Many thanks Martin, I add your reponse to my knowledge. >> //1// I only use series in analysis, >> so series with severe variables are very rare for me. >> >> There are some >> >> sum(in x^i) sum(in y^j) ... x^i y^j = sum(in y^j) sum(in x^j) ... >> >> but this use of axiom isn't so fine because it's impossible to find the >> function from this serie in axiom. >> I forgot that the series functions in axiom neither compute general term of a serie from its function nor the limit of the sum.
In fact I look a link to this exercice with perhaps x=y. I give you the first line, you tell me the last one or the other way. But the series function in axiom seems to have no interest for this exercice : A= sum ((x^i*y^k) / (2^i * k!), k=0..infty and i=0..infty) = sum (sum (y^k/k!, k=0..infty) * x^i / 2^i, i=0..infty) = sum (e^y * x^i / 2^i, i=0..infty) = 1/(1-x/2) * e^y =B < infty. It's in fact two limit commands if we want to compute from A to B. And series only gives the first coefficients of A from B of corse ! not the general formula. > Sorry, you need to be more explicit. Am I a bit more clear ? This exercice seems out of your improvement. >> //4// Sometime I compute the recip of y=1+x+x^3 arround x=0. So I get a serie >> x=1-y+... Do you know if there is this revert function in axiom. >> It's not recip that is equal to 1/... >> My eyes didn't see the revert function in my emacs buffer in )sh UTS. shame for me. In fact the right place of this revert may be in a package because the revert of the serie f(x) for x arround a is a serie arround b=f(a)... and even a Puiseux serie if y=f(x)=x^2+x^3. > The series expansion of the root of 1+ x + x^3 itself is ugly, and revert > expects that the constant term vanishes... > Are you sure ? z = revert (x+x^3) is easy. With a graph there is a translation between f(x)=x+x^3 and g(x)=1+x+x^3 by the vector [0,1]. For the revert function the translation use the vector [1,0] and the UTS is arround 1. I feel that the formula is almost the same... The real difficult case is for revert (x^2+x^3). I get an awfull NIL/NIL result with fricas. >> //5// I use the coefficient function with polynom. >> By example M(a,0) and N(0,b) are points. Give a vector of (MN) line. >> >> The equation of the line is >> eqline := determinant matrix [[a,0,1],[0,b,1],[x,y,1]] (equal 0) >> >> A vector is vector [coefficient (eqline, y, 1), - coefficient (eqline, x, >> 1)]. >> >> I don't know if this method may be used for series. >> > > *What* do you want to do with series? > This command shows an example where coefficient : POLY -> POLY is useful. I ignore if there are analogous exemple for series. The common test I do about Bernoulli polynomial is right : sr := series (t*exp(t*x)/(exp t -1), t=0) coefficient (sr, 5) --- and not coefficient (st, t, 5) there is only one variable. >> //8// I quickly try to test the silly series (x^a-a^x) and (x^x - a^a) for a >> fixed a when x is arround a but I can't with axiom. >> > > (67) -> series(x^a-a^x, x=a) > I persisted to search this serie with a silly coerce command. UUTS ==> UTS (Fraction Integer ? ? ?) UATS ==> UTS (Expression Integer, x, a) seems right with a lot of elementry coerce... I will remind the questions I ask to my students for the x=tan x equation. Thanks you again for this axiom course ! Francois --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "FriCAS - computer algebra system" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/fricas-devel?hl=en -~----------~----~----~----~------~----~------~--~---
