20040910, 14:18  #1 
Jul 2004
16_{8} Posts 
General Solution to Polynomial Equations
What are the implications for factoring/primality testing (if any) of the newly announced general solution to polynomial equations? This piqued my interest because I remembered something about FFT being based on representing numbers as polynomial equations (sorry if I'm incorrect).
http://arxiv.org/ftp/math/papers/0408/0408264.pdf 
20040910, 19:15  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
years old. It is a lot of media ignorance and hype. Methods have been around since the time of Hermite (or earlier). Since polynomials are 11 in the close neighborhood of any root, you can always find an infinite series representation to the root (by the Inverse Function Thm). This socalled "new" method is nothing more than an infinite series solution. But it has been known for a long time that roots could be expressed in terms of Elliptic or Hypergeometric or Theta functions. (2) It has ZERO implications for factoring/prime testing. I am curious as to why you think there MIGHT be a connection? 

20040910, 19:27  #3 
"Mike"
Aug 2002
2^{2}×7×281 Posts 
Apparently this is hitting the news big time...
http://science.slashdot.org/science/.../1229221.shtml 
20040911, 12:51  #4  
Jul 2004
2×7 Posts 
Quote:
57829 = 5x^4 + 7x^3 + 8x^2 + 2x + 9 I wasn't sure if the roots of such an equation were useful for factoring, etc, since the article made it seem like it was one of nature's unsolved mysteries (or something) to have a general equation for numerically solving polynomial equations. 

20040911, 14:10  #5 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Hmm, I thought that the NewtonRhapson method already provided a method for finding the roots of any polynomial numerically to arbitrarily high accuracy.
Last fiddled with by jinydu on 20040911 at 14:12 
20040913, 11:36  #6  
Nov 2003
2^{2}×5×373 Posts 
Quote:
What makes you think that there is something special here? 

20040913, 11:38  #7  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Your premise is not correct. I suggest that you apply Newton's method to (say) x^3  3x^2 + 4. 

20040913, 20:49  #8  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
289E_{16} Posts 
Quote:
For extra points, explain why it all goes badly wrong. Paul 

20040914, 02:26  #9  
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Quote:
Let f(x) = x^3  3x^2 + 4 Then f '(x) = 3x^2  6x My first guess will be x = 1 f(1) = 2 f '(1) = 3 Next guess = 1  (2/3) = 5/3 ~= 1.6666666666666666666 Second guess is x = 5/3 f(5/3) = 8/27 f '(5/3) = 5/3 Next guess = (5/3)  ([8/27]/[5/3]) = 83/45 ~= 1.8444444444444444 The next two iterations give: 21563/11205 ~= 1.92440875 1422643403/724840245 ~= 1.96269925 The method seems to be working, or did I just get lucky with my starting guess? 

20040914, 13:00  #10  
Nov 2003
2^{2}·5·373 Posts 
Quote:
gets slower and slower and slower and....... Ask yourself: what is special about the polynomial I gave? I did not choose it randomly.... 

20040914, 14:27  #11  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×3×1,733 Posts 
Quote:
That's why I put a smiley after quoting your example, and why I asked him to explain why it all goes terribly wrong. I'm well aware, of course, why you chose that particular polynomial. Paul 

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