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Let $A_k=\{x \mid 2^k < x^k +x^{k+1} < 2^{k+1}\},\, k=1, 2, 3, \ldots$.

Find the largest positive integer $n$ such that $$ A_1\cap A_2\cap\cdots \cap A_n \ne \emptyset. $$

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$A_1\cap A_2\cap A_3\neq \emptyset$, but $A_1\cap A_4=\emptyset$. I have no analytic proof, just a calculation on wolframalpha:

$A_4\approx(1.58,1.83)$, while $A_1\approx (-2.56,-2)\cup (1,1.56)$

Also $A_2\approx (1.31, 1.72)$ and $A_3\approx(-2.31,-2)\cup(1.48,1.79)$, so $A_1\cap A_2\cap A_3\approx(1.48,1.56)$

Proof that $A_1\cap A_2\cap A_3\neq \emptyset$: Take $x=1.5$. $1.5+1.5^2=3.75$ and $2<3.75<4$. $1.5^2+1.5^3=5.625$ and $4<5.625<8$. $1.5^3+1.5^4=8.4375$ and $8<8.4375<16$.

Proof that $A_1\cap A_4=\emptyset$: Take $x=1.57$. On the interval $(0,+\infty)$, $f(x)=x+x^2$ is increasing and $f(1.57)=4.0349>4$. But also $g(x)=x^4+x^5$ is increasing on the same interval, and $g(1.57)=15.6146312657<16$. Hence $1.57$ is strictly above $A_1$ and strictly below $A_4$.

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  • $\begingroup$ The answer is right, thank you. I need an analytic proof. $\endgroup$ – Chung. J May 23 '13 at 23:07
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Computers these days can do exact arithmetic, so there's no reason you can't have a computer do an "analytic" proof - or at least have it do the relevant computations.

The positive portions of your sets

\begin{array}{l} 2<x^2+x<4 \\ 4<x^3+x^2<8 \\ 8<x^4+x^3<16 \\ 16<x^5+x^4<32 \\ \end{array}

look something like so

enter image description here

To show that the intersection of the first and last is empty, it suffices to find $x_0$ such that $x_0^2+x_0>4$ (i.e., $x_0$ lies to the right of the first interval) and $x_0^5+x_0^4<16$ (i.e., $x_0$ lies to the left of the last interval). I propose that you choose $$x_0 = \frac{196377467}{125086164}.$$ Then $$x_0^2+x_0 - 4 = \frac{542019891463093}{15646548424234896} > 0$$ and $$x_0^5+x_0^4 -16= -\frac{11888789705707430179748265315192519808033}{30622903893638169223375610368116647629824} < 0.$$

I obtained the number $x_0$ by simply rationalizing the average of numerical approximations to the two relevant endpoints, but the exact computation with rational numbers proves that $x_0$ has the desired properties.

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  • $\begingroup$ Thank you for the smart answer. $\endgroup$ – Chung. J May 24 '13 at 3:15

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