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Partial fractions question

  1. Jun 11, 2013 #1
  2. jcsd
  3. Jun 11, 2013 #2
    Please use the template as provided next time.

    You can substitute those values to find the values of [itex]A[/itex], [itex]B[/itex] and [itex]C[/itex] since we want to find those values that equal to both sides, selecting a "root" for one of the factors. We want to make both sides have same coefficients.
     
  4. Jun 11, 2013 #3
    I'm not sure what you mean, I know this method, I'm aware of how to do it, I'm also aware of how to use other methods.

    What I don't get (still after your post) is why it's allowed to use values which the question specifies is not allowed to be used (as f(x) would be undefined). Thanks for your reply, I'll be sure to use the template next time.
     
  5. Jun 11, 2013 #4

    CAF123

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    You are writing the fraction as: $$\frac{15 - 17x}{(2+x)(1-3x)^2} = \frac{A(1-3x)^2 + B(2+x)(1-3x) + C(2-x)}{(2+x)(1-3x)^2}.$$For this to hold, the denominator must be the same on both sides. This is true by inspection. Similarly, the numerator on LHS must be the same as the numerator on the RHS. This gives: $$15 - 17x = A(1-3x)^2 + B(2+x)(1-3x) + C(2-x)$$ which has to be satisifed by all ##x##.

    The condition on x not being equal to 1/3 and -2 is so that we don't end up with f(x) being undefined. The exercise above is simply to determine A,B and C. The reason we choose 1/3 and -2 is because that simplies things greatly.
     
  6. Jun 12, 2013 #5

    Ray Vickson

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    As CAF123 has pointed out, you have
    [tex]\frac{15-17x}{(2+x)(1-3x)^2} =
    \frac{A(1-3x)^2 + B(2+x)(1-3x)+C(2+x)}{(2+x)(1-3x)^2} \; x \neq -2,\, 1/3[/tex]
    (although CAF123 wrote ##C(2-x)## instead of ##C(2+x)##---probably a typo).
    Therefore, we have
    [tex] 15-17x = A(1-3x)^2 + B(2+x)(1-3x)+C(2+x) [/tex]
    for all ##x \neq -2, \, 1/3.## The reason we had to exclude -2 and 1/3 was so that we would not be dividing by zero. However, when we eliminate the denominators and just look at the numerators, we are no longer prevented from setting x to -2 or 1/3, because both numerators are perfectly well-defined at those points. In fact, we have an equation of the form
    [tex] \text{polynomial 1}(x) = \text{polynomial 2}(x) [/tex] for all x different from -2 and 1/3. But, since polynomials are continuous everywhere, the equation also holds at the points -2 and 1/3. The reason that is important to note is that when x = -2 or x = 1/3, the right-hand-side is very easy to evaluate, so we can get immediately the numbers A and C. Since the two sides must be equal for all x, so are their derivatives; that allows us to get B as well.

    An alternative would be to note that since the two polynomials are equal for all x, their ##x^n## coefficients must be equal. So, if you expand out the right-hand-side, you can get three equations for the three parameters A, B and C.
     
  7. Jun 12, 2013 #6

    CAF123

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    Hmm didn't notice that, indeed a typo.
     
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