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In preparation for Lagrange's Interpolation Theorem, I'm curious to know why a rational function can be put in the following form.

Suppose $Q$ is a polynomial with distinct roots $a_1,\dots,a_n$, and let $P$ be a polynomial with $\deg(P)<n$. Then why does $$ \frac{P(x)}{Q(x)}=\sum_{i=1}^n\frac{P(a_i)}{Q'(a_i)(x-a_i)}? $$ I tried expanding the right hand sum over a common denominator, but it got horribly messy quickly. I also tried induction, assuming $Q$ has just one distinct root, and thus $Q=C(x-a_1)^k $. But then $\deg(P)<1$, so $P$ is constant, say $P=A$. But then $$ \frac{P(a_1)}{Q'(a_1)(x-a_1)}=\frac{A}{kC(a_1-a_1)^{k-1}(x-a_i)} $$ which seems problematic if $k>1$. What's the slicker way to conclude this equality? Thanks,

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    $\begingroup$ You are not interpreting "$Q$ is a polynomial with distinct roots" in the intended way. It is intended to mean that the polynomial has degree say $n$, and all of its $n$ complex roots are different from each other. So $Q(x)=(x-1)^2$ is not covered. Indeed, the result is in general false for this $Q$, as you may remember from partial fractions. $\endgroup$ – André Nicolas Jan 24 '12 at 4:15
  • $\begingroup$ That clears it up, thanks. $\endgroup$ – Gemma Jan 24 '12 at 4:17
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Multiply through by $Q(x)$ and note that both sides are polynomials of degree less than $n$ agreeing at the $n$ points $x=a_i$.

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  • $\begingroup$ How can you be sure the right hand side has degree less than $n$? Each summand $\displaystyle\frac{P(a_i)Q(x)}{Q'(a_i)(x-a_1)}$ has degree $\deg(Q)-1$, but what if $\deg(Q)>n$? $\endgroup$ – Gemma Jan 24 '12 at 3:58
  • $\begingroup$ @Gemma, the original question was somwhat imprecise on this point, but I think that what was meant in it is that the listed roots are all of them, and they are simple. $\endgroup$ – Lubin Jan 24 '12 at 4:08
  • $\begingroup$ Thanks @lubin, I see that I misinterpreted the question. $\endgroup$ – Gemma Jan 24 '12 at 4:18

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