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I have the following question:


$f_1(a), f_2(a),\ldots, f_n(a)$ and $g_1(a), g_2(a),\ldots, g_n(a)$ are strictly increasing positive "polynomial" functions of $a$.

It is also known that

$$\frac{f_1(a)}{g_1(a)}, \frac{f_2(a)}{g_2(a)},\ldots,\frac{f_n(a)}{g_n(a)}$$ are strictly increasing functions of $a$.


\begin{equation} \frac{f_1(a)+f_2(a)+\cdots+f_n(a)}{g_1(a)+g_2(a)+\cdots+g_n(a)} \end{equation}

EVENTUALLY increases with $a$?

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Are you just talking about positive $a$? There is no polynomial on $\mathbb R$ that is both strictly increasing and positive. – Robert Israel Apr 30 '12 at 18:58
Yes $a$ should be positive. Sorry for the mistake. – Seyhmus Güngören Apr 30 '12 at 20:00
I would also check out the theory of Padé approximants: – deoxygerbe Apr 30 '12 at 21:00
up vote 0 down vote accepted

No, take for counter-example these four function: $$f_1(a) = a + 0.5$$ $$g_1(a) = 2a + 1.01$$ $$f_2(a) = 1.01a + 2$$ $$g_2(a) = 0.5a + 1$$

Here all the function and both $f_i(a)/g_i(a)$ are increasing for each $a \in \mathbb{R_+}$, but $\sum f_i / \sum g_i$ decrease for each $a \in \mathbb{R_+}$.

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Interesting example. What's really going on here is that two $2 \times 2$ matrices with positive entries and positive determinant can have a sum with negative determinant. – Robert Israel Apr 30 '12 at 21:43

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