{
  "index": "1994-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "For any real number $\\alpha$, define the function $f_{\\alpha}(x)\n= \\lfloor \\alpha x \\rfloor$. Let $n$ be a positive integer. Show that\nthere exists an $\\alpha$ such that for $1 \\leq k \\leq n$,\n\\[\nf_\\alpha^k(n^2) = n^2 - k = f_{\\alpha^k}(n^2).\n\\]",
  "solution": "Solution 1. We will show that \\( \\alpha \\) satisfies the conditions of the problem if and only if\n\\[\n1-\\frac{1}{n^{2}} \\leq \\alpha<\\left(\\frac{n^{2}-n+1}{n^{2}}\\right)^{1 / n}\n\\]\nand then show that this interval is nonempty.\nWe have \\( f_{\\alpha}^{k}\\left(n^{2}\\right)=n^{2}-k \\) for \\( k=1, \\ldots, n \\) if and only if \\( \\left\\lfloor\\alpha\\left(n^{2}-k+1\\right)\\right\\rfloor=n^{2}-k \\) for \\( k=1, \\ldots, n \\), which holds if and only if\n\\[\n\\frac{n^{2}-k}{n^{2}-k+1} \\leq \\alpha<1 \\quad \\text { for } k=1, \\ldots, n\n\\]\n\nSince\n\\[\n\\frac{n^{2}-k}{n^{2}-k+1}=1-\\frac{1}{n^{2}-k+1}\n\\]\ndecreases with \\( k \\), these hold if and only if \\( 1-\\frac{1}{n^{2}} \\leq \\alpha<1 \\). We assume this from now on.\n\nNext we consider the conditions \\( f_{\\alpha^{k}}\\left(n^{2}\\right)=n^{2}-k \\) for \\( k=1, \\ldots, n \\). Since \\( f_{\\alpha}^{k}\\left(n^{2}\\right) \\) is an integer less than \\( \\alpha^{k} n^{2} \\), we have \\( f_{\\alpha}^{k}\\left(n^{2}\\right) \\leq f_{\\alpha^{k}}\\left(n^{2}\\right) \\). We have already arranged that \\( f_{\\alpha}^{k}\\left(n^{2}\\right)=n^{2}-k \\), so \\( f_{\\alpha^{k}}\\left(n^{2}\\right)=n^{2}-k \\) if and only if \\( \\alpha^{k} n^{2}<n^{2}-k+1 \\). Moreover, if the latter holds for \\( k=n \\), then it holds for \\( k=1, \\ldots, n \\) by reverse induction, since multiplying \\( \\alpha^{k} n^{2}<n^{2}-k+1 \\) by\n\\[\n\\alpha^{-1} \\leq \\frac{n^{2}}{n^{2}-1}=1+\\frac{1}{n^{2}-1} \\leq 1+\\frac{1}{n^{2}-k+1} \\leq \\frac{n^{2}-(k-1)+1}{n^{2}-k+1}\n\\]\nyields the inductive step. Hence all the conditions hold if and only if\n\\[\n1-\\frac{1}{n^{2}} \\leq \\alpha<\\left(\\frac{n^{2}-n+1}{n^{2}}\\right)^{1 / n}\n\\]\n\nIt remains to show that this interval is nonempty, or equivalently, that\n\\[\n\\left(1-\\frac{1}{n^{2}}\\right)^{n}<1-\\frac{1}{n}+\\frac{1}{n^{2}} .\n\\]\n\nFirst we will prove\n\\[\n(1-x)^{n} \\leq 1-n x+\\binom{n}{2} x^{2} \\quad \\text { for } 0 \\leq x \\leq 1\n\\]\n\nThe two sides are equal at \\( x=0 \\), so it suffices to show that their derivatives satisfy\n\\[\n-n(1-x)^{n-1} \\leq-n+n(n-1) x \\quad \\text { for } 0 \\leq x \\leq 1 .\n\\]\n\nAgain the two sides are equal at \\( x=0 \\), so, differentiating again, it suffices to show\n\\[\nn(n-1)(1-x)^{n-2} \\leq n(n-1) \\quad \\text { for } 0 \\leq x \\leq 1 .\n\\]\n\nThis is obvious. Taking \\( x=1 / n^{2} \\) in (2) yields\n\\[\n\\left(1-\\frac{1}{n^{2}}\\right)^{n} \\leq 1-\\frac{1}{n}+\\frac{n(n-1) / 2}{n^{4}}<1-\\frac{1}{n}+\\frac{1}{n^{2}} .\n\\]\n\nRemark. Inequality (2) is a special case of the fact that for real \\( a>b>0 \\) and integers \\( 0<k<n \\), the partial binomial expansion\n\\[\na^{n}-\\binom{n}{1} a^{n-1} b+\\cdots+(-1)^{k}\\binom{n}{k} a^{n-k} b^{k}\n\\]\nis greater than or less than \\( (a-b)^{n} \\), according to whether \\( k \\) is even or odd. It can be proved by the same argument used to prove (2). This is sometimes called the Inclusion-Exclusion Inequality, because if \\( a, b \\) are sizes of sets \\( A \\supsetneq B \\supsetneq \\emptyset \\), then (3) is the overestimate or underestimate for \\( \\#(A-B)^{n} \\) obtained by terminating the inclusion-exclusion argument prematurely.\n\nSolution 2 (Dave Rusin). Let \\( \\alpha=e^{-1 / n^{2}} \\). The same method used to prove (2) shows that\n\\[\n1-r+r^{2} / 2>e^{-r}>1-r\n\\]\n\nSubstituting \\( r=k / n^{2}(0<k \\leq n) \\) and simplifying, we find\n\\[\nn^{2}-k+\\frac{1}{2}\\left(\\frac{k}{n}\\right)^{2}>\\alpha^{k} n^{2}>n^{2}-k .\n\\]\n\nThe right side is an integer, and the left side is at most \\( 1 / 2 \\) more, so \\( \\left\\lfloor\\alpha^{k} n^{2}\\right\\rfloor=n^{2}-k \\). Since \\( \\alpha>1-1 / n^{2} \\) (again by (4)), \\( f_{\\alpha}^{k}\\left(n^{2}\\right)=n^{2}-k \\) for \\( 1 \\leq k \\leq n \\) by the same argument as in the previous solution.",
  "vars": [
    "n",
    "k",
    "x",
    "r",
    "a",
    "b",
    "A",
    "B"
  ],
  "params": [
    "\\\\alpha",
    "f_\\\\alpha"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "squaresize",
        "k": "iterindex",
        "x": "realinput",
