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Math Help - series..?

  1. #1
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    series..?

    I know these are very very basic series(if they are.. ) but I have no idea of what they actually are, I found them in my way and need to understand why the result is that.

    If I have the ? [from 1 to 6] of (1/(1.3)^ (s/12)) how do I solve it and why?

    Thanks for all the help.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by 0123 View Post
    I know these are very very basic series(if they are.. ) but I have no idea of what they actually are, I found them in my way and need to understand why the result is that.

    If I have the ? [from 1 to 6] of (1/(1.3)^ (s/12)) how do I solve it and why?

    Thanks for all the help.
    I'm guessing this is the problem you're trying to solve:

    \sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}

    (I replaced "s" with "k" to make it easier to see.)

    If I'm correct, then here's how you would solve it:

    First, let S_6=\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}

    [1] S_6=\left(\frac{1}{1.3}\right)^\frac{1}{12}+\left(  \frac{1}{1.3}\right)^\frac{2}{12}+\left(\frac{1}{1  .3}\right)^\frac{3}{12}+\left(\frac{1}{1.3}\right)  ^\frac{4}{12}+\left(\frac{1}{1.3}\right)^\frac{5}{  12}+\left(\frac{1}{1.3}\right)^\frac{6}{12}

    Now multiply S_6 by a negative \left(\frac{1}{1.3}\right)^\frac{1}{12}. This will add 1/12 to the exponent of every term on the right side of the equation, to get:

    [2] -\left(\frac{1}{1.3}\right)^\frac{1}{12}S_6=-\left(\frac{1}{1.3}\right)^\frac{2}{12}-\left(\frac{1}{1.3}\right)^\frac{3}{12}-\left(\frac{1}{1.3}\right)^\frac{4}{12}-\left(\frac{1}{1.3}\right)^\frac{5}{12}-\left(\frac{1}{1.3}\right)^\frac{6}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    Now we can add both sides of equations [1] and [2] to get, which elliminates everything but the 1st and 7th terms:

    S_6-\left(\frac{1}{1.3}\right)^\frac{1}{12}S_6=\left(\  frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    \left(1-\left(\frac{1}{1.3}\right)^\frac{1}{12}\right)S_6=  \left(\frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    S_6=\frac{1}{1-\left(\frac{1}{1.3}\right)^\frac{1}{12}}\left(\lef  t(\frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)\app  rox 5.56186547919

    So instead of calculating what all of the terms equalled, you need only to find what these equal to find the total sum. Hope that helps.

    ----------------------------

    Thank you CaptainBlack for checking my work.
    Last edited by ecMathGeek; May 26th 2007 at 11:29 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by 0123 View Post
    I know these are very very basic series(if they are.. ) but I have no idea of what they actually are, I found them in my way and need to understand why the result is that.

    If I have the ? [from 1 to 6] of (1/(1.3)^ (s/12)) how do I solve it and why?

    Thanks for all the help.
    This is a finite geometric series:

    <br />
\sum_{s=1}^6 \frac{1}{1.3^{s/12}} = \sum_{s=1}^6 (1/1.3)^{s/12}=\sum_{s=1}^6 ((1/1.3)^{1/12})^{s}<br />

    Now you could look up the formula for the sum of a finite geometric series or
    derive it form the observation that:

    <br />
(1-b) \left[ \sum_{s=0}^n b^s \right] = 1 - b^{n+1}<br />

    since if you multiply out the brackets on the left all but the first and last
    terms in the result cancel out.

    So we have:

    <br />
\left[ \sum_{s=0}^n b^s \right] = \frac{1 - b^{n+1}}{1-b}<br />

    or:

    <br />
\left[ \sum_{s=1}^n b^s \right] +1 = \frac{1 - b^{n+1}}{1-b}<br />

    <br />
\left[ \sum_{s=1}^n b^s \right] = \frac{1 - b^{n+1}}{1-b} -1<br />

    for any b and n, and so in particular for our
    series, where n=6, and b=(1/1.3)^{1/12}.

    Hence:

    <br />
\sum_{s=1}^6 \frac{1}{1.3^{s/12}} = \sum_{s=1}^6 (1/1.3)^{s/12}=\sum_{s=1}^6 ((1/1.3)^{1/12})^{s} = \frac{1 - ((1/1.3)^{1/12})^{7}}{1-(1/1.3)^{1/12}}-1 <br />
= \frac{1 - 1/1.3^{7/12}}{1-1/1.3^{1/12}}-1 \approx 5.56187

    Reality check:

    Code:
    >
    >.. bute force sum of series:
    >
    >k=1:6;
    >S=(1/1.3)^(k/12);
    >SS=sum(S)
          5.56187 
    >
    >
    >.. CaptainBlack's formula for sum of series:
    >
    >S6 = (1-1/1.3^(7/12)) / (1-1/1.3^(1/12))-1
          5.56187 
    >
    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    S_6=\frac{13}{3}\left(\left(\frac{1}{1.3}\right)^\  frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)
    Reality check:

    Code:
    >S6 = (13/3)*((1/1.3)^(1/12)-(1/1.3)^(7/12))
         0.521227
    which can't be right as every term in the sum is positive, and the first one
    is bigger than this.

    So there must be a mistake in your derivation of the result (or in my
    evaluation of your formula ).

    Checking through your working (as if done correctly it should give the
    correct answer) I can't see where it has gone wrong.

    RonL
    Last edited by CaptainBlack; May 26th 2007 at 10:07 PM.
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I'm guessing this is the problem you're trying to solve:

    \sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}

    (I replaced "s" with "k" to make it easier to see.)

    If I'm correct, then here's how you would solve it:

    First, let S_6=\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}

    [1] S_6=\left(\frac{1}{1.3}\right)^\frac{1}{12}+\left(  \frac{1}{1.3}\right)^\frac{2}{12}+\left(\frac{1}{1  .3}\right)^\frac{3}{12}+\left(\frac{1}{1.3}\right)  ^\frac{4}{12}+\left(\frac{1}{1.3}\right)^\frac{5}{  12}+\left(\frac{1}{1.3}\right)^\frac{6}{12}

    Now multiply S_6 by a negative \frac{1}{1.3}. This will add 1 to the exponent of every term on the right side of the equation, to get: (This is where I made my mistake.)

    [2] -\frac{1}{1.3}S_6=-\left(\frac{1}{1.3}\right)^\frac{2}{12}-\left(\frac{1}{1.3}\right)^\frac{3}{12}-\left(\frac{1}{1.3}\right)^\frac{4}{12}-\left(\frac{1}{1.3}\right)^\frac{5}{12}-\left(\frac{1}{1.3}\right)^\frac{6}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    Now we can add both sides of equations [1] and [2] to get, which elliminates everything but the 1st and 7th terms:

    S_6-\frac{1}{1.3}S_6=\left(\frac{1}{1.3}\right)^\frac{  1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    \frac{3}{13}S_6=\left(\frac{1}{1.3}\right)^\frac{1  }{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}

    S_6=\frac{13}{3}\left(\left(\frac{1}{1.3}\right)^\  frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)

    So instead of calculating what all of the terms equalled, you need only to find what these equal to find the total sum. Hope that helps.
    I found my mistake. I'll go back and correct it.
    Last edited by ecMathGeek; May 26th 2007 at 11:31 PM.
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