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Math Help - Derive a relation using Cauchy Schwarz inequality

  1. #1
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    Derive a relation using Cauchy Schwarz inequality

    So I am trying to derive the following relation using Cauchy Schwarz inequality for any collection of N real numbers:

    \left(\frac{a_1+a_2+\cdots+a_N}{N}\right)^2 =< \frac{a^2_1+a^2_2+\cdots+a^2_N}{N}.

    This says that the square of the average is less or equal than the average of the squares.

    So I want to use |v \cdot w|^2 =(v \cdot v)*(w \cdot w) to derive this.

    So v \cdot w needs to be \frac{a_1+a_2+\cdots+a_N}{N} therefore, I have v \cdot v as a^2_1+a^2_2+\cdots+a^2_N and w \cdot w as \frac{1}{N^2} providing that:

    v: a_1+a_2+\cdots+a_N

    w: \frac{1}{N}\\frac
    {1}{N}" alt="{1}{N}" />

    Am I going in the right direction? Can anyone provide me some guidance?
    Last edited by MichaelH; October 23rd 2013 at 08:44 AM.
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  2. #2
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    Re: Derive a relation using Cauchy Schwarz inequality

    Quote Originally Posted by MichaelH View Post
    So I am trying to derive the following relation using Cauchy Schwarz inequality for any collection of N real numbers:

    \left(\frac{a_1+a_2+\cdots+a_N}{N}\right)^2 =< \frac{a^2_1+a^2_2+\cdots+a^2_N}{N}.

    This says that the square of the average is less or equal than the average of the squares.

    So I want to use |v \cdot w|^2 =(v \cdot v)*(w \cdot w) to derive this.

    So v \cdot w needs to be \frac{a_1+a_2+\cdots+a_N}{N} therefore, I have v \cdot v as a^2_1+a^2_2+\cdots+a^2_N and w \cdot w as \frac{1}{N^2} providing that:

    v: a_1+a_2+\cdots+a_N

    w: \frac{1}{N}

    Am I going in the right direction? Can anyone provide me some guidance?
    No. Currently, you have v,w \in \mathbb{R}^1, so v\cdot v = (a_1+a_2+\cdots+a_N)^2 \neq a_1^2+a_2^2+\cdots + a_N^2

    Instead, try v = \left(a_1,a_2,\ldots,a_N\right), w = \left(\underbrace{\dfrac{1}{N},\dfrac{1}{N},\ldots  ,\dfrac{1}{N}}_{N\text{ times}}\right). Now, v\cdot w and v\cdot v are as you wanted, but w\cdot w = \dfrac{1}{N}, not \dfrac{1}{N^2}. Fortunately, the inequality you are given has \dfrac{1}{N} on the RHS, not \dfrac{1}{N^2}.
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  3. #3
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    Re: Derive a relation using Cauchy Schwarz inequality

    I do apologise, it was a typo on my end and I forgot to add that I was using 1/N n amount of times.

    But was my calculation of v dot product v incorrect then?
    Last edited by MichaelH; October 23rd 2013 at 08:37 AM.
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    Re: Derive a relation using Cauchy Schwarz inequality

    If you choose v = a_1+a_2+\cdots + a_N and w = \dfrac{1}{N}, then those are real numbers. So, yes, your calculation of v\cdot v is incorrect, but your calculation of w\cdot w was correct. You would get \dfrac{1}{N^2}. You cannot take it N times because you chose w to be a real number, not a number repeated N times.
    Last edited by SlipEternal; October 23rd 2013 at 02:51 PM.
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    Re: Derive a relation using Cauchy Schwarz inequality

    { v : \left(a_1,a_2,\ldots,a_N\right) | v \in R^N} and { w : \left(\underbrace{\dfrac{1}{N},\dfrac{1}{N},\ldots  ,\dfrac{1}{N}}_{N\text{ times}}\right) | w \in R^N}

    Ok so I am still a little confused with whether I have this right or not now? And if this is correct for v and w then where do I go with v \cdot v, w \cdot w and v \cdot w in order to completely derive the inequality?
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  6. #6
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    Re: Derive a relation using Cauchy Schwarz inequality

    Yes, you have it right now.

    v\cdot w = \dfrac{a_1}{N} + \dfrac{a_2}{N} + \cdots + \dfrac{a_N}{N} = \dfrac{a_1+a_2+\cdots + a_N}{N}

    v\cdot v = a_1^2+a_2^2 + \cdots + a_N^2

    w\cdot w = \underbrace{\dfrac{1}{N^2} + \dfrac{1}{N^2} + \cdots + \dfrac{1}{N^2}}_{N\text{ times}} = \dfrac{N}{N^2} = \dfrac{1}{N}
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    Re: Derive a relation using Cauchy Schwarz inequality

    Great. So now how do I show that the RHS is greater than or equal to the LHS using these facts?
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  8. #8
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    Re: Derive a relation using Cauchy Schwarz inequality

    To quote you:

    Quote Originally Posted by MichaelH View Post
    So I want to use |v \cdot w|^2 =(v \cdot v)*(w \cdot w) to derive this.
    Edit: Although, that should be \le, not =.
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  9. #9
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    Re: Derive a relation using Cauchy Schwarz inequality

    Quote Originally Posted by SlipEternal View Post
    To quote you:



    Edit: Although, that should be \le, not =.
    Ahh that's why I couldn't figure it out! Haha, my fault, should have double checked it. I was staring at = thinking... what am I supposed to do! Cheers mate.
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