Author
Message
arp
Poincare Conjecture
Offline Joined: 31 Dec 2004 Posts: 173
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Zeros of a polynomial
How do you find the zeros of a polynomial of even degree above 4 whose graph lies above the x-axis?
Posted: Fri Feb 18, 2005 9:13 pm
EFuzzy
Riemann Hypothesis
Offline Joined: 31 May 2004 Posts: 389 Location: Maryland, USA
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
There's no fool proof method for finding the (exact) complex zeroes, only some tricks for special cases.
_________________
Go KSEG! Free Geometry!
http://www.mit.edu/~ibaran/kseg.html
Posted: Sat Feb 19, 2005 5:47 am
arp
Poincare Conjecture
Offline Joined: 31 Dec 2004 Posts: 173
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
hm could you give me an example of some of these tricks?
Posted: Sat Feb 19, 2005 7:33 am
p_adic
New Member
Offline Joined: 08 Dec 2003 Posts: 10 Location: Swarthmore, PA
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Polynomials
Examples:
P(x) = x^6 + x^4 + x^2 + 1 is always positive, for real x to find complex zeros note that
P(x) = (x^8 -1)/(x^2 - 1) and hence P(x) is zero when x^8 - 1 = 0 and x^2 - 1 is nonzero.
Q(x) = x^8 + 2x^4 + 1 factors as (x^4 + 1)^2 and so the zeros are the same as those of x^4 + 1 = 0.
R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:
x^4 + 5(x + 1/x) + 1/x^4 = 0
Let u = x + 1/x
Then u^2 = x^2 + 1/x^2 + 2,
(u^2 - 2)^2 = x^4 + 1/x^4 + 2
So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is
(u^2 -2)^2 - 2 + 5u = 0
which can be solved for u by the quartic formula and then for u.
Posted: Sat Feb 19, 2005 6:17 pm
goodyfresh741
Hodge Conjecture
Offline Joined: 02 Oct 2004 Posts: 55 Location: Maryland
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Unsolvability of quintic
Don't go around looking for a solution (by algebraic methods) to the generalized quintic equation. In the 1800's (I'm not sure exactly when) Abel proved that it's impossible to find an algebraic solution to a general equation of degree >=5. In other words, choose any equation of degree >=5, and (with almost 100 % likelihood) it is impossible through a finite number of algebraic operations (addition and subtraction of polynomials from both sides, plus root-taking operations and such) to solve the equation. That can only be done in special cases. If you're interested in reading more about it, there is a "general solution" to the quintic equation using something called Jacobian theta functions (you can also use Weirstrassian theta functions for that purpose) but it's quite high-level math and won't do anything for you on a high school math contest.
_________________ There is an infinite set A that is not too big-- John Von Neumann
Posted: Mon Feb 21, 2005 12:49 pm
jli
Navier-Stokes Equations
Offline Joined: 17 Nov 2004 Posts: 2341 Location: california
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
find all the factors of the coeficient of the one with the largest power
find the factors of the number at the end
divide them in all the possible ways and put a plus/minus in fron of the fractions and those are the possible roots
Posted: Tue Feb 22, 2005 1:30 pm
goodyfresh741
Hodge Conjecture
Offline Joined: 02 Oct 2004 Posts: 55 Location: Maryland
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Um, that's not completely true
jli wrote:
find all the factors of the coeficient of the one with the largest power
find the factors of the number at the end
divide them in all the possible ways and put a plus/minus in fron of the fractions and those are the possible roots
That's partially correct, but the roots you get as a result are all the possible RATIONAL roots, and some or all of the numbers you get that way may not be roots at all. The process helps not one bit in finding the irrational roots. However, on math contests, the so-called "rational zero theorem" helps ALOT, because many times, the roots of polynomials in math contests problems are integers or fractions.
_________________ There is an infinite set A that is not too big-- John Von Neumann
Posted: Tue Feb 22, 2005 8:43 pm
Scrambled
Navier-Stokes Equations
Offline Joined: 06 Jan 2004 Posts: 1006 Location: Svalbard, Norway
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Re: Polynomials
p_adic wrote:
R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:
x^4 + 5(x + 1/x) + 1/x^4 = 0
Let u = x + 1/x
Then u^2 = x^2 + 1/x^2 + 2,
(u^2 - 2)^2 = x^4 + 1/x^4 + 2
So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is
(u^2 -2)^2 - 2 + 5u = 0
which can be solved for u by the quartic formula and then for u.
that equation is symmetric, (having the same terms on either side), and a symmetric equations can be solved this way in generla
and i dont think you need that quartic?
_________________ Lynbrook High!
SKY HIGH!
Posted: Wed Feb 23, 2005 11:02 am
ZetaX
Birch & Swinnerton Dyer
Offline Joined: 21 Dec 2004 Posts: 6141 Location: München
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
goodyfresh741: can you please give me a link or something like that to a solution for quintic equations using Thetafunctions¿
Posted: Wed Feb 23, 2005 11:33 am
bénabar
Hodge Conjecture
Offline Joined: 10 Feb 2005 Posts: 65
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Re: Unsolvability of quintic
goodyfresh741 wrote:
In the 1800's (I'm not sure exactly when) Abel proved that it's impossible to find an algebraic solution to a general equation of degree >=5.
Isn't it Galois? Of course, It is not very important...
Posted: Wed Feb 23, 2005 11:36 am
bénabar
Hodge Conjecture
Offline Joined: 10 Feb 2005 Posts: 65
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
...but Galois was french .
Posted: Wed Feb 23, 2005 11:54 am
ZetaX
Birch & Swinnerton Dyer
Offline Joined: 21 Dec 2004 Posts: 6141 Location: München
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
I think Abel proofed it for rational polynomials, Galois in a more general case for fields. (But's possible that I'm wrong)
Posted: Wed Feb 23, 2005 12:09 pm
goodyfresh741
Hodge Conjecture
Offline Joined: 02 Oct 2004 Posts: 55 Location: Maryland
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
Abel and Galois
Yes, Abel proved it first, but Galois proved it for a more general class of polynomials over fields.
_________________ There is an infinite set A that is not too big-- John Von Neumann
Posted: Fri Feb 25, 2005 2:59 pm
AntonioMainenti
Riemann Hypothesis
Offline Joined: 12 Sep 2004 Posts: 433
Not_yet_rated
Poor (Spam)
Poor (Spam)
Below average
Below average
Average
Average
Good
Good
Very good
Very good
Excellent
To rate posts you must be logged in
This guy (http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ruffini.html) proved that the quintic is unsolvable by radicals around 1800 and sent it to Lagrange and others but no one seemed to care or believe him. I know about him because he was a northern Italian whose last name was Ruffini. My father is a northern Italian whose last name is Ruffini We could be related..
_________________ "i already made like infinity of those at scout camp" - napoleon d.
Posted: Fri Feb 25, 2005 7:02 pm
Display posts from previous: All Posts 1 Day 7 Days 2 Weeks 1 Month 3 Months 6 Months 1 Year Sort by: Post Time Post Subject Author Ascending Descending