        "r": "ratioamt",
        "a": "scalarone",
        "b": "scalartwo",
        "A": "superset",
        "B": "subsetbb",
        "\\alpha": "scalaralpha",
        "f_\\alpha": "alphafunc"
      },
      "question": "For any real number $scalaralpha$, define the function $alphafunc(realinput)=\\lfloor scalaralpha\\,realinput\\rfloor$. Let $squaresize$ be a positive integer. Show that there exists a $scalaralpha$ such that for $1 \\leq iterindex \\leq squaresize$,\\[\nalphafunc^{iterindex}(squaresize^2)=squaresize^2-iterindex=f_{scalaralpha^{iterindex}}(squaresize^2).\n\\]",
      "solution": "Solution 1. We will show that $scalaralpha$ satisfies the conditions of the problem if and only if\n\\[\n1-\\frac{1}{squaresize^{2}}\\le scalaralpha<\\left(\\frac{squaresize^{2}-squaresize+1}{squaresize^{2}}\\right)^{1/squaresize}\n\\]\nand then show that this interval is non-empty.\n\nWe have $alphafunc^{iterindex}(squaresize^{2})=squaresize^{2}-iterindex$ for $iterindex=1,\\ldots ,squaresize$ if and only if $\\lfloor scalaralpha\\,(squaresize^{2}-iterindex+1)\\rfloor=squaresize^{2}-iterindex$ for $iterindex=1,\\ldots ,squaresize$, which holds if and only if\n\\[\n\\frac{squaresize^{2}-iterindex}{squaresize^{2}-iterindex+1}\\le scalaralpha<1\\qquad( iterindex=1,\\ldots ,squaresize).\n\\]\nSince\n\\[\n\\frac{squaresize^{2}-iterindex}{squaresize^{2}-iterindex+1}=1-\\frac{1}{squaresize^{2}-iterindex+1}\n\\]\ndecreases with $iterindex$, these inequalities hold precisely when $1-\\frac{1}{squaresize^{2}}\\le scalaralpha<1$. We assume this from now on.\n\nNext we consider the conditions $f_{scalaralpha^{iterindex}}(squaresize^{2})=squaresize^{2}-iterindex$ for $iterindex=1,\\ldots ,squaresize$.  Because $alphafunc^{iterindex}(squaresize^{2})$ is an integer less than $scalaralpha^{iterindex}squaresize^{2}$, we have $alphafunc^{iterindex}(squaresize^{2})\\le f_{scalaralpha^{iterindex}}(squaresize^{2})$.  We have already arranged that $alphafunc^{iterindex}(squaresize^{2})=squaresize^{2}-iterindex$, so $f_{scalaralpha^{iterindex}}(squaresize^{2})=squaresize^{2}-iterindex$ occurs exactly when\n\\[\nscalaralpha^{iterindex}squaresize^{2}<squaresize^{2}-iterindex+1.\n\\]\nIf this holds for $iterindex=squaresize$ it also holds for $iterindex=1,\\ldots ,squaresize$ by reverse induction, since multiplying $scalaralpha^{iterindex}squaresize^{2}<squaresize^{2}-iterindex+1$ by\n\\[\nscalaralpha^{-1}\\le\\frac{squaresize^{2}}{squaresize^{2}-1}=1+\\frac{1}{squaresize^{2}-1}\\le1+\\frac{1}{squaresize^{2}-iterindex+1}\\le\\frac{squaresize^{2}-(iterindex-1)+1}{squaresize^{2}-iterindex+1}\n\\]\nyields the inductive step.  Hence all the conditions hold exactly when\n\\[\n1-\\frac{1}{squaresize^{2}}\\le scalaralpha<\\left(\\frac{squaresize^{2}-squaresize+1}{squaresize^{2}}\\right)^{1/squaresize}.\n\\]\n\nIt remains to show that this interval is non-empty, i.e.\n\\[\n\\left(1-\\frac{1}{squaresize^{2}}\\right)^{squaresize}<1-\\frac{1}{squaresize}+\\frac{1}{squaresize^{2}}.\n\\]\n\nFirst we prove\n\\[\n(1-realinput)^{squaresize}\\le1-squaresize\\,realinput+\\binom{squaresize}{2}realinput^{2}\\qquad(0\\le realinput\\le1).\n\\]\nThe two sides agree at $realinput=0$, so it suffices to check the derivatives:\n\\[\n-squaresize(1-realinput)^{squaresize-1}\\le -squaresize+squaresize(squaresize-1)realinput\\qquad(0\\le realinput\\le1).\n\\]\nAgain they coincide at $realinput=0$; differentiating once more we need only verify\n\\[\nsquaresize(squaresize-1)(1-realinput)^{squaresize-2}\\le squaresize(squaresize-1)\\qquad(0\\le realinput\\le1),\n\\]\nwhich is clear.  Taking $realinput=1/squaresize^{2}$ gives\n\\[\n\\left(1-\\frac{1}{squaresize^{2}}\\right)^{squaresize}\\le1-\\frac{1}{squaresize}+\\frac{squaresize(squaresize-1)/2}{squaresize^{4}}<1-\\frac{1}{squaresize}+\\frac{1}{squaresize^{2}}.\n\\]\n\nRemark.  The inequality just proved is a special case of the fact that for real $scalarone>scalartwo>0$ and integers $0<iterindex<squaresize$, the partial binomial sum\n\\[\nscalarone^{squaresize}-\\binom{squaresize}{1}scalarone^{squaresize-1}scalartwo+\\cdots+(-1)^{iterindex}\\binom{squaresize}{iterindex}scalarone^{squaresize-iterindex}scalartwo^{iterindex}\n\\]\nis greater or less than $(scalarone-scalartwo)^{squaresize}$ according as $iterindex$ is even or odd.  This is sometimes called the Inclusion–Exclusion Inequality; if $scalarone,scalartwo$ are the sizes of sets $superset\\supsetneq subsetbb\\supsetneq\\emptyset$, the inequality gives the over- or under-estimate for $\\#(superset\\setminus subsetbb)^{squaresize}$ obtained by truncating the inclusion–exclusion expansion.