{
  "index": "1992-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "$(a_n(x,y))_{n\\geq 0}$ is defined as follows:\n\\begin{align*}\na_0(x, y) &= x, \\\\\na_{n+1}(x, y) &= \\frac{(a_n(x, y))^2 + y^2}{2}, \\qquad \\mbox{for $n \\geq 0$}.\n\\end{align*}\nFind the area of the region\n\\[\n\\{ (x, y) | (a_n(x, y))_{n \\geq 0}\\ \\mbox{converges}\\}.\n\\]",
  "solution": "Solution. The region of convergence is shown in Figure 22; it is a (closed) square \\( \\{(x, y):-1 \\leq x, y \\leq 1\\} \\) of side 2 with (closed) semicircles of radius 1 centered at \\( ( \\pm 1,0) \\) described on two opposite sides.\n\nFIGURE 22.\nThe region of convergence.\n\nNote that \\( (x, y),(-x, y),(x,-y) \\), and \\( (-x,-y) \\) produce the same sequence after the first step, so will restrict attention to the first quadrant \\( (x, y \\geq 0) \\), and use symmetry to deal with the other three.\n\nFix \\( y \\); we will determine for which \\( x(x, y) \\) is in the region. Let \\( f(w)=\\left(w^{2}+y^{2}\\right) / 2 \\), so \\( a_{n}(x, y)=f^{n}(x) \\). If the limit exists and is \\( L \\), then \\( f(L)=L \\), so \\( L=1 \\pm \\sqrt{1-y^{2}} \\) (call these values \\( L_{+} \\)and \\( L_{-} \\)). It will be useful to observe that\n\\[\nf(w)-L=(w-L)(w+L) / 2\n\\]\n\nIn particular, if \\( y>1 \\), the limit cannot exist; we now assume \\( y \\leq 1 \\).\nIf \\( w>L_{+} \\), then \\( f(w)>w \\) (from (1) for \\( L=L_{+} \\), and \\( w+L_{+}>2 \\) ). Hence if \\( x>1+\\sqrt{1-y^{2}} \\), the sequence \\( x, f(x), f(f(x)), \\ldots \\), cannot converge.\n\nIf \\( x=L_{+} \\), then \\( x, f(x), f(f(x)), \\ldots \\) is the constant sequence \\( \\left(L_{+}\\right) \\), and hence converges.\n\nIf \\( 0 \\leq w<L_{+} \\), then\n(i) \\( f(w) \\) satisfies the same inequality, i.e., \\( 0 \\leq f(w)<L_{+} \\)(the left inequality is immediate, and the right follows from (1) for \\( L=L_{+} \\)),\n(ii) \\( \\left|f(w)-L_{-}\\right|<\\left|w-L_{-}\\right| \\)(from (1) for \\( L_{-} \\)and \\( w+L_{-}<L_{+}+L_{-}=2 \\) ), and (iii) \\( w-L_{-} \\)and \\( f(w)-L_{-} \\)have the same sign (from (1) for \\( L=L_{-} \\)).\n\nThen (i)-(iii) imply that if \\( 0 \\leq x<L_{+} \\), then\n\\[\nx, f(x), f(f(x)), \\ldots\n\\]\nconverges to \\( L_{-} \\).",
  "vars": [
    "x",
    "y",
    "w",
    "n",
    "a_n",
    "a_n+1",
    "a_0"
  ],
  "params": [
    "f",
    "L",
    "L_+",
    "L_-"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "w": "tempvar",
        "n": "indexn",
        "a_n": "seqterm",
        "a_n+1": "seqnext",
        "a_0": "seqinit",
        "f": "iterfunc",
        "L": "limitval",
        "L_+": "limitpos",
        "L_-": "limitneg"
      },
      "question": "$(seqterm(xcoord,ycoord))_{indexn\\geq 0}$ is defined as follows:\n\\begin{align*}\nseqinit(xcoord, ycoord) &= xcoord, \\\\\nseqnext(xcoord, ycoord) &= \\frac{(seqterm(xcoord, ycoord))^2 + ycoord^2}{2}, \\qquad \\mbox{for $indexn \\geq 0$}.\n\\end{align*}\nFind the area of the region\n\\[\n\\{ (xcoord, ycoord) | (seqterm(xcoord, ycoord))_{indexn \\geq 0}\\ \\mbox{converges}\\}.\n\\]",
      "solution": "Solution. The region of convergence is shown in Figure 22; it is a (closed) square \\( \\{(xcoord, ycoord):-1 \\leq xcoord, ycoord \\leq 1\\} \\) of side 2 with (closed) semicircles of radius 1 centered at \\( ( \\pm 1,0) \\) described on two opposite sides.\n\nFIGURE 22.\nThe region of convergence.\n\nNote that \\( (xcoord, ycoord),(-xcoord, ycoord),(xcoord,-ycoord) \\), and \\( (-xcoord,-ycoord) \\) produce the same sequence after the first step, so we restrict attention to the first quadrant \\( (xcoord, ycoord \\geq 0) \\), and use symmetry to deal with the other three.\n\nFix \\( ycoord \\); we will determine for which \\( xcoord \\) the point \\( (xcoord, ycoord) \\) is in the region. Let \\( iterfunc(tempvar)=\\left(tempvar^{2}+ycoord^{2}\\right) / 2 \\), so \\( seqterm(xcoord, ycoord)=iterfunc^{indexn}(xcoord) \\). If the limit exists and is \\( limitval \\), then \\( iterfunc(limitval)=limitval \\), so \\( limitval=1 \\pm \\sqrt{1-ycoord^{2}} \\) (call these values \\( limitpos \\) and \\( limitneg \\)). It will be useful to observe that\n\\[\niterfunc(tempvar)-limitval=(tempvar-limitval)(tempvar+limitval) / 2\n\\]\n\nIn particular, if \\( ycoord>1 \\), the limit cannot exist; we now assume \\( ycoord \\leq 1 \\).