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Math Help - Power Series Representation

  1. #1
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    Power Series Representation

    I understand fully part a and b.

    But for part c they just plug in 1 , why is that? Why didnt they plug in 2 or 3 because those are between -5 and 5?
    Then how do they calculate the sum by plugging in 1?

    Thanks
    Attached Thumbnails Attached Thumbnails Power Series Representation-123.png  
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  2. #2
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    Re: Power Series Representation

    Hey minneola24.

    I think they are doing that in b) to answer c) since 1^anything = 1 for anything.
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    Re: Power Series Representation

    But then where do they get 5/16 as the sum?
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  4. #4
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    Re: Power Series Representation

    They evaluated the derivative at x = 1 since the derivative and the power series are the same thing.
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    Re: Power Series Representation

    Ok, but I still dont see where they got the 5/16 from?

    How do you evaluate the derivative when x=1 and there is still another variable (n)
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  6. #6
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    Re: Power Series Representation

    The n is a dummy variable for the power series: the function f'(x) has no other variables other than x.
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    Re: Power Series Representation

    Quote Originally Posted by minneola24 View Post
    Ok, but I still dont see where they got the 5/16 from?

    How do you evaluate the derivative when x=1 and there is still another variable (n)
    Ok. Then where is the 5/16 coming from? If anyone is understanding that part please enlighten me!
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  8. #8
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    Re: Power Series Representation

    Quote Originally Posted by minneola24 View Post
    Ok. Then where is the 5/16 coming from? If anyone is understanding that part please enlighten me!
    You have \frac{1}{16}=\frac{1}{5}Y so Y=\frac{5}{16}.
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  9. #9
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    Re: Power Series Representation

    I'm confused where the 1/16 is coming tho!
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  10. #10
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    Re: Power Series Representation

    Do the arithmetic! You have already shown that the derivative of \frac{1}{5- x} is \frac{1}{(5- x)^2} and, that by differentiating the power series for \frac{1}{5- x}, \frac{1}{(5- x)^2}= \frac{1}{5}\sum \frac{n}{5^{n+1}}x^{n-1}.

    Setting x= 1 in both of those you get \frac{1}{4^2}= \frac{1}{5}\sum \frac{n}{5^{n+1}}.

    Multiply both sides of that equation by 5.
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