\n\nSolution 2 (Dave Rusin).  Let $scalaralpha=e^{-1/squaresize^{2}}$.  The same argument used above shows\n\\[\n1-ratioamt+\\frac{ratioamt^{2}}{2}>e^{-ratioamt}>1-ratioamt.\n\\]\nSubstituting $ratioamt=iterindex/squaresize^{2}$ $(0<iterindex\\le squaresize)$ we obtain\n\\[\nsquaresize^{2}-iterindex+\\frac12\\left(\\frac{iterindex}{squaresize}\\right)^{2}>scalaralpha^{iterindex}squaresize^{2}>squaresize^{2}-iterindex.\n\\]\nThe right side is an integer and the left is at most $\\tfrac12$ larger, so $\\lfloor scalaralpha^{iterindex}squaresize^{2}\\rfloor=squaresize^{2}-iterindex$.  Since $scalaralpha>1-1/squaresize^{2}$, we also have $alphafunc^{iterindex}(squaresize^{2})=squaresize^{2}-iterindex$ for $1\\le iterindex\\le squaresize$ by the argument of Solution 1."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "k": "flashcard",
        "x": "batteries",
        "r": "raincloud",
        "a": "drumstick",
        "b": "lighthouse",
        "A": "sandstone",
        "B": "butterfly",
        "\\\\alpha": "marshmallow",
        "f_\\\\alpha": "cheesecake"
      },
      "question": "For any real number $marshmallow$, define the function $cheesecake(batteries)\n= \\lfloor marshmallow\\, batteries \\rfloor$. Let pineapple be a positive integer. Show that\nthere exists an $marshmallow$ such that for $1 \\leq flashcard \\leq pineapple$,\n\\[\ncheesecake^{flashcard}(pineapple^{2}) = pineapple^{2} - flashcard = f_{marshmallow^{flashcard}}(pineapple^{2}).\n\\]",
      "solution": "Solution 1. We will show that \\( marshmallow \\) satisfies the conditions of the problem if and only if\n\\[\n1-\\frac{1}{pineapple^{2}} \\leq marshmallow<\\left(\\frac{pineapple^{2}-pineapple+1}{pineapple^{2}}\\right)^{1 / pineapple}\n\\]\nand then show that this interval is nonempty.\nWe have \\( cheesecake^{flashcard}\\left(pineapple^{2}\\right)=pineapple^{2}-flashcard \\) for \\( flashcard=1, \\ldots, pineapple \\) if and only if \\( \\left\\lfloor marshmallow\\left(pineapple^{2}-flashcard+1\\right)\\right\\rfloor=pineapple^{2}-flashcard \\) for \\( flashcard=1, \\ldots, pineapple \\), which holds if and only if\n\\[\n\\frac{pineapple^{2}-flashcard}{pineapple^{2}-flashcard+1} \\leq marshmallow<1 \\quad \\text { for } flashcard=1, \\ldots, pineapple\n\\]\nSince\n\\[\n\\frac{pineapple^{2}-flashcard}{pineapple^{2}-flashcard+1}=1-\\frac{1}{pineapple^{2}-flashcard+1}\n\\]\ndecreases with \\( flashcard \\), these hold if and only if \\( 1-\\frac{1}{pineapple^{2}} \\leq marshmallow<1 \\). We assume this from now on.\n\nNext we consider the conditions \\( f_{marshmallow^{flashcard}}\\left(pineapple^{2}\\right)=pineapple^{2}-flashcard \\) for \\( flashcard=1, \\ldots, pineapple \\). Since \\( cheesecake^{flashcard}\\left(pineapple^{2}\\right) \\) is an integer less than \\( marshmallow^{flashcard} \\, pineapple^{2} \\), we have \\( cheesecake^{flashcard}\\left(pineapple^{2}\\right) \\leq f_{marshmallow^{flashcard}}\\left(pineapple^{2}\\right) \\). We have already arranged that \\( cheesecake^{flashcard}\\left(pineapple^{2}\\right)=pineapple^{2}-flashcard \\), so \\( f_{marshmallow^{flashcard}}\\left(pineapple^{2}\\right)=pineapple^{2}-flashcard \\) if and only if \\( marshmallow^{flashcard} \\, pineapple^{2}<pineapple^{2}-flashcard+1 \\). Moreover, if the latter holds for \\( flashcard=pineapple \\), then it holds for \\( flashcard=1, \\ldots, pineapple \\) by reverse induction, since multiplying \\( marshmallow^{flashcard} \\, pineapple^{2}<pineapple^{2}-flashcard+1 \\) by\n\\[\nmarshmallow^{-1} \\leq \\frac{pineapple^{2}}{pineapple^{2}-1}=1+\\frac{1}{pineapple^{2}-1} \\leq 1+\\frac{1}{pineapple^{2}-flashcard+1} \\leq \\frac{pineapple^{2}-(flashcard-1)+1}{pineapple^{2}-flashcard+1}\n\\]\nyields the inductive step. Hence all the conditions hold if and only if\n\\[\n1-\\frac{1}{pineapple^{2}} \\leq marshmallow<\\left(\\frac{pineapple^{2}-pineapple+1}{pineapple^{2}}\\right)^{1 / pineapple}\n\\]\nIt remains to show that this interval is nonempty, or equivalently, that\n\\[\n\\left(1-\\frac{1}{pineapple^{2}}\\right)^{pineapple}<1-\\frac{1}{pineapple}+\\frac{1}{pineapple^{2}} .\n\\]\nFirst we will prove\n\\[\n(1-batteries)^{pineapple} \\leq 1-pineapple\\, batteries+\\binom{pineapple}{2} batteries^{2} \\quad \\text { for } 0 \\leq batteries \\leq 1\n\\]\nThe two sides are equal at \\( batteries=0 \\), so it suffices to show that their derivatives satisfy\n\\[\n-pineapple(1-batteries)^{pineapple-1} \\leq-pineapple+pineapple(pineapple-1) batteries \\quad \\text { for } 0 \\leq batteries \\leq 1 .