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ganeshie8
 one year ago
ganeshie8
 one year ago

This Question is Closed

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.0i saw tat question but i couldt help ;/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4there is really a very simple and cute solution, give it a try again :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2inb4 Astrophysics tags me

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2These are perfect squares. The only way we have it is that each one is zero.

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.0I couldnt find a way but now that i see the correct answer it clears up the fog in my mind, good job @ParthKohli :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4please finish it off @ParthKohli i think @praxer is still looking for a solution

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Oh, sorry. OS is acting up for me again. I type out things and it removes them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this suddenly look like ultimate troll question after you explain it lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Haha if you don't like the simplicity, there is a complicated solution, which is quite enlightening too :) Familiar with CauchySchwarz inequality ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2IT REMOVED IT AGAIN! Yes, I actually solved the same question on my test.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no im not at that stage of mathematics yet

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2The OP isn't actually too far from directly solving it. You can just expand and complete the square.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4what do you mean by directly expand and complete the square ? i thought we will have to use cauchyschwarz inequality

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah same, can you show it parth

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh I think you mean what op was already doing

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ohkiee, I'll finish off your solution using geometry : dw:1438963298665:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] then discriminant is given by : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^24\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4thats what OP had too, so we're on same page :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh right, but whats the justification for \(\le 0\) ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well I think since \[\sum_{i=1}^{n} (a_ix+b_i)^2 = 0\] is only way we can get real roots, so each term in the sum will have to be 0, and I don't really know the exact proof of it goes, but if we want what you have above \[\left(2\sum_{i=1}^{n}a_ib_i\right)^24\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right)\] this would need to be <= 0, for the roots of \[x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] exist

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That looks good! since the quadratic \(\sum_{i=1}^{n} (a_ix+b_i)^2 \) is nonnegative, its determinant must be \(\le 0\) : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^24\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \le 0\] We just proved CauchySchwarz inequality!

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.0Why is this still up and viewing xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4one of the cool things in math in general is TMTOWTDI ;p
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