## Problema do mês :: Problem of the month #3

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Seja $P(x)$ um polinómio real de grau $n\geq 2$. Suponha que o coeficiente do termo de maior grau de $P$ é igual a $1$. Prove que $\dfrac{P^{\prime \prime }(x)}{P^{\prime }(x)}=\displaystyle\sum_{k=1}^{n-1}\dfrac{1}{x-w_{k}}$, em que $w_{1},w_{2},\ldots ,w_{n-1}$ são as raízes de $P^{\prime }(x)$.

• O prazo limite para apresentação das resoluções é 10.02.2010, através de email acltavares@sapo.pt ou comentando no blogue.

Problem Statement

Let $P(x)$ be a polynomial of degree $n\geq 2$. Assume that the leading coefficient of $P$ is equal to $1$. Prove that $\dfrac{P^{\prime \prime }(x)}{P^{\prime }(x)}=\displaystyle\sum_{k=1}^{n-1}\dfrac{1}{x-w_{k}}$, where $w_{1},w_{2},\ldots ,w_{n-1}$ are the roots of $P^{\prime }(x)$.

• The deadline for submitting solutions is February 10, 2010 either via e-mail  acltavares@sapo.pt or comment box.

Anúncios

## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Matemática, Math, Problem, Problem Of The Month, Problema do mês com as etiquetas , , , . ligação permanente.

### 3 respostas a Problema do mês :: Problem of the month #3

1. MathOMan diz:

P'(x) = n (x – w_1)…(x – w_{n-1})

P”(x) = n sum{k=1…n-1} (x-w_1)…(x-w_{k-1})(x-w_{k+1})…(x-w_{n-1})

= sum{k=1…n-1} P'(x)/(x-w_k)

The result follows by dividing by P'(x). I don’t know why the problem is stated about P’ and P”. It is clearer to state it about P and P’ .

[Latex edited version by AT:

$P'(x)=n (x-w_1)\cdots (x - w_{n-1})$

$P''(x)=n\displaystyle\sum_{k=1,\dots ,n-1}(x-w_1)\cdots (x-w_{k-1})(x-w_{k+1})\cdots (x-w_{n-1})$

$=\displaystyle\sum_{k=1,\dots ,n-1}P'(x)/(x-w_k)$

The result follows by dividing by $P'(x)$. I don’t know why the problem is stated about $P'$ and $P''$. It is clearer to state it about $P$ and $P'$ . ]

• You are right. There is also this result (for $P'$ and $P''$ )

2. Obtive no Gaussianos http://gaussianos.com/ estas respostas:

1 – De Dani
(em http://gaussianos.com/formando-2010/#comment-32303)

« Parece bastante inmediato, no? Si $P$ es mónico, será el coeficiente de $x^{n-1}$ en $P^{\prime}$ igual a $n$. Por tanto podemos escribirlo de esta forma:

$P^{\prime }(x)=n(x-w_{1})(x-w_{1})\cdots (x-w_{n-1})=n\displaystyle\prod_{k=1}^{n-1}\left( x-w_{k}\right)$

Luego derivando término a término obtenemos

$P^{\prime \prime }(x)=n\displaystyle\sum_{k=1}^{n-1}\left( \displaystyle\prod_{j\neq k}\left( x-w_{j}\right) \right)$

de donde al dividir por $P^{\prime}$ nos da

$\dfrac{P^{\prime \prime }(x)}{P^{\prime }(x)}$ $=\dfrac{n\displaystyle\sum_{k=1}^{n-1}\left( \displaystyle\prod_{j\neq k}\left( x-w_{j}\right) \right) }{n\displaystyle\prod_{k=1}^{n-1}\left( x-w_{k}\right) }$ $=\displaystyle\sum_{k=1}^{n-1}\left( \dfrac{n\left( \displaystyle\prod_{j\neq k}\left( x-w_{j}\right) \right) }{n\displaystyle\prod_{k=1}^{n-1}\left( x-w_{k}\right) }\right)$ $=\displaystyle\sum_{k=1}^{n-1}\dfrac{1}{x-w_{k}}$

como queríamos demostrar. »

2 – De M ( em http://gaussianos.com/formando-2010/#comment-32311):
« Si $p(x)=\displaystyle\prod_{i=1}^{n}\left( x-a_{i}\right)$, entonces

$\log \left\vert p(x)\right\vert =\displaystyle\sum_{i=1}^{n}\log \left\vert x-a_{i}\right\vert$ (para $x\neq a_{i}$). Derivando (cuidando bien el signo entre las raíces):

$\dfrac{p^{\prime }(x)}{p(x)}=\displaystyle\sum_{i=1}^{n}\dfrac{1}{x-a_{i}}$. Tu caso se particulariza con $\left\{ p^{\prime },p^{\prime \prime }\right\}$. »

3 – E ainda outra resolução de M para o caso de $p(x)$ ter raízes simples.