\n\\]\nAgain the two sides are equal at \\( batteries=0 \\), so, differentiating again, it suffices to show\n\\[\npineapple(pineapple-1)(1-batteries)^{pineapple-2} \\leq pineapple(pineapple-1) \\quad \\text { for } 0 \\leq batteries \\leq 1 .\n\\]\nThis is obvious. Taking \\( batteries=1 / pineapple^{2} \\) in (2) yields\n\\[\n\\left(1-\\frac{1}{pineapple^{2}}\\right)^{pineapple} \\leq 1-\\frac{1}{pineapple}+\\frac{pineapple(pineapple-1) / 2}{pineapple^{4}}<1-\\frac{1}{pineapple}+\\frac{1}{pineapple^{2}} .\n\\]\nRemark. Inequality (2) is a special case of the fact that for real \\( drumstick>lighthouse>0 \\) and integers \\( 0<flashcard<pineapple \\), the partial binomial expansion\n\\[\ndrumstick^{pineapple}-\\binom{pineapple}{1} drumstick^{pineapple-1} lighthouse+\\cdots+(-1)^{flashcard}\\binom{pineapple}{flashcard} drumstick^{pineapple-flashcard} lighthouse^{flashcard}\n\\]\nis greater than or less than \\( (drumstick-lighthouse)^{pineapple} \\), according to whether \\( flashcard \\) is even or odd. It can be proved by the same argument used to prove (2). This is sometimes called the Inclusion-Exclusion Inequality, because if \\( drumstick, lighthouse \\) are sizes of sets \\( sandstone \\supsetneq butterfly \\supsetneq \\emptyset \\), then (3) is the overestimate or underestimate for \\( \\#(sandstone-lighthouse)^{pineapple} \\) obtained by terminating the inclusion-exclusion argument prematurely.\n\nSolution 2 (Dave Rusin). Let \\( marshmallow=e^{-1 / pineapple^{2}} \\). The same method used to prove (2) shows that\n\\[\n1-raincloud+raincloud^{2} / 2>e^{-raincloud}>1-raincloud\n\\]\nSubstituting \\( raincloud=flashcard / pineapple^{2}(0<flashcard \\leq pineapple) \\) and simplifying, we find\n\\[\npineapple^{2}-flashcard+\\frac{1}{2}\\left(\\frac{flashcard}{pineapple}\\right)^{2}>marshmallow^{flashcard} \\, pineapple^{2}>pineapple^{2}-flashcard .\n\\]\nThe right side is an integer, and the left side is at most \\( 1 / 2 \\) more, so \\( \\left\\lfloor marshmallow^{flashcard} \\, pineapple^{2}\\right\\rfloor=pineapple^{2}-flashcard \\). Since \\( marshmallow>1-1 / pineapple^{2} \\) (again by (4)), \\( cheesecake^{flashcard}\\left(pineapple^{2}\\right)=pineapple^{2}-flashcard \\) for \\( 1 \\leq flashcard \\leq pineapple \\) by the same argument as in the previous solution."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "uncountable",
        "k": "constantvalue",
        "x": "fixednumber",
        "r": "largescale",
        "a": "minimumval",
        "b": "maximalval",
        "A": "subsetset",
        "B": "supersetset",
        "\\alpha": "omegaparam",
        "f_\\alpha": "fixedfunction"
      },
      "question": "For any real number $omegaparam$, define the function fixedfunction(fixednumber)\n= \\lfloor omegaparam fixednumber \\rfloor$. Let uncountable be a positive integer. Show that\nthere exists an omegaparam such that for $1 \\leq constantvalue \\leq uncountable$,\n\\[\nfixedfunction^{constantvalue}(uncountable^{2}) = uncountable^{2} - constantvalue = f_{omegaparam^{constantvalue}}(uncountable^{2}).\n\\]",
      "solution": "Solution 1. We will show that \\( omegaparam \\) satisfies the conditions of the problem if and only if\n\\[\n1-\\frac{1}{uncountable^{2}} \\leq omegaparam<\\left(\\frac{uncountable^{2}-uncountable+1}{uncountable^{2}}\\right)^{1 / uncountable}\n\\]\nand then show that this interval is nonempty.\nWe have fixedfunction^{constantvalue}\\left(uncountable^{2}\\right)=uncountable^{2}-constantvalue for constantvalue=1, \\ldots, uncountable if and only if \\( \\left\\lfloor omegaparam\\left(uncountable^{2}-constantvalue+1\\right)\\right\\rfloor=uncountable^{2}-constantvalue \\) for constantvalue=1, \\ldots, uncountable, which holds if and only if\n\\[\n\\frac{uncountable^{2}-constantvalue}{uncountable^{2}-constantvalue+1} \\leq omegaparam<1 \\quad \\text { for } constantvalue=1, \\ldots, uncountable\n\\]\nSince\n\\[\n\\frac{uncountable^{2}-constantvalue}{uncountable^{2}-constantvalue+1}=1-\\frac{1}{uncountable^{2}-constantvalue+1}\n\\]\ndecreases with \\( constantvalue \\), these hold if and only if \\( 1-\\frac{1}{uncountable^{2}} \\leq omegaparam<1 \\). We assume this from now on.