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Math Help - Partial fractions

  1. #1
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    Partial fractions

    I'm stuck on b) iii) of this question. I'll show you what i have got so far.

    the function f is given by f(x)= \frac{9}{(1+2x)(4 - x)}

    a) express f(x) as a partial fraction.

    I got \frac{2}{(1+2x)} + \frac{1}{(4 -x)}

    b) i) Find the first three terms in the expansion of \frac{1}{4 -x}

    I got \frac{1}{4} + \frac{x}{16} + \frac{x^2}{64}

    b) ii) Obtain a similar expansion for \frac{1}{1+2x}

    I got 1 - 2x + 4x^2

    where i am stuck on is b) iii)

    Hence, or otherwise, obtain the first three terms in the expansion of f(x) in ascending powers of x
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  2. #2
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    Hello Beard
    Quote Originally Posted by Beard View Post
    I'm stuck on b) iii) of this question. I'll show you what i have got so far.

    the function f is given by f(x)= \frac{9}{(1+2x)(4 - x)}

    a) express f(x) as a partial fraction.

    I got \frac{2}{(1+2x)} + \frac{1}{(4 -x)}

    b) i) Find the first three terms in the expansion of \frac{1}{4 -x}

    I got \frac{1}{4} + \frac{x}{16} + \frac{x^2}{64}

    b) ii) Obtain a similar expansion for \frac{1}{1+2x}

    I got 1 - 2x + 4x^2

    where i am stuck on is b) iii)

    Hence, or otherwise, obtain the first three terms in the expansion of f(x) in ascending powers of x
    But you've done all the hard work! - and you've done it right so far. You're going to kick yourself, I think, because all you need to do is to use your Partial Fractions answer:

    f(x) = 2(1+2x)^{-1} + (4-x)^{-1}

    together with your answers to b(i) and b(ii) to write the answer down. Can you see it now?

    Grandad
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  3. #3
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    I don't feel too bad considering I haven't done this thing for ages but it was a nice little reminder. Thanks.
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