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shubhamsrg
 3 years ago
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..
shubhamsrg
 3 years ago
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..

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1The 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)

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0try making use of complex conjugates.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0though not sure if this will work.

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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1Actually its: p_1=(a+b+c...k)

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0multiply those ... and put back x=1

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0can euler's formulla help ?

experimentX
 3 years ago
Best 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0looks 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0looks promising though looks rigorous.

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0on 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 \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (1)^k (i)^{j+k} (1)^j\]

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0i'd give this a try will come back to you @experimentX

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0yeah sure ... still looks pretty rigorous.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the 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.

siddhantsharan
 3 years ago
Best ResponseYou've already chosen the best response.4Its 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.

siddhantsharan
 3 years ago
Best ResponseYou've already chosen the best response.4Is it fine guys?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0yeah seem simpler that way.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0im late ... lol nice :)

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0cool! B thanks a lot @siddhantsharan
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