resolvent

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These are replies submitted by resolvent

Please let me know if my file NahayCubicResolvent.PDF makes any sense to you. Thanks. Please replace this text with the link to your file. The link can be found in the File Manager
Yes, I tried it. That works! Thank you, Mariner!
Yes, I tried it. That works! Thank you, Mariner!
I've found a use, if one can call it that, for these so-called "useless" closed-form solutions. Specifically, I am examining the structure of differential resolvents of polynomials. In April 1999 I found a way to factor certain terms of my powersum formula for a differential resolvent. I have wrestled with how to factor the remaining terms. By factor, in my case, I mean to give a matrix determinantal formula, e.g. my formula for the k-th term of a differential resolvent is given as det(A(k)) of a certain matrix A(k). Since 1999 I have tried to factor the A(k)'s, i.e determine the entries of matrices B(k) such that A(k) = M*B(k) where M is the matrix which is known to factor out of A(k) for SOME of the k (but I can't figure out how it factors out for the others). I started a paper this year, and am still writing it, for joint differential resolvents of polynomials. In all cases, I run up to computational difficulty. Namely, enormous intermediate blowup problems. For example, just last night, I calculated that my powersum formula for a 6-th order joint differential resolvent of two polynomials would consist of computing the determinant of roughly a 684 x 684 matrix with polynomial entries. These closed-form solutions of polynomials may give me insights, even if I never use them for direct computations. Thank you for informing me of Lauricella functions. I have never heard of them. I am just furious and frustrated at always hitting a mathematical complexity wall no matter which approach I take in any math paper I start to write.
Thank you, Robert Israel! Yes, the Mumford book "Tata Lectures on Theta II". In the appendix is Umemura's formula - as much as you wish to call it a formula. I've seen it nowhere else. Many of the polynomials with which I deal ARE highly structured. Or, if the polynomials themselves are not highly structured, then the specific things I do WITH them have a lot of structure - for example, computing differential resolvents of such polynomials. Ideally, at the end of the day (i.e. at the end of my lifetime) I would like some sort of "advanced category-theory type description" of all this computational work I've done with polynomials and differential equations.
You are half correct. Thank you for catching my error. I didn't care about complex vs real solutions or initial conditions of my artificial differential equation. I just want to test Maple's powers. I'll worry about all that later after I get the main solution-generating algorithm going. (-5)^(2/5) = ((-5)^(1/5))^2 = (- (5^(1/5)))^2 = (5^(1/5))^2 = 5^(2/5) One of the fifth roots of a negative number is a real negative number. Though it certainly IS very sloppy math to ignore this fact in a published math paper (or in a math class), often in practical applications, one might come across this situation for which one DOES seek the real negative solution. (-1)^5 = -1 implies (in a universe without complex numbers!) -1 = (-1)^(1/5) x^5+1=0 has 1 real solution x=-1 and 4 complex solutions {-1*h, -1*h^2, -1*h^3, -1*h^4} where h = exp(2*pi*sqrt(-1)/5)
You are half correct. Thank you for catching my error. I didn't care about complex vs real solutions or initial conditions of my artificial differential equation. I just want to test Maple's powers. I'll worry about all that later after I get the main solution-generating algorithm going. (-5)^(2/5) = ((-5)^(1/5))^2 = (- (5^(1/5)))^2 = (5^(1/5))^2 = 5^(2/5) One of the fifth roots of a negative number is a real negative number. Though it certainly IS very sloppy math to ignore this fact in a published math paper (or in a math class), often in practical applications, one might come across this situation for which one DOES seek the real negative solution. (-1)^5 = -1 implies (in a universe without complex numbers!) -1 = (-1)^(1/5) x^5+1=0 has 1 real solution x=-1 and 4 complex solutions {-1*h, -1*h^2, -1*h^3, -1*h^4} where h = exp(2*pi*sqrt(-1)/5)
Thank you, Robert Israel. I meant to ask - where in the online help glossary is the syntax x:='x' explained? I must make a habit of asking this, since this is what I am often really after when I post a question on this board. Obviously, I now know to look up 'eval' and 'unassign'.
How could "nops" be considered anywhere NEAR "perfectly suited"? In C++ and Java, probably the two most common computer programming languages on earth, "length" or "listlength" or something very similar to these and obvious is used to return the length of a list or an array.
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