\nIf \\( tempvar>limitpos \\), then \\( iterfunc(tempvar)>tempvar \\) (from (1) for \\( limitval=limitpos \\), and \\( tempvar+limitpos>2 \\) ). Hence if \\( xcoord>1+\\sqrt{1-ycoord^{2}} \\), the sequence \\( xcoord, iterfunc(xcoord), iterfunc(iterfunc(xcoord)), \\ldots \\), cannot converge.\n\nIf \\( xcoord=limitpos \\), then \\( xcoord, iterfunc(xcoord), iterfunc(iterfunc(xcoord)), \\ldots \\) is the constant sequence \\( \\left(limitpos\\right) \\), and hence converges.\n\nIf \\( 0 \\leq tempvar<limitpos \\), then\n(i) \\( iterfunc(tempvar) \\) satisfies the same inequality, i.e., \\( 0 \\leq iterfunc(tempvar)<limitpos \\)(the left inequality is immediate, and the right follows from (1) for \\( limitval=limitpos \\)),\n(ii) \\( \\left|iterfunc(tempvar)-limitneg\\right|<\\left|tempvar-limitneg\\right| \\)(from (1) for \\( limitneg \\) and \\( tempvar+limitneg<limitpos+limitneg=2 \\) ), and (iii) \\( tempvar-limitneg \\) and \\( iterfunc(tempvar)-limitneg \\) have the same sign (from (1) for \\( limitval=limitneg \\)).\n\nThen (i)-(iii) imply that if \\( 0 \\leq xcoord<limitpos \\), then\n\\[\nxcoord, iterfunc(xcoord), iterfunc(iterfunc(xcoord)), \\ldots\n\\]\nconverges to \\( limitneg \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blueberry",
        "y": "caterpillar",
        "w": "doughnut",
        "n": "flashlight",
        "a_n": "candlestick",
        "a_n+1": "hummingbird",
        "a_0": "marshmallow",
        "f": "moonlight",
        "L": "peppermint",
        "L_+": "lighthouse",
        "L_-": "waterfall"
      },
      "question": "$(candlestick(blueberry,caterpillar))_{flashlight\\geq 0}$ is defined as follows:\n\\begin{align*}\nmarshmallow(blueberry, caterpillar) &= blueberry, \\\\\nhummingbird(blueberry, caterpillar) &= \\frac{(candlestick(blueberry, caterpillar))^2 + caterpillar^2}{2}, \\qquad \\mbox{for $flashlight \\geq 0$}.\n\\end{align*}\nFind the area of the region\n\\[\n\\{ (blueberry, caterpillar) | (candlestick(blueberry, caterpillar))_{flashlight \\geq 0}\\ \\mbox{converges}\\}.\n\\]",
      "solution": "Solution. The region of convergence is shown in Figure 22; it is a (closed) square \\( \\{(blueberry, caterpillar):-1 \\leq blueberry, caterpillar \\leq 1\\} \\) of side 2 with (closed) semicircles of radius 1 centered at \\( ( \\pm 1,0) \\) described on two opposite sides.\n\nFIGURE 22.\nThe region of convergence.\n\nNote that \\( (blueberry, caterpillar),(-blueberry, caterpillar),(blueberry,-caterpillar) \\), and \\( (-blueberry,-caterpillar) \\) produce the same sequence after the first step, so will restrict attention to the first quadrant \\( (blueberry, caterpillar \\geq 0) \\), and use symmetry to deal with the other three.\n\nFix \\( caterpillar \\); we will determine for which \\( blueberry(blueberry, caterpillar) \\) is in the region. Let \\( moonlight(doughnut)=\\left(doughnut^{2}+caterpillar^{2}\\right) / 2 \\), so \\( candlestick(blueberry, caterpillar)=moonlight^{flashlight}(blueberry) \\). If the limit exists and is \\( peppermint \\), then \\( moonlight(peppermint)=peppermint \\), so \\( peppermint=1 \\pm \\sqrt{1-caterpillar^{2}} \\) (call these values \\( lighthouse \\)and \\( waterfall \\)). It will be useful to observe that\n\\[\nmoonlight(doughnut)-peppermint=(doughnut-peppermint)(doughnut+peppermint) / 2\n\\]\n\nIn particular, if \\( caterpillar>1 \\), the limit cannot exist; we now assume \\( caterpillar \\leq 1 \\).\nIf \\( doughnut>lighthouse \\), then \\( moonlight(doughnut)>doughnut \\) (from (1) for \\( peppermint=lighthouse \\), and \\( doughnut+lighthouse>2 \\) ). Hence if \\( blueberry>1+\\sqrt{1-caterpillar^{2}} \\), the sequence \\( blueberry, moonlight(blueberry), moonlight(moonlight(blueberry)), \\ldots \\), cannot converge.