Results 1 to 5 of 5

Math Help - Help with proof of Arithmetic mean

  1. #1
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6

    Help with proof of Arithmetic mean

    x = \frac{1}{n} \sum^{n}_{i = 1} x_{i}

    x = \frac{1}{n} \sum^{n}_{i = 1} x_{i} - a + a

    x = \frac{1}{n} \left( \sum^{n}_{i = 1} x_{i} - na \right) + a <- What the heck happened with that na thing? The transformation from the previous step, to this one, and the next step?

    x = \frac{1}{n} \left[ \sum^{n}_{i = 1} x_{i} - a \right] + a

    x = \frac{ \sum^{n}_{i = 1} (x_{i} - a) }{n} + a

    By the way on the x on the left hand side there should be a little bar on top, indicating average.

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    Is this what you are missing?
    \begin{gathered}<br />
  \sum\limits_{k = 1}^n {\left( {x_k  - a} \right)}  = \left( {\sum\limits_{k = 1}^n {x_k } } \right) - na \hfill \\<br />
  \frac{1}<br />
{n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1}<br />
{n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - a \hfill \\ <br />
\end{gathered} <br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Plato View Post
    Is this what you are missing?
    \begin{gathered}<br />
  \sum\limits_{k = 1}^n {\left( {x_k  - a} \right)}  = \left( {\sum\limits_{k = 1}^n {x_k } } \right) - na \hfill \\<br />
  \frac{1}<br />
{n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1}<br />
{n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - a \hfill \\ <br />
\end{gathered} <br />
    But didn't that n of na get factored out somehow?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    Is this what you are missing?
    \frac{1}<br />
{n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1}<br />
{n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - \frac{1}<br />
{n}\left( {na} \right)<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Plato View Post
    Is this what you are missing?
    \frac{1}<br />
{n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1}<br />
{n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - \frac{1}<br />
{n}\left( {na} \right)<br />
    There we go, that works for me. Thanks Plato.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof Using Fundamental Thm of Arithmetic
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 10th 2011, 03:52 PM
  2. Arithmetic proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 5th 2010, 08:31 PM
  3. Arithmetic proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 1st 2010, 06:04 PM
  4. Proof with arithmetic functions
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: November 24th 2009, 05:58 AM
  5. modulo arithmetic proof
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: February 17th 2008, 12:33 PM

Search Tags


/mathhelpforum @mathhelpforum