floor function?

achen29 wrote: floor function? yes

The condition is equivalent to $\displaystyle {x_i}^{1/\lfloor x_i\rfloor}={x_j}^{1/\lfloor x_j\rfloor}$ for all $i$ and $j$, and hence to $\displaystyle {x_i}^{1/\lfloor x_i\rfloor}=$const. Hence let us study the function $f(x)=x^{1/\lfloor x\rfloor}$. $\bullet$ For $1\le x<2$, we have $f(x)=x$ and $f$ attains all values in the interval $[1,2)$. $\bullet$ For $2\le x<3$, we have $f(x)=\sqrt{x}$ and $f$ attains all values in the interval $[\sqrt2,\sqrt3)$. $\bullet$ For $3\le x<4$, we have $f(x)=x^{1/3}$ and $f$ attains all values in the interval $[3^{1/3},4^{1/3})$. $\bullet$ And so on. $\bullet$ For $n\le x<n+1$, we see that $f$ attains all values in the interval $[n^{1/n},(n+1)^{1/n})$. Hence let us denote interval $I_n=[n^{1/n},(n+1)^{1/n})$. $\bullet$ It is easy to see that $10/7\approx1.428$ lies in the four intervals $I_1$, $I_2$, $I_3$ and $I_4$. $\bullet$ It is easy to show that for $n\ge3$, the left endpoint of $I_n$ is strictly larger than the right endpoint of $I_{n+2}$, $I_{n+3}$, and all further intervals. $\bullet$ Hence the maximum possible value is $k=4$.

Solved with nukelauncher and Th3Numb3rThr33. The goal is to determine over constants \(a\ge1\) the maximum number of \(x\) for which \[x^{1/\lfloor x\rfloor}=a\]holds. I claim the maximum number of \(x\) is \(k=4\). First solution, by piecewise argument The claimed answer \(k=4\) is achieved by \(a=3^{1/3}\) and \(x\in\{1,2,3,4\}\). Now we will show \(k=4\) is maximal. Note that for \(x\in[n,n+1)\), \(f\) is increasing and outputs every number in the interval \(I_n=[n^{1/n},(n+1)^{1/n})\) exactly once. Thus we want to show each \(C\) lies on at most four of these intervals \(I_n\). However, Claim: \(n^{1/n}\ge(n+k+1)^{1/(n+k)}\) for integers \(n\ge3\), \(k\ge2\). Proof. The inequality is equivalent to \(n^{n+k}\ge(n+k+1)^n\). Manually verify \(n=3\) and \((n,k)=(4,2)\) by hand. For remaining \((n,k)\), the final inequality holds in \[\left(1+\frac{k+1}n\right)^n\le e^{k+1}\le n^k,\]so the claim is proven \(\blacksquare\) Now fix the constant value \(a\), and let \(N\) be the smallest \(N\ge3\) with \(a\in I_N\); it follows that \(a\) is not in \(I_{N+2}\), \(I_{N+3}\), \ldots, so the only intervals \(a\) could lie in are \(I_1\), \(I_2\), \(I_N\), \(I_{N+1}\), the end. Second solution, by function analysis Set \(n=\lfloor x_i\rfloor\). Note that \(\lfloor a^n\rfloor=\lfloor a^{\lfloor x_i\rfloor}\rfloor=\lfloor x_i\rfloor=n\), so by letting \(n=\lfloor x_i\rfloor\) vary, we are counting the number of integers \(n\) with \[a^n-n\in[0,1).\]We will show at most four such \(n\) exist. (This finishes the problem, since each \(n\) yields at most one working \(x_i\).) Four is achieved by \(a=e^{1/e}\), as will be shown later. Claim: The number of intersections between \(y=a^x\) and \(y=x\) over \(x>0\) is zero for \(a>e^{1/e}\); one for \(a=e^{1/e}\); and two for \(a<e^{1/e}\). Proof. Evidently the second bullet point implies the other two. Since \(a^x\) is convex, it suffices to find a tangency point. I claim \(x=e\) is the desired tangency point; indeed \(a^e=e\) and \[\frac d{dx}a^x\bigg\rvert_{x=e}=\ln a\cdot a^e=1,\]as needed. \(\blacksquare\) Now observe that if \(a>e^{1/e}\), but \(y=a^x\) intersects \(y=x+1\) at two points \(x=0\) and \(x=s\), then the number of \(n\) is the number of positive integers in the interval \((0,s)\). But \(s\) strictly decreases as \(a\) increases, so the number of \(n\) is at most the number of \(n\) when \(a=e^{1/e}\). For \(e^{1/e}\) we can manually verify \(n=1,2,3,4\) work, but no other \(n\) work. Finally let \(a<e^{1/e}\). Then \(y=a^x\) intersects \(y=x+1\) at two points \(x=0\) and \(x=s\) and \(y=x\) at two points \(x=r_1\) and \(x=r_2\). The number of \(n\) is the number of positive integers in the set \((0,r_1]\cup[r_2,s)\). Since \(a^e<e\), we have \(r_1<e<r_2\) Then \((0,r_1]\) always contains at most two positive integers. If \(s<5\) then \([r_2,s)\) contains at most two positive integers. Let \(t=s+1\), and assume for contradiction \(r_2\le t-3\). Then \(a^{t-3}>t-3\) and \(a^{t-1}=t\). Thus \[\sqrt[t-1]t=a<\sqrt[t-3]{t-3}\implies(t-3)^{t-1}<t^{t-3},\]which fails for \(t\ge6\). Therefore \(s\ge5\), and \([r_2,s)\) contains at most two positive integers. It follows that at most four \(n\) exist, as desired.