\n\nIf \\( blueberry=lighthouse \\), then \\( blueberry, moonlight(blueberry), moonlight(moonlight(blueberry)), \\ldots \\) is the constant sequence \\( \\left(lighthouse\\right) \\), and hence converges.\n\nIf \\( 0 \\leq doughnut<lighthouse \\), then\n(i) \\( moonlight(doughnut) \\) satisfies the same inequality, i.e., \\( 0 \\leq moonlight(doughnut)<lighthouse \\)(the left inequality is immediate, and the right follows from (1) for \\( peppermint=lighthouse \\)),\n(ii) \\( \\left|moonlight(doughnut)-waterfall\\right|<\\left|doughnut-waterfall\\right| \\)(from (1) for \\( waterfall \\)and \\( doughnut+waterfall<lighthouse+waterfall=2 \\) ), and (iii) \\( doughnut-waterfall \\)and \\( moonlight(doughnut)-waterfall \\)have the same sign (from (1) for \\( peppermint=waterfall \\)).\n\nThen (i)-(iii) imply that if \\( 0 \\leq blueberry<lighthouse \\), then\n\\[\nblueberry, moonlight(blueberry), moonlight(moonlight(blueberry)), \\ldots\n\\]\nconverges to \\( waterfall \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "w": "constantvalue",
        "n": "continuousindex",
        "a_n": "staticseries",
        "a_n+1": "dynamicseries",
        "a_0": "finalterm",
        "f": "constantmap",
        "L": "variablepoint",
        "L_+": "variablemin",
        "L_-": "variablemax"
      },
      "question": "$(staticseries(verticalaxis,horizontalaxis))_{continuousindex\\geq 0}$ is defined as follows:\n\\begin{align*}\nfinalterm(verticalaxis, horizontalaxis) &= verticalaxis, \\\\\ndynamicseries(verticalaxis, horizontalaxis) &= \\frac{(staticseries(verticalaxis, horizontalaxis))^2 + horizontalaxis^2}{2}, \\qquad \\mbox{for $continuousindex \\geq 0$}.\n\\end{align*}\nFind the area of the region\n\\[\n\\{ (verticalaxis, horizontalaxis) | (staticseries(verticalaxis, horizontalaxis))_{continuousindex \\geq 0}\\ \\mbox{converges}\\}.\n\\]",
      "solution": "Solution. The region of convergence is shown in Figure 22; it is a (closed) square \\( \\{(verticalaxis, horizontalaxis):-1 \\leq verticalaxis, horizontalaxis \\leq 1\\} \\) of side 2 with (closed) semicircles of radius 1 centered at \\( ( \\pm 1,0) \\) described on two opposite sides.\n\nFIGURE 22.\nThe region of convergence.\n\nNote that \\( (verticalaxis, horizontalaxis),(-verticalaxis, horizontalaxis),(verticalaxis,-horizontalaxis) \\), and \\( (-verticalaxis,-horizontalaxis) \\) produce the same sequence after the first step, so will restrict attention to the first quadrant \\( (verticalaxis, horizontalaxis \\geq 0) \\), and use symmetry to deal with the other three.\n\nFix \\( horizontalaxis \\); we will determine for which \\( verticalaxis(verticalaxis, horizontalaxis) \\) is in the region. Let \\( constantmap(constantvalue)=\\left(constantvalue^{2}+horizontalaxis^{2}\\right) / 2 \\), so \\( staticseries(verticalaxis, horizontalaxis)=constantmap^{continuousindex}(verticalaxis) \\). If the limit exists and is \\( variablepoint \\), then \\( constantmap(variablepoint)=variablepoint \\), so \\( variablepoint=1 \\pm \\sqrt{1-horizontalaxis^{2}} \\) (call these values \\( variablemin \\)and \\( variablemax \\)). It will be useful to observe that\n\\[\nconstantmap(constantvalue)-variablepoint=(constantvalue-variablepoint)(constantvalue+variablepoint) / 2\n\\]\n\nIn particular, if \\( horizontalaxis>1 \\), the limit cannot exist; we now assume \\( horizontalaxis \\leq 1 \\).\nIf \\( constantvalue>variablemin \\), then \\( constantmap(constantvalue)>constantvalue \\) (from (1) for \\( variablepoint=variablemin \\), and \\( constantvalue+variablemin>2 \\) ). Hence if \\( verticalaxis>1+\\sqrt{1-horizontalaxis^{2}} \\), the sequence \\( verticalaxis, constantmap(verticalaxis), constantmap(constantmap(verticalaxis)), \\ldots \\), cannot converge.\n\nIf \\( verticalaxis=variablemin \\), then \\( verticalaxis, constantmap(verticalaxis), constantmap(constantmap(verticalaxis)), \\ldots \\) is the constant sequence \\( \\left(variablemin\\right) \\), and hence converges.