20200319, 15:08  #1 
Aug 2019
11 Posts 
Solving Quartic Equation with a Coefficient of 1 MB Space
I have an equation of 4 degree (Quartic equation)and a coefficient of this equation takes 1 megabyte space in a text file. I want to solve this Quartic equation using computer. If the the equation has rational solution, I want to get rational solution with the exact numerator and the exact denominator (not the approximation). Is it possible?
There are are programming languages (e.g. MAGMA), computer algebra systems (e.g. PARI/GP, SageMath etc, here PARI is C library, can be called from a highlevel language application ,for instance, written in C, C++, Pascal, Fortran, Perl, or Python). If possible, then which programming language or computer algebra systems or library or softaware will be best to solve the Quartic equation as described above? What are the additional issues (configurations of RAM, Processor)? 
20200319, 20:22  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Is the lead coefficient a prime? [or monic]? If not, are these coefficients special in some way? Please explain why you need/want to do this. Also, please show us that you have some understanding of the problem by explaining why I asked what I did. Why are these questions important? Also, tell us what you want to do if there are no rational solutions. A full solution via Ferrari's formula will be one total cluster f*ck. Last fiddled with by R.D. Silverman on 20200319 at 20:26 

20200320, 06:09  #3  
Romulan Interpreter
Jun 2011
Thailand
9153_{10} Posts 
Quote:
Everything is possible if you have the knowhow, the resources, and the patience. Are all the coefficients integers? (if rationals, amplify to the LCM of denominators). How many coefficients? (i.e. can you reduce to degree 2 or 3 by substitutions?) Can you solve the equation in \(β\) (to know approximations of solution?) Can you apply Newton method (googlegoogle) in the vicinity of the solutions to get to the rational solution? (or as better as needed rational approximation?) (you may keep those lines in the rational domain by some tricks, you can start rational, and store the numerator and denominator separate, like, in fact, pari does). Edit: Ha!? It seems like AMS symbols don't work in matjax? I had to cheat that R by using UTF8 symbol Last fiddled with by LaurV on 20200320 at 06:34 

20200320, 08:19  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:
` Quote:
(2) You are wrong in your assertions. Reduction to lower degree requires that the equation has rational solutions. The O.P. did not say that it does. Indeed. Determining whether it does was part of the problem. (3) You totally` missed the point of my questions!!!!! (4) We don't even know if it has solutions in R. I would not want to compute a Sturm sequence. 

20200320, 10:26  #5 
Romulan Interpreter
Jun 2011
Thailand
3^{4}×113 Posts 
It depends how you look at it. I still assert that everything is possible, if you have enough resources, know how, and patience.
Sorry not to wait for the OP to reply to your questions, I know you had very good reasons to ask them (and I know a part of those reasons, indeed not all of them), but I just went a step ahead and I assumed the OP wants to make a factorization. I went that way long time ago (and banged my head on the wall, because, of course, I am not enough clever nor I know enough math. etc etc). About your point 2, well, that is false. Reduction to lower degree does NOT require the solutions to be rational. Think about X^42=0, it has no rational solution, and still it can be written as Y^22=0, by substituting X^2=Y. Also think about how people few centuries ago used to solve (a subset of) cubic equations, making clever substitutions to reduce them to quadratics (I could point you to some Mathologer/Numberphile videos, hihi, but I am sure you know that already better than me, you just brainfarted or made a typo ). Last fiddled with by LaurV on 20200320 at 10:36 
20200320, 12:06  #6 
Feb 2017
Nowhere
2^{4}·3^{2}·29 Posts 
It is not clear to me that the question is even being posed seriously. I will assume, for the sake of discussion, that it is.
I wonder where the coefficients came from. Are they integers? Exact fractions? What? The clause "a coefficient of this equation takes 1 megabyte space in a text file" is ambiguous. Does it only apply to one of the coefficients, or do theyall require that much space? Or is, perhaps, the polynomial monic? Assuming the coefficients are integers, one could also ask what is actually known about them. Are they of some special form? In addition to LaurV's suggestion of approximate solution, there is also (assuming integer coefficients) the possibility of reducing the polynomial (mod p) for small prime numbers p. Assuming the coefficients are integers, and the reduction is not the zero polynomial in F_{p}[x], it can easily be factored in F_{p}[x]. If any of these mod p factorizations have no linear factors, you have the answer to your rational zeroes question  there aren't any. 
20200320, 13:23  #7  
Nov 2003
2^{2}×5×373 Posts 
Quote:
reducible runs into the same questions that I asked, as well as making any linear transforms. Last fiddled with by R.D. Silverman on 20200320 at 13:31 

20200320, 15:08  #8 
Aug 2006
3·1,987 Posts 
There's a closed algebraic form for the solutions of a quartic, so you can just substitute away and be done. The solutions mightl be, say, 20 MB each before simplification, but that shouldn't be a problem. You can then simplify and check if they're rational as desired.

20200320, 18:01  #9  
Nov 2003
16444_{8} Posts 
Quote:
contains 6th degree terms. The solution contains many terms expressed in terms of the resolvent cubic which itself has many terms in its solution. I am also unsure whether it CAN be "simplified". Casus irreducibilis says that Cardano's formula for the resolvent cubic can not be simplified. I may, of course, be missing something. 

20200320, 18:16  #10  
"Robert Gerbicz"
Oct 2005
Hungary
1,429 Posts 
Quote:
To find "only" the rational roots, you could (fully) factorize the polynom, an easier/faster way (could be well known, found yesterday) : Let the polynom in Z[x]: a(n)*x^n+...+a(1)*x+a(0)=0 approximate the (real) roots with better than 1/(2*abs(a(n))) absolute difference. and for each x0 approximated root test the r=g/a(n) rational number as a root, where g=round(x0*a(n)). You'll get all rational solutions, and this works for every n. (you can simplify r by computing gcd(g,a(n)) , Ok, small (trivial) example: Code:
f(x)=10*x^2  17*x  63 let x0=1.801 approx. solution, then a(2)=10, g=round(1.801*10)=18, and r=18/10=9/5 is indeed a rational root. (and the other root of f(x) is also rational Last fiddled with by R. Gerbicz on 20200320 at 18:36 Reason: typos 

20200320, 18:18  #11  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{2}·37·71 Posts 
Quote:
Bob doubtless remembers the last time I cracked this gag. Something to do with running the LL test backwards. 

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