A community for students.
Here's the question you clicked on:
 0 viewing
 2 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..
 2 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..

This Question is Closed

hartnn
 2 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
 2 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
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0try making use of complex conjugates.

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

sauravshakya
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1Actually its: p_1=(a+b+c...k)

experimentX
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0multiply those ... and put back x=1

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

experimentX
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0looks promising though looks rigorous.

experimentX
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0i'd give this a try will come back to you @experimentX

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

experimentX
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.4Is it fine guys?

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

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

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0cool! B thanks a lot @siddhantsharan
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.