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The Philosophy of Repeat Linear Fractions?

  1. Jun 11, 2010 #1
    Hello, I'm wondering what the reason for repeat linear factors in partial fractions is?

    I can't find an explanation online, they all just say do it!*

    I kind of understand why

    [tex]\frac{A}{x + 2} + \frac{B}{(x + 2)^2}[/tex]

    can turn into ;

    [tex] \frac{6x + 7}{(x + 2)^2}[/tex]

    Is there any reasonably easy way to rigorously understand this, nearly everything in calc I've found has a really easy way of understanding it, this has to join the group! :cool:

    because of the common factor thing, but that sketchy notion isn't enough anymore.





    *more or less :tongue2:
     
  2. jcsd
  3. Jun 12, 2010 #2

    Mark44

    Staff: Mentor

    I'm not sure that there's any deep explanation, other than this is what works. If there is some sort of explanation like that, I don't remember seeing it.

    A naive approach would be something like this.
    [tex]\frac{A}{x + 2} + \frac{B}{x + 2} = \frac{6x + 7}{(x + 2)^2}[/tex]

    This seems to be a reasonable approach, as it mimics what we would do if the factors in the denominator on the right side weren't repeated.

    Multiplying both sides of the equation above by (x + 2)2, we get
    A(x + 2) + B(x + 2) = 6x + 7
    ==> (A + B)x + (2A + 2B) = 6x + 7

    The equations above have to be identically true (i.e., true for all values of x), with the exception of x = -2 in the first equation. The only way the equations can be identically true is for the coefficients of x to be equal and the constant terms to be equal.

    Equating the coefficients of x and the constant term gives
    A + B = 6 and 2A + 2B = 7

    A little thought shows that if A + B = 6, then 2A + 2B = 2(A + B) must equal 12, not 7. We conclude that there are no constants that make the equation an identity, so this decomposition is not valid.

    Since that didn't work, someone must have come up with the bright idea of seeing if constants A and B could be found so that
    [tex]\frac{A}{x + 2} + \frac{B}{(x + 2)^2} = \frac{6x + 7}{(x + 2)^2}[/tex]
    is an identity.
     
  4. Jun 12, 2010 #3
    Thanks, I understand the procedure when you get a simple fraction like this, what about those ones that have huge powers & many factors?

    When you have a fraction like this;

    [tex] \frac{A}{x + 2} + \frac{B}{x + 2} + \frac{C}{x + 1} = \frac{6x + 7}{(x + 2)^2(x + 1)}[/tex]

    \frac{A}{x + 2} + \frac{B}{x + 2} + \frac{C}{x + 1} = \frac{6x + 7}{(x + 2)^2(x + 1)}

    (if it doesn't work)

    see the change, If that's correct? :redface:

    The problem is, I can rederive the method when it's a simple equation like a quadratic in the denominator, but if you've got some crazy equation then there are all sorts of random things you have to put into the equation like repeat linear factors, an Ax + B in some term for some reason, etc...

    It just makes no sense unless you memorize it, which I refuse to do :tongue2:

    There has to be an easy & logical way to rederive everything from scratch on the back of an envelope after you've put away your calculator @ that dinner party :cool:
     
  5. Jun 13, 2010 #4
    Alright, properly now;

    [tex] \frac{P(x)}{(x - 1)(x + 2)^3} \ = \ \frac{A}{x - 1} \ + \ \frac{B}{x + 2} \ + \ \frac{C}{(x + 2)^2} \ + \ \frac{D}{(x + 2)^3} [/tex]

    I can understand this in a sense, but;

    [tex] \frac{P(x)}{(ax^2 + bx + c)^3(x - 1)(x + 2)^3} \ = \frac{A_1x + B_1}{ax^2 + bx + c}\ \ + \ \frac{A_2x + B_2}{(ax^2 + bx + c)^2}\ \ + \ \frac{A_3x + B_3}{(ax^2 + bx + c)^3}\ \ + \ \frac{C}{x - 1} \ + \ \frac{D}{x + 2} \ + \ \frac{E}{(x + 2)^2} \ + \ \frac{F}{(x + 2)^3} [/tex]


    (If that isn't correct, please let me know!)

