You realize that you cannot express roots of polynomials in terms of radicals generally?
Maybe you should figure what you are doing with apart(). If it is integration of rational functions, maybe you should find another path. RJF On Thursday, August 21, 2014 1:05:42 PM UTC-7, Mateusz Paprocki wrote: > > Hi, > > On 21 August 2014 16:26, clemens novak <[email protected] <javascript:>> > wrote: > > Am Mittwoch, 20. August 2014 23:45:55 UTC+2 schrieb Mateusz Paprocki: > >> > >> Hi, > >> > >> On 20 August 2014 21:28, Ondřej Čertík <[email protected]> wrote: > >> > Hi Clemens, > >> > > >> > On Wed, Aug 20, 2014 at 5:15 AM, clemens novak <[email protected]> > >> > wrote: > >> >> I have a question regarding the apart function. I want to obtain the > >> >> partial > >> >> fractions for eq = z / (z**2-z-1) . > >> >> > >> >> The denominator has real roots solve(denom(eq)) yields [1/2 + > sqrt(5)/2 > >> >> , > >> >> 1/2- sqrt(5)/2], but apart does not return the partial fractions; > i,e. > >> >> apart(eq, z) = z/(z**2 - z - 1) . > >> >> > >> >> Using apart(eq, z, full=True) yields RootSum(_w**2 - _w - 1, > Lambda(_a, > >> >> (_a/5 + 2/5)/(-_a + z))) which doesn't seem to be of much help (at > >> >> least for > >> >> me). > >> >> > >> >> Using apart with a fraction with "simpler" roots produces the > desired > >> >> partial fractions; e.g. apart (z / (z**2+z-2), z) yields 2/(3*(z + > 2)) > >> >> + > >> >> 1/(3*(z - 1)) > >> >> > >> >> Are there any additional options I can provide to apart to obtain > the > >> >> desired result? > >> > > >> > Good point. I checked Mathematica and it produces the same result as > >> > SymPy. > >> > However, clearly this can be decomposed, so I don't quite understand > >> > if this is a bug in Mathematica as well, > >> > or whether perhaps there is some reason not to do the decomposition > >> > for this case. > >> > >> There is no bug in this case. apart() uses polynomial factorization > >> routines to "decompose" the denominator. By default, sympy (but also > >> many other computer algebra systems) work over the rationals and z**2 > >> - z - 1 doesn't have rational zeros. If you invoke apart(f, z), then > >> you will get partial fraction decomposition over rationals. To get the > >> expected result, you have set the domain of computation properly, e.g. > >> apart(f, z, extension=sqrt(5)) is helpful in this case. apart() > >> accepts all options that factor() and other polynomial manipulation > >> functions accept. > >> > >> In [1]: apart(z/(z**2 - z - 1), z, extension=sqrt(5)) > >> Out[1]: > >> ___ ___ > >> ╲╱ 5 + 5 -5 + ╲╱ 5 > >> ─────────────────── - ─────────────────── > >> ⎛ ___ ⎞ ⎛ ___⎞ > >> 5⋅⎝2⋅z - ╲╱ 5 - 1⎠ 5⋅⎝2⋅z - 1 + ╲╱ 5 ⎠ > >> > >> In [2]: simplify(_) > >> Out[2]: > >> z > >> ────────── > >> 2 > >> z - z - 1 > >> > >> Another option is to use full=True (no need for extension, because > >> this method doesn't use factorization at all). The result seems > >> useless at first but you can use .doit() on the resulting RootSum to > >> get more familiar result: > >> > >> In [1]: apart(z/(z**2 - z - 1), z, full=True) > >> Out[1]: > >> ⎛ a 2 ⎞ > >> ⎜ ─ + ─ ⎟ > >> ⎜ 2 5 5 ⎟ > >> RootSum⎜w - w - 1, a ↦ ──────⎟ > >> ⎝ -a + z⎠ > >> > >> In [2]: _.doit() > >> Out[2]: > >> ___ ___ > >> ╲╱ 5 1 ╲╱ 5 1 > >> ───── + ─ - ───── + ─ > >> 10 2 10 2 > >> ───────────── + ───────────── > >> ___ ___ > >> ╲╱ 5 1 1 ╲╱ 5 > >> z - ───── - ─ z - ─ + ───── > >> 2 2 2 2 > >> > >> In [3]: simplify(_) > >> Out[3]: > >> z > >> ────────── > >> 2 > >> z - z - 1 > >> > >> It would be good to document this in apart()'s docstring (the current > >> one is pretty weak), because it's a common misconception that apart() > >> uses either roots() or solve() to decompose the denominator. > >> > >> Mateusz > >> > >> > Ondrej > >> > > >> >> > >> >> Thanks & regards - Clemens > >> >> > >> >> -- > >> > > > > Hello, > > > > thanks for quick answer and support. Both suggestions (apart with > full=True > > + doit() and apart with extension) work for me. > > > > I read factor's documentation > > ( > http://docs.sympy.org/latest/modules/polys/reference.html#symbolic-root-finding-algorithms) > > > > which says: > > > > "By default, the factorization is computed over the rationals. To factor > > over other domain, e.g. an lgebraic or finite field, use appropriate > > options: > > ``extension``, ``modulus`` or ``domain``." > > > > I don't quite get it: > > > > "extension" allows providing a list of irrational numbers which shall be > > considered during factorization: > > > > eq = x**2 - sqrt(2)*x + sqrt(3)*x - sqrt(6) > > factor(eq, extension=[sqrt(2), sqrt(3)]) > > > > This works for complex numbers as well > > > > eq = x**2+x+5 > > factor(eq, extension=sqrt(-19)) > > > > > > "modulus allows factorization over a finite field (there is an example > in > > the "docstring) > > > > However, I don't get the use of "domain" - from the name I would assume > that > > you can restrict the domain (e.g. integer, ration, irrational, > complex...) > > from which the factors are obtained? What domains does sympy provide > (and > > where can I find that in the documentation)? > > Domain is the ultimate way to set the domain of computation. > extension, modulus, etc. are helpers, that, in the end, will generate > a domain. For example: > > In [1]: Poly(x**2 + sqrt(2), extension=sqrt(2)) > Out[1]: Poly(x**2 + sqrt(2), x, domain='QQ<sqrt(2)>') > > You can see that extension=sqrt(2) really means an algebraic domain > over sqrt(2). In case of x**2 + sqrt(2) you could use extension=True > as well (equivalent of Mathematica's auto): > > In [2]: Poly(x**2 + sqrt(2), extension=True) > Out[2]: Poly(x**2 + sqrt(2), x, domain='QQ<sqrt(2)>') > > Ultimately you can construct domain yourself: > > In [3]: Poly(x**2 + sqrt(2), domain=QQ.algebraic_field(sqrt(2))) > Out[3]: Poly(x**2 + sqrt(2), x, domain='QQ<sqrt(2)>') > > extension allow you to type less and gives you Mathematica-like feel. > Domains can be useful on their own e.g.: > > In [4]: x**2 + sqrt(2) in QQ[x] > Out[4]: False > > In [5]: x**2 + sqrt(2) in QQ.algebraic_field(sqrt(2))[x] > Out[5]: True > > Mateusz > > > If you want I can try to improve the apart (maybe also the factor) > docstring > > and send a corresponding PR (probably in the next few days). > > > > Regards - Clemens > > > > -- > > You received this message because you are subscribed to the Google > Groups > > "sympy" group. > > To unsubscribe from this group and stop receiving emails from it, send > an > > email to [email protected] <javascript:>. > > To post to this group, send email to [email protected] > <javascript:>. > > Visit this group at http://groups.google.com/group/sympy. > > To view this discussion on the web visit > > > https://groups.google.com/d/msgid/sympy/ed1fcf30-233e-4551-8b34-040e047e42c5%40googlegroups.com. > > > > > > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "sympy" group. 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