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Math Help - Infinite series

  1. #1
    sps
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    Infinite series

    Think about we have a infinite series
    summation (j from 0 to infinity) (0.5)^2j=4/3

    How did we get this answer?Please help me.
    thanks.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by sps View Post
    Think about we have a infinite series
    summation (j from 0 to infinity) (0.5)^2j=4/3

    How did we get this answer?Please help me.
    thanks.
    (0.5)^{2j}=[(0.5)^2]^j=\left(\frac 14\right)^j

    And this is a geometric series
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  3. #3
    sps
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    Thanks Moo
    I understood.
    I have one more question:

    summation (j from 0 to infinity) j*(0.5)^2j=100/225
    summation (j from 0 to infinity) j^2*(0.5)^2j=100/135

    How did we get those answers? Could you please help me again?
    thanks in advance.
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  4. #4
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    I can't get line breaks to parse properly on this forum...

    There are several ways to do both questions. Here's just an example

    View the first one as a geometric series of a geometric series

    the second is a little tricker...
    Let's try to find a general form for  x + 4x^2 + 9x^3 + \ldots assuming |x| is less than 1

    there are probably much easier ways to do this
    Attached Thumbnails Attached Thumbnails Infinite series-sum-1.gif   Infinite series-sum-3.gif  
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by sps View Post
    Think about we have a infinite series
    summation (j from 0 to infinity) (0.5)^2j=4/3

    How did we get this answer?Please help me.
    thanks.
    Geometric series are of the form ar^n, but we have the form ar^{2n}, so all we need to do is change the base from r \rightarrow r^2

    Let r = 0.5^2 or \frac{1}{4}

    We now have the sum, \sum_0^{\infty} ar^j, where a=1,r=0.25

    The sum of a geometric series such as that is \frac{a}{1-r}. Therefore, the sum is \frac{1}{1-0.25} = \frac{1}{0.75} \Rightarrow \frac{1}{\frac{3}{4}} = \frac{4}{3}.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jjzshen View Post
    I can't get line breaks to parse properly on this forum...

    There are several ways to do both questions. Here's just an example

    View the first one as a geometric series of a geometric series

    the second is a little tricker...
    Let's try to find a general form for  x + 4x^2 + 9x^3 + \ldots assuming |x| is less than 1

    there are probably much easier ways to do this
    There is a much easier way

    Notice that we have a summation of the form

    \sum_{n=0}^{\infty}nx^n

    Where x=\frac{1}{4}

    Well

    \sum_{n=0}^{\infty}nx^n=x\cdot\sum_{n=1}^{\infty}n  x^{n-1}=x\cdot\frac{d}{dx}\bigg[\sum_{n=0}^{\infty}x^n\bigg]

    Now we know that

    \sum_{n=0}^{\infty}x^n=\frac{1}{1-x},\forall{x}\in[-1,1]

    So we have

    x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\frac{x}{(x-1)^2}

    So now we can see that

    \sum_{n=0}^{\infty}n\left(\frac{1}{4}\right)^n=\fr  ac{\frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}=\frac{4}{9}

    The second is the same, except you must consider the second derivative
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