\n\nNext we consider the conditions \\( f_{omegaparam^{constantvalue}}\\left(uncountable^{2}\\right)=uncountable^{2}-constantvalue \\) for constantvalue=1, \\ldots, uncountable. Since fixedfunction^{constantvalue}\\left(uncountable^{2}\\right) is an integer less than \\( omegaparam^{constantvalue}uncountable^{2} \\), we have fixedfunction^{constantvalue}\\left(uncountable^{2}\\right) \\leq f_{omegaparam^{constantvalue}}\\left(uncountable^{2}\\right). We have already arranged that fixedfunction^{constantvalue}\\left(uncountable^{2}\\right)=uncountable^{2}-constantvalue, so \\( f_{omegaparam^{constantvalue}}\\left(uncountable^{2}\\right)=uncountable^{2}-constantvalue \\) if and only if \\( omegaparam^{constantvalue}uncountable^{2}<uncountable^{2}-constantvalue+1 \\). Moreover, if the latter holds for constantvalue=uncountable, then it holds for constantvalue=1, \\ldots, uncountable by reverse induction, since multiplying \\( omegaparam^{constantvalue}uncountable^{2}<uncountable^{2}-constantvalue+1 \\) by\n\\[\nomegaparam^{-1} \\leq \\frac{uncountable^{2}}{uncountable^{2}-1}=1+\\frac{1}{uncountable^{2}-1} \\leq 1+\\frac{1}{uncountable^{2}-constantvalue+1} \\leq \\frac{uncountable^{2}-(constantvalue-1)+1}{uncountable^{2}-constantvalue+1}\n\\]\nyields the inductive step. Hence all the conditions hold if and only if\n\\[\n1-\\frac{1}{uncountable^{2}} \\leq omegaparam<\\left(\\frac{uncountable^{2}-uncountable+1}{uncountable^{2}}\\right)^{1 / uncountable}\n\\]\n\nIt remains to show that this interval is nonempty, or equivalently, that\n\\[\n\\left(1-\\frac{1}{uncountable^{2}}\\right)^{uncountable}<1-\\frac{1}{uncountable}+\\frac{1}{uncountable^{2}} .\n\\]\n\nFirst we will prove\n\\[\n(1-fixednumber)^{uncountable} \\leq 1-uncountable fixednumber+\\binom{uncountable}{2} fixednumber^{2} \\quad \\text { for } 0 \\leq fixednumber \\leq 1\n\\]\n\nThe two sides are equal at \\( fixednumber=0 \\), so it suffices to show that their derivatives satisfy\n\\[\n-uncountable(1-fixednumber)^{uncountable-1} \\leq-uncountable+uncountable(uncountable-1) fixednumber \\quad \\text { for } 0 \\leq fixednumber \\leq 1 .\n\\]\n\nAgain the two sides are equal at \\( fixednumber=0 \\), so, differentiating again, it suffices to show\n\\[\nuncountable(uncountable-1)(1-fixednumber)^{uncountable-2} \\leq uncountable(uncountable-1) \\quad \\text { for } 0 \\leq fixednumber \\leq 1 .\n\\]\n\nThis is obvious. Taking \\( fixednumber=1 / uncountable^{2} \\) in (2) yields\n\\[\n\\left(1-\\frac{1}{uncountable^{2}}\\right)^{uncountable} \\leq 1-\\frac{1}{uncountable}+\\frac{uncountable(uncountable-1) / 2}{uncountable^{4}}<1-\\frac{1}{uncountable}+\\frac{1}{uncountable^{2}} .\n\\]\n\nRemark. Inequality (2) is a special case of the fact that for real \\( minimumval>maximalval>0 \\) and integers \\( 0<constantvalue<uncountable \\), the partial binomial expansion\n\\[\nminimumval^{uncountable}-\\binom{uncountable}{1} minimumval^{uncountable-1} maximalval+\\cdots+(-1)^{constantvalue}\\binom{uncountable}{constantvalue} minimumval^{uncountable-constantvalue} maximalval^{constantvalue}\n\\]\nis greater than or less than \\( (minimumval-maximalval)^{uncountable} \\), according to whether \\( constantvalue \\) is even or odd. It can be proved by the same argument used to prove (2). This is sometimes called the Inclusion-Exclusion Inequality, because if \\( minimumval, maximalval \\) are sizes of sets \\( subsetset \\supsetneq supersetset \\supsetneq \\emptyset \\), then (3) is the overestimate or underestimate for \\#(subsetset-supersetset)^{uncountable} obtained by terminating the inclusion-exclusion argument prematurely.\n\nSolution 2 (Dave Rusin). Let \\( omegaparam=e^{-1 / uncountable^{2}} \\). The same method used to prove (2) shows that\n\\[\n1-largescale+largescale^{2} / 2>e^{-largescale}>1-largescale\n\\]\n\nSubstituting \\( largescale=constantvalue / uncountable^{2}(0<constantvalue \\leq uncountable) \\) and simplifying, we find\n\\[\nuncountable^{2}-constantvalue+\\frac{1}{2}\\left(\\frac{constantvalue}{uncountable}\\right)^{2}>omegaparam^{constantvalue} uncountable^{2}>uncountable^{2}-constantvalue .\n\\]\n\nThe right side is an integer, and the left side is at most \\( 1 / 2 \\) more, so \\( \\left\\lfloor omegaparam^{constantvalue} uncountable^{2}\\right\\rfloor=uncountable^{2}-constantvalue \\). Since \\( omegaparam>1-1 / uncountable^{2} \\) (again by (4)), fixedfunction^{constantvalue}\\left(uncountable^{2}\\right)=uncountable^{2}-constantvalue for \\( 1 \\leq constantvalue \\leq uncountable \\) by the same argument as in the previous solution."
