Let $P$ be a polynomial of degree $n$ , $n \geq 2$, $P(X)=\prod_{i=1}^n \left(X-z_i \right)^{m_i}$

Question

Let $P$ be a polynomial of degree $n$ , $n \geq 2$. $$P(X)=\prod_{i=1}^n \left(X-z_i \right)^{m_i}$$ with zeros $z_1,z_2,\ldots,z_n$

• Decompose the rational fraction $\dfrac{P'}{P}$

My thoughts:

• $P(X)=\prod_{i=1}^n \left(X-z_i \right)^{m_i}$
• $P'(X)=\sum_{i=1}^{n}m_i\left(X-z_i \right)^{m_i-1}\prod_{?}^{?}\left(X-z_i \right)^{m_?}$

I don't know with what i should fill the ?

Then:

\begin{align*} \dfrac{P'}{P}&=\dfrac{\sum_{i=1}^{n}m_i\left(X-z_i \right)^{m_i-1}\prod_{?}^{?}\left(X-z_i \right)^{m_i}}{\prod_{i=1}^n \left(X-z_i \right)^{m_i}} \\ &= ? \end{align*}

I'm stuck here

Answer 1

The product rule generalizes such that the result is the sum of things where each thing is a derivative of one of the factors times the rest of the original products. For example $(fgh)' = f'gh+fg'h+fgh'$. So for your product, you want the product from $j=1$ to $i-1$ and then from $ j=i+1$ to $n$.

Answer 2

You can act as follows:$$P(X)=\prod_{i=1}^n \left(X-z_i \right)^{m_i}\Rightarrow P'(X)=\sum_i^n m_i(X-z_i)^{m_i-1}\prod_{j\ne i}(X-z_j)^{m_j}=\sum_i^n m_i(X-z_i)^{m_i-1}\prod_{j\ne i}(X-z_j)^{m_j}\cdot\left( \frac{X-z_i}{X-z_i}\right)=\sum_i^n m_i(X-z_i)^{m_i-2}\prod_{i=1}^{n}(X-z_i)^{m_i}$$ Hence $$\frac{P'(X)}{P(X)}=\sum_{i=1}^n m_i(X-z_i)^{m_i-2}$$