Show $P(z) $ is a polynomial of degree $n-1$ interpolating an Analytic function.

Question

Let $C$ be a regular curve enclosing the distinct points $ω_1,ω_2,...ω_n$ and let $p(ω) = (ω −ω_1)(ω −ω_2) \cdots (ω −ω_n)$. Suppose that $f (ω)$ is analytic in a region that includes $C$. Show that

$$P(z) = \frac{1}{2 \pi i} \int_C \frac{f(ω)}{p(ω)} . \frac{p(ω) -p(z)}{ω -z} dω$$

is a polynomial of degree $n-1$ and $P(ω_i) = f(ω_i)$.

Answer 1

This directly follows from the residue theorem: $$ P(z)=\sum_i \frac{f(\omega_i)}{\prod_{j\ne i}(\omega_i-\omega_j)}\frac{-p(z)}{\omega_i-z} $$ which can be written as $$ P(z)=\sum_i \frac{f(\omega_i)}{\prod_{j\ne i}(\omega_i-\omega_j)}\prod_{j\ne i} (z-\omega_j) $$ The last product $\prod_{j\ne i} (z-\omega_j)$ is a polynomial of degree $n-1$ which cancels at $z=\omega_k$ for $k\ne i$. $P(z)$ is thus a polynomial of degree $n-1$ with $P(\omega_i)=f(\omega_i)$.

Answer 2

As $\omega - z$ divides $p(\omega) - p(z)$, this quotient is actually a polynomial: $$\frac{p(\omega) - p(z)}{\omega - z} = R(\omega,z) = \sum_{k=0}^{n-1} R_k(\omega)z^k.$$ Then, $$P(z) = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − p(z)}{\omega − z}d\omega = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\left(\sum_{k=0}^{n-1} R_k(\omega)z^k\right)d\omega = $$ $$ \sum_{k=0}^{n-1}\int_C\left(\frac1{2\pi i}\frac{f(\omega)}{p(\omega)} R_k(\omega)d\omega\right)z^k. $$ ($P$ is a polynomial)

On the other hand, $$P(\omega_k) = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − p(\omega_k)}{\omega − \omega_k}d\omega = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − 0}{\omega − \omega_k}d\omega = $$ $$ = \frac1{2\pi i}\int_C\frac{f(\omega)}{\omega − \omega_k}d\omega = f(\omega_k) $$ by the Cauchy integral formula.

Answer 3

Re-writing $P$ give the following form $$P(z) = f(z)- \frac{1}{2i\pi}\int_{C}\prod_{j=0}^{n}\left(\frac{z-\omega_j}{\omega-\omega_j}\right)\frac{f(\omega)d\omega}{\omega-z}$$ Howver, we have, $$\prod_{j=0}^{n}\left(\frac{1}{\omega-\omega_j}\right) = \sum_{i=0}^{n} \frac{A_i}{\omega-\omega_i}~~~\text{with}~~~A_i =\prod_{\stackrel{i\neq j }{j=0}}^{n}\left(\frac{1}{\omega-\omega_j}\right)$$

Hence, $$\prod_{j=0}^{n}\left(\frac{z-\omega_j}{\omega-\omega_j}\right) = \sum_{i=0}^{n} \prod_{\stackrel{i\neq j }{j=0}}^{n}\left(\frac{z-\omega_j}{\omega-\omega_j}\right)\frac{z-\omega_i}{\omega-\omega_i}=\sum_{i=0}^{n} \ell_i(z)\frac{z-\omega_i}{\omega-\omega_i}~~ ~$$

With $$\color{blue}{\ell_i(z) =\prod_{\stackrel{i\neq j }{j=0}}^{n}\left(\frac{z-\omega_j}{\omega-\omega_j}\right)\in\Bbb C_{n-1}[X]. } $$ Therefore, $$P(z) = f(z)- \frac{1}{2i\pi}\int_{C}\prod_{j=0}^{n}\left(\frac{z-\omega_j}{\omega-\omega_j}\right)\frac{f(\omega)d\omega}{\omega-z} \\= f(z)- \frac{1}{2i\pi}\int_{C}\sum_{i=0}^{n} \ell_i(z)\frac{z-\omega_i}{\omega-\omega_i}~\frac{f(\omega)d\omega}{\omega-z}\\=f(z)- \sum_{i=0}^{n} \ell_i(z)\frac{1}{2i\pi}\int_{C}\frac{z-\omega_i}{\omega-\omega_i}~\frac{f(\omega)d\omega}{\omega-z} $$

On the other hand, by **Cauchy formula we have, ** $$\frac{1}{2i\pi}\int_{C}\frac{z-\omega_i}{\omega-\omega_i}~\frac{f(\omega)d\omega}{\omega-z} =\frac{1}{2i\pi}\int_{C}~\frac{f(\omega)d\omega}{\omega-z}- \frac{1}{2i\pi}\int_{C}\frac{f(\omega)d\omega}{\omega-\omega_i}\\= f(z) -f(\omega_i)$$ Finally we have,

$$P(z) = f(z)- \sum_{i=0}^{n} \ell_i(z)\frac{1}{2i\pi}\int_{C}\frac{z-\omega_i}{\omega-\omega_i}~\frac{f(\omega)d\omega}{\omega-z} \\=f(z)- \sum_{i=0}^{n} \ell_i(z)(f(z) -f(\omega_i)) =\sum_{i=0}^{n} \ell_i(z)f(\omega_i) \in \Bbb C_n[X] $$

Since, By Lagrange interpolation we can see that $$\sum_{i=0}^{n} \ell_i(z) =1$$

Thus,

$$P(z) =\sum_{i=0}^{n} \ell_i(z)f(\omega_i) \in \Bbb C_{n-1}[X] $$ By Lagrange interpolation again $P(z)$ is the only polynomial of degree $n$ interpolating $f$ at $\omega_i,~~i= 0,\cdots,n$