{
  "index": "1950-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "GEO"
  ],
  "difficulty": "",
  "question": "5. Answer either (i) or (ii).\n(i) Given that the sequence whose \\( n \\)th term is \\( \\left(s_{n}+2 s_{n+1}\\right) \\) converges, show that the sequence \\( \\left\\{s_{n}\\right\\} \\) converges also.\n(page 306)\n(ii) A plane varies so that it includes a cone of constant volume equal to \\( \\pi a^{3} / 3 \\) with the surface the equation of which in rectangular coordinates is \\( 2 x y=z^{2} \\). Find the equation of the envelope of the various positions of this plane.\n\nState the result so that it applies to a general cone (that is, conic surface) of the second order.",
  "solution": "Solution. Suppose \\( \\lim \\left(s_{n}+2 s_{n+1}\\right)=3 L \\). Then\n\\[\n\\lim \\left[\\left(s_{n}-L\\right)+2\\left(s_{n+1}-L\\right)\\right]=0\n\\]\n\nPut \\( t_{n}=s_{n}-L \\); then \\( \\lim \\left(t_{n}+2 t_{n+1}\\right)=0 \\). We shall prove that \\( \\lim t_{n}=0 \\). This will\nsequence. Given \\( \\epsilon> \\)\n\n0 , choose \\( k \\) so that\n\\[\n\\left|t_{n}+2 t_{n+1}\\right|<\\epsilon \\quad \\text { for all } n \\geq k .\n\\]\n\nBy induction on \\( p \\) we find that\n\\[\nt_{k}-(-2)^{p} t_{k+p}=\\sum_{i=0}^{p-1}(-2)^{i}\\left(t_{k+i}+2 t_{k+i+1}\\right)\n\\]\n\nTherefore\n\\[\n\\left|t_{k}-(-2)^{p} t_{k+p}\\right| \\leq \\sum_{i=0}^{p-1} 2^{i}\\left|t_{k+i}+2 t_{k+i+1}\\right|<2^{p} \\epsilon,\n\\]\nprovided \\( p \\geq 1 \\). Dividing by \\( 2^{p} \\), we get\nso\n\\[\n\\left|t_{k+p}\\right|<\\epsilon+\\frac{1}{2^{j}}\\left|t_{k}\\right| .\n\\]\n\nHence\n\\( \\limsup _{p \\rightarrow-\\infty}\\left|t_{k+p}\\right| \\leq \\epsilon \\).\nTherefore\n\\[\n\\limsup _{n \\rightarrow \\infty}\\left|t_{n}\\right| \\leq \\epsilon .\n\\]\n\nSince \\( \\epsilon \\) was arbitrary, this implies \\( \\lim t_{n}=0 \\). As we remarked before, this proves that the sequence \\( \\left\\{s_{n}\\right\\} \\) converges to \\( L \\).\n\nRemark. The argument generalizes to prove that if \\( s_{n}-\\lambda s_{n+1} \\rightarrow A \\),\nwhere \\( |\\lambda|>1 \\), then \\( s_{n} \\rightarrow A /(1-\\lambda) \\). Solution. Make the change of coordinates\n\\[\n\\begin{array}{l}\nx=\\frac{1}{2} \\sqrt{2}(u+v) \\\\\ny=\\frac{1}{2} \\sqrt{2}(u-v)\n\\end{array}\n\\]\n\nThen the \\( u v \\)-axes are orthogonal and rotated by \\( \\pi / 4 \\) radians from the \\( x y \\) axes. The equation of the given surface in the new coordinates is\n\\[\nz^{2}+v^{2}=u^{2}\n\\]\nwhich is a right circular cone with axis the \\( u \\)-axis.\nNext we find the volume of the conical region cut off from the solid cone\nby a plane. Because of rotational symmetry we need only consider planes of the form \\( u=m v+b \\). In order that the plane should cut off a bounded\nregion it is necessary that \\( |m|<1 \\). The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base.\nThe altitude is the distance from the origin to the plane, namely\n\\[\n\\frac{|b|}{\\sqrt{1+m^{2}}}\n\\]\n\nThe area of the base is \\( \\sqrt{1+m^{2}} \\) times the area of the ellipse obtained by projecting it orthogonally onto the \\( \\boldsymbol{v z} \\)-plane. To find the equation of the\nprojected ellipse we eliminate \\( u \\) between the equations of the cone and the projected ellipse we eliminate \\( u \\) between the equations of the cone and the plane, getting\n\\[\nz^{2}+v^{2}=(m v+b)^{2}\n\\]\n\nCollecting the \\( v \\) terms and completing the square we get\n\\[\nz^{2}+\\left(1-m^{2}\\right)\\left(v-\\frac{m b}{1-m^{2}}\\right)^{2}=b^{2}\\left(\\frac{1}{1-m^{2}}\\right)\n\\]\n[Note that this is indeed an ellipse because \\( |m|<1 \\) ]. The area of this ellipse is\n\\[\nA=\\pi b^{2} \\frac{1}{\\left(1-m^{2}\\right)^{3 / 2}} .\n\\]\n\nThe volume of the conical region is therefore\n\\[\n\\begin{aligned}\n\\frac{1}{3} \\text { base } \\times \\text { altitude } & =\\frac{1}{3}\\left(\\sqrt{1+m^{2}} \\pi b^{2} \\frac{1}{\\left(1-m^{2}\\right)^{3 / 2}}\\right)\\left(\\frac{|b|}{\\sqrt{1+m^{2}}}\\right) \\\\\n& =\\frac{1}{3} \\pi|b|^{3} \\frac{1}{\\left(1-m^{2}\\right)^{3 / 2}} .