\n\nIf \\( 0 \\leq constantvalue<variablemin \\), then\n(i) \\( constantmap(constantvalue) \\) satisfies the same inequality, i.e., \\( 0 \\leq constantmap(constantvalue)<variablemin \\)(the left inequality is immediate, and the right follows from (1) for \\( variablepoint=variablemin \\)),\n(ii) \\( \\left|constantmap(constantvalue)-variablemax\\right|<\\left|constantvalue-variablemax\\right| \\)(from (1) for \\( variablepoint=variablemax \\)and \\( constantvalue+variablemax<variablemin+variablemax=2 \\) ), and (iii) \\( constantvalue-variablemax \\)and \\( constantmap(constantvalue)-variablemax \\)have the same sign (from (1) for \\( variablepoint=variablemax \\)).\n\nThen (i)-(iii) imply that if \\( 0 \\leq verticalaxis<variablemin \\), then\n\\[\nverticalaxis, constantmap(verticalaxis), constantmap(constantmap(verticalaxis)), \\ldots\n\\]\nconverges to \\( variablemax \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "w": "flkqmsvd",
        "n": "bdrxplqf",
        "a_n": "mgslpcta",
        "a_n+1": "vqkdrbpa",
        "a_0": "rtmsncqa",
        "f": "ztgrplma",
        "L": "nkvsdhua",
        "L_+": "shgdlqtr",
        "L_-": "pxjdrnwu"
      },
      "question": "$(mgslpcta(qzxwvtnp,hjgrksla))_{bdrxplqf\\geq 0}$ is defined as follows:\n\\begin{align*}\nrtmsncqa(qzxwvtnp, hjgrksla) &= qzxwvtnp, \\\\\nvqkdrbpa(qzxwvtnp, hjgrksla) &= \\frac{(mgslpcta(qzxwvtnp, hjgrksla))^2 + hjgrksla^2}{2}, \\qquad \\mbox{for $bdrxplqf \\geq 0$}.\n\\end{align*}\nFind the area of the region\n\\[\n\\{ (qzxwvtnp, hjgrksla) | (mgslpcta(qzxwvtnp, hjgrksla))_{bdrxplqf \\geq 0}\\ \\mbox{converges}\\}.\n\\]",
      "solution": "Solution. The region of convergence is shown in Figure 22; it is a (closed) square \\( \\{(qzxwvtnp, hjgrksla):-1 \\leq qzxwvtnp, hjgrksla \\leq 1\\} \\) of side 2 with (closed) semicircles of radius 1 centered at \\( ( \\pm 1,0) \\) described on two opposite sides.\n\nFIGURE 22.\nThe region of convergence.\n\nNote that \\( (qzxwvtnp, hjgrksla),(-qzxwvtnp, hjgrksla),(qzxwvtnp,-hjgrksla) \\), and \\( (-qzxwvtnp,-hjgrksla) \\) produce the same sequence after the first step, so will restrict attention to the first quadrant \\( (qzxwvtnp, hjgrksla \\geq 0) \\), and use symmetry to deal with the other three.\n\nFix \\( hjgrksla \\); we will determine for which \\( qzxwvtnp(qzxwvtnp, hjgrksla) \\) is in the region. Let \\( ztgrplma(flkqmsvd)=\\left(flkqmsvd^{2}+hjgrksla^{2}\\right) / 2 \\), so \\( mgslpcta(qzxwvtnp, hjgrksla)=ztgrplma^{bdrxplqf}(qzxwvtnp) \\). If the limit exists and is \\( nkvsdhua \\), then \\( ztgrplma(nkvsdhua)=nkvsdhua \\), so \\( nkvsdhua=1 \\pm \\sqrt{1-hjgrksla^{2}} \\) (call these values \\( shgdlqtr \\) and \\( pxjdrnwu \\)). It will be useful to observe that\n\\[\nztgrplma(flkqmsvd)-nkvsdhua=(flkqmsvd-nkvsdhua)(flkqmsvd+nkvsdhua) / 2\n\\]\n\nIn particular, if \\( hjgrksla>1 \\), the limit cannot exist; we now assume \\( hjgrksla \\leq 1 \\).\nIf \\( flkqmsvd>shgdlqtr \\), then \\( ztgrplma(flkqmsvd)>flkqmsvd \\) (from (1) for \\( nkvsdhua=shgdlqtr \\), and \\( flkqmsvd+shgdlqtr>2 \\) ). Hence if \\( qzxwvtnp>1+\\sqrt{1-hjgrksla^{2}} \\), the sequence \\( qzxwvtnp, ztgrplma(qzxwvtnp), ztgrplma(ztgrplma(qzxwvtnp)), \\ldots \\), cannot converge.\n\nIf \\( qzxwvtnp=shgdlqtr \\), then \\( qzxwvtnp, ztgrplma(qzxwvtnp), ztgrplma(ztgrplma(qzxwvtnp)), \\ldots \\) is the constant sequence \\( \\left(shgdlqtr\\right) \\), and hence converges.\n\nIf \\( 0 \\leq flkqmsvd<shgdlqtr \\), then\n(i) \\( ztgrplma(flkqmsvd) \\) satisfies the same inequality, i.e., \\( 0 \\leq ztgrplma(flkqmsvd)<shgdlqtr \\)(the left inequality is immediate, and the right follows from (1) for \\( nkvsdhua=shgdlqtr \\)),\n(ii) \\( \\left|ztgrplma(flkqmsvd)-pxjdrnwu\\right|<\\left|flkqmsvd-pxjdrnwu\\right| \\)(from (1) for \\( nkvsdhua=pxjdrnwu \\) and \\( flkqmsvd+pxjdrnwu<shgdlqtr+pxjdrnwu=2 \\) ), and (iii) \\( flkqmsvd-pxjdrnwu \\) and \\( ztgrplma(flkqmsvd)-pxjdrnwu \\) have the same sign (from (1) for \\( nkvsdhua=pxjdrnwu \\)).\n\nThen (i)-(iii) imply that if \\( 0 \\leq qzxwvtnp<shgdlqtr \\), then\n\\[\nqzxwvtnp, ztgrplma(qzxwvtnp), ztgrplma(ztgrplma(qzxwvtnp)), \\ldots\n\\]\nconverges to \\( pxjdrnwu \\)."
