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Re-Expressing a Quotient of Polynomials

  1. Aug 8, 2012 #1
    This is a basic question but I don't think I've ever seen anything like it before.

    If [tex]Q(x) \ = \ b_0(x \ - \ a_1)(x \ - \ a_2)\cdot \ . \ . \ . \ \cdot (x \ - \ a_s)[/tex]

    then

    [tex]\frac{P(x)}{Q(x)} \ = \ R(x) \ + \ \sum_{i=1}^s \frac{P(a_i)}{(x \ - \ a_i)Q'(a_i)}[/tex]


    I just don't understand where the P(aᵢ) & Q'(aᵢ) come from, maybe it's just notation masking something obvious but idk... I can't really make sense of it.

    If this doesn't make sense it's from page 12 of this.
     
  2. jcsd
  3. Aug 9, 2012 #2

    gabbagabbahey

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    This should really be in the Calculus and beyond forums (due to the fact that it involves a derivative). Anyways, what have you tried? If you are completely stuck, what do you get when you multiply your equation by [itex](x-a_n)[/itex] and take the limit as [itex]x \to a_n[/itex]?
     
    Last edited: Aug 9, 2012
  4. Aug 9, 2012 #3

    HallsofIvy

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    You might start by looking at a very simple example. Suppose [itex]Q(x)= (x- 1)(x- 2)= x^2- 3x+ 2[/itex] and [itex]P(x)= x^3[/itex]. Then
    [tex]\frac{P(x)}{Q(x)}= x+ 3+\frac{7x- 1}{(x- 1)(x- 2)}[/tex]
    Q'(x), the derivative of Q(x), is 2x- 3.

    Now, [itex]a_1= 1[/itex], [itex]P(a_1)= P(1)= 1^3= 1[/itex] and [itex]Q'(a_1)= Q'(1)= 2(1)- 3= -1[/itex] while [itex]a_2= 2[/itex], [itex]P(a_2)= P(2)= 2^3= 8[/itex] and [itex]Q'(a_2)= Q'(2)= 2(2)- 2= 1[/itex]. So
    [tex]\sum \frac{P(a_i)}{(x- a_i)Q'(a_i)}= \frac{1}{(x- 1)(-1)}+ \frac{8}{(x- 2)(1)}= \frac{-(x-2)}{(x-1)(x-2)}+ \frac{8(x- 1)}{(x-1)(x-2)}[/tex]
    [tex]= \frac{-x+ 2+ 8x- 8}{(x-1)(x- 2)}= \frac{7x- 6}{(x-1)(x-2)}[/tex]
    just as it should.
     
  5. Aug 9, 2012 #4
    Aargh I should have just ignored the P(aᵢ) & Q'(aᵢ) inventing fantasy reasons why the Q' term is in the denominator etc... & worked from first principles!
    Okay so working from
    [tex]\frac{P(x)}{Q(x)} \ = \ R(x) \ + \ \sum_{i=1}^n \frac{d_i}{(x \ - \ a_i)}[/tex]

    [tex](x \ - \ a_j) \cdot \ \frac{P(x)}{Q(x)} \ = \ (x \ - \ a_j) \cdot \ R(x) \ + \ (x \ - \ a_j) \cdot \ \sum_{i=1}^n \frac{d_i}{(x \ - \ a_i)}[/tex]

    [tex] \frac{P(x)}{Q'(x)} \ = \ (x \ - \ a_j) \cdot \ R(x) \ + \ (x \ - \ a_j) \cdot \ \sum_{i=1,i\neq j}^n \frac{d_i}{(x \ - \ a_i)} \ + d_j[/tex]
    (Where Q'(x) = Q(x)/(x - aj) )

    [tex] \lim_{x \rightarrow a_j} \ \frac{P(x)}{Q'(x)} \ = \ \lim_{x \rightarrow a_j} (x \ - \ a_j) \cdot \ R(x) \ + \ \lim_{x \rightarrow a_j} \ (x \ - \ a_j) \cdot \ \sum_{i=1,i\neq j}^n \frac{d_i}{(x \ - \ a_i)} \ + \lim_{x \rightarrow a_j} \ d_j[/tex]

    [tex] \frac{P(a_j)}{Q'(a_j)} \ = \ 0 \cdot \ R(a) \ + \ 0 \cdot \ \sum_{i=1,i\neq j}^n \frac{d_i}{(a_j \ - \ a_i)} \ + \ d_j[/tex]

    [tex] \frac{P(a_j)}{Q'(a_j)} \ = \ d_j[/tex]

    So I don't see what the derivative has to do with anything :confused: The book is nearly a hundred years old so I didn't really think it was fair to assume Q' means the derivative unless he specified this, which he does when he actually uses derivatives a page or so later, so thanks a lot!
     
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