\n\\end{aligned}\n\\]\n\nThe problem restricts consideration to those planes which cut off a vol-\nme \\( \\frac{1}{3} \\pi a^{3} \\), that is, to planes for which ume \\( \\frac{1}{3} \\pi a^{3} \\), that is, to planes for which\n\\[\n|b|=a \\sqrt{1-m^{2}}\n\\]\nwhere \\( m \\) is the tangent of the angle between the plane and the \\( z v \\)-plane. We are to find the envelope \\( E \\) of all such planes. Clearly \\( E \\) must share\nthe rotational symmetry of the entire configuration, so it suffices to find the rotational symmetry of the entire co\nthe intersection \\( I \\) of \\( E \\) with the \\( u v \\)-plane.\nConsider the plane \\( P \\) tangent to \\( E \\) at a point of \\( I \\). Reflection in the \\( u v \\) plane preserves \\( P \\), since it preserves \\( E \\) and fixes each point of \\( I \\). Therefore \\( P \\) is perpendicular to the \\( u v \\)-plane, so it has an equation of the form \\( u= \\)\n\\( m v+b \\) where, as we have seen, \\( |b|=a \\sqrt{1-m^{2}} \\). The line \\( l \\) in which \\( P \\) meets the \\( u v \\)-plane is clearly tangent to \\( I \\). Hence \\( I \\) is the envelope of all lines \\( l \\) in the \\( u v \\)-plane having equations of the form\n(1)\n\\[\nu=m v \\pm a \\sqrt{1-m^{2}} .\n\\]\n\nThe envelope problem is thus reduced to two dimensions.\nTo find the envelope,\nbetween the equation\n(2)\n\\( \\qquad \\)\nand the equation obtained by differentiating (2) witl\n(3)\n\\[\n0=v-\\frac{a m}{\\sqrt{1-m^{2}}} .\n\\]\n\nFrom (3) we find\n\\[\nm^{2}=\\frac{v^{2}}{v^{2}+a^{2}}, \\quad \\text { so } \\quad 1-m^{2}=\\frac{a^{2}}{v^{2}+a^{2}} .\n\\]\n\nEquation (2) now becomes\n\\[\nu=\\sqrt{1-m^{2}}\\left(\\frac{m}{\\sqrt{1-m^{2}}} v+a\\right)=\\sqrt{1-m^{2}}\\left(\\frac{p^{2}+a^{2}}{a}\\right),\n\\]\ngiving finally\n(4)\n\\[\nu^{2}=v^{2}+a^{2}\n\\]\nas the equation of the envelope \\( I \\). The same equation results if we start with the negative sign.\n[Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the\nfamily of tangent lines to the hyperbola (4). This examination reveals that family of tangent lines to the hyperbola (4). This examination reveas that\nthe tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).]\n\\( u \\) axis; hence its equation is\n\\[\nu^{2}=v^{2}+z^{2}+a^{2} .\n\\]\n\nTransforming back to the original coordinates its equation is\n\\[\n2 x y=z^{2}+a^{2} .\n\\]\n\nGeneralization. Any non-degenerate quadratic cone is affinely equivalent to a right circular cone, and since ratios of volumes are preserved under an affine transformation, the calculations above lead to the conclusion that if all planes of a family cut off solids of the same fixed\nvolume from a quadratic cone, the planes must be tangent to a hyperboloid which is asymptotic to the cone.\n\nRemark. We have seen that all planes tangent to a hyperboloid of two sheets cut off the same volume from the asymptotic cone. This is analogous to the fact that all lines tangent to a hyperbola form with the asymptotes Consider \\( H \\) the \"hyperboloid of two sheets\" in \\( \\mathbf{R}^{n} \\) given by\n\\[\nx_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}=x_{n}{ }^{2}-a^{2} .\n\\]\n\nIts asymptotic cone \\( C \\) has equation\n\\[\nx_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}=x_{n}{ }^{2} .\n\\]\n\nIf \\( P \\) is any hyperplane tangent to \\( H \\), then \\( P \\) and \\( C \\) divide \\( \\mathbf{R}^{n} \\) into five regions (seven, if \\( n=2 \\) ) of which only one is bounded. The \\( n \\)-dimensional\nvolume of this bounded region \\( B_{P} \\) is the same for all choices of \\( P \\). To or all choices of \\( P \\).\npreserve the quadratic form \\( x_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}-x_{n}{ }^{2} \\). \\( G \\) clearly preserves \\( H \\) and \\( C \\), and any hyperplane tangent to \\( H \\) is mapped by any element of \\( G \\) to a hyperplane tangent to \\( H \\). Since \\( G \\) acts transitively on \\( H \\), it acts transitively on the set of hyperplanes tangent to \\( H \\), and hence on the are all volume-preserving, and thus all the regions \\( B_{P} \\) have the same\nvolume.\n\nAs in the case of three dimensions, all \"hyperboloids of two sheets\" (i.e.,\nnon-degenerate, disconnected quadric surfaces) in \\( \\mathbf{R}^{n} \\) are equivalent to \\( H \\)\nunder a linear transformation.",
  "vars": [
    "n",
    "s_n",
    "s_n+1",
    "t_n",
    "t_n+1",
    "t_k+p",
    "t_k+i",
    "t_k+i+1",
    "t_k",
    "p",
    "i",
    "k",
    "x",
    "y",
