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    We learnt this method in maths today, and my friend and I were wondering why it works. Can anyone explain?

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    (Original post by pineneedles)
    We learnt this method in maths today, and my friend and I were wondering why it works. Can anyone explain?

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    Well, note that you could start with \frac{A}{x-1} + \frac{Bx+C}{(2x+1)^2}; that gives you enough unknowns to uniquely specify the coefficients of the constant term and the terms in x, x^2.

    But note also that \frac{B}{2x+1}+\frac{C}{(2x+1)^2  } = \frac{Dx+E}{(2x+1)^2} for some values of D,E.

    So the two expressions are equivalent.
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    (Original post by pineneedles)
    We learnt this method in maths today, and my friend and I were wondering why it works. Can anyone explain?

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    because the LCM of
    (x-1), (2x+1) and (2x+1)2 is (x-1)(2x+1)2
    (x-1) and (2x+1)2 is still (x-1)(2x+1)2
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    (Original post by pineneedles)
    We learnt this method in maths today, and my friend and I were wondering why it works. Can anyone explain?

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    (i) Partial fractions are essentially the decomposition of rational expressions.
    (ii) when asked to express a rational expression as the sum of its partial fractions, what you are really being asked is " What fractions could have been added together in order to give you the rational expression"
    (iii) Given that the denominator of your rational expression is (x-1)(2x+1)^2 then what denominators could have been used to give rise to this denominator?
    (iV) well it could have been the sum of A/(x-1) + B/(2x-1)+C/(2x+1)^2 + D/(x-1)(2x+1)^2+E/(x-1)(2x+1)
    v) However, you do not need to bother with the last 2 parts as they can always be further reduced using partial fractions.
    (vi) you need to consider A/(x-1) + B/(2x+1)+C/(2x+1)^2

    (vii) try it with numbers to visualize ... e.g. what three non reducible fractions add to give 13/12 (hint think of the denominator as 3 x 2^2 (i.e. prime factorized)
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    (Original post by pineneedles)
    We learnt this method in maths today, and my friend and I were wondering why it works. Can anyone explain?

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Views: 91
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    The previous posts have given you some good explanations of why this is needed.

    At a practical level, you can also reassure yourself that those denominators are needed by adding together the fractions on the RHS and convincing yourself that you can get to a composite fraction with the required denominator and any numerator of the form ax + b for some constants a and b.

    Alternatively, are you happy as to why we can write

    \dfrac{5x + 1}{(x-1)(2x+1)} = \dfrac{a}{x-1} + \dfrac{b}{2x+1} (*)

    for some constants a and b?

    If you are, then your original fraction is just 1/(2x+1) times the fraction on the LHS of (*) above, and therefore it is also equal to 1/(2x+1) times the RHS of (*).

    So we now know that your original fraction can be written as

    \dfrac{a}{(x-1)(2x+1)} +\dfrac{b}{(2x+1)^2} (**)

    The 2nd term in (**) is one of the three terms that you're looking for, whilst the 1st term in (**) can be reduced to partial fractions by reapplying the rule that we are already happy with, i.e.

    \dfrac{a}{(x-1)(2x+1)} = \dfrac{k}{x-1} + \dfrac{l}{2x+1}

    for some constants k and l. So overall we end up with 3 fractions in the partial fraction expansion, which match up with the rule you've been given for decomposing your original fraction.

    Hope this helps
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    (Original post by atsruser)
    Well, note that you could start with \frac{A}{x-1} + \frac{Bx+C}{(2x+1)^2}; that gives you enough unknowns to uniquely specify the coefficients of the constant term and the terms in x, x^2.

    But note also that \frac{B}{2x+1}+\frac{C}{(2x+1)^2  } = \frac{Dx+E}{(2x+1)^2} for some values of D,E.

    So the two expressions are equivalent.
    (Original post by TeeEm)
    because the LCM of
    (x-1), (2x+1) and (2x+1)2 is (x-1)(2x+1)2
    (x-1) and (2x+1)2 is still (x-1)(2x+1)2
    (Original post by dpm)
    (i) Partial fractions are essentially the decomposition of rational expressions.
    (ii) when asked to express a rational expression as the sum of its partial fractions, what you are really being asked is " What fractions could have been added together in order to give you the rational expression"
    (iii) Given that the denominator of your rational expression is (x-1)(2x+1)^2 then what denominators could have been used to give rise to this denominator?
    (iV) well it could have been the sum of A/(x-1) + B/(2x-1)+C/(2x+1)^2 + D/(x-1)(2x+1)^2+E/(x-1)(2x+1)
    v) However, you do not need to bother with the last 2 parts as they can always be further reduced using partial fractions.
    (vi) you need to consider A/(x-1) + B/(2x+1)+C/(2x+1)^2

    (vii) try it with numbers to visualize ... e.g. what three non reducible fractions add to give 13/12 (hint think of the denominator as 3 x 2^2 (i.e. prime factorized)
    (Original post by davros)
    The previous posts have given you some good explanations of why this is needed.

    At a practical level, you can also reassure yourself that those denominators are needed by adding together the fractions on the RHS and convincing yourself that you can get to a composite fraction with the required denominator and any numerator of the form ax + b for some constants a and b.

    Alternatively, are you happy as to why we can write

    \dfrac{5x + 1}{(x-1)(2x+1)} = \dfrac{a}{x-1} + \dfrac{b}{2x+1} (*)

    for some constants a and b?

    If you are, then your original fraction is just 1/(2x+1) times the fraction on the LHS of (*) above, and therefore it is also equal to 1/(2x+1) times the RHS of (*).

    So we now know that your original fraction can be written as

    \dfrac{a}{(x-1)(2x+1)} +\dfrac{b}{(2x+1)^2} (**)

    The 2nd term in (**) is one of the three terms that you're looking for, whilst the 1st term in (**) can be reduced to partial fractions by reapplying the rule that we are already happy with, i.e.

    \dfrac{a}{(x-1)(2x+1)} = \dfrac{k}{x-1} + \dfrac{l}{2x+1}

    for some constants k and l. So overall we end up with 3 fractions in the partial fraction expansion, which match up with the rule you've been given for decomposing your original fraction.

    Hope this helps
    Thanks everyone for taking the time to explain this, it's really helped to have everyone's different take on it. I don't think I'll forget this method now I understand! (hopefully) O.O
 
 
 
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