shubhamsrg
  • shubhamsrg
if a,b,c...k are the roots of some n degree equation, x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(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..
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hartnn
  • hartnn
simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?
anonymous
  • anonymous
The first step would be this one: x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p_n =(x-a)(x-b)(x-c)...(x-k)
shubhamsrg
  • shubhamsrg
yep i got it for 2 nd degree,,same problem,,how to generalize @hartnn 2nd step ? @sauravshakya

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
But I dont this will lead to the solution or not: p_0=1 p_1=a+b+c...k
experimentX
  • experimentX
try making use of complex conjugates.
experimentX
  • experimentX
though not sure if this will work.
anonymous
  • anonymous
(x-a)(x-b)(x-c)...(x-k) =x^n-(a+b+c+...+k)x^n-1 +...+(abc...k)
anonymous
  • anonymous
Actually its: p_1=-(a+b+c...k)
experimentX
  • experimentX
\[ (x - ia_1)(x-ia_2)(x-ia_3) .... (x-ia_n) = x^n + i(p_1)(x^{n-}1) - (p_2)(x^{n-2})-ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n\]
experimentX
  • experimentX
multiply those ... and put back x=1
shubhamsrg
  • shubhamsrg
can euler's formulla help ?
experimentX
  • experimentX
\[ (x + ia_1)(x+ia_2)(x+ia_3) .... (x+ia_n) = x^n - i(p_1)(x^{n-}1) + (p_2)(x^{n-2})+ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n \] looks like awful generalization.
experimentX
  • experimentX
looks like the general term would be \[ \huge \prod_{i=1}^{n}(x-ia_0) = \sum_{j=0}^n x^{n-j} p_j i^j\\ \huge \prod_{i=1}^{n}(x+ia_0) = \sum_{k=0}^n x^{n-k} p_k (-i)^k\] where p_0 = 1 try multiplying those two and put x=1
experimentX
  • experimentX
looks promising though looks rigorous.
experimentX
  • experimentX
on the putting x=1, you would naturally get (1+ a^2) (1 +b^2)....(1 +k^2) on LHS side.
experimentX
  • experimentX
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 \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k (i)^{j+k} (1)^j\]
shubhamsrg
  • shubhamsrg
i'd give this a try will come back to you @experimentX
experimentX
  • experimentX
yeah sure ... still looks pretty rigorous.
experimentX
  • experimentX
the other way would be \[ (x+a^2)(x+b^2) ... = x^n + (a^2+b^2 ..)x^{n-1} + ... \\ x^n + ((a+b+...)^2 - 2(ab +bc +...))x^{n-1} + ... \\ x^n + (-p_1^2 - 2p_2)x^{n-1} + ... \] this looks equally rigorous.
anonymous
  • anonymous
Its pretty simple. Write P(x) = (x-a)(x-b)......(x-k) = x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p Put=> x = i in this identity. => (i - a)(i-b).........(i-k) = (1 - p_2 + p_4 -p_6 ....)i + (p_1 - p_3 + p_5 - .........) OR it can also be: => (i - a)(i-b).........(i-k) = (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.
anonymous
  • anonymous
Is it fine guys?
experimentX
  • experimentX
yeah seem simpler that way.
anonymous
  • anonymous
im late ... lol nice :)
shubhamsrg
  • shubhamsrg
cool! B| thanks a lot @siddhantsharan

Looking for something else?

Not the answer you are looking for? Search for more explanations.