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Thread: 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:

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

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

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

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

    $\displaystyle v: a_1+a_2+\cdots+a_N$

    $\displaystyle w: \frac{1}{N}\\frac$
    $\displaystyle {1}{N}$

    Am I going in the right direction? Can anyone provide me some guidance?
    Last edited by MichaelH; Oct 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:

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

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

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

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

    $\displaystyle v: a_1+a_2+\cdots+a_N$

    $\displaystyle w: \frac{1}{N}$

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

    Instead, try $\displaystyle v = \left(a_1,a_2,\ldots,a_N\right), w = \left(\underbrace{\dfrac{1}{N},\dfrac{1}{N},\ldots ,\dfrac{1}{N}}_{N\text{ times}}\right)$. Now, $\displaystyle v\cdot w$ and $\displaystyle v\cdot v$ are as you wanted, but $\displaystyle w\cdot w = \dfrac{1}{N}$, not $\displaystyle \dfrac{1}{N^2}$. Fortunately, the inequality you are given has $\displaystyle \dfrac{1}{N}$ on the RHS, not $\displaystyle \dfrac{1}{N^2}$.
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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 $\displaystyle v$ dot product $\displaystyle v$ incorrect then?
    Last edited by MichaelH; Oct 23rd 2013 at 08:37 AM.
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    Re: Derive a relation using Cauchy Schwarz inequality

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

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

    Ok so I am still a little confused with whether I have this right or not now? And if this is correct for $\displaystyle v$ and $\displaystyle w$ then where do I go with $\displaystyle v \cdot v$, $\displaystyle w \cdot w$ and $\displaystyle v \cdot w$ 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.

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

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

    $\displaystyle w\cdot w = \underbrace{\dfrac{1}{N^2} + \dfrac{1}{N^2} + \cdots + \dfrac{1}{N^2}}_{N\text{ times}} = \dfrac{N}{N^2} = \dfrac{1}{N}$
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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 $\displaystyle |v \cdot w|^2 =(v \cdot v)*(w \cdot w)$ to derive this.
    Edit: Although, that should be $\displaystyle \le$, not $\displaystyle =$.
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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 $\displaystyle \le$, not $\displaystyle =$.
    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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