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I'm having some trouble deriving certain identities. If

$$S(z) = \prod_{i=1}^n (z-z_i)$$

then how can I write

$$\frac{1}{S(z)}\frac{d^2S}{dz^2} = \sum_{i=1}^n\frac{1}{z-z_i}\sum_{j\neq i}^n\frac{2}{z_i-z_j}$$

and $$ \frac{1}{S(z)}\frac{dS}{dz}= \sum_{i=1}^n\frac{1}{z-z_i} $$

Truth be told, I'm having some trouble writing the derivative of $S(z)$ in a neat, compact form.

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2 Answers 2

Hint for the second: $${\frac {{ \frac {d}{dz}}S \left( z \right) }{S \left( z \right) }}={\frac {d}{dz}}\ln \left( S \left( z \right) \right) $$

Hint A for the first: $${\frac {{\frac {d^{2}}{d{z}^{2}}}S \left( z \right) }{S \left( z \right) }}={\frac {d^{2}}{d{z}^{2}}}\ln \left( S \left( z \right) \right) + \left( {\frac {d}{dz}}\ln \left( S \left( z \right) \right) \right) ^{2}$$

Hint B for the first:

$${\frac {1}{ \left( z-z_{{i}} \right) \left( z-z_{{j}} \right) }}={ \frac {1}{ \left( z_{{i}}-z_{{j}} \right) \left( z-z_{{i}} \right) } }+{\frac {1}{ \left( z_{{j}}-z_{{i}} \right) \left( z-z_{{j}} \right) }}$$ Where the identities hold for any differentiable function (proof: chain rule).

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$\frac { dS } { dz} =\sum _ { i = 1 } ^ n\Pi_{j\neq i}(z-z_j)$

So $ \frac { 1 } {S ( z)}\frac { dS } { dz}=\sum _ { i = 1 } ^ n\Pi_{j\neq i}(z-z_j)/(\Pi_{i=1}^n(z-z_j))=\sum _ { i = 1 } ^ n\frac { 1 } {z - z_i}$

Now can you do the second derivative?

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