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Thread: Geometric progression?

  1. #1
    Super Member fardeen_gen's Avatar
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    Geometric progression?

    If $\displaystyle a_{i}\in\mathbb{R}, i = 1,2,3,...N$and all $\displaystyle a_{i}\ 's$ are distinct such that:
    $\displaystyle \left(\sum_{i = 1}^{n - 1} a_{i}^2\right)x^2 + 2\left(\sum_{i = 1}^{n - 1} a_{i}a_{i + 1}\right)x + \sum_{i = 2}^{n} a_{i}^2\leq 0$
    ,then show that $\displaystyle a_{1},a_{2},a_{3}...$ are in G.P.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If $\displaystyle a_{i}\in\mathbb{R}, i = 1,2,3,...N$and all $\displaystyle a_{i}\ 's$ are distinct such that:
    $\displaystyle \left(\sum_{i = 1}^{n - 1} a_{i}^2\right)x^2 + 2\left(\sum_{i = 1}^{n - 1} a_{i}a_{i + 1}\right)x + \sum_{i = 2}^{n} a_{i}^2\leq 0$
    ,then show that $\displaystyle a_{1},a_{2},a_{3}...$ are in G.P.
    Hi fardeen_gen.

    The discriminant is

    $\displaystyle 4\left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)^2-4\left(\sum_{i\,=\,1}^{n-1}a_i^2\right)\left(\sum_{i\,=\,2}^na_i^2\right)$

    which is never positive by the Cauchy–Schwarz inequality. Hence, since the coefficient of $\displaystyle x^2$ is never negative, we have

    $\displaystyle \left(\sum_{i\,=\,1}^{n-1}a_i^2\right)x^2+2\left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)x+\left(\sum_{i\,=\,2}^{n-1}a_i^2\right)\ \ge\ 0$

    Combining this with the given condition, we find that the quadratic expression is equal to zero. Hence the discriminant is zero, so we have equality for Cauchy–Schwarz:

    $\displaystyle \left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)^2\ =\ \left(\sum_{i\,=\,1}^{n-1}a_i^2\right)\left(\sum_{i\,=\,2}^na_i^2\right)$

    This means that the vectors $\displaystyle \left(a_1,a_2,\ldots,a_{n-1}\right)$ and $\displaystyle \left(a_2,a_3,\ldots,a_n\right)$ are linearly dependent, i.e. $\displaystyle \exists\,r$ such that $\displaystyle r\left(a_1,a_2,\ldots,a_{n-1}\right)=\left(a_2,a_3,\ldots,a_n\right).$ Hence the $\displaystyle a_i$’s are in geometric progression.
    Last edited by TheAbstractionist; May 2nd 2009 at 05:48 AM.
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