Friday, February 08, 2002
Credibility check: Alternate approach to polynomial
Yesterday I talked about a way to analyze the roots of the polynomial
W^3 -3vW/(v^3+1) - 2/(v^3+1) = 0,
to show that two of them have the "operator" of 1/sqrt(v+1), while one does not.
I realized that many of you find it easier to dismiss a long post with new ideas than to check to see if it actually works, so here's a short post with an alternate approach using more traditional methods.
I post it deliberately to show you why many of the people posting on the newsgroup should no longer be considered credible by you. In fact, I think the people still making negative posts are doing so to preserve their perception of a social order, without regard for the truth. Or maybe they believe their social order IS the truth, as if titles and academic positions determine that.
Ok, I assume you know that you can solve the polynomial given, and you also should know that solution has no validity at v=-1 because you have divide by zero errors all over the place.
However, you can consider the roots as v approaches -1.
In that case, two of your roots start going to infinity, while one approaches 2/3.
That's easy to see by multiplying through by v^3+1 to get
(v^3+1)W^3 - 3vW - 2 = 0, where in the limit as v approaches -1,
the polynomial approaches 3W - 2 = 0.
Oh, and using this approach it's also possible to consider factors of sqrt(v+1).
For instance, in the limit as v approaches -1, sqrt((sqrt(v)-i))/sqrt(v+i) equals 1/sqrt(sqrt(v)+i) = 1/sqrt(2i), which is clearly nowhere to be seen in the result 3W - 2 = 0.
So, it's clear that the current system, which posters have been heatedly claiming can't differentiate between the two roots that go to infinity and the one that goes to 2/3 without something indirect like this, is flawed.
That is, while I can say, given 1/3 + 2/3 - 1 = 0, that two of the numbers are fractions while one is not, you've seen a lot of posts—attacking me and trying to use this to cast doubt on my proof of Fermat's Last Theorem—claiming that there's no way to distinguish the roots of the polynomial I've given in the same way.
When I talked of factors of the roots of 1/sqrt(v+1), with one root not having 1/sqrt(v+1), I was attacked.
When I tried to say some of the roots of the polynomial have a denominator of sqrt(v+1) while one does not, I was attacked.
So, yes, you can continue to doubt your own judgement against that of the mathematical establishment, and you can continue to believe that just because mathematicians say so, there's no way to talk of two of the roots being different from the third with regard to 1/sqrt(v+1), but that will say something about you as a person, and nothing about the truth.
I know that most of you have been trained all of your lives that there are certain experts who cannot be challenged. While I was trained to be a scientist while at Vanderbilt University for four years earning my bachelors degree in Physics, which means I was trained to challenge the experts.
Are there any of you who will go by the truth versus your need to believe in the infallibility of a group of human beings?
What? Do you think that they are just too "smart" to be wrong?
W^3 -3vW/(v^3+1) - 2/(v^3+1) = 0,
to show that two of them have the "operator" of 1/sqrt(v+1), while one does not.
I realized that many of you find it easier to dismiss a long post with new ideas than to check to see if it actually works, so here's a short post with an alternate approach using more traditional methods.
I post it deliberately to show you why many of the people posting on the newsgroup should no longer be considered credible by you. In fact, I think the people still making negative posts are doing so to preserve their perception of a social order, without regard for the truth. Or maybe they believe their social order IS the truth, as if titles and academic positions determine that.
Ok, I assume you know that you can solve the polynomial given, and you also should know that solution has no validity at v=-1 because you have divide by zero errors all over the place.
However, you can consider the roots as v approaches -1.
In that case, two of your roots start going to infinity, while one approaches 2/3.
That's easy to see by multiplying through by v^3+1 to get
(v^3+1)W^3 - 3vW - 2 = 0, where in the limit as v approaches -1,
the polynomial approaches 3W - 2 = 0.
Oh, and using this approach it's also possible to consider factors of sqrt(v+1).
For instance, in the limit as v approaches -1, sqrt((sqrt(v)-i))/sqrt(v+i) equals 1/sqrt(sqrt(v)+i) = 1/sqrt(2i), which is clearly nowhere to be seen in the result 3W - 2 = 0.
So, it's clear that the current system, which posters have been heatedly claiming can't differentiate between the two roots that go to infinity and the one that goes to 2/3 without something indirect like this, is flawed.
That is, while I can say, given 1/3 + 2/3 - 1 = 0, that two of the numbers are fractions while one is not, you've seen a lot of posts—attacking me and trying to use this to cast doubt on my proof of Fermat's Last Theorem—claiming that there's no way to distinguish the roots of the polynomial I've given in the same way.
When I talked of factors of the roots of 1/sqrt(v+1), with one root not having 1/sqrt(v+1), I was attacked.
When I tried to say some of the roots of the polynomial have a denominator of sqrt(v+1) while one does not, I was attacked.
So, yes, you can continue to doubt your own judgement against that of the mathematical establishment, and you can continue to believe that just because mathematicians say so, there's no way to talk of two of the roots being different from the third with regard to 1/sqrt(v+1), but that will say something about you as a person, and nothing about the truth.
I know that most of you have been trained all of your lives that there are certain experts who cannot be challenged. While I was trained to be a scientist while at Vanderbilt University for four years earning my bachelors degree in Physics, which means I was trained to challenge the experts.
Are there any of you who will go by the truth versus your need to believe in the infallibility of a group of human beings?
What? Do you think that they are just too "smart" to be wrong?
# posted by James Harris @ 11:48 