    "z",
    "u",
    "v"
  ],
  "params": [
    "a",
    "m",
    "b",
    "L",
    "\\\\lambda",
    "A",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexnumb",
        "s_n": "seriescurr",
        "s_n+1": "seriesnext",
        "t_n": "shiftedcurr",
        "t_n+1": "shiftednext",
        "t_k+p": "shiftkplusp",
        "t_k+i": "shiftkplusi",
        "t_k+i+1": "shiftkiplusone",
        "t_k": "shiftatk",
        "p": "stepcount",
        "i": "summindex",
        "k": "startindex",
        "x": "firstcoord",
        "y": "secondcoord",
        "z": "thirdcoord",
        "u": "rotcoordu",
        "v": "rotcoordv",
        "a": "fixedscalar",
        "m": "slopeparam",
        "b": "planeoffset",
        "L": "limitvalue",
        "\\lambda": "lambdapar",
        "A": "areavalue",
        "\\epsilon": "epsilonval"
      },
      "question": "5. Answer either (i) or (ii).\n(i) Given that the sequence whose \\( indexnumb \\)th term is \\( \\left(seriescurr+2 seriesnext\\right) \\) converges, show that the sequence \\( \\left\\{seriescurr\\right\\} \\) converges also.\n(page 306)\n(ii) A plane varies so that it includes a cone of constant volume equal to \\( \\pi fixedscalar^{3} / 3 \\) with the surface the equation of which in rectangular coordinates is \\( 2 firstcoord secondcoord=thirdcoord^{2} \\). Find the equation of the envelope of the various positions of this plane.\n\nState the result so that it applies to a general cone (that is, conic surface) of the second order.",
      "solution": "Solution. Suppose \\( \\lim \\left(seriescurr+2 seriesnext\\right)=3 limitvalue \\). Then\n\\[\n\\lim \\left[\\left(seriescurr-limitvalue\\right)+2\\left(seriesnext-limitvalue\\right)\\right]=0\n\\]\n\nPut \\( shiftedcurr=seriescurr-limitvalue \\); then \\( \\lim \\left(shiftedcurr+2 shiftednext\\right)=0 \\). We shall prove that \\( \\lim shiftedcurr=0 \\). This will\nsequence. Given \\( epsilonval>0 \\), choose \\( startindex \\) so that\n\\[\n\\left|shiftedcurr+2 shiftednext\\right|<epsilonval \\quad \\text { for all } indexnumb \\geq startindex .\n\\]\n\nBy induction on \\( stepcount \\) we find that\n\\[\nshiftatk-(-2)^{stepcount} shiftkplusp=\\sum_{summindex=0}^{stepcount-1}(-2)^{summindex}\\left(shiftkplusi+2 shiftkiplusone\\right)\n\\]\n\nTherefore\n\\[\n\\left|shiftatk-(-2)^{stepcount} shiftkplusp\\right| \\leq \\sum_{summindex=0}^{stepcount-1} 2^{summindex}\\left|shiftkplusi+2 shiftkiplusone\\right|<2^{stepcount} epsilonval,\n\\]\nprovided \\( stepcount \\geq 1 \\). Dividing by \\( 2^{stepcount} \\), we get so\n\\[\n\\left|shiftkplusp\\right|<epsilonval+\\frac{1}{2^{j}}\\left|shiftatk\\right| .\n\\]\n\nHence \\( \\limsup _{stepcount \\rightarrow \\infty}\\left|shiftkplusp\\right| \\leq epsilonval \\).\nTherefore\n\\[\n\\limsup _{indexnumb \\rightarrow \\infty}\\left|shiftedcurr\\right| \\leq epsilonval .\n\\]\n\nSince \\( epsilonval \\) was arbitrary, this implies \\( \\lim shiftedcurr=0 \\). As we remarked before, this proves that the sequence \\( \\left\\{seriescurr\\right\\} \\) converges to \\( limitvalue \\).\n\nRemark. The argument generalizes to prove that if \\( seriescurr-lambdapar seriesnext \\rightarrow areavalue \\),\nwhere \\( |lambdapar|>1 \\), then \\( seriescurr \\rightarrow areavalue /(1-lambdapar) \\).\n\nSolution. Make the change of coordinates\n\\[\n\\begin{array}{l}\nfirstcoord=\\frac{1}{2} \\sqrt{2}(rotcoordu+rotcoordv) \\\nsecondcoord=\\frac{1}{2} \\sqrt{2}(rotcoordu-rotcoordv)\n\\end{array}\n\\]\n\nThen the \\( rotcoordu rotcoordv \\)-axes are orthogonal and rotated by \\( \\pi / 4 \\) radians from the \\( firstcoord secondcoord \\) axes. The equation of the given surface in the new coordinates is\n\\[\nthirdcoord^{2}+rotcoordv^{2}=rotcoordu^{2}\n\\]\nwhich is a right circular cone with axis the \\( rotcoordu \\)-axis.\nNext we find the volume of the conical region cut off from the solid cone\nby a plane. Because of rotational symmetry we need only consider planes of the form \\( rotcoordu=slopeparam rotcoordv+planeoffset \\). In order that the plane should cut off a bounded\nregion it is necessary that \\( |slopeparam|<1 \\). The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base.