    },
    "garbled_string": {
      "map": {
        "n": "kibuvazx",
        "k": "gxqlameu",
        "x": "odnukype",
        "r": "jifyteda",
        "a": "wupelori",
        "b": "madosire",
        "A": "zicogavu",
        "B": "tepuyola",
        "\\alpha": "yorunepa",
        "f_\\alpha": "vutahime"
      },
      "question": "For any real number $yorunepa$, define the function $vutahime(odnukype)\n= \\lfloor yorunepa odnukype \\rfloor$. Let $kibuvazx$ be a positive integer. Show that\nthere exists an $yorunepa$ such that for $1 \\leq gxqlameu \\leq kibuvazx$,\n\\[\nvutahime^{gxqlameu}(kibuvazx^2) = kibuvazx^2 - gxqlameu = f_{yorunepa^{gxqlameu}}(kibuvazx^2).\n\\]",
      "solution": "Solution 1. We will show that \\( yorunepa \\) satisfies the conditions of the problem if and only if\n\\[\n1-\\frac{1}{kibuvazx^{2}} \\leq yorunepa<\\left(\\frac{kibuvazx^{2}-kibuvazx+1}{kibuvazx^{2}}\\right)^{1 / kibuvazx}\n\\]\nand then show that this interval is nonempty.\nWe have \\( vutahime^{gxqlameu}\\left(kibuvazx^{2}\\right)=kibuvazx^{2}-gxqlameu \\) for \\( gxqlameu=1, \\ldots, kibuvazx \\) if and only if \\( \\left\\lfloor yorunepa\\left(kibuvazx^{2}-gxqlameu+1\\right)\\right\\rfloor=kibuvazx^{2}-gxqlameu \\) for \\( gxqlameu=1, \\ldots, kibuvazx \\), which holds if and only if\n\\[\n\\frac{kibuvazx^{2}-gxqlameu}{kibuvazx^{2}-gxqlameu+1} \\leq yorunepa<1 \\quad \\text { for } gxqlameu=1, \\ldots, kibuvazx\n\\]\n\nSince\n\\[\n\\frac{kibuvazx^{2}-gxqlameu}{kibuvazx^{2}-gxqlameu+1}=1-\\frac{1}{kibuvazx^{2}-gxqlameu+1}\n\\]\ndecreases with \\( gxqlameu \\), these hold if and only if \\( 1-\\frac{1}{kibuvazx^{2}} \\leq yorunepa<1 \\). We assume this from now on.\n\nNext we consider the conditions \\( f_{yorunepa^{gxqlameu}}\\left(kibuvazx^{2}\\right)=kibuvazx^{2}-gxqlameu \\) for \\( gxqlameu=1, \\ldots, kibuvazx \\). Since \\( vutahime^{gxqlameu}\\left(kibuvazx^{2}\\right) \\) is an integer less than \\( yorunepa^{gxqlameu} kibuvazx^{2} \\), we have \\( vutahime^{gxqlameu}\\left(kibuvazx^{2}\\right) \\leq f_{yorunepa^{gxqlameu}}\\left(kibuvazx^{2}\\right) \\). We have already arranged that \\( vutahime^{gxqlameu}\\left(kibuvazx^{2}\\right)=kibuvazx^{2}-gxqlameu \\), so \\( f_{yorunepa^{gxqlameu}}\\left(kibuvazx^{2}\\right)=kibuvazx^{2}-gxqlameu \\) if and only if \\( yorunepa^{gxqlameu} kibuvazx^{2}<kibuvazx^{2}-gxqlameu+1 \\). Moreover, if the latter holds for \\( gxqlameu=kibuvazx \\), then it holds for \\( gxqlameu=1, \\ldots, kibuvazx \\) by reverse induction, since multiplying \\( yorunepa^{gxqlameu} kibuvazx^{2}<kibuvazx^{2}-gxqlameu+1 \\) by\n\\[\nyorunepa^{-1} \\leq \\frac{kibuvazx^{2}}{kibuvazx^{2}-1}=1+\\frac{1}{kibuvazx^{2}-1} \\leq 1+\\frac{1}{kibuvazx^{2}-gxqlameu+1} \\leq \\frac{kibuvazx^{2}-(gxqlameu-1)+1}{kibuvazx^{2}-gxqlameu+1}\n\\]\nyields the inductive step. Hence all the conditions hold if and only if\n\\[\n1-\\frac{1}{kibuvazx^{2}} \\leq yorunepa<\\left(\\frac{kibuvazx^{2}-kibuvazx+1}{kibuvazx^{2}}\\right)^{1 / kibuvazx}\n\\]\n\nIt remains to show that this interval is nonempty, or equivalently, that\n\\[\n\\left(1-\\frac{1}{kibuvazx^{2}}\\right)^{kibuvazx}<1-\\frac{1}{kibuvazx}+\\frac{1}{kibuvazx^{2}} .\n\\]\n\nFirst we will prove\n\\[\n(1-odnukype)^{kibuvazx} \\leq 1-kibuvazx \\, odnukype+\\binom{kibuvazx}{2} \\, odnukype^{2} \\quad \\text { for } 0 \\leq odnukype \\leq 1\n\\]\n\nThe two sides are equal at \\( odnukype=0 \\), so it suffices to show that their derivatives satisfy\n\\[\n-kibuvazx(1-odnukype)^{kibuvazx-1} \\leq-kibuvazx+kibuvazx(kibuvazx-1) \\, odnukype \\quad \\text { for } 0 \\leq odnukype \\leq 1 .\n\\]\n\nAgain the two sides are equal at \\( odnukype=0 \\), so, differentiating again, it suffices to show\n\\[\nkibuvazx(kibuvazx-1)(1-odnukype)^{kibuvazx-2} \\leq kibuvazx(kibuvazx-1) \\quad \\text { for } 0 \\leq odnukype \\leq 1 .\n\\]\n\nThis is obvious. Taking \\( odnukype=1 / kibuvazx^{2} \\) in (2) yields\n\\[\n\\left(1-\\frac{1}{kibuvazx^{2}}\\right)^{kibuvazx} \\leq 1-\\frac{1}{kibuvazx}+\\frac{kibuvazx(kibuvazx-1) / 2}{kibuvazx^{4}}<1-\\frac{1}{kibuvazx}+\\frac{1}{kibuvazx^{2}} .