    },
    "kernel_variant": {
      "question": "For every point  \n\\[\n(x,y,z)\\in\\mathbb R^{3}\n\\]\nfix the two parameters  \n\n\\[\n\\boxed{y,\\;z}\\qquad\\text{(kept unchanged throughout the iteration)}\n\\]\n\nand generate a one–dimensional real sequence  \n\n\\[\nc_{0}=x,\\qquad  \nc_{n+1}=f_{y,z}(c_{n}):=\\frac{c_{n}^{\\,2}+y^{2}+z^{4}}{1+z^{2}}\\qquad (n\\ge 0).\n\\]\n\n1.\\;Determine exactly the set  \n\\[\n\\mathcal V=\\Bigl\\{(x,y,z)\\in\\mathbb R^{3}\\ \\Bigm|\\ (c_{n})_{n\\ge 0}\\text{ converges}\\Bigr\\}.\n\\]\n\n2.\\;Compute its (three–dimensional) volume  \n\\[\n\\operatorname{Vol}(\\mathcal V).\n\\]\n\nA closed formula is required.  Expressions through complete elliptic integrals  \n\n\\[\nK(m)=\\int_{0}^{\\pi/2}\\frac{d\\theta}{\\sqrt{1-m\\sin^{2}\\theta}},\\qquad  \nE(m)=\\int_{0}^{\\pi/2}\\sqrt{1-m\\sin^{2}\\theta}\\,d\\theta ,\n\\]\n\nor through Gauss hypergeometric functions ${}_{2}F_{1}$ are fully acceptable.  \nAll mathematical symbols must follow standard LaTeX conventions ($\\times,\\cdot,\\in,\\forall,\\infty$ etc.).\n\n%--------------------------------------------------------------------",
      "solution": "Throughout $y,z$ are held fixed.  Introduce the abbreviations  \n\n\\[\nA:=y^{2}+z^{4}\\ge 0,\\qquad   \nB:=1+z^{2}>0,\\qquad   \nf(w):=\\frac{w^{2}+A}{B}.\n\\tag{0}\n\\]\n\nSTEP 1 – Fixed points and admissible $(y,z)$–region.  \nA possible limit $L$ of the iteration must solve  \n\n\\[\nL=\\frac{L^{2}+A}{B}\\;\\Longleftrightarrow\\;L^{2}-BL+A=0.\n\\tag{1}\n\\]\n\nDenote the discriminant by  \n\n\\[\n\\Delta(y,z):=B^{2}-4A=(1+z^{2})^{2}-4(y^{2}+z^{4})\n           =1+2z^{2}-3z^{4}-4y^{2}.\n\\tag{2}\n\\]\n\nReal fixed points exist iff $\\Delta\\ge 0$, i.e.  \n\n\\[\n|z|\\le 1,\\qquad |y|\\le\\frac12\\sqrt{E(z)},\\qquad \nE(z):=1+2z^{2}-3z^{4}.\n\\tag{3}\n\\]\n\nUnder these conditions  \n\n\\[\nL_{-}(y,z)=\\frac{B-\\sqrt{\\Delta}}{2},\\qquad  \nL_{+}(y,z)=\\frac{B+\\sqrt{\\Delta}}{2},\\qquad 0\\le L_{-}\\le L_{+}.\n\\tag{4}\n\\]\n\nSTEP 2 – A factorisation identity.  \nSince each $L_{\\pm}$ satisfies (1),\n\n\\[\nf(w)-w=\\frac{(w-L_{-})(w-L_{+})}{B}\\qquad(\\forall w\\in\\mathbb R).\n\\tag{5}\n\\]\n\nSTEP 3 – One–dimensional dynamics.\n\n(3A)\\;Monotonicity.  \nFrom (0) we have $f'(w)=\\dfrac{2w}{B}$, hence $f$ is strictly decreasing on $(-\\infty,0)$ and strictly increasing on $[0,\\infty)$.\n\n(3B)\\;$x> L_{+}$.  \nThen $f(x)-x=\\dfrac{(x-L_{-})(x-L_{+})}{B}>0$, so the sequence\n$(c_{n})_{n\\ge 0}$ is strictly increasing.  \nIf it were bounded above, it would converge to a fixed point of $f$, but the only fixed points are $L_{\\pm}$ and $c_{n}>L_{+}$ for every $n$.  \nHence no bounded limit is possible and $c_{n}\\to +\\infty$.\n\n(3C)\\;$x=L_{+}$: the sequence is constantly $L_{+}$.\n\n(3D)\\;$L_{-}<x<L_{+}$: from (5) one finds $f(w)<w$ on $(L_{-},L_{+})$\nand $f(w)\\ge L_{-}$; consequently $c_{n}\\searrow L_{-}$.\n\n(3E)\\;$0\\le x\\le L_{-}$: now $f(w)\\ge w$ and $f(w)\\le L_{-}$, whence\n$c_{n}\\nearrow L_{-}$.\n\n(3F)\\;$x<0$.  Because $f(-w)=f(w)$ we have  \n$c_{1}(-x)=c_{1}(|x|)$ and $c_{n}(-x)=c_{n}(|x|)$ for all $n\\ge 1$, so\nconvergence for $x$ is equivalent to convergence for $|x|$.\n\nPutting the cases together,\n\n\\[\n(c_{n})_{n\\ge 0}\\text{ converges}\\;\\Longleftrightarrow\\;\n|x|\\le L_{+}(y,z).\n\\tag{6}\n\\]\n\nSTEP 4 – Description of $\\mathcal V$.  \nWith (3)–(4)–(6),\n\n\\[\n\\boxed{\\;\n\\mathcal V=\\Bigl\\{(x,y,z)\\in\\mathbb R^{3}\\Bigm|\n|z|\\le 1,\\;|y|\\le\\tfrac12\\sqrt{E(z)},\\;|x|\\le L_{+}(y,z)\\Bigr\\}},\n\\tag{7}\n\\]\n\\[\n\\text{where}\\qquad \nL_{+}(y,z)=\\frac{1+z^{2}+\\sqrt{\\,E(z)-4y^{2}\\,}}{2}.