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shubhamsrg
Group Title
if a,b,c...k are the roots of some n degree equation,
x^n + (p_1)(x^n1) + (p_2)(x^n2)....(p_n1)(x) + p_n = 0
p_i would mean " i " is in the subscript ..
then how to prove that :
(1+ a^2) (1 +b^2)....(1 +k^2) = (1  p_2 + p_4 p_6 ....)^2 + (p_1  p_3 + p_5  .........)^2
hope i could write it clearly..
 one year ago
 one year ago
shubhamsrg Group Title
if a,b,c...k are the roots of some n degree equation, x^n + (p_1)(x^n1) + (p_2)(x^n2)....(p_n1)(x) + p_n = 0 p_i would mean " i " is in the subscript .. then how to prove that : (1+ a^2) (1 +b^2)....(1 +k^2) = (1  p_2 + p_4 p_6 ....)^2 + (p_1  p_3 + p_5  .........)^2 hope i could write it clearly..
 one year ago
 one year ago

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hartnn Group TitleBest ResponseYou've already chosen the best response.0
simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
The first step would be this one: x^n + (p_1)(x^n1) + (p_2)(x^n2)....(p_n1)(x) + p_n =(xa)(xb)(xc)...(xk)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yep i got it for 2 nd degree,,same problem,,how to generalize @hartnn 2nd step ? @sauravshakya
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
But I dont this will lead to the solution or not: p_0=1 p_1=a+b+c...k
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
try making use of complex conjugates.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
though not sure if this will work.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
(xa)(xb)(xc)...(xk) =x^n(a+b+c+...+k)x^n1 +...+(abc...k)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Actually its: p_1=(a+b+c...k)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ (x  ia_1)(xia_2)(xia_3) .... (xia_n) = x^n + i(p_1)(x^{n}1)  (p_2)(x^{n2})ip_3 x^{n3} .... \\ (p_n1)(x) + p_n\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
multiply those ... and put back x=1
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
can euler's formulla help ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ (x + ia_1)(x+ia_2)(x+ia_3) .... (x+ia_n) = x^n  i(p_1)(x^{n}1) + (p_2)(x^{n2})+ip_3 x^{n3} .... \\ (p_n1)(x) + p_n \] looks like awful generalization.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
looks like the general term would be \[ \huge \prod_{i=1}^{n}(xia_0) = \sum_{j=0}^n x^{nj} p_j i^j\\ \huge \prod_{i=1}^{n}(x+ia_0) = \sum_{k=0}^n x^{nk} p_k (i)^k\] where p_0 = 1 try multiplying those two and put x=1
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
looks promising though looks rigorous.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
on the putting x=1, you would naturally get (1+ a^2) (1 +b^2)....(1 +k^2) on LHS side.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
on the right side you would be \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (1)^k (i)^{j+k} \] the imaginary terms would cancel out because it only happens if j+k is odd ... so either j is odd or k is odd. ... if that's the case then the other variable will be even and we have \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (1)^k (i)^{j+k} (1)^j\]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i'd give this a try will come back to you @experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yeah sure ... still looks pretty rigorous.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the other way would be \[ (x+a^2)(x+b^2) ... = x^n + (a^2+b^2 ..)x^{n1} + ... \\ x^n + ((a+b+...)^2  2(ab +bc +...))x^{n1} + ... \\ x^n + (p_1^2  2p_2)x^{n1} + ... \] this looks equally rigorous.
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.4
Its pretty simple. Write P(x) = (xa)(xb)......(xk) = x^n + (p_1)(x^n1) + (p_2)(x^n2)....(p_n1)(x) + p Put=> x = i in this identity. => (i  a)(ib).........(ik) = (1  p_2 + p_4 p_6 ....)i + (p_1  p_3 + p_5  .........) OR it can also be: => (i  a)(ib).........(ik) = (1  p_2 + p_4 p_6 ....) + (p_1  p_3 + p_5  .........)i In either case take modulus both sides to prove the required.
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.4
Is it fine guys?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yeah seem simpler that way.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
im late ... lol nice :)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
cool! B thanks a lot @siddhantsharan
 one year ago
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