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Thread: Series help

  1. #1
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    Series help

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

    How do I convert this to a series with $\displaystyle \sum$

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

    How do I convert this to a series with $\displaystyle \sum$

    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: $\displaystyle (c_n)^2 \:= \:\frac{1}{2}(c_{n-1})^2$

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

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

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

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

    Then: .$\displaystyle \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; Jun 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 $\displaystyle c_1 = a$, the first term.

    Then crank out the first few terms:

    $\displaystyle c_1\;=\;a$

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

    But someone check my work . . . please!
    Hello Soroban!
    It should be
    $\displaystyle (c_2)^2\;=\;\frac{a^2}{2}$
    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:
    $\displaystyle {c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

    How do I convert this to a series with $\displaystyle \sum$

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

    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)
    $\displaystyle c_1 = 1$

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

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

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

    $\displaystyle c_n = ?$

    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 $\displaystyle n_{10}$ 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,
    $\displaystyle c_1=a$
    then,
    $\displaystyle c_n=(1/\sqrt{2})^{n-1}a$
    or,
    $\displaystyle c_n=(-1/\sqrt{2})^{n-1}a$
    Where $\displaystyle a$ is the initial term.
    Note, it is not possible to find $\displaystyle a$, 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)

    $\displaystyle c_1 = 1$

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

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

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

    $\displaystyle c_n = \:?$

    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 $\displaystyle n_{10}$
    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: .$\displaystyle 1,\;\frac{1}{2},\;\frac{1}{4},\;\frac{1}{8},\; \frac{1}{16},\;\hdots$

    How long does it take for you to see that the denominators are doubled each time?
    With a little thought, we see that the $\displaystyle n^{th}$ denominator is: $\displaystyle 2^{n-1}$
    So the general term is: .$\displaystyle 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: .$\displaystyle 1,\;\frac{1}{2},\;\frac{1}{4},\;\frac{1}{8},\; \frac{1}{16},\;\hdots$

    How long does it take for you to see that the denominators are doubled each time?
    With a little thought, we see that the $\displaystyle n^{th}$ denominator is: $\displaystyle 2^{n-1}$
    So the general term is: .$\displaystyle 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:
    $\displaystyle {c_n}^2 = \frac{1}{2} {c_{n-1}}^2 - 2$

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

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