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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Mar 2008 15:09:51 IST
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Find all  , polynomials in two variables with real or complex coefficients, that satisfy this condition:  .
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8 Mar 2008 15:19:18 IST
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all polynomials of the form
F(x-y)...
as the difference remains constant
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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8 Mar 2008 15:35:08 IST
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I think I can wait for more answers. can you make your answer more general?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
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8 Mar 2008 15:41:29 IST
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i mean all polynomials reducible to the form:
P(x,y)=a(x-y)n+b(x-y)n-1+c(x-y)n-2+....+z(x-y)+q
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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8 Mar 2008 16:27:50 IST
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excellent akhil! I have a nice proof in mind, but the way you "logically" reasoned out the answer is simply awesome! The question is still open, can anyone prove it?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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9 Mar 2008 15:33:42 IST
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Can we have ur proof Abhijith?
we all have a feeling its gonna be brilliant...or jst give us the general direction..
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
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10 Mar 2008 21:26:29 IST
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Ok here goes.. Otherwise, suppose  be any term of P(x,y), where b is a constant. Then:  
Now because P(x,y) is a sum of such terms we'll have this:
where  are polynomials in y with constant coefficients. Therefore we must have: Because this equation is true for every x and y, we must have: so if  then we must have: Conversely, such polynomials obviously satisfy Q.E.D.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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10 Mar 2008 21:39:11 IST
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a0,a1....an and b0 till bn rep coeff
P(x,y)= (a0y^0 + a1y^1+ a2y^2.....any^n)+(b0x^0 + b1x^1+ b2x^2.....bnx^n)
p(x+1,y+1)={a0(y+1)^0......an(y+1)^n}+{b0(x+1)^0......bn(x+1)^n}
p(x,y)=p(x+1,y+1)
so subtract the two
v get:
a1+ a2((y+1)^2-y^2)....an((y+1)^n-y^n)+ b1+ b2((x+1)^2-x^2)....bn((x+1)^n-x^n)
this shud also be valid for P(x,x)=P(x+1,x+1)
as it is valid for all x and y
so puttin x=y..
we get:
a1+ a2((x+1)^2-x^2)....an((x+1)^n-x^n)+ b1+ b2((x+1)^2-x^2)....bn((x+1)^n-x^n)
this is same as:
(a1+b1) +(a2+b2)((x+1)^2-x^2)...(an+bn)((x+1)^n-x^n)
the variable terms r all positive.. as (x+1)^n always > than x^n..
hence the numeric coeff must all reduce to 0 or a1=-b1...an=-bn u may say it is possibe tht a1 + b1 may cancel out wid all other terms but tht is not possibe as foe diff values of x those terms keep changin but a1+b1 is a constant..
so the soln is
P(x,y)=(a0y^0 + a1y^1+ a2y^2.....any^n)+(b0x^0 + b1x^1+ b2x^2.....bnx^n)=a1(y-x) + a2(y^2-x^2)...an(y^n-x^n)
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10 Mar 2008 22:11:55 IST
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Almost everything seems right, but clear this: You have taken  Why not  ?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
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10 Mar 2008 22:27:03 IST
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yeah il think abt tht..it din strike me initially..
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Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
, then 
and for every 
which is not possible unless
. 
.
, for every y and 
.So