\nThe altitude is the distance from the origin to the plane, namely\n\\[\n\\frac{|planeoffset|}{\\sqrt{1+slopeparam^{2}}}\n\\]\n\nThe area of the base is \\( \\sqrt{1+slopeparam^{2}} \\) times the area of the ellipse obtained by projecting it orthogonally onto the \\( \\boldsymbol{rotcoordv\\, thirdcoord} \\)-plane. To find the equation of the\nprojected ellipse we eliminate \\( rotcoordu \\) between the equations of the cone and the plane, getting\n\\[\nthirdcoord^{2}+rotcoordv^{2}=(slopeparam rotcoordv+planeoffset)^{2}\n\\]\n\nCollecting the \\( rotcoordv \\) terms and completing the square we get\n\\[\nthirdcoord^{2}+\\left(1-slopeparam^{2}\\right)\\left(rotcoordv-\\frac{slopeparam planeoffset}{1-slopeparam^{2}}\\right)^{2}=planeoffset^{2}\\left(\\frac{1}{1-slopeparam^{2}}\\right)\n\\]\n[Note that this is indeed an ellipse because \\( |slopeparam|<1 \\) ]. The area of this ellipse is\n\\[\nareavalue=\\pi planeoffset^{2} \\frac{1}{\\left(1-slopeparam^{2}\\right)^{3 / 2}} .\n\\]\n\nThe volume of the conical region is therefore\n\\[\n\\begin{aligned}\n\\frac{1}{3} \\text { base } \\times \\text { altitude } & =\\frac{1}{3}\\left(\\sqrt{1+slopeparam^{2}} \\pi planeoffset^{2} \\frac{1}{\\left(1-slopeparam^{2}\\right)^{3 / 2}}\\right)\\left(\\frac{|planeoffset|}{\\sqrt{1+slopeparam^{2}}}\\right) \\\\\n& =\\frac{1}{3} \\pi|planeoffset|^{3} \\frac{1}{\\left(1-slopeparam^{2}\\right)^{3 / 2}} .\n\\end{aligned}\n\\]\n\nThe problem restricts consideration to those planes which cut off a vol-\nume \\( \\frac{1}{3} \\pi fixedscalar^{3} \\), that is, to planes for which\n\\[\n|planeoffset|=fixedscalar \\sqrt{1-slopeparam^{2}}\n\\]\nwhere \\( slopeparam \\) is the tangent of the angle between the plane and the \\( thirdcoord rotcoordv \\)-plane. We are to find the envelope \\( E \\) of all such planes. Clearly \\( E \\) must share\nthe rotational symmetry of the entire configuration, so it suffices to find the\nintersection \\( I \\) of \\( E \\) with the \\( rotcoordu rotcoordv \\)-plane.\nConsider the plane \\( P \\) tangent to \\( E \\) at a point of \\( I \\). Reflection in the \\( rotcoordu rotcoordv \\) plane preserves \\( P \\), since it preserves \\( E \\) and fixes each point of \\( I \\). Therefore \\( P \\) is perpendicular to the \\( rotcoordu rotcoordv \\)-plane, so it has an equation of the form \\( rotcoordu=slopeparam rotcoordv+planeoffset \\) where, as we have seen, \\( |planeoffset|=fixedscalar \\sqrt{1-slopeparam^{2}} \\). The line \\( l \\) in which \\( P \\) meets the \\( rotcoordu rotcoordv \\)-plane is clearly tangent to \\( I \\). Hence \\( I \\) is the envelope of all lines \\( l \\) in the \\( rotcoordu rotcoordv \\)-plane having equations of the form\n(1)\n\\[\nrotcoordu=slopeparam rotcoordv \\pm fixedscalar \\sqrt{1-slopeparam^{2}} .\n\\]\n\nThe envelope problem is thus reduced to two dimensions.\nTo find the envelope, differentiate the relation\n(2)\n\\[\nrotcoordu=slopeparam rotcoordv \\pm fixedscalar \\sqrt{1-slopeparam^{2}}\n\\]\nwith respect to \\( slopeparam \\) and eliminate \\( slopeparam \\) between (2) and\n(3)\n\\[\n0=rotcoordv-\\frac{fixedscalar slopeparam}{\\sqrt{1-slopeparam^{2}}} .\n\\]\n\nFrom (3) we find\n\\[\nslopeparam^{2}=\\frac{rotcoordv^{2}}{rotcoordv^{2}+fixedscalar^{2}}, \\quad \\text { so } \\quad 1-slopeparam^{2}=\\frac{fixedscalar^{2}}{rotcoordv^{2}+fixedscalar^{2}} .\n\\]\n\nEquation (2) now becomes\n\\[\nrotcoordu=\\sqrt{1-slopeparam^{2}}\\left(\\frac{slopeparam}{\\sqrt{1-slopeparam^{2}}} rotcoordv+fixedscalar\\right)=\\sqrt{1-slopeparam^{2}}\\left(\\frac{rotcoordv^{2}+fixedscalar^{2}}{fixedscalar}\\right),\n\\]\ngiving finally\n(4)\n\\[\nrotcoordu^{2}=rotcoordv^{2}+fixedscalar^{2}\n\\]\nas the equation of the envelope \\( I \\). The same equation results if we start with the negative sign.\n[Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the\nfamily of tangent lines to the hyperbola (4). This examination reveals that\nthe tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).]\nThe envelope surface shares the axis of the original cone, namely the \\( rotcoordu \\) axis; hence its equation is\n\\[\nrotcoordu^{2}=rotcoordv^{2}+thirdcoord^{2}+fixedscalar^{2} .\n\\]\n\nTransforming back to the original coordinates its equation is\n\\[\n2 firstcoord secondcoord=thirdcoord^{2}+fixedscalar^{2} .