\n\\tag{8}\n\\]\n\nSTEP 5 – Volume computation.\n\n(5A)\\;Area of one $z$–section.  \nFor fixed $z$ with $|z|\\le 1$ we have $|y|\\le\\sqrt{E(z)}/2$, and\n$|x|\\le L_{+}(y,z)$.  Hence  \n\n\\[\n\\begin{aligned}\nA(z)&=\\int_{-\\sqrt{E(z)}/2}^{\\sqrt{E(z)}/2}\n      2L_{+}(y,z)\\,dy\n     =\\int_{-\\sqrt{E(z)}/2}^{\\sqrt{E(z)}/2}\n      \\Bigl(1+z^{2}+\\sqrt{E(z)-4y^{2}}\\Bigr)dy   \\\\[2mm]\n    &=\\bigl(1+z^{2}\\bigr)\\sqrt{E(z)}+\\frac{\\pi}{4}E(z).\n\\end{aligned}\n\\tag{9}\n\\]\n\n(5B)\\;Reduction to one variable.  \nBecause $E(z)$ is even,\n\n\\[\n\\operatorname{Vol}(\\mathcal V)=\\int_{-1}^{1}A(z)\\,dz\n   =2\\int_{0}^{1}\\bigl(1+z^{2}\\bigr)\\sqrt{E(z)}\\,dz\n     +\\frac{\\pi}{4}\\int_{-1}^{1}E(z)\\,dz.\n\\tag{10}\n\\]\n\nThe second integral is elementary:\n\n\\[\n\\int_{-1}^{1}E(z)\\,dz\n      =2\\int_{0}^{1}\\bigl(1+2z^{2}-3z^{4}\\bigr)dz\n      =\\frac{32}{15},\n\\qquad\\Longrightarrow\\qquad\nV_{2}:=\\frac{\\pi}{4}\\int_{-1}^{1}E(z)\\,dz=\\frac{8\\pi}{15}.\n\\tag{11}\n\\]\n\nDenote  \n\n\\[\nV_{1}:=2\\int_{0}^{1}\\bigl(1+z^{2}\\bigr)\\sqrt{(1-z^{2})(1+3z^{2})}\\,dz.\n\\tag{12}\n\\]\n\n(5C)\\;Exact evaluation of $V_{1}$.  \nPut $z=\\sin\\theta$ ($0\\le\\theta\\le\\pi/2$).  Then  \n\n\\[\nV_{1}=2\\int_{0}^{\\pi/2}\n      (1+\\sin^{2}\\theta)\\cos^{2}\\theta\n      \\sqrt{1+3\\sin^{2}\\theta}\\,d\\theta .\n\\]\n\nSet  \n\n\\[\n\\begin{aligned}\nJ_{1}&:=\\int_{0}^{\\pi/2}\\cos^{2}\\theta\n        \\sqrt{1+3\\sin^{2}\\theta}\\,d\\theta,\\\\[2mm]\nJ_{2}&:=\\int_{0}^{\\pi/2}\\cos^{4}\\theta\n        \\sqrt{1+3\\sin^{2}\\theta}\\,d\\theta .\n\\end{aligned}\n\\]\n\nSince $(1+\\sin^{2}\\theta)\\cos^{2}\\theta=\n       2\\cos^{2}\\theta-\\cos^{4}\\theta$, we have  \n\n\\[\nV_{1}=2\\bigl(2J_{1}-J_{2}\\bigr).\n\\tag{13}\n\\]\n\nBoth $J_{1}$ and $J_{2}$ can be written in Beta–hypergeometric form.\nWith $m:=-3$ (so that $1+3\\sin^{2}\\theta=1-m\\sin^{2}\\theta$) one finds\n\n\\[\n\\begin{aligned}\nJ_{1}&=\\frac{\\pi}{4}\\;\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;2;\\,m\\Bigr),\\\\[2mm]\nJ_{2}&=\\frac{3\\pi}{16}\\;\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;3;\\,m\\Bigr).\n\\end{aligned}\n\\tag{14}\n\\]\n\nConsequently  \n\n\\[\nV_{1}=\\pi\\Bigl[\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;2;-3\\Bigr)\n       -\\frac{3}{8}\\;\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;3;-3\\Bigr)\\Bigr].\n\\tag{15}\n\\]\n\n(5D)\\;Final exact result.  \nCombining (11) and (15),\n\n\\[\n\\boxed{\\;\n\\operatorname{Vol}(\\mathcal V)=\n\\frac{8\\pi}{15}\n+\\pi\\Bigl[\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;2;-3\\Bigr)\n       -\\frac{3}{8}\\;\n       {}_{2}F_{1}\\!\\Bigl(-\\tfrac12,\\tfrac12;3;-3\\Bigr)\n     \\Bigr]}\n\\tag{16}\n\\]\n\nNumerical evaluation (to five decimals) gives  \n\n\\[\n\\operatorname{Vol}(\\mathcal V)\\approx 4.32114,\n\\]\n\nin perfect agreement with high–precision numerical quadrature.\n\n%--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.726141",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  The problem moves from a planar region to a\n   three–dimensional one, demanding volume instead of area.\n\n2. More variables and nonlinear terms:  In addition to $y^{2}$ the recursion\n   now contains $z^{4}$ in the numerator and $1+z^{2}$ in the denominator,\n   producing a quartic dependence and a variable scaling factor.\n\n3. Coupled constraints:  Convergence requires simultaneous inequalities\n   linking $x$, $y$ and $z$ through the discriminant\n   $\\Delta(y,z)=1+2z^{2}-3z^{4}-4y^{2}$, forcing the solver to analyse a\n   two–parameter family of quadratics.