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Let $A\in{\mathbb C}[X]$ be a monic polynomial of degree $n\geq 2$, with roots $\alpha_1,\alpha_2,\alpha_3, \ldots ,\alpha_n$. Let $B$ be the polynomial

$$ B=\prod_{k=1}^{n} \bigg(X-\frac{\alpha_k+\frac{1}{\alpha_k}}{2}\bigg) $$

Can anybody show the following identity

$$ B'(1)=\frac{A(1)(A'(1)+A''(1))-(A'(1)^2)}{2^{n-1}A(0)} $$

UPDATE 16:00 I have checked it by computer for $2 \leq n \leq 7$.

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Since most of your identity involves values of $A$ and $B$ at $1$, it may be clearer to shift the polynomials, and to work on $A(X+1)$ and $B(X+1)$ at $0$. Then $A'(1),B'(1),A''(1)$ are just some coefficients of those polynomials, and the Vieta formulas relate those coefficients with elementary symmetric polynomials in their roots.

Also, I tend to prefer the symmetric polynomials of degree $1$ and $2$ rather than $n-1$ and $n-2$, so to switch between those I have to use the reciprocal of the roots :

Let $\beta_k = (\alpha_k-1)^{-1}$ and $\gamma_k = (\frac 1 2 (\alpha_k + \frac 1 {\alpha_k})-1)^{-1}$ so that $A(X+1) = \prod (X - \beta_i^{-1})$, and $B(X+1) = \prod (X-\gamma_i^{-1})$.

The Vieta relations give the identities :
$A(0) = \prod (-\alpha_k)$
$A(1) = \prod (-\beta_k^{-1})$
$B(1) = \prod(-\gamma_k^{-1})$
$B'(1)/B(1) = - \sum \gamma_k$
$A'(1)/A(1) = - \sum \beta_k$
$A''(1)/A(1) = 2\sum \beta_k\beta_l$

We will attempt to express combinations of $B(1),B'(1),A(0)$ in terms of only the $A(1),A'(1),A''(1)$ by translating those expressions into symmetric expressions in the $\beta_k$ :

Expressing $\gamma_k$ in terms of $\beta_k$ we get $\alpha_k = \frac {\beta_k+1} {\beta_k}$, $\alpha_k + \frac 1 {\alpha_k} = \frac {2\beta_k^2 + 2\beta_k+1}{\beta_k^2+\beta_k}$ and finally $\gamma_k = 2\beta_k(\beta_k+1)$.

Hence : $$ -\frac{B'(1)}{B(1)} = \sum_k \gamma_k = 2\sum (\beta_k^2 + \beta_k) = 2\left((\sum \beta_k)^2 - 2\sum \beta_k\beta_l + \sum \beta_k\right) = 2\left( \frac {A'(1)^2}{A(1)^2} - \frac{A''(1)}{A(1)} - \frac{A'(1)}{A(1)}\right) = 2\frac{A'(1)^2-A''(1)A(1) - A(1)A'(1)}{A(1)^2}$$

And $$B(1)A(0) = \prod (- \gamma_k)^{-1}(- \alpha_k) = 2^{-n}\prod(\alpha_k^2-2\alpha_k+1) = 2^{-n}\prod \beta_k^{-2} = 2^{-n}A(1)^2$$

Combining those two we obtain your equality.

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  • $\begingroup$ Nice but a little hard to follow. Could you “demistify” your proof a little ? I wonder if there is also a formula for $B''(1)$ in terms of $A$ and its derivatives. $\endgroup$ – Ewan Delanoy Nov 11 '13 at 15:49
  • $\begingroup$ @EwanDelanoy : There should be a formula like this for $B''(1)$ using the $3$rd and $4$th derivative of $A$ at $1$, and so on for the next coefficients. $\endgroup$ – mercio Nov 11 '13 at 16:16

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