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Partial Fractions

  1. Apr 22, 2012 #1
    In partial fractions, why

    [itex]\frac{3x+5}{(1-2x)^2}[/itex] = [itex]\frac{A}{(1-2x)^2}[/itex] + [itex]\frac{B}{(1-2x)}[/itex]

    and not
    [itex]\frac{3x+5}{(1-2x)^2}[/itex] = [itex]\frac{A}{(1-2x)}[/itex] + [itex]\frac{B}{(1-2x)}[/itex]

    Why exists the exponent on the denominator in the right hand side of the equation?
     
  2. jcsd
  3. Apr 22, 2012 #2


    Well, [itex]\frac{A}{(1-2x)}[/itex] + [itex]\frac{B}{(1-2x)}=\frac{A+B}{1+2x}=\frac{C}{1+2x}[/itex] , with C a consant, and this clearly cannot equal the original

    expression since then we don't have a square in the denominator...

    DonAntonio
     
  4. Apr 22, 2012 #3
    1 - So, why [itex]\frac{B}{1-2x}[/itex] doesn't have an exponent of 2 in the denominator ([itex]\frac{B}{(1-2x)^2}[/itex])?

    2 - I also have a basic question related to your reply. Why
    [itex]\frac{A}{(1-2x)} + \frac{B}{(1-2x)} = \frac{A+B}{(1+2x)}[/itex] and not [itex]\frac{A+B}{(1-2x)}[/itex]
     
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4

    That was only a mistake of mine: I wrote + instead of - .

    DonAntonio
     
  6. Apr 22, 2012 #5
    I'm trying to solve the equation, but I get different values:

    [itex]\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A(1-2x)}{(1-2x)^3} + \frac{B(1-2x)^2}{(1-2x)^3} = \frac{(1-2x)(A+B(1-2x))}{(1-2x)^3} = \frac{A+B(1-2x)}{(1-2x)^2}[/itex]

    from this point forward I can't put this fraction to be like [itex]\frac{A+B}{(1-2x)}[/itex], I think...
     
  7. Apr 22, 2012 #6


    Of course you can't. What makes you think you could? The original expression is a rational function

    with denominator's degree equal to 2, and in the last line above you wrote a rational fraction with denominator's

    degree equal to 1...of course they can't be equal.

    DonAntonio
     
  8. Apr 22, 2012 #7
    So, how can both expressions be equal?

    [itex]\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A+B}{(1-2x)}[/itex]?

    And, how do I know that
    [itex]\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{3x+5}{(1-2x)^2}[/itex] ?
     
    Last edited: Apr 22, 2012
  9. Apr 22, 2012 #8


    You can use the procedure to decompose in partial fractions: you have to know that it is possible to write

    [itex]\displaystyle{\frac{3x+5}{(1-2x)^2} =\frac{A}{1-2x}+\frac{B}{(1-2x)^2}\,,\,\,A\,,\,B\,\,}[/itex] constants. Now multiply through by the common denominator:

    [itex]\displaystyle{ 3x+5=A(1-2x)+B\,\,}[/itex] . This is a polynomial identity so you can compare particular values of x in both sides and/or

    coefficients of powers of x to find out the values of the constants A,B. For example:

    The coefficient in both sides of [itex]\,\,\displaystyle{x: 3=-2A\Longrightarrow\,A=-\frac{3}{2}}[/itex] , and now the free coefficient

    in both sides: [itex]\displaystyle{5=A+B\Longrightarrow B=5-A=5-\left(-\frac{3}{2}\right)=\frac{13}{2}\,\,}[/itex] , so

    [itex]\displaystyle{\frac{3x+5}{(1-2x)^2}=-\frac{3}{2(1-2x)}+\frac{13}{2(1-2x)^2}}[/itex] , which you can easily check...

    DonAntonio
     
  10. Apr 22, 2012 #9

    HallsofIvy

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    Have you ever actually taken an algebra course? If so you should be able to do the sum on the left and see that it does, in fact, give the right side.
     
  11. Apr 22, 2012 #10
    As DonAntonio pointed out, the reason it's not the second one is because the second one doesn't work. The next section on Wikipedia describes the general decomposition:

    http://en.wikipedia.org/wiki/Partial_fraction#Over_the_reals

    In particular, if you work it out you should be able to see that a repeated factor in the denominator results in the higher powers being seen in the decomposition.
     
  12. Apr 23, 2012 #11
    Thanks, now I got it.
     
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