Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $ $$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$ For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$. I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?
Show that $\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$
-
0Hint: move $Q(z)$ to the other side, then prove that the equality holds for each $z=\alpha_k\,$. Though, in fairness, that won't come easy if you really know nothing about derivatives and what $Q'$ means. – 2017-02-27
-
0Well, I know as much about derivatives of functions with complex coefficients as is presented (that I can understand without the aid of a classroom) in the analytic functions and polynomials sections in Ahlfors, which are the two sections preceding this one. There is a sentence which says that $P(\alpha) = P'(\alpha) = \cdots = P^{(h-1)}(\alpha) = 0$, so I suppose then that $Q'(a_k) = 0$ given that $n>k$? I am still unsure how to relate $\alpha_k$ to $P$. Thank you for the hint. – 2017-02-27
-
0Elaborated some more in the posted answer. `Q′(a_k)=0 given that n>k?` No, $Q′(\alpha_k)$ is the value of the derivative $Q'(z)$ at $z=\alpha_k$. `still unsure how to relate α_k to P` They are not directly "related" except by the given identity. $P'(\alpha_k)$ is just the value $P(z)$ takes when $z=\alpha_k\,$. – 2017-02-27
2 Answers
Hint (without L'Hôpital): assume WLOG $Q$ is monic (otherwise cancel out the leading term between the two sides). Then $Q(z)=\prod_{k=1}^n(z-\alpha_k)$ and, by the product rule of differentiation, $\,Q'(z)=\sum_{k=1}^n \prod_{j \ne k}(z-\alpha_j)\,$ so in particular $\,Q'(\alpha_k)=\prod_{j \ne k}(\alpha_k-\alpha_j)\,$.
Multiplying by $Q(z)$ and using the above, the equality to prove becomes:
$$ \begin{align} P(z) = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} & = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} \\[5px] & = \sum_{k=1}^{n} \frac{P(\alpha_k)\,\prod_{j \ne k}(z-\alpha_j)}{\prod_{j \ne k}(\alpha_k-\alpha_j)} \\[5px] \end{align} $$
It follows that the equality holds for all $\,z=\alpha_k\,$ and, since the two sides of the equality are polynomials of degree $\le n-1\,$ that are equal at $n$ points, it further follows that the equality holds in general, for all $z$.
-
1This is the [Lagrange Polynomial](https://en.wikipedia.org/wiki/Lagrange_polynomial) matching at the $a_k$. – 2017-02-27
-
0@robjohn It is indeed, thanks for pointing out. The above is (quite obviously) a more painful derivation than yours. Yet, given the little context provided, I thought a purely algebraic alternative might be worth mentioning. – 2017-02-27
-
0Thank you for this! I was, surprisingly, able to follow everything, but I still must think about why the conclusion follows. I apologize for my lack of knowledge (and hence the little context) on the topic. My only source is this textbook, and it's challenging for me. I have one last question: following up from your comment: if $a_k$ is a root for $Q$, why does $Q'(a_k) \neq 0$? – 2017-02-27
-
0@playitright $Q'(\alpha_k) \ne 0$ because all roots are assumed to be simple. If you ask why it's allowed to multiply/divide by $Q(z)$ despite it being zero at the $n$ points $\alpha_k\,$, then the answer is a bit more complicated. If taken as abstract "fractions" of polynomials, then there is no problem. Else if taken as complex *functions*, note that both sides of the original equality are technically undefined for $z=\alpha_k\,$. The identity can be extended to the entire $\mathbb{C}$ by continuity, though, and the same argument could be used with the proof above. – 2017-02-27
-
1@playitright: If $Q(x)=Q'(x)=0$, then $x$ is a double root of $Q$. The question stated that $Q$ has distinct roots. – 2017-02-27
Use Partial Fractions and L'Hôpital: $$ \frac{P(x)}{Q(x)}=\sum_{k=1}^n\frac{A_k}{x-a_k}\\ $$ where $A_k=\lim\limits_{x\to a_k}(x-a_k)\frac{P(x)}{Q(x)}=\frac{P(a_k)}{Q'(a_k)}$.
-
0Curious that the zeroes of $P$ are implicitly embedded in the expansion. It might be worthwhile mentioning this. – 2017-02-27