ganeshie8
  • ganeshie8
http://assets.openstudy.com/updates/attachments/55c4a0c1e4b0c7f4a978d9c7-praxer-1438949589754-screenshot_58.png
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ali2x2
  • ali2x2
i saw tat question but i couldt help ;/
ali2x2
  • ali2x2
thats cheap asto
ganeshie8
  • ganeshie8
there is really a very simple and cute solution, give it a try again :)

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ali2x2
  • ali2x2
ok :)
ParthKohli
  • ParthKohli
inb4 Astrophysics tags me
ParthKohli
  • ParthKohli
These are perfect squares. The only way we have it is that each one is zero.
ganeshie8
  • ganeshie8
Thats it!
ali2x2
  • ali2x2
I couldnt find a way but now that i see the correct answer it clears up the fog in my mind, good job @ParthKohli :)
ganeshie8
  • ganeshie8
please finish it off @ParthKohli i think @praxer is still looking for a solution
ParthKohli
  • ParthKohli
Oh, sorry. OS is acting up for me again. I type out things and it removes them.
anonymous
  • anonymous
this suddenly look like ultimate troll question after you explain it lol
ganeshie8
  • ganeshie8
Haha if you don't like the simplicity, there is a complicated solution, which is quite enlightening too :) Familiar with Cauchy-Schwarz inequality ?
ParthKohli
  • ParthKohli
IT REMOVED IT AGAIN! Yes, I actually solved the same question on my test.
anonymous
  • anonymous
no im not at that stage of mathematics yet
Astrophysics
  • Astrophysics
\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\]
ParthKohli
  • ParthKohli
The OP isn't actually too far from directly solving it. You can just expand and complete the square.
ganeshie8
  • ganeshie8
what do you mean by directly expand and complete the square ? i thought we will have to use cauchy-schwarz inequality
Astrophysics
  • Astrophysics
Yeah same, can you show it parth
Astrophysics
  • Astrophysics
Oh I think you mean what op was already doing
ganeshie8
  • ganeshie8
Ohkiee, I'll finish off your solution using geometry : |dw:1438963298665:dw|
ganeshie8
  • ganeshie8
\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] then discriminant is given by : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \]
ganeshie8
  • ganeshie8
thats what OP had too, so we're on same page :)
Astrophysics
  • Astrophysics
perfect
Astrophysics
  • Astrophysics
\[\le 0\]
ganeshie8
  • ganeshie8
Ahh right, but whats the justification for \(\le 0\) ?
Astrophysics
  • Astrophysics
Well 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 \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right)\] 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
  • ganeshie8
That looks good! since the quadratic \(\sum_{i=1}^{n} (a_ix+b_i)^2 \) is nonnegative, its determinant must be \(\le 0\) : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \le 0\] We just proved Cauchy-Schwarz inequality!
Astrophysics
  • Astrophysics
Haha, xD nice!
ali2x2
  • ali2x2
Why is this still up and viewing xD
ganeshie8
  • ganeshie8
one of the cool things in math in general is TMTOWTDI ;p

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