\n\\]\n\nGeneralization. Any non-degenerate quadratic cone is affinely equivalent to a right circular cone, and since ratios of volumes are preserved under an affine transformation, the calculations above lead to the conclusion that if all planes of a family cut off solids of the same fixed\nvolume from a quadratic cone, the planes must be tangent to a hyperboloid which is asymptotic to the cone.\n\nRemark. We have seen that all planes tangent to a hyperboloid of two sheets cut off the same volume from the asymptotic cone. This is analogous to the fact that all lines tangent to a hyperbola form with the asymptotes. Consider \\( H \\) the \"hyperboloid of two sheets\" in \\( \\mathbf{R}^{indexnumb} \\) given by\n\\[\nfirstcoord_{1}{ }^{2}+firstcoord_{2}{ }^{2}+\\cdots+firstcoord_{indexnumb-1}{ }^{2}=firstcoord_{indexnumb}{ }^{2}-fixedscalar^{2} .\n\\]\n\nIts asymptotic cone \\( C \\) has equation\n\\[\nfirstcoord_{1}{ }^{2}+firstcoord_{2}{ }^{2}+\\cdots+firstcoord_{indexnumb-1}{ }^{2}=firstcoord_{indexnumb}{ }^{2} .\n\\]\n\nIf \\( P \\) is any hyperplane tangent to \\( H \\), then \\( P \\) and \\( C \\) divide \\( \\mathbf{R}^{indexnumb} \\) into five regions (seven, if \\( indexnumb=2 \\)) of which only one is bounded. The \\( indexnumb \\)-dimensional\nvolume of this bounded region \\( B_{P} \\) is the same for all choices of \\( P \\). The group of linear transformations preserving the quadratic form \\( firstcoord_{1}{ }^{2}+firstcoord_{2}{ }^{2}+\\cdots+firstcoord_{indexnumb-1}{ }^{2}-firstcoord_{indexnumb}{ }^{2} \\) acts transitively on the set of hyperplanes tangent to \\( H \\) and is volume-preserving; hence all the regions \\( B_{P} \\) have the same\nvolume.\n\nAs in the case of three dimensions, all \"hyperboloids of two sheets\" (i.e.,\nnon-degenerate, disconnected quadric surfaces) in \\( \\mathbf{R}^{indexnumb} \\) are equivalent to \\( H \\)\nunder a linear transformation."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "batteryage",
        "s_n": "compassdust",
        "s_n+1": "gardenloom",
        "t_n": "hammockrice",
        "t_n+1": "lanternfoil",
        "t_k+p": "marbletwig",
        "t_k+i": "napkinfate",
        "t_k+i+1": "orchardveil",
        "t_k": "pocketdrum",
        "p": "quartzseed",
        "i": "ribbonflax",
        "k": "saddlemug",
        "x": "tabletfern",
        "y": "umbrellaoak",
        "z": "violinrope",
        "u": "walnutbeam",
        "v": "yarncrown",
        "a": "zeppelinjar",
        "m": "anchorplume",
        "b": "blizzardink",
        "L": "cactusmoss",
        "\\\\lambda": "dolphinreed",
        "A": "emeraldsock",
        "\\\\epsilon": "flamingoice"
      },
      "question": "5. Answer either (i) or (ii).\n(i) Given that the sequence whose \\( batteryage \\)th term is \\( \\left(compassdust+2 gardenloom\\right) \\) converges, show that the sequence \\( \\left\\{compassdust\\right\\} \\) converges also.\n(page 306)\n(ii) A plane varies so that it includes a cone of constant volume equal to \\( \\pi zeppelinjar^{3} / 3 \\) with the surface the equation of which in rectangular coordinates is \\( 2 tabletfern umbrellaoak=violinrope^{2} \\). Find the equation of the envelope of the various positions of this plane.\n\nState the result so that it applies to a general cone (that is, conic surface) of the second order.",
      "solution": "Solution. Suppose \\( \\lim \\left(compassdust+2 gardenloom\\right)=3 cactusmoss \\). Then\n\\[\n\\lim \\left[\\left(compassdust-cactusmoss\\right)+2\\left(gardenloom-cactusmoss\\right)\\right]=0\n\\]\n\nPut \\( hammockrice=compassdust-cactusmoss \\); then \\( \\lim \\left(hammockrice+2 lanternfoil\\right)=0 \\). We shall prove that \\( \\lim hammockrice=0 \\). This will\nsequence. Given \\( flamingoice> 0 \\), choose \\( saddlemug \\) so that\n\\[\n\\left|hammockrice+2 lanternfoil\\right|<flamingoice \\quad \\text { for all } batteryage \\geq saddlemug .\n\\]\n\nBy induction on \\( quartzseed \\) we find that\n\\[\npocketdrum-(-2)^{quartzseed} marbletwig=\\sum_{ribbonflax=0}^{quartzseed-1}(-2)^{ribbonflax}\\left(napkinfate+2 orchardveil\\right)\n\\]\n\nTherefore\n\\[\n\\left|pocketdrum-(-2)^{quartzseed} marbletwig\\right| \\leq \\sum_{ribbonflax=0}^{quartzseed-1} 2^{ribbonflax}\\left|napkinfate+2 orchardveil\\right|<2^{quartzseed} flamingoice,\n\\]\nprovided \\( quartzseed \\geq 1 \\). Dividing by \\( 2^{quartzseed} \\), we get\nso\n\\[\n\\left|marbletwig\\right|<flamingoice+\\frac{1}{2^{j}}\\left|pocketdrum\\right| .