We claim that the answer is $k=4$. \[x_i^{ \lfloor x_j\rfloor}=x_j^{ \lfloor x_i\rfloor} \iff(x_i^{\frac{1}{\lfloor x_i\rfloor}})^{\lfloor x_j\rfloor.\lfloor x_i\rfloor}=(x_j^{\frac{1}{\lfloor x_j\rfloor}})^{\lfloor x_i\rfloor.\lfloor x_j\rfloor}\iff x_i^{\frac{1}{\lfloor x_j\rfloor}}=x_j^{\frac{1}{\lfloor x_j\rfloor}}\]Hence $x_i^{\frac{1}{\lfloor x_i\rfloor}}$ is a constant. Let this constant be $c$. For all $1\leq i\leq k$, we get $x_i^{\frac{1}{\lfloor x_i\rfloor}}=c$. Since $x_i\neq x_j$, we observe that $\lfloor x_i\rfloor\neq \lfloor x_j\rfloor$. Let $x_i=m_i+r_i$ where $0\leq r_i<1$. Note that $m_i\geq 1$. Since $m_i+1>c^{m_i}\geq m_i$ for all $1\leq i\leq k$, \[\min\{(m_1+1)^{\frac{1}{m_1}},\dots, (m_k+1)^{\frac{1}{m_k}}\}>\max\{m_1^{\frac{1}{m_1}},\dots,m_k^{\frac{1}{m_k}}\}\]For $k=4$, choose $5^{\frac{1}{4}}>c>3^{\frac{1}{3}}$. Now, suppose that $k\geq 5$. Claim: $(n+1)^\frac{1}{n}\geq (n+2)^{\frac{1}{n+1}}\iff (n+1)^{n+1}\geq (n+2)^n$ for each positive integer. Proof: This holds for $n=1$ and $n=2$. Assume that it holds for $n-1$, we will show that $n+1$ also satisfies. \[(n+1)^{n+1}\geq (n+1)^{n+1}.\frac{(n+1)^{n-1}}{n^n}=[\frac{(n+1)^2}{n}]^n>(n+2)^n\]As we have expected.$\square$ Claim: $n^{\frac{1}{n}}\geq (n+3)^{\frac{1}{n+2}}\iff n^{n+2}\geq (n+3)^n$ for $n\geq 3$ positive integers. Proof: It holds for $n=3$. Assume that it's true for $n-1$. \[n^{n+2}\geq n^{n+2}.\frac{(n+2)^{n-1}}{(n-1)^{n+1}}\overset{?}{\geq} (n+3)^n\implies n^3.(n^2+2n)^{n-1}\overset{?}{\geq}(n^2-4n+3)^n(n-1)\]We see that $n^2+2n>n^2-4n+3$ and $n^3>(n^2-4n+3)(n-1)\iff n^3>n^3-4n^2+3n-n^2+4n-3$ or $5n^2-7n+3>0$ which is true.$\square$ If $m_2\geq 3$, then $m_2^{\frac{1}{m_2}}\geq (m_2+3)^{\frac{1}{m_2+2}}\geq (m_5+1)^{\frac{1}{m_5}}$ which results in a contradiction. Thus, $m_1=1$ and $m_2=2$. If $m_3=3$, then $3^{\frac{1}{3}}>6^{\frac{1}{5}}\geq (m_5+1)^{\frac{1}{m_5}}>3^{\frac{1}{3}}$ which is impossible again. Hence $m_3\geq 4$. However, $2^{\frac{1}{2}}>7^{\frac{1}{6}}\geq (m_5+1)^{\frac{1}{m_5}}$ So we get a contradiction once more. We see that $k\geq 5$ fails as desired.$\blacksquare$