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Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$,

$\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|$

The given inequality seemed too much similar to limits so I applied R tending to $\sqrt[3]{2}$. And then I tried solving the equation I get however I am unable to solve for integers. Any help please

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  • $\begingroup$ I don't understand the description of your computation. Could you post the details, please? $\endgroup$ – saulspatz Aug 22 '18 at 8:03
  • $\begingroup$ @saulspatz I am not confident about my computation. However as R tends to cube root of 2, the RHS of the inequality tends to zero, hence as I apply the same limit on the LHS it should tend to zero as well. $\endgroup$ – saisanjeev Aug 22 '18 at 12:11
  • $\begingroup$ You say you got an equation you couldn't solve. You should show the equation and the work that led up to it. Perhaps someone can solve the equation; perhaps someone can point out a mistake in the derivation. That way, you can get some actual help. As it is, the only thing anyone can do is solve the problem from scratch, which won't help you near as much. $\endgroup$ – saulspatz Aug 22 '18 at 16:06
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(Too long for a comment.) The question is missing a lot of context, so it is hard to guess if the following even applies at all. But FWIW the given form reminisces of an iterated approximation steadily approaching $\,\sqrt[3]{2}\,$.

Newton-Raphson would come to mind first, but the canonical $\,f(x) = x^3-2\,$ doesn't give $\,x - f(x) / f'(x)\,$ in the prescribed form of a ratio of two quadratics.

Another way to "guess" an answer could be to start from the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, so $\,\sqrt[3]{2}=1+1 \big/ \left(\left(\sqrt[3]{2}\right)^2+\sqrt[3]{2}+1\right)\,$ i.e. $\,\sqrt[3]{2}$ is a fixed point of $\,f(x)=(x^2+x+2)/(x^2+x+1)\,$ which corresponds to $\,a=b=d=e=f=1, c=2\,$ and does indeed appear to converge as required. Formally proving that convergence would take some more work, though.


[ EDIT ]   Following up on OP's insight that...

The given inequality seemed too much similar to limits so I applied R tending to $\sqrt[3]{2}$.

Formalizing the above, let $\,f(x)=\dfrac{ax^2+bx+c}{dx^2+ex+f}\,$ and consider the sequence defined by $\,r_{n+1}=f(r_n)\,$ for some $\,r_0 \in \mathbb{R}\,$. Furthermore, suppose that $\,r_n \to r = \sqrt[3]{2}\,$, then passing the recurrence relation to the limit gives:

$$r=f(r) \;\;\iff\;\; dr^3+(e-a)r^2+(f-b)r-c = 0$$

But the minimal polynomial of $\,r=\sqrt[3]{2}\,$ over $\,\Bbb Z[X]\,$ is $x^3 - 2$, so it follows that $\,\begin{align}\begin{cases}2d &= c \\ e&=a \\f&=b\end{cases}\end{align}\,$ and therefore $\,f(x) = \dfrac{ax^2+bx+2d}{dx^2+ax+b}\,$ or, choosing WLOG $\,d=1\,$, $\,f(x) = \dfrac{ax^2+bx+2}{x^2+ax+b}\,$ .

The case given in the original answer $\,f(x)=\dfrac{x^2+x+2}{x^2+x+1}\,$ corresponds to $\,a=b=1\,$ above.

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