\n\\]\n\nHence\n\\( \\limsup _{quartzseed \\rightarrow-\\infty}\\left|marbletwig\\right| \\leq flamingoice \\).\nTherefore\n\\[\n\\limsup _{batteryage \\rightarrow \\infty}\\left|hammockrice\\right| \\leq flamingoice .\n\\]\n\nSince \\( flamingoice \\) was arbitrary, this implies \\( \\lim hammockrice=0 \\). As we remarked before, this proves that the sequence \\( \\left\\{compassdust\\right\\} \\) converges to \\( cactusmoss \\).\n\nRemark. The argument generalizes to prove that if \\( compassdust-dolphinreed gardenloom \\rightarrow emeraldsock \\), where \\( |dolphinreed|>1 \\), then \\( compassdust \\rightarrow emeraldsock /(1-dolphinreed) \\).\n\nSolution. Make the change of coordinates\n\\[\n\\begin{array}{l}\ntabletfern=\\frac{1}{2} \\sqrt{2}(walnutbeam+yarncrown) \\\\\numbrellaoak=\\frac{1}{2} \\sqrt{2}(walnutbeam-yarncrown)\n\\end{array}\n\\]\n\nThen the \\( walnutbeam yarncrown \\)-axes are orthogonal and rotated by \\( \\pi / 4 \\) radians from the \\( tabletfern umbrellaoak \\) axes. The equation of the given surface in the new coordinates is\n\\[\nviolinrope^{2}+yarncrown^{2}=walnutbeam^{2}\n\\]\nwhich is a right circular cone with axis the \\( walnutbeam \\)-axis.\nNext we find the volume of the conical region cut off from the solid cone by a plane. Because of rotational symmetry we need only consider planes of the form \\( walnutbeam=anchorplume yarncrown+blizzardink \\). In order that the plane should cut off a bounded region it is necessary that \\( |anchorplume|<1 \\). The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base.\nThe altitude is the distance from the origin to the plane, namely\n\\[\n\\frac{|blizzardink|}{\\sqrt{1+anchorplume^{2}}}\n\\]\n\nThe area of the base is \\( \\sqrt{1+anchorplume^{2}} \\) times the area of the ellipse obtained by projecting it orthogonally onto the \\( \\boldsymbol{yarncrown violinrope} \\)-plane. To find the equation of the projected ellipse we eliminate \\( walnutbeam \\) between the equations of the cone and the plane, getting\n\\[\nviolinrope^{2}+yarncrown^{2}=(anchorplume yarncrown+blizzardink)^{2}\n\\]\n\nCollecting the \\( yarncrown \\) terms and completing the square we get\n\\[\nviolinrope^{2}+\\left(1-anchorplume^{2}\\right)\\left(yarncrown-\\frac{anchorplume blizzardink}{1-anchorplume^{2}}\\right)^{2}=blizzardink^{2}\\left(\\frac{1}{1-anchorplume^{2}}\\right)\n\\]\n[Note that this is indeed an ellipse because \\( |anchorplume|<1 \\) ]. The area of this ellipse is\n\\[\nemeraldsock=\\pi blizzardink^{2} \\frac{1}{\\left(1-anchorplume^{2}\\right)^{3 / 2}} .\n\\]\n\nThe volume of the conical region is therefore\n\\[\n\\begin{aligned}\n\\frac{1}{3} \\text { base } \\times \\text { altitude } & =\\frac{1}{3}\\left(\\sqrt{1+anchorplume^{2}} \\pi blizzardink^{2} \\frac{1}{\\left(1-anchorplume^{2}\\right)^{3 / 2}}\\right)\\left(\\frac{|blizzardink|}{\\sqrt{1+anchorplume^{2}}}\\right) \\\\\n& =\\frac{1}{3} \\pi|blizzardink|^{3} \\frac{1}{\\left(1-anchorplume^{2}\\right)^{3 / 2}} .\n\\end{aligned}\n\\]\n\nThe problem restricts consideration to those planes which cut off a volume \\( \\frac{1}{3} \\pi zeppelinjar^{3} \\), that is, to planes for which\n\\[\n|blizzardink|=zeppelinjar \\sqrt{1-anchorplume^{2}}\n\\]\nwhere \\( anchorplume \\) is the tangent of the angle between the plane and the \\( violinrope yarncrown \\)-plane. We are to find the envelope \\( E \\) of all such planes. Clearly \\( E \\) must share the rotational symmetry of the entire configuration, so it suffices to find the intersection \\( I \\) of \\( E \\) with the \\( walnutbeam yarncrown \\)-plane.\nConsider the plane \\( P \\) tangent to \\( E \\) at a point of \\( I \\). Reflection in the \\( walnutbeam yarncrown \\)-plane preserves \\( P \\), since it preserves \\( E \\) and fixes each point of \\( I \\). Therefore \\( P \\) is perpendicular to the \\( walnutbeam yarncrown \\)-plane, so it has an equation of the form \\( walnutbeam=anchorplume yarncrown+blizzardink \\) where, as we have seen, \\( |blizzardink|=zeppelinjar \\sqrt{1-anchorplume^{2}} \\). The line \\( l \\) in which \\( P \\) meets the \\( walnutbeam yarncrown \\)-plane is clearly tangent to \\( I \\). Hence \\( I \\) is the envelope of all lines \\( l \\) in the \\( walnutbeam yarncrown \\)-plane having equations of the form\n(1)\n\\[\nwalnutbeam=anchorplume yarncrown \\pm zeppelinjar \\sqrt{1-anchorplume^{2}} .