Re-Expressing a Quotient of Polynomials

[sponsoredwalk]

This is a basic question but I don't think I've ever seen anything like it before.

If Q(x) \ = \ b_0(x \ - \ a_1)(x \ - \ a_2)\cdot \ . \ . \ . \ \cdot (x \ - \ a_s)

then

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

 

[gabbagabbahey]

This is a basic question but I don't think I've ever seen anything like it before.

If Q(x) \ = \ b_0(x \ - \ a_1)(x \ - \ a_2)\cdot \ . \ . \ . \ \cdot (x \ - \ a_s)

then

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

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 (x-a_n) and take the limit as x \to a_n?

 

[HallsofIvy]

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

Now, a_1= 1, P(a_1)= P(1)= 1^3= 1 and Q'(a_1)= Q'(1)= 2(1)- 3= -1 while a_2= 2, P(a_2)= P(2)= 2^3= 8 and Q'(a_2)= Q'(2)= 2(2)- 2= 1. So
\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)}
= \frac{-x+ 2+ 8x- 8}{(x-1)(x- 2)}= \frac{7x- 6}{(x-1)(x-2)}
just as it should.

 

[sponsoredwalk]

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
\frac{P(x)}{Q(x)} \ = \ R(x) \ + \ \sum_{i=1}^n \frac{d_i}{(x \ - \ a_i)}

(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)}

\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
(Where Q'(x) = Q(x)/(x - a_j) )

\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

\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

\frac{P(a_j)}{Q'(a_j)} \ = \ d_j

So I don't see what the derivative has to do with anything 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!