    The only justification I have for this is the idea of standard factoring, i.e. when you get a common denominator it will be the highest power in the denominator, but that does nothing to explain the numerator's weird shape...

    How are you supposed to generalize this idea? There is no way I could magically know how to do this for weirder shaped equations...

    I have a seperate side issue too that may explain something;

    I read a kind of explanation online for the kind of equation I have at the top of this latest post, it says;

    If you let:

    [tex] \frac{P(x)}{(x - 1)(x + 2)^2} \ = \ \frac{1}{x + 2}(\frac{P(x)}{(x - 1)(x + 2)}) [/tex]

    then you have:

    [tex] \frac{1}{x + 2}(\frac{P}{(x - 1)} + \frac{Q}{(x + 2}) = \frac{P}{(x - 1)(x + 2)} + \frac{Q}{(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} \ + \ \frac{Q}{(x + 2)^2} [/tex]

    which satisfied the person who was asking the question but it just escapes me...

    After then you have: the person just broke up the fraction magically inside the brackets and then magically the P becomes an A & a B, it's coming out of nowhere...

    I couldn't follow this shady logic for my equation which is degree 3.
     
  6. Jun 13, 2010 #5

    Mark44

    Staff: Mentor

    Fixed your LaTeX.
    And it won't work. Since you have a repeated factor in (x + 2)2, the decomposition needs to look like this:
    [tex]\frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{x + 1} = \frac{6x + 7}{(x + 2)^2(x + 1)}[/tex]
     
  7. Jun 13, 2010 #6

    Mark44

    Staff: Mentor

    It is correct.
    Makes sense to me. I've never seen this before, but I like it!
    Your difficulty in understanding seems to come from this:
    [tex] \frac{P}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}[/tex]

    Suppose you had
    [tex] \frac{5}{(x - 1)(x + 2)} [/tex]

    How would you decompose it to the sum of two fractions? Wouldn't you do this exactly like the right side of the equation above?
     
  8. Jun 13, 2010 #7
    [tex]\frac{5}{(x - 1)(x + 2)} = \frac{A}{x - 1} \ + \ \frac{B}{x + 2}[/tex]

    Multiplying both sides by (x - 1)(x + 2)

    [tex][(x - 1)(x + 2)] \cdot \frac{5}{(x - 1)(x + 2)} = [(x - 1)(x + 2)] \cdot [\frac{A}{x - 1} \ + \ \frac{B}{x + 2}][/tex]

    EDIT: MAde a mistake with the minus sign, but you get the idea anyway!

    RE-EDIT: Fixed it:

    [tex] \frac{5}{(x - 1)(x + 2)} = \frac{\frac{5}{3}}{x - 1} \ - \ \frac{\frac{5}{3}}{x + 2} [/tex]

    (it all multiplies back out fine!)


    That's how I would do it, I get this but when you've got repeat factors due to powers in the denominator, and then weird formula's because of a quadratic in the denominator, I mean where does any of that come from???
     
    Last edited: Jun 13, 2010
  9. Jun 13, 2010 #8
    Jeesh I don't know how I didn't realise what that person was doing by taking the [tex] \frac{1}{x + 2} [/tex] outside of the whole fraction.

    It's a brilliant idea!!! I only realised you'd use partial fractions inside the brackets before multiplying back in the above fraction, explaining the powers!!!!

    This could just be iterated a few times, albeit slowly, but it explains things massively!!!

    :biggrin:

    Still though, what's up with the quadratic equation?

    I can't find an explanation for this & I really don't understand the numerator's shape?
     
  10. Jun 13, 2010 #9

    Mark44

    Staff: Mentor

    Discounting repeated factors, you want the numerator to be of degree 1 less than the degree of the denominator. If one denominator is ax^2 + bx + c (it's assumed that this quadratic is irreducible over the reals), the form of the numerator would be Ax + B, a polynomial of degree 1.

    If you have repeated irreducible quadratic factors, what you do is similar for what you do with repeated linear factors, except that each numerator will now be a linear polynomial instead of a constant. Just like what you did in post #4.
     
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