\n\\]\n\nThe envelope problem is thus reduced to two dimensions.\nTo find the envelope, between the equation\n(2)\n\\( \\qquad \\) and the equation obtained by differentiating (2) witl\n(3)\n\\[\n0=yarncrown-\\frac{zeppelinjar anchorplume}{\\sqrt{1-anchorplume^{2}}} .\n\\]\n\nFrom (3) we find\n\\[\nanchorplume^{2}=\\frac{yarncrown^{2}}{yarncrown^{2}+zeppelinjar^{2}}, \\quad \\text { so } \\quad 1-anchorplume^{2}=\\frac{zeppelinjar^{2}}{yarncrown^{2}+zeppelinjar^{2}} .\n\\]\n\nEquation (2) now becomes\n\\[\nwalnutbeam=\\sqrt{1-anchorplume^{2}}\\left(\\frac{anchorplume}{\\sqrt{1-anchorplume^{2}}} yarncrown+zeppelinjar\\right)=\\sqrt{1-anchorplume^{2}}\\left(\\frac{p^{2}+zeppelinjar^{2}}{zeppelinjar}\\right),\n\\]\ngiving finally\n(4)\n\\[\nwalnutbeam^{2}=yarncrown^{2}+zeppelinjar^{2}\n\\]\nas the equation of the envelope \\( I \\). The same equation results if we start with the negative sign.\n[Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the family of tangent lines to the hyperbola (4). This examination reveals that the tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).]\n\\( walnutbeam \\) axis; hence its equation is\n\\[\nwalnutbeam^{2}=yarncrown^{2}+violinrope^{2}+zeppelinjar^{2} .\n\\]\n\nTransforming back to the original coordinates its equation is\n\\[\n2 tabletfern umbrellaoak=violinrope^{2}+zeppelinjar^{2} .\n\\]\n\nGeneralization. Any non-degenerate quadratic cone is affinely equivalent to a right circular cone, and since ratios of volumes are preserved under an affine transformation, the calculations above lead to the conclusion that if all planes of a family cut off solids of the same fixed volume from a quadratic cone, the planes must be tangent to a hyperboloid which is asymptotic to the cone.\n\nRemark. We have seen that all planes tangent to a hyperboloid of two sheets cut off the same volume from the asymptotic cone. This is analogous to the fact that all lines tangent to a hyperbola form with the asymptotes Consider \\( H \\) the \\\"hyperboloid of two sheets\\\" in \\( \\mathbf{R}^{n} \\) given by\n\\[\nx_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}=x_{n}{ }^{2}-zeppelinjar^{2} .\n\\]\n\nIts asymptotic cone \\( C \\) has equation\n\\[\nx_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}=x_{n}{ }^{2} .\n\\]\n\nIf \\( P \\) is any hyperplane tangent to \\( H \\), then \\( P \\) and \\( C \\) divide \\( \\mathbf{R}^{n} \\) into five regions (seven, if \\( n=2 \\) ) of which only one is bounded. The \\( n \\)-dimensional volume of this bounded region \\( B_{P} \\) is the same for all choices of \\( P \\). To or all choices of \\( P \\).\npreserve the quadratic form \\( x_{1}{ }^{2}+x_{2}{ }^{2}+\\cdots+x_{n-1}{ }^{2}-x_{n}{ }^{2} \\). \\( G \\) clearly preserves \\( H \\) and \\( C \\), and any hyperplane tangent to \\( H \\) is mapped by any element of \\( G \\) to a hyperplane tangent to \\( H \\). Since \\( G \\) acts transitively on \\( H \\), it acts transitively on the set of hyperplanes tangent to \\( H \\), and hence on the are all volume-preserving, and thus all the regions \\( B_{P} \\) have the same volume.\n\nAs in the case of three dimensions, all \\\"hyperboloids of two sheets\\\" (i.e., non-degenerate, disconnected quadric surfaces) in \\( \\mathbf{R}^{n} \\) are equivalent to \\( H \\) under a linear transformation."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "continuous",
        "s_n": "divergent",
        "s_n+1": "divergentnext",
        "t_n": "constant",
        "t_n+1": "constantnext",
        "t_k+p": "constantoffset",
        "t_k+i": "constantinner",
        "t_k+i+1": "constantinnernext",
        "t_k": "constantbase",
        "p": "aggregate",
        "i": "totality",
        "k": "selector",
        "x": "knownvalue",
        "y": "certainty",
        "z": "horizontal",
        "u": "orthogonal",
        "v": "stationary",
        "a": "variable",
        "m": "flatness",
        "b": "originpoint",
        "L": "startpoint",
        "\\lambda": "stability",
        "A": "emptiness",
        "\\epsilon": "gigantic"
      },
