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Math Help - Series help

  1. #1
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    Series help

    I have this formula:
    {c_n}^2 = \frac{1}{2}{c_{n-1}}^2

    How do I convert this to a series with \sum

    From n = 1 to n = 10
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  2. #2
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    Quote Originally Posted by chancey
    I have this formula:
    {c_n}^2 = \frac{1}{2}{c_{n-1}}^2

    How do I convert this to a series with \sum

    From n = 1 to n = 10
    Not really sure what you are asking.
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  3. #3
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    Hello, Malay!

    Edit: You're right! . . . *blush*
    . . . . I'll correct it now.


    I have this formula: (c_n)^2 \:= \:\frac{1}{2}(c_{n-1})^2

    How do I convert this to a series with \sum^{10}_{n=1}

    We have: . c_n\;=\;\frac{1}{\sqrt{2}}c_{n-1}

    We have a geometric series with first term c_1 = a and common ratio r = \frac{1}{\sqrt{2}}

    The k^{th} term is: . c_n\:=\:\frac{a}{2^{\frac{n-1}{2}}}

    Then: . \sum^{10}_{k=1}\frac{a}{2^{\frac{n-1}{2}}} \;=\;a \cdot \frac{1 - \frac{1}{(\sqrt{2})^{10}}} {1 - \frac{1}{\sqrt{2}}}

    Last edited by Soroban; June 21st 2006 at 12:11 PM.
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, chancey!

    I had to baby-talk my way through it . . .


    [size=3]
    Let c_1 = a, the first term.

    Then crank out the first few terms:

    c_1\;=\;a

    c_2\;=\;\frac{a^2}{2}

    But someone check my work . . . please!
    Hello Soroban!
    It should be
    (c_2)^2\;=\;\frac{a^2}{2}
    Keep Smiling

    Malay
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by chancey
    I have this formula:
    {c_n}^2 = \frac{1}{2}{c_{n-1}}^2

    How do I convert this to a series with \sum

    From n = 1 to n = 10
    {c_n}^2 = \frac{1}{2}{c_{n-1}}^2
    I observed that
    \frac{c_n}{c_{n-1}}=\frac{1}{\sqrt{2}}
    Hence the given sequence is a G.P. with common ratio \frac{1}{\sqrt{2}}

    Keep Smiling
    Malay
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  6. #6
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    I need the series to do this:

    Example (not the actual formula, changed it so the numbers were easier)
    c_1 = 1

    c_2 = \frac{1}{2} 1 = 0.5

    c_3 = \frac{1}{2} 0.5 = 0.25

    c_4 = \frac{1}{2} 0.25 = 0.125

    c_n = ?

    Where each number in the series has to be calculated with the previous number. My question is, can that be put into a series so that i can run a formula once to find n_{10} or do I have to do it 10 times
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  7. #7
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    As explained in the previous discussions this sequence is geometric, thus, if,
    c_1=a
    then,
    c_n=(1/\sqrt{2})^{n-1}a
    or,
    c_n=(-1/\sqrt{2})^{n-1}a
    Where a is the initial term.
    Note, it is not possible to find a, it can be anything number.
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  8. #8
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    Hello, chancey!

    I need the series to do this:

    Example (not the actual formula, changed it so the numbers were easier)

    c_1 = 1

    c_2 = \frac{1}{2}\cdot 1 = \frac{1}{2} . Don't use decimals!

    c_3 = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}

    c_4 = \frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}

    c_n = \:?

    Where each number in the series has to be calculated with the previous number.
    My question is, can that be put into a series so that i can run a formula once to find n_{10}
    or do I have to do it 10 times
    You're expected to be familiar with Arithmetic Series and Geometric Series by now
    . . and to be able to "eyeball" a sequence and determine its general form.

    This sequence is: . 1,\;\frac{1}{2},\;\frac{1}{4},\;\frac{1}{8},\; \frac{1}{16},\;\hdots

    How long does it take for you to see that the denominators are doubled each time?
    With a little thought, we see that the n^{th} denominator is: 2^{n-1}
    So the general term is: . c_n\:= \:\frac{1}{2^{n-1}}
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  9. #9
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    Quote Originally Posted by Soroban
    Hello, chancey!


    You're expected to be familiar with Arithmetic Series and Geometric Series by now
    . . and to be able to "eyeball" a sequence and determine its general form.

    This sequence is: . 1,\;\frac{1}{2},\;\frac{1}{4},\;\frac{1}{8},\; \frac{1}{16},\;\hdots

    How long does it take for you to see that the denominators are doubled each time?
    With a little thought, we see that the n^{th} denominator is: 2^{n-1}
    So the general term is: . c_n\:= \:\frac{1}{2^{n-1}}
    Of course I can see that, that was just an example, im not really that interested in the answer, but the process is what I want to know.

    So:
    {c_n}^2 = \frac{1}{2} {c_{n-1}}^2 - 2

    {c_n} = \sqrt{\frac{1}{2} {c_{n-1}}^2 - 2}

    {c_n} = (\sqrt{\frac{1}{2} a^2 - 2} )^{n-1}
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