\n\\]\n\nRemark. Inequality (2) is a special case of the fact that for real \\( wupelori>madosire>0 \\) and integers \\( 0<gxqlameu<kibuvazx \\), the partial binomial expansion\n\\[\nwupelori^{kibuvazx}-\\binom{kibuvazx}{1} wupelori^{kibuvazx-1} madosire+\\cdots+(-1)^{gxqlameu}\\binom{kibuvazx}{gxqlameu} wupelori^{kibuvazx-gxqlameu} madosire^{gxqlameu}\n\\]\nis greater than or less than \\( (wupelori-madosire)^{kibuvazx} \\), according to whether \\( gxqlameu \\) is even or odd. It can be proved by the same argument used to prove (2). This is sometimes called the Inclusion-Exclusion Inequality, because if \\( wupelori, madosire \\) are sizes of sets \\( zicogavu \\supsetneq tepuyola \\supsetneq \\emptyset \\), then (3) is the overestimate or underestimate for \\( \\#(zicogavu-tepuyola)^{kibuvazx} \\) obtained by terminating the inclusion-exclusion argument prematurely.\n\nSolution 2 (Dave Rusin). Let \\( yorunepa=e^{-1 / kibuvazx^{2}} \\). The same method used to prove (2) shows that\n\\[\n1-jifyteda+jifyteda^{2} / 2>e^{-jifyteda}>1-jifyteda\n\\]\n\nSubstituting \\( jifyteda=gxqlameu / kibuvazx^{2}(0<gxqlameu \\leq kibuvazx) \\) and simplifying, we find\n\\[\nkibuvazx^{2}-gxqlameu+\\frac{1}{2}\\left(\\frac{gxqlameu}{kibuvazx}\\right)^{2}>yorunepa^{gxqlameu} kibuvazx^{2}>kibuvazx^{2}-gxqlameu .\n\\]\n\nThe right side is an integer, and the left side is at most \\( 1 / 2 \\) more, so \\( \\left\\lfloor yorunepa^{gxqlameu} kibuvazx^{2}\\right\\rfloor=kibuvazx^{2}-gxqlameu \\). Since \\( yorunepa>1-1 / kibuvazx^{2} \\) (again by (4)), \\( vutahime^{gxqlameu}\\left(kibuvazx^{2}\\right)=kibuvazx^{2}-gxqlameu \\) for \\( 1 \\leq gxqlameu \\leq kibuvazx \\) by the same argument as in the previous solution."
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 be an integer and put N := n^3.  For a real number α define the map\n        f_α : ℝ → ℝ , f_α(x) = ⌊αx⌋ .\nFor k ≥ 1 write  f_α^{∘ k} for the k–fold iterate of f_α (composition of k copies of f_α).\n\nProve that there exists a real number α such that, simultaneously for every k = 1,2,…,n,\n        f_α^{∘ k}(N) = N − 2k = f_{α^{k}}(N).\nIn other words, the same number N – 2k is produced both by k-fold iteration of f_α and by applying the single floor map with parameter α^k.",
      "solution": "Throughout we fix an integer n ≥ 2 and write N := n^3.  All floor symbols ⌊·⌋ refer to the ordinary order on ℝ.\n\n0.  A convenient interval for α\n--------------------------------\nSet\n        L := 1 − 2 ⁄ N and U := ((N − 2n + 1)/N)^{1/n}.     (⋆)\nWe shall prove\n(i)   L < U (hence the open interval (L , U) is non–empty), and\n(ii)  every α ∈ (L , U) satisfies for all k = 1,…,n\n        f_α^{∘ k}(N) = N − 2k = f_{α^{k}}(N).                 (1)\n\nThe proof splits into two parts, one for each equality in (1).\n\n1.  The iterates f_α^{∘ k}(N)\n-----------------------------\nFor 1 ≤ k ≤ n define\n        l_k := (N − 2k)/(N − 2k + 2), u_k := (N − 2k + 1)/(N − 2k + 2).\n\nLemma 1 (monotonicity of the bounds).\nThe sequences (l_k)_{k=1}^n and (u_k)_{k=1}^n are strictly decreasing.\n\nProof.  Put A := N − 2k (>0).  Then\n        l_{k+1} = (A−2)/A, l_k = A/(A+2), and l_{k+1}−l_k = −4/(A(A+2)) < 0.\nA similar calculation yields u_{k+1} − u_k = −2/(A(A+2)) < 0. ∎\n\nLemma 2 (one–step criterion).\nIf α satisfies l_j ≤ α < u_j for every j = 1,…,k, then\n        f_α^{∘ j}(N) = N − 2j for all j = 1,…,k.             (2)\n\nProof.  Induction over j.  The base j = 1 is immediate from l_1 ≤ α < u_1.\nAssume (2) holds for some j < k.  Then\n        f_α^{∘(j+1)}(N) = ⌊α · f_α^{∘ j}(N)⌋ = ⌊α (N − 2j)⌋.\nBecause α ≥ l_{j+1} and α < u_{j+1}, the real number α (N − 2j) lies in the half–open unit interval [N − 2(j+1), N − 2(j+1)+1), whose only integer is N − 2(j+1), giving the claim. ∎\n\nNext we relate the bounds l_k , u_k with L , U.\nLemma 1 gives l_k ≥ l_n, and since N > N − 2n + 2 we have L = 1 − 2/N > 1 − 2/(N − 2n + 2) = l_n.  Hence\n        α ≥ L  ⇒  α ≥ l_k for every k ≤ n.                (3)\nLemma 3 below shows U ≤ u_n ≤ u_k, so\n        α < U  ⇒  α < u_k for every k ≤ n.                (4)\nCombining (3), (4) with Lemma 2 (taking k = n) we obtain the first equality in (1).\n\nLemma 3.  With L, U as in (⋆) one has U ≤ u_n.\n\nProof.  Put D := N − 2n + 1 (>0).  Then u_n = D/(D+1) and U = (D/(D+2n−1))^{1/n}.  We must show\n        (D/(D+2n−1))^{1/n} ≤ D/(D+1).           (5)\nLet x := 1/D, 0 < x ≤ 1/(2n−1) (because N = n^3 ≥ 2n+1).  Inequality (5) is equivalent to\n        ln(1+(2n−1)x) ≥ n ln(1+x).               (6)\nDefine h(x) := ln(1+(2n−1)x) − n ln(1+x).  Then\n        h′(x) = (2n−1)/(1+(2n−1)x) − n/(1+x)\n               = (n−1)(1−(2n−1)x) / [(1+(2n−1)x)(1+x)] ≥ 0,\nbecause x ≤ 1/(2n−1).  Thus h is non-decreasing and h(0)=0, whence h(x) ≥ 0, proving (6) and therefore (5). ∎\n\n2.  The single–step values f_{α^{k}}(N)\n---------------------------------------\nFor any real y we have ⌊y⌋ ≤ y < ⌊y⌋+1.  Consequently\n        f_{α^{k}}(N) = N − 2k ⇔ N − 2k ≤ α^{k} N < N − 2k + 1.       (7)\n\nLower bound in (7).\nBecause α ≥ L = 1 − 2/N and Bernoulli’s inequality (1−x)^k ≥ 1−kx (for x ≥ 0),\n        α^{k} ≥ (1 − 2/N)^k ≥ 1 − 2k/N,\nso α^{k} N ≥ N − 2k.\n\nUpper bound in (7).\nSince α < U we have α^{k} < U^{k}.  Put β := U.  By definition β^{n}N = N − 2n + 1.  For 1 ≤ k ≤ n we estimate β^{k}N via the elementary convexity inequality (1−t)^θ ≤ 1 − θt for 0 ≤ θ ≤ 1 and 0 ≤ t ≤ 1.  Here t := (2n−1)/N (<1) and β^{n} = 1 − t, whence\n        β^{k} ≤ 1 − (k/n)t.\nMultiplying by N and noting Nt = 2n−1 gives\n        β^{k} N ≤ N − (k/n)(2n−1) = N − 2k + k/n < N − 2k + 1   (for k ≤ n).     (8)\nCombining α^{k}N < β^{k}N with (8) yields the strict upper inequality in (7).  Hence (7) is fulfilled for every k and we obtain the second equality in (1).\n\n3.  The interval (L , U) is non-empty\n-------------------------------------\nIt remains to check L < U, i.e.\n        1 − 2/N < (1 − (2n−1)/N)^{1/n}.                        (9)\nRaising both sides to the power n gives\n        (1 − 2/N)^{n} < 1 − (2n−1)/N.                          (10)\nPut x := 2/N.  Note 0 < x ≤ 1/4 (since n ≥ 2).  For 0 ≤ x ≤ 1 the quadratic truncation of the binomial theorem yields\n        (1 − x)^{n} ≤ 1 − nx + n(n−1)x²/2.                    (11)\nWith x = 2/N and N = n^3, inequality (11) becomes\n        (1 − 2/N)^{n} ≤ 1 − 2/n² + n(n−1)/(2n^6)\n                       < 1 − 2/n² + 1/n³\n                       = 1 − (2n−1)/N,\nwhich is exactly (10).  Therefore L < U.\n\n4.  Conclusion\n--------------\nChoose any α ∈ (L , U).  Section 1 showed f_α^{∘ k}(N) = N − 2k, and Section 2 showed f_{α^{k}}(N) = N − 2k, for every k = 1,…,n.  Thus such an α exists, completing the proof. ∎",
      "_meta": {
        "core_steps": [
          "Convert the equalities f_α^k(n^2)=n^2−k into the double inequality (n^2−k)/(n^2−k+1)≤α<1, giving the lower bound α≥1−1/n² by monotonicity.",
          "Rewrite f_{α^k}(n^2)=n^2−k as α^k n² < n²−k+1; a reverse–induction argument shows it suffices to check the case k=n, yielding the upper bound α<((n²−n+1)/n²)^{1/n}.",
          "Intersect the two bounds to obtain a single interval for α: 1−1/n² ≤ α < ((n²−n+1)/n²)^{1/n}.",
          "Prove this interval is non–empty by the truncated–binomial/Taylor inequality (1−x)^n ≤ 1−nx+ (n choose 2)x² applied to x=1/n², giving (1−1/n²)^n < 1−1/n+1/n²."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The initial integer on which the iterates are evaluated (currently n²).  Any sufficiently large integer N (e.g. N≥n) can replace n²; all bounds then use N in place of n².",
            "original": "n^2"
          },
          "slot2": {
            "description": "The size of the fixed decrement between successive target values (currently 1, so targets are n²−k).  A constant positive step d could be used instead, i.e. require f_α^k(N)=N−d k; the same chain of inequalities goes through with +d replacing +1 wherever it appears.",
            "original": "1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}