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| Science Forum Index » Mathematics Forum » Explaining Alain Verghote's Identity Generator... |
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| Achava Nakhash, the Loving Snake... |
 Posted: Sun Nov 01, 2009 5:08 pm |
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I have been unable to completely follow Alain's explanations, but I
used them freely to come up with a method that is probably completely
equivalent, so I thought I would share it with the group.
His opening salvo was:
4(a^3 + b^3) = (a+b)^3 + 3((a+b)(a-b)^2
which he explained as equating event parts of a power of g(s), where g
(s) is some function, and calculating it by applying the
decomposition after taking the power or the decomposition of g(s)
before taking the power. The functioin he suggested was either as +
b, or bs + a, I don't remember, but it really doesn't matter. I did
notice that there were a lot of (a+b) and (a-b) terms which are simply
f(1) anad f(-1) which would occur in the decompostion into even and
odd parts.
Here is what I figured out that is probably the same thing. Notice
that
2a = (a+b) - (a-b), 2b = (a+b) + (a-b)
8(a^3 + b^3) = (a+b)^3 - 3(a+b)^2(a-b) + 3(a+b)(a-b)^2 - (a-b)^3
+ (a+b)^3 + 3(a+b)^2(a-b) + 3(a+b)(a-b)^2 + (a-b)^3
= 2(a+b)^3 + 6(a+b)(a-b)^2
from which his identity follows after dividing both sides by 2.
This idea can be generalized to probably an arbitrary number of
variables by changing the signs of
a_1 + a_2 + ... + a_n, taking n choices of signs and doing a little
linear algebra to isolate each varaible. I suspect that one will be
able to generate all of identities in this fashion, because suspect
that this is essentially what he is doing. Am I right, Alain?
Regards,
Achava |
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| alainverghote at (no spam) gmail.com... |
| Posted: Mon Nov 02, 2009 12:34 am |
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| alainverghote at (no spam) gmail.com... |
| Posted: Mon Nov 02, 2009 1:26 am |
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[quote]Dear Achava
[/quote]
Thanks for your nice answer.
Sure, there many a way to generate identies.
The decomposition you figured out:
2a =(a+b)-(a-b) ,2b=(a+b)+(a-b) works
like the even/odd splitting
of a polynomial:g(s)=(g(s)+g(-s))/2+(g(s)-g(-s))/2
g(s)=(a+b) , g(-s)=a-b
I do think the palying with signs or using simple
algebra can't drive us very far.
The method I follow is like a waltz
three times:1) n roots of unity
2) additive splitting of a polynomial
3) links between parts of a power and power of parts,
Example:a three additive splitting
1) x^3 = 1 roots 1, j,j^2 , j =(-1+I*sqrt(3))/2 ,1+j+j^2=0
2) x^(3n)terms {p(x)+p(jx)+p(j^2.x)]/3
x^(3n+2)terms {p(x)+j.p(jx)+j^2.p(j^2.x)]/3
x^(3n+1)terms {p(x)+j^2.p(jx)+j.p(j^2.x)]/3
3)comparison of x^(3n+2)terms inside p(x)^2
{p(x)^2+j.p(jx)^2+j^2.p(j^2.x)^2]/3
and directly square of x^(3n+1)terms plus
product of x^(3n)and x^(3n+1)terms ,
With p(x)=a,p(jx)=b ,p(j^2.x)=c
You've got
3(a^2+j^2.b^2+jc^2)
=(a+j^2.b+j.c)^2+2*(a+b+c)*(a+jb+j^2.b)
What about building a 16 square terms identity?
Amicalement,
Alain |
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