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Math Help - Annoying Partial Fractions...

  1. #1
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    Annoying Partial Fractions...

    Hi guys,

    Hopefully somebody can help me out

    Basically, I've been plodding through a load of these questions and have managed to get to grips with it...except for one scenario...

    x
    (x-2)^2

    I would REALLY appriciate any help!! The sooner the better

    Sean

    (PS: Virtual Mars bar in it )
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sean Woodcraft View Post
    Hi guys,

    Hopefully somebody can help me out

    Basically, I've been plodding through a load of these questions and have managed to get to grips with it...except for one scenario...

    x
    (x-2)^2

    I would REALLY appriciate any help!! The sooner the better

    Sean

    (PS: Virtual Mars bar in it )
    Remember, for partial fractions when given a ratonal functon of the form f(x)/x^2 you put it in partial fractions in the form A/x + B/x^2, similarly

    ......x..........=.......A......+........B
    ----------.......---------.....--------
    (x - 2)^2.........(x - 2)........(x - 2)^2

    now we multiply through by (x - 2)^2. we obtain:

    x = A(x - 2) + B
    => x = Ax + (B - 2A)
    now we equate the coefficients of like powers of x
    => 1 = A
    => 0 = B - 2A
    => 0 = B - 2
    => B = 2

    so

    ......x..........=.......1......+........2
    ----------.......---------.....--------
    (x - 2)^2.........(x - 2)........(x - 2)^2
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Remember, for partial fractions when given a ratonal functon of the form f(x)/x^2 you put it in partial fractions in the form A/x + B/x^2, similarly

    ......x..........=.......A......+........B
    ----------.......---------.....--------
    (x - 2)^2.........(x - 2)........(x - 2)^2

    now we multiply through by (x - 2)^2. we obtain:

    x = A(x - 2) + B
    => x = Ax + (B - 2A)
    now we equate the coefficients of like powers of x
    => 1 = A
    => 0 = B - 2A
    => 0 = B - 2
    => B = 2

    so

    ......x..........=.......1......+........2
    ----------.......---------.....--------
    (x - 2)^2.........(x - 2)........(x - 2)^2
    Hmm...looks soo simple when you put it that way

    Thanks

    (As promised:
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sean Woodcraft View Post
    Hmm...looks soo simple when you put it that way

    Thanks

    (As promised:
    Wow, thanks, I've never had a mars bar before. Just thought i'd let you know about another protocol for partial fractions (just in case you dont know).

    The thing we did for the (x - 1)^2 in the last problem expands as we have higher powers. that is,

    if we had f(x)/(x - 1)^3
    then f(x)/(x - 1)^3 = A/(x - 1) + B/(x - 1)^2 + C/(x - 1)^3

    if we had f(x)/(x - 1)^4
    then f(x)/(x - 1) = A/(x - 1) + B/(x - 1)^2 + C/(x - 1)^3 + D/(x - 1)^4

    so in general, we expand as much as is necessary:

    f(x)/g(x)^n = A/g(x) + B/g(x)^2 + C/g(x)^3 + ... + D/g(x)^n

    ..ok, take a deep breath and wait for that to sink in...
    **deep breath** ...and exhale, ok. now to more complicated things.

    if you have a function, f(x)/g(x) and g(x) is a function that does not foil into anything useful, then f(x)/g(x) = [Ax + B]/g(x). If g(x) is raised to a power, then we also have to apply what we did above. Confused? Let's do an example (i won't complete the problem, you can if you want to):

    (2x^2 + 5)/(x^2 + 1)^2

    ok, we notice that the function under the bottom is squared, so we are going to have to split it into two functions, one of which is the regular function, the other one is a square. but we also notice we have an x^2 on the inside, we ascertain further that x^2 + 1 does not foil into anything nice like (x + a)(x + b) or so. so we have to compensate for that by following the preceeding rule.

    (2x^2 + 5)/(x^2 + 1)^2 = [Ax + B]/(x^2 + 1) + [Cx + D]/(x^2 + 1)^2

    yeah, its messed up i know, but those are the rules. and they make sense if you think about it. if you're interested i can tell you why, but maybe you're not.

    and like our previous rule, this rule expands with the power that we are given, except here, we compensate in the numerator. let's see what would happen if we had (x^3 + 1)^2, then:

    (2x^2 + 5)/(x^3 + 1)^2 = [Ax^2 + Bx + C]/(x^3 + 1) + [Dx^2 + Ex + F]/(x^3 + 1)^2

    so in general:

    f(x)/(x^n + c)^m = [Ax^(n-1) +...+B]/(x^n + c) + [Cx^(n-1) + ... +D]/(x^n + c)^2 + .......... + [Ex^(n-1) + ... + F]/(x^n + c)^m

    so don't trouble your brain with this stuff unless you come across problems that look like this. Problems like the first example (see below) i did are actually common though


    (2x^2 + 5)/(x^2 + 1)^2
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  5. #5
    Junior Member frenzy's Avatar
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    just for fun...

    x/(x-2)^2


    let u=x-2

    then x=u+2

    x/(x-2)^2=(u+2)/u^2

    =u/u^2+2/u^2

    =1/u+2/u^2

    now go back to x

    1/(x-2)+2/(x-2)^2
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by frenzy View Post
    just for fun...

    x/(x-2)^2


    let u=x-2

    then x=u+2

    x/(x-2)^2=(u+2)/u^2

    =u/u^2+2/u^2

    =1/u+2/u^2

    now go back to x

    1/(x-2)+2/(x-2)^2
    that's nice! i've never seen it done that way before. thank you for that
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