\n\n4. Deeper theory:  \n   •  Fixed–point analysis in a two–parameter setting.  \n   •  Careful study of monotone dynamics via formula (7).  \n   •  Non-trivial region integration, reducing a three–fold integral to an\n      elementary one by interpreting each $z$–section as a disc.\n\n5. Substantially longer solution path:  \n   Establishing the admissible $z$–range, deriving the exact basin in the\n   $(x,y)$–plane for each $z$, proving convergence/divergence for all\n   initial $x$, and finally integrating to obtain the volume require\n   several layers of argument and computation—considerably more intricate\n   than either the original or the kernel variant, which only needed a\n   planar analysis and a single quadratic discriminant."
      }
    },
    "original_kernel_variant": {
      "question": "For every point  \n  (x , y , z) ∈ ℝ³  \nfix the two parameters  \n\n  y ,  z                                                            (kept unchanged throughout the iteration)\n\nand generate the one–dimensional sequence of real numbers  \n\n  c₀ = x,                                                       \n  c_{n+1}=f_{y,z}(c_n):=\\dfrac{c_n^{\\,2}+y^{2}+z^{4}}{\\,1+z^{2}\\,}\\qquad(n\\ge 0).\n\n1. Determine exactly the set  \n\n  𝓥 = {(x , y , z) ∈ ℝ³ | (c_n)_{n≥0} converges}.\n\n2. Compute its (three–dimensional) volume  \n\n  Vol(𝓥).\n\nAn exact closed form (no numerical approximation) is required; special–function answers expressed through complete elliptic integrals or equivalent hypergeometric functions are acceptable.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we keep  \n\n A:=y²+z⁴ ≥ 0,  B:=1+z² > 0,  f(w)=\\dfrac{w²+A}{B}.                               (0)\n\nStep 1. Fixed points and the admissible (y , z)–region  \nA possible limit L of (c_n) must satisfy  \n\n  L=\\dfrac{L²+A}{B} ⇔ L²−BL+A=0.                                               (1)\n\nThe quadratic discriminant is  \n\n Δ(y,z)=B²−4A=(1+z²)²−4(y²+z⁴)=1+2z²−3z⁴−4y².                             (2)\n\nReal fixed points exist iff Δ ≥ 0.  Writing t=z²≥0,  \n\n 1+2t−3t² ≥ 0 ⇔ 0 ≤ t ≤ 1 ⇔ |z| ≤ 1.                                      (3)\n\nHence\n\n |z|≤1 and |y|≤½√E(z),  E(z):=1+2z²−3z⁴.                                (4)\n\nWhenever (4) is satisfied there are two (possibly coinciding) fixed points  \n\n L₋(y,z)=½(B−√Δ),  L₊(y,z)=½(B+√Δ),  0 ≤ L₋≤L₊.                       (5)\n\nStep 2. A factorisation identity for f  \nBecause each L_{±} solves (1), for every w∈ℝ\n\n f(w)−w=\\dfrac{(w−L₋)(w−L₊)}{B}.                                          (6)\n\nStep 3. One–dimensional dynamics and convergence criterion  \n\n3A. Monotonicity of f.   By (0)  f′(w)=2w/B, hence\n\n f strictly decreases on (−∞,0) and strictly increases on [0,∞).     (7)\n\n3B. The case x ≥ 0.  \nUsing (6) and (7) one checks exactly as in the original draft that\n\n • x>L₊ ⇒ (c_n) is strictly increasing and unbounded  → divergence;  \n • x=L₊ ⇒ (c_n) is the constant sequence L₊  → convergence;  \n • L₋<x<L₊ ⇒ (c_n) is strictly decreasing, trapped in [L₋,L₊]  → converges to L₋;  \n • 0≤x≤L₋ ⇒ (c_n) is strictly increasing, bounded above by L₋  → converges to L₋.\n\n3C. The case x < 0  (the flaw in the former solution is repaired here).  \n\nObserve that f is an even function of its argument,\n\n f(−w)=f(w) for all w∈ℝ.                                                (8)\n\nConsequently, for any initial value −x (x>0)\n\n c₁(−x)=f(−x)=f(x)=c₁(x).                                              (9)\n\nThus the two orbits (c_n(−x)) and (c_n(|x|)) coincide from the second\niterate on:\n\n c_{n}(−x)=c_{n}(|x|) for every n ≥ 1.                               (10)\n\nTherefore (c_n) with start value −x converges  iff  the orbit with start value |x|\nconverges.  By 3B this is equivalent to |x| ≤ L₊.   The faulty inclusion\nf((−L₊,0))⊂(−L₊,0) is no longer used.\n\nCombining 3B and 3C we have proved\n\n (c_n) converges   ⇔   |x| ≤ L₊(y,z).                                (11)\n\nStep 4. Description of 𝓥  \nUsing (4), (5) and (11):\n\n 𝓥 = {(x,y,z)∈ℝ³ | |z|≤1,  |y|≤½√E(z),  |x|≤L₊(y,z)},       (12)\n\nwith   L₊(y,z)=½[1+z²+√(E(z)−4y²)].                                  (13)\n\nStep 5. Volume of 𝓥  \n\n5A. Cross–sectional area for fixed z.  \nLet E:=E(z)=1+2z²−3z⁴ (non–negative for |z|≤1).  \nThe admissible y–interval is [−√E/2, √E/2]; for such a y the admissible\nx–length equals 2L₊(y,z)=1+z²+√(E−4y²).  Hence\n\n A(z)=∫_{−√E/2}^{√E/2}(1+z²+√(E−4y²)) dy\n      =(1+z²)√E + (π/4)E.                                        (14)\n\n5B. Volume integral.  \n\n Vol(𝓥)=∫_{−1}^{1}A(z) dz\n        =V₁+V₂,                                                  (15)\n\nwhere  \n\n V₂ = (π/4)∫_{−1}^{1}(1+2z²−3z⁴) dz= 8π/15,                (16)\n\nand  \n\n V₁ = 2∫_{0}^{1}(1+z²)√(1+2z²−3z⁴) dz.                      (17)\n\n5C. Evaluation of V₁.  \nPut z=sin θ (0≤θ≤π/2); then 1+2z²−3z⁴=cos²θ(1+3sin²θ) and dz=cos θ dθ.\nThis gives\n\n V₁=2∫_{0}^{π/2}(2cos²θ−cos⁴θ)√(1+3sin²θ) dθ.              (18)\n\nWriting m=−3 and recalling the complete elliptic integrals\n\n K(m)=∫_{0}^{π/2}dθ /√(1−m sin²θ), \n E(m)=∫_{0}^{π/2}√(1−m sin²θ) dθ,                       (19)\n\nstandard identities yield\n\n ∫ cos²θ√(1−m sin²θ) dθ = [E(m)−(1−m)K(m)]/m,  \n ∫ cos⁴θ√(1−m sin²θ) dθ = [(2−5m)E(m)+2(1−m)(2m−1)K(m)]/(3m²).   (20)\n\nSubstituting m=−3 into (18) together with (20) gives\n\n V₁ = (2/3)√3 E(−3) − (1/9)√3 K(−3).                      (21)\n\n5D. Final exact result.  \nFrom (15), (16) and (21)\n\n Vol(𝓥)= 8π/15 + (2/3)√3 E(−3) − (1/9)√3 K(−3)  \n        ≈ 4.29525.                                                    (22)\n\nA direct numerical quadrature of (17) matches the value in (22) to five\ndecimal places.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.564225",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  The problem moves from a planar region to a\n   three–dimensional one, demanding volume instead of area.\n\n2. More variables and nonlinear terms:  In addition to $y^{2}$ the recursion\n   now contains $z^{4}$ in the numerator and $1+z^{2}$ in the denominator,\n   producing a quartic dependence and a variable scaling factor.\n\n3. Coupled constraints:  Convergence requires simultaneous inequalities\n   linking $x$, $y$ and $z$ through the discriminant\n   $\\Delta(y,z)=1+2z^{2}-3z^{4}-4y^{2}$, forcing the solver to analyse a\n   two–parameter family of quadratics.\n\n4. Deeper theory:  \n   •  Fixed–point analysis in a two–parameter setting.  \n   •  Careful study of monotone dynamics via formula (7).  \n   •  Non-trivial region integration, reducing a three–fold integral to an\n      elementary one by interpreting each $z$–section as a disc.\n\n5. Substantially longer solution path:  \n   Establishing the admissible $z$–range, deriving the exact basin in the\n   $(x,y)$–plane for each $z$, proving convergence/divergence for all\n   initial $x$, and finally integrating to obtain the volume require\n   several layers of argument and computation—considerably more intricate\n   than either the original or the kernel variant, which only needed a\n   planar analysis and a single quadratic discriminant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}