Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

hartnnBest 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

sauravshakyaBest 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

shubhamsrgBest 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

sauravshakyaBest 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

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

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

sauravshakyaBest 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

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

experimentXBest 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

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

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

experimentXBest 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

experimentXBest 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

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

experimentXBest 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

experimentXBest 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

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

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

experimentXBest 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

siddhantsharanBest 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

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

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

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

shubhamsrgBest ResponseYou've already chosen the best response.0
cool! B thanks a lot @siddhantsharan
 one year ago
See more questions >>>
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.