>> One last question. I´m trying to expand that expression up to n=2000,
>> but I get a memory error. Besides buying more physical RAM, can you
>> recommend any other way of doing it?
>>
...
>>
>> This polynomial has specific structure, so it should be possible to
>> derive a formula for the terms of `f_n`. SymPy should help you with
>> experimentation with `f_n` for small `n` (as you did already).
>
This is where python shines, I think. If you make some little helper functions
it can make the task easier.
Not wanting to spoil the fun of finding the terms of your original expression,
let's say you wanted to investigate the terms of (x+y)**n as n increased. One
way to order terms would be by the exponent that is on the x, so we write a
function to tell us that:
h[19] >>> def xpow(t):
... xpart = t.as_independent(x)[1]
... if xpart.has(x):
... return xpart.as_base_exp()[1]
... return 0
...
We test it
h[19] >>> xpow(S(1)) # S(1) gives a sympy 1 so we don't get an error
0
h[19] >>> xpow(x)
1
h[19] >>> xpow(x**2)
2
h[20] >>> xpow(y*x**3)
3
h[20] >>> p=list(((x+y)**5).expand().args)
h[20] >>> p
[x**5, 10*x**2*y**3, 5*y*x**4, 5*x*y**4, y**5, 10*x**3*y**2]
h[21] >>> p.sort(key=lambda a: xpow(a))
h[21] >>> p
[y**5, 5*x*y**4, 10*x**2*y**3, 10*x**3*y**2, 5*y*x**4, x**5]
Now let's see if we can see a pattern as we change n:
h[21] >>> for n in range(1, 5):
... p = list(((x+y)**n).expand().args)
... p.sort(key=lambda a: xpow(a))
... print n,p
...
1 [y, x]
2 [y**2, 2*x*y, x**2]
3 [y**3, 3*x*y**2, 3*y*x**2, x**3]
4 [y**4, 4*x*y**3, 6*x**2*y**2, 4*y*x**3, x**4]
So if each term (x and y) is independent of n we get the binomial coefficients.
What if y has a coefficient of n?
h[35] >>> for nn in range(1,7):
... nn,sorted(list(product(x+n*y,
(n,1,nn)).expand().args),key=lambda l:
xpow(l))
...
(1, [y, x])
(2, [2*y**2, 3*x*y, x**2])
(3, [6*y**3, 11*x*y**2, 6*y*x**2, x**3])
(4, [24*y**4, 50*x*y**3, 35*x**2*y**2, 10*y*x**3, x**4])
(5, [120*y**5, 274*x*y**4, 225*x**2*y**3, 85*x**3*y**2, 15*y*x**4, x**5])
(6, [720*y**6, 1764*x*y**5, 1624*x**2*y**4, 735*x**3*y**3, 175*x**4*y**2,
21*y*x**5, x**6])
If you arrange these coefficients into a "Pascal's triangle", can you see any
pattern?
/c
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