      "question": "5. Answer either (i) or (ii).\n(i) Given that the sequence whose \\( continuous \\)th term is \\( \\left(divergent+2 divergentnext\\right) \\) converges, show that the sequence \\( \\left\\{divergent\\right\\} \\) converges also.\n(page 306)\n(ii) A plane varies so that it includes a cone of constant volume equal to \\( \\pi variable^{3} / 3 \\) with the surface the equation of which in rectangular coordinates is \\( 2 knownvalue certainty=horizontal^{2} \\). Find the equation of the envelope of the various positions of this plane.\n\nState the result so that it applies to a general cone (that is, conic surface) of the second order.",
      "solution": "Solution. Suppose \\( \\lim \\left(divergent+2 divergentnext\\right)=3 startpoint \\). Then\n\\[\\lim \\left[\\left(divergent-startpoint\\right)+2\\left(divergentnext-startpoint\\right)\\right]=0\\]\n\nPut \\( constant=divergent-startpoint \\); then \\( \\lim \\left(constant+2 constantnext\\right)=0 \\). We shall prove that \\( \\lim constant=0 \\). This will sequence. Given \\( gigantic>0 \\), choose \\( selector \\) so that\n\\[\\left|constant+2 constantnext\\right|<gigantic \\quad \\text { for all } continuous \\geq selector .\\]\n\nBy induction on \\( aggregate \\) we find that\n\\[constantbase-(-2)^{aggregate} constantoffset=\\sum_{totality=0}^{aggregate-1}(-2)^{totality}\\left(constantinner+2 constantinnernext\\right)\\]\n\nTherefore\n\\[\\left|constantbase-(-2)^{aggregate} constantoffset\\right| \\leq \\sum_{totality=0}^{aggregate-1} 2^{totality}\\left|constantinner+2 constantinnernext\\right|<2^{aggregate} gigantic,\\]\nprovided \\( aggregate \\geq 1 \\). Dividing by \\( 2^{aggregate} \\), we get so\n\\[\\left|constantoffset\\right|<gigantic+\\frac{1}{2^{j}}\\left|constantbase\\right| .\\]\n\nHence \\( \\limsup _{aggregate \\rightarrow-\\infty}\\left|constantoffset\\right| \\leq gigantic \\). Therefore\n\\[\\limsup _{continuous \\rightarrow \\infty}\\left|constant\\right| \\leq gigantic .\\]\n\nSince \\( gigantic \\) was arbitrary, this implies \\( \\lim constant=0 \\). As we remarked before, this proves that the sequence \\( \\left\\{divergent\\right\\} \\) converges to \\( startpoint \\).\n\nRemark. The argument generalizes to prove that if \\( divergent-stability divergentnext \\rightarrow emptiness \\), where \\( |stability|>1 \\), then \\( divergent \\rightarrow emptiness /(1-stability) \\).\n\nSolution. Make the change of coordinates\n\\[\\begin{array}{l}\nknownvalue=\\frac{1}{2} \\sqrt{2}(orthogonal+stationary) \\\\\ncertainty=\\frac{1}{2} \\sqrt{2}(orthogonal-stationary)\n\\end{array}\\]\n\nThen the \\( orthogonal stationary \\)-axes are orthogonal and rotated by \\( \\pi / 4 \\) radians from the \\( knownvalue certainty \\) axes. The equation of the given surface in the new coordinates is\n\\[horizontal^{2}+stationary^{2}=orthogonal^{2}\\]\nwhich is a right circular cone with axis the \\( orthogonal \\)-axis.\n\nNext we find the volume of the conical region cut off from the solid cone by a plane. Because of rotational symmetry we need only consider planes of the form \\( orthogonal=flatness stationary+originpoint \\). In order that the plane should cut off a bounded region it is necessary that \\( |flatness|<1 \\). The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base.\nThe altitude is the distance from the origin to the plane, namely\n\\[\\frac{|originpoint|}{\\sqrt{1+flatness^{2}}}\\]\n\nThe area of the base is \\( \\sqrt{1+flatness^{2}} \\) times the area of the ellipse obtained by projecting it orthogonally onto the \\( stationary horizontal \\)-plane. To find the equation of the projected ellipse we eliminate \\( orthogonal \\) between the equations of the cone and the plane, getting\n\\[horizontal^{2}+stationary^{2}=(flatness stationary+originpoint)^{2}\\]\n\nCollecting the \\( stationary \\) terms and completing the square we get\n\\[horizontal^{2}+\\left(1-flatness^{2}\\right)\\left(stationary-\\frac{flatness originpoint}{1-flatness^{2}}\\right)^{2}=originpoint^{2}\\left(\\frac{1}{1-flatness^{2}}\\right)\\]\n[Note that this is indeed an ellipse because \\( |flatness|<1 \\).] The area of this ellipse is\n\\[A=\\pi originpoint^{2} \\frac{1}{\\left(1-flatness^{2}\\right)^{3 / 2}} .\\]\n\nThe volume of the conical region is therefore\n\\[\\begin{aligned}\n\\frac{1}{3} \\text { base } \\times \\text { altitude } &