2020-10-01, 20:46 | #23 |
"Robert Gerbicz"
Oct 2005
Hungary
10110000011_{2} Posts |
Thanks. More examples:
polynom: x^5 - 1615*x^4 + 861790*x^3 - 174419170*x^2 + 14998420705*x - 465948627343 Irreducible polynomial factors The 4 factors are: x − 653 (x − 103)2 x2 − 756x + 67259 Roots The 5 roots are: x1 = 653 x2 = x3 = 103 x4 = 103 x5 = 653 What is wrong x2-756+67259 is reducible. The correct answer should be as the roots are displayed: (x-653)^2*(x-103)^3 . But you haven't grouped even correctly the roots in that section. For evaluation part: polynom: (x+1)/x*(x+1) Your polynomial x + 1 This is wrong. polynom: x/2*2 your answer: Polynomial division is not integer This could be still ok, if you require that all subresults should be in Z[x]. Last fiddled with by R. Gerbicz on 2020-10-01 at 20:47 |
2020-10-02, 01:18 | #24 | |||
Aug 2002
Buenos Aires, Argentina
31·43 Posts |
Quote:
Quote:
This means that when you enter (x+1)/x*(x+1), parsing left to right, the program calculates (x+1)/x = 1 (the remainder 1 of the polynomial division is discarded) and then the program multiplies the previous result 1 by x+1, so the result is x+1. Quote:
Last fiddled with by alpertron on 2020-10-02 at 01:20 |
|||
2020-10-02, 16:23 | #25 |
"Robert Gerbicz"
Oct 2005
Hungary
2603_{8} Posts |
OK, more inputs:
(3*x^2)/(2*x+1) Your polynomial 3x2 Clearly wrong. Following your way it would mean that 3*x^2=(2*x+1)*3*x^2+R(x) and in this case the remainder would be 3rd degree? Meaningless. x^5 - 4161938199135571*x^4 + 6750425908270471236962285227630*x^3 - 5309242550166291213723686988859597734543042314*x^2 + 2016699400841707878060752590276325131391118358776183137438993*x - 295975920745818133920126480489914719398004640082403818556796416884133107491 Irreducible polynomial factors The polynomial is irreducible Roots The 5 roots are: The quintic equation cannot be expressed with radicands. Wrong, because p(x)=(x-505340926559057)^2*(x-1050418782005819)^3 Other such polynoms where you find the roots but display the wrong factors and grouping problems: x^5 - 569676319*x^4 + 105098371422759466*x^3 - 7011576352652045449773950*x^2 + 195375348220798308339573946630661*x - 1952243687462206905824707435700917386803 and x^5 - 2876590967309*x^4 + 3229067054667107663190970*x^3 - 1768676572192568691887072530348224746*x^2 + 473795689206607977454622626698064450356613773013*x - 49793205560537089066888851381785438595315265096906181547929 |
2020-10-02, 19:42 | #26 |
Aug 2002
Buenos Aires, Argentina
31·43 Posts |
All these examples are working now. Thanks for finding the errors.
Please refresh the page to get the current version. |
2020-10-04, 21:20 | #27 |
Aug 2002
Buenos Aires, Argentina
2465_{8} Posts |
When the irreducible polynomial has degree >= 5, now my code performs the factorization of this polynomials with prime modulus up to 100.
If the conditions of Keith Conrad's paper that you can read at https://kconrad.math.uconn.edu/blurb...galoisSnAn.pdf are true, the Galois group is A_{n} or S_{n}, so my application indicates that the roots of the polynomial cannot be expressed as radical expressions along with the conditions found. Example of new output from https://www.alpertron.com.ar/POLFACT.HTM Your polynomial x^{12} + 45x^{7} − 23 Irreducible polynomial factors The polynomial is irreducible Roots The 12 roots are: x1 to x12 : The roots of the polynomial cannot be expressed by radicals. The degrees of the factors of polynomial modulo 7 are 1, 2 and 9 (the Galois group contains a cycle of length 2) and the degrees of the factors of polynomial modulo 17 are 1 and 11 (the Galois group contains a cycle of prime length greater than half the degree of polynomial) Last fiddled with by alpertron on 2020-10-04 at 21:35 |
Thread Tools | |
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Where can I find a arbitrary precision Calculator online that can handle this # | ONeil | Information & Answers | 9 | 2018-04-17 18:18 |
On polynomials without roots modulo small p | fivemack | Computer Science & Computational Number Theory | 2 | 2015-09-18 12:54 |
How to find values of polynomials with nice factorization? | Drdmitry | Computer Science & Computational Number Theory | 18 | 2015-09-10 12:23 |
How much ECM does it take to find a given factor? | geoff | Factoring | 5 | 2004-09-29 20:14 |
How large a factor can P-1 testing find ? | dsouza123 | Software | 3 | 2003-12-11 00:48 |