was my approach:
def estPi(terms):
outPut = 0.0
for i in range (1, (2 * terms), 4):
outPut = (outPut + (1/i) - (1/(i+2)))
return 4 * outPut
I take in the number of terms the user wants, then in the for loop I double it to account for only using odds.
at 100 terms I get 3.1315929035585537
at 1000 terms I get 3.140592653839794
at 10000 terms I get 3.1414926535900345
at 100000 terms I get 3.1415826535897198
at 1000000 terms I get 3.1415916535897743
at 10000000 terms I get 3.1415925535897915
at 100000000 terms I get 3.141592643589326
at 1000000000 terms I get 3.1415926525880504
Actual Pi is 3.1415926535897932
Got to love a convergent series.

def myPi(iters):
pi = 0
sign = 1
denominator = 1
for i in range(iters):
pi = pi + (sign/denominator)
# alternating between negative and positive
sign = sign * -1
denominator = denominator + 2
pi = pi * 4.0
return pi
pi_approx = myPi(10000)
print(pi_approx)

old thread, but i wanted to stuff around with this and coincidentally i came up with pretty much the same as user3220980
# gregory-leibnitz
# pi acurate to 8 dp in around 80 sec
# pi to 5 dp in .06 seconds
import time
start_time = time.time()
pi = 4 # start at 4
times = 100000000
for i in range(3,times,4):
pi -= (4/i) + (4/(i + 2))
print(pi)
print("{} seconds".format(time.time() - start_time))

Python: "long int too large to convert to float" when calculating pi

I get this error when using a python script that calculates pi using the Gauss-Legendre algorithm. You can only use up to 1024 iterations before getting this:
C:\Users\myUsernameHere>python Desktop/piWriter.py
End iteration: 1025
Traceback (most recent call last):
File "Desktop/piWriter.py", line 15, in <module>
vars()['t' + str(sub)] = vars()['t' + str(i)] - vars()['p' + str(i)] * math.
pow((vars()['a' + str(i)] - vars()['a' + str(sub)]), 2)
OverflowError: long int too large to convert to float
Here is my code:
import math
a0 = 1
b0 = 1/math.sqrt(2)
t0 = .25
p0 = 1
finalIter = input('End iteration: ')
finalIter = int(finalIter)
for i in range(0, finalIter):
sub = i + 1
vars()['a' + str(sub)] = (vars()['a' + str(i)] + vars()['b' + str(i)])/ 2
vars()['b' + str(sub)] = math.sqrt((vars()['a' + str(i)] * vars()['b' + str(i)]))
vars()['t' + str(sub)] = vars()['t' + str(i)] - vars()['p' + str(i)] * math.pow((vars()['a' + str(i)] - vars()['a' + str(sub)]), 2)
vars()['p' + str(sub)] = 2 * vars()['p' + str(i)]
n = i
pi = math.pow((vars()['a' + str(n)] + vars()['b' + str(n)]), 2) / (4 * vars()['t' + str(n)])
print(pi)
Ideally, I want to be able to plug in a very large number as the iteration value and come back a while later to see the result.
Any help appreciated!
Thanks!

Floats can only represent numbers up to sys.float_info.max, or 1.7976931348623157e+308. Once you have an int with more than 308 digits (or so), you are stuck. Your iteration fails when p1024 has 309 digits:
179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137216L
You'll have to find a different algorithm for pi, one that doesn't require such large values.
Actually, you'll have to be careful with floats all around, since they are only approximations. If you modify your program to print the successive approximations of pi, it looks like this:
2.914213562373094923430016933707520365715026855468750000000000
3.140579250522168575088244324433617293834686279296875000000000
3.141592646213542838751209274050779640674591064453125000000000
3.141592653589794004176383168669417500495910644531250000000000
3.141592653589794004176383168669417500495910644531250000000000
3.141592653589794004176383168669417500495910644531250000000000
3.141592653589794004176383168669417500495910644531250000000000
In other words, after only 4 iterations, your approximation has stopped getting better. This is due to inaccuracies in the floats you are using, perhaps starting with 1/math.sqrt(2). Computing many digits of pi requires a very careful understanding of the numeric representation.

As noted in previous answer, the float type has an upper bound on number size. In typical implementations, sys.float_info.max is 1.7976931348623157e+308, which reflects the use of 10 bits plus sign for the exponent field in a 64-bit floating point number. (Note that 1024*math.log(2)/math.log(10) is about 308.2547155599.)
You can add another half dozen decades to the exponent size by using the Decimal number type. Here is an example (snipped from an ipython interpreter session):
In [48]: import decimal, math
In [49]: g=decimal.Decimal('1e12345')
In [50]: g.sqrt()
Out[50]: Decimal('3.162277660168379331998893544E+6172')
In [51]: math.sqrt(g)
Out[51]: inf
This illustrates that decimal's sqrt() function performs correctly with larger numbers than does math.sqrt().

As noted above, getting lots of digits is going to be tricky, but looking at all those vars hurts my eyes. So here's a version of your code after (1) replacing your use of vars with dictionaries, and (2) using ** instead of the math functions:
a, b, t, p = {}, {}, {}, {}
a[0] = 1
b[0] = 2**-0.5
t[0] = 0.25
p[0] = 1
finalIter = 4
for i in range(finalIter):
sub = i + 1
a[sub] = (a[i] + b[i]) / 2
b[sub] = (a[i] * b[i])**0.5
t[sub] = t[i] - p[i] * (a[i] - a[sub])**2
p[sub] = 2 * p[i]
n = i
pi_approx = (a[n] + b[n])**2 / (4 * t[n])
Instead of playing games with vars, I've used dictionaries to store the values (the link there is to the official Python tutorial) which makes your code much more readable. You can probably even see an optimization or two now.
As noted in the comments, you really don't need to store all the values, only the last, but I think it's more important that you see how to do things without dynamically creating variables. Instead of a dict, you could also have simply appended the values to a list, but lists are always zero-indexed and you can't easily "skip ahead" and set values at arbitrary indices. That can occasionally be confusing when working with algorithms, so let's start simple.
Anyway, the above gives me
>>> print(pi_approx)
3.141592653589794
>>> print(pi_approx-math.pi)
8.881784197001252e-16

A simple solution is to install and use the arbitrary-precisionmpmath module which now supports Python 3. However, since I completely agree with DSM that your use ofvars()to create variables on the fly is an undesirable way to implement the algorithm, I've based my answer on his rewrite of your code and [trivially] modified it to make use ofmpmath to do the calculations.
If you insist on usingvars(), you could probably do something similar -- although I suspect it might be more difficult and the result would definitely harder to read, understand, and modify.
from mpmath import mpf # arbitrary-precision float type
a, b, t, p = {}, {}, {}, {}
a[0] = mpf(1)
b[0] = mpf(2**-0.5)
t[0] = mpf(0.25)
p[0] = mpf(1)
finalIter = 10000
for i in range(finalIter):
sub = i + 1
a[sub] = (a[i] + b[i]) / 2
b[sub] = (a[i] * b[i])**0.5
t[sub] = t[i] - p[i] * (a[i] - a[sub])**2
p[sub] = 2 * p[i]
n = i
pi_approx = (a[n] + b[n])**2 / (4 * t[n])
print(pi_approx) # 3.14159265358979