Results 1 to 15 of 15

Math Help - breaking this expressions into areas..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    breaking this expressions into areas..

    \sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}= \sum_{n=1}^{\infty}e^{inx}+ r^{0}e^{0}+ \sum_{n=-\infty}^{\infty}r^{n}e^{-inx}

    if we look at the left side
    i have been told that it was broken by intervals
    the left most is for n=1.. infinity the central is for n=0
    the right most for n=-1 ..-infinity

    i know the definition of the |x| function
    for some values its x for other its -x

    but here the power of e changes too
    why?
    Last edited by transgalactic; December 25th 2009 at 11:53 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i made a typing mistake
    there is no absolute value in the expression

    and i still cant see the logic in this separation
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post
    i made a typing mistake
    there is no absolute value in the expression

    and i still cant see the logic in this separation
    Still does not look right.

    C B
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    ok how would you break the exression
    the left most is for n=1.. infinity the central is for n=0
    the right most for n=-1 ..-infinity
    how it should look like?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post
    ok how would you break the exression
    the left most is for n=1.. infinity the central is for n=0
    the right most for n=-1 ..-infinity
    how it should look like?
    \sum_{n=-\infty}^{\infty}r^ne^{inx}=\sum_{n=-\infty}^{-1} r^ne^{inx}+r^0e^0 +\sum_{n=1}^{\infty}r^ne^{inx}

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    but the power of should be -n on the minus case
    correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post
    but the power of should be -n on the minus case
    correct?
    No, I can change it to minus, but then I will have to change a few more things.

    <br />
\sum_{n=-\infty}^{\infty}r^ne^{inx}=\sum_{n=1}^{\infty} r^{-n}e^{-inx}+r^0e^0 +\sum_{n=1}^{\infty}r^ne^{inx}<br />

    or we could have (if you want the absolute value in the exponent of r on the left):

    <br />
\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}=\sum_{n=1}^{\infty} r^{n}e^{-inx}+r^0e^0 +\sum_{n=1}^{\infty}r^ne^{inx}<br />

    Now if you could clarify what the sum on the left is supposed to be and what you are trying to do, that would help. At the moment I feel I am wasting my time as you have not ma
    de it clear what you are trying to do or why.

    CB
    Last edited by CaptainBlack; December 26th 2009 at 06:15 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    but its absolute value
    its minus for negative numbers
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,603
    Thanks
    1421
    Yes, it is- but notice that Captain Black has switched indices also.

    The sum in question is \sum_{n=-\infty}^1 r^{|n|}e^{inx}. Since n is negative, that is the same as \sum_{n= -\infty}^1 r^{-n}e^{inx}. Now let k= -n so that n= -k, when n= -1, k= 1 and when n= -\infty, k= \infty. In terms of index k the sum is \sum_{k=1}^\infty r^{k}e^{-ikx}.

    Finally, since this is a "dummy index" anyway, it doesn't matter whether you call it "k" or "n". That is exactly the same as \sum_{n=1}^\infty r^n e^{-inx}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    if you say that n=-k
    than you cant say that k=n because that gives us
    n=-n wich is true only for n=0
    not for n=-1 to - infinty
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,603
    Thanks
    1421
    Quote Originally Posted by transgalactic View Post
    if you say that n=-k
    than you cant say that k=n because that gives us
    I didn't say that k=n! I said that k= -n.

    n=-n wich is true only for n=0
    not for n=-1 to - infinty
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    you said -n=k
    <br />
       \sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}<br />
    and then you said k is just a dummy and we can call k as n
    which gives us
    <br />
        \sum_{n=1}^\infty r^n e^{-inx}<br />

    but saying that contradicts the logic of
    n=-n wich is true only for n=0
    not for n=-1 to - infinty
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    thanks i got the idea
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,603
    Thanks
    1421
    Quote Originally Posted by transgalactic View Post
    you said -n=k
    <br />
       \sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}<br />
    and then you said k is just a dummy and we can call k as n
    which gives us
    <br />
        \sum_{n=1}^\infty r^n e^{-inx}<br />

    but saying that contradicts the logic of
    n=-n wich is true only for n=0
    not for n=-1 to - infinty
    The two "n"s have nothing to do with each other.

    If n+ 3= 6, I can replace n by "-k" and get -k+ 3= 6.

    I could then replace k by n and get -n+ 3= 6. The two equations "n+3= 6" and "-n+ 3= 6" [b]would[b] contradict each other if they were the same n.

    But in sums we have, as I said "dummy" indices, if A= \sum_{n=1}^4 \FRAC{1}{n}, I can replace n by -k and have A= \sum_{k= -1}^{-4}\frac{1}{-k} and then replace k by n to get <br />
A= \sum_{n= -1}\frac{1}{-n}.

    Those all say the same thing: A= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    thanks i understand this thing now
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Breaking a code.
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 29th 2010, 06:16 PM
  2. code needs breaking
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: October 20th 2009, 02:21 PM
  3. Need help breaking a matrix code
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: March 2nd 2009, 02:49 PM
  4. Breaking a stick.
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 26th 2009, 02:40 AM
  5. Logarithms and Line Breaking
    Posted in the LaTeX Help Forum
    Replies: 0
    Last Post: February 23rd 2009, 04:36 PM

Search Tags


/mathhelpforum @mathhelpforum