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Math Help - Find the power series representation of this function

  1. #1
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    Find the power series representation of this function

    f(x)=\frac{3}{2x-1} , c=2

    This is what I did:

    \frac{a}{1-r} = \frac{-3}{1-2x} = -3\sum^\infty_ {n=0} (2x)^n

    Now, what do I do next since its centered at c=2??
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    Quote Originally Posted by saiyanmx89 View Post
    f(x)=\frac{3}{2x-1} , c=2

    This is what I did:

    \frac{a}{1-r} = \frac{-3}{1-2x} = -3\sum^\infty_ {n=0} (2x)^n

    Now, what do I do next since its centered at c=2??
    Hint: you need to write it in the form \sum_{n = 0}^\infty C_n (x - 2)^n

    can you figure that out? you do that manipulation before you even get to the first series.
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    -3 \sum^ \infty_ {n=0}2(x-2)^n ???
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    Quote Originally Posted by saiyanmx89 View Post
    -3 \sum^ \infty_ {n=0}2(x-2)^n ???
    all you did was replace the x in your series with (x - 2)

    i said you make the change BEFORE getting to that series.

    \frac 3{2x - 1} = \frac {-3}{1 - 2x} = \frac {-3}{4 - 3 - 2x} = \frac {-3}{-3 -2(x - 2)}

    now go on to forming your series
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    (-3)(x-2)^n ?
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    Quote Originally Posted by saiyanmx89 View Post
    (-3)(x-2)^n ?
    question: what form for the power series are you trying to use? you are trying to use the fact that \frac 1{1 - x} = \sum_{n = 0}^\infty x^n, for |x| < 1, am i right?

    now look at what we have. does it look like it is in that form? is the constant in the denominator a 1? is it 1 - (some function of x) in the denominator? no it isn't. get it into that form before you make the power series
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    Could this work:

    \frac{-3}{1(-3)-2(x-2)}

    But the -3 still needs to be taken out right??
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    Quote Originally Posted by saiyanmx89 View Post
    Could this work:

    \frac{-3}{1(-3)-2(x-2)}

    But the -3 still needs to be taken out right??
    yes, you need to factor -1/3 out of the fraction
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    I'm sorry for asking such a mundane question but, how would I take out the (1/3)?
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    Quote Originally Posted by saiyanmx89 View Post
    I'm sorry for asking such a mundane question but, how would I take out the (1/3)?
    you could start by factoring out -3 from the denominator and then you can just remove it and write -1/3 in front of the fraction. you should review factoring

    \frac {-3}{-3 -2(x - 2)} = \frac {-3}{-3 \left[ 1 + \frac {2(x - 2)}3\right]} = \frac 1{1 + \frac {2(x - 2)}3} = \frac 1{1 - \left[ - \frac {2(x - 2)}3\right]}

    now we have the required form.
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    Why is this form acceptable then for c=0??

    \frac{-3}{1-2x} , c=0

    I understand that \frac{-3}{4-3-2x} is the same thing, but is this just something you have to do when the function isn't centered at 0?
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    Quote Originally Posted by saiyanmx89 View Post
    Why is this form acceptable then for c=0??

    \frac{-3}{1-2x} , c=0

    I understand that \frac{-3}{4-3-2x} is the same thing, but is this just something you have to do when the function isn't centered at 0?
    if c = 0 it means you want your function part to look like (x - 0) which happens to just be x. thus, having that x there (possibly times some constant) is good enough. then, you have the constant in front being a "+1", so we have the form. you would NOT change it to 4 - 3 - 2x, because that is getting away from the form you want, which is \frac 1{1 - f(x)}, where f(x) is some function in x. in general, you would want in this case, \frac 1{1 - f(x - c)}
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