Question

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..

Answer 1

simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?

Answer 2

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)

Answer 3

yep i got it for 2 nd degree,,same problem,,how to generalize @hartnn
2nd step ? @sauravshakya

Answer 4

But I dont this will lead to the solution or not:
p_0=1
p_1=a+b+c...k

Answer 5

try making use of complex conjugates.

Answer 6

though not sure if this will work.

Answer 7

(x-a)(x-b)(x-c)...(x-k)
=x^n-(a+b+c+...+k)x^n-1 +...+(abc...k)

Answer 8

Actually its:
p_1=-(a+b+c...k)

Answer 9

\[ (x - ia_1)(x-ia_2)(x-ia_3) .... (x-ia_n) = x^n + i(p_1)(x^{n-}1) - (p_2)(x^{n-2})-ip_3 x^{n-3} .... \\
(p_n-1)(x) + p_n\]

Answer 10

multiply those ... and put back x=1

Answer 11

can euler's formulla help ?

Answer 12

\[ (x + ia_1)(x+ia_2)(x+ia_3) .... (x+ia_n) = x^n - i(p_1)(x^{n-}1) + (p_2)(x^{n-2})+ip_3 x^{n-3} .... \\
(p_n-1)(x) + p_n \]

looks like awful generalization.

Answer 13

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

Answer 14

looks promising though looks rigorous.

Answer 15

on the putting x=1, you would naturally get (1+ a^2) (1 +b^2)....(1 +k^2) on LHS side.

Answer 16

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

\[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k (i)^{j+k} (1)^j\]

Answer 17

i'd give this a try will come back to you @experimentX

Answer 18

yeah sure ... still looks pretty rigorous.

Answer 19

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.

Answer 20

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.

Answer 21

Is it fine guys?

Answer 22

yeah seem simpler that way.

Answer 23

im late ... lol
nice :)

Answer 24

cool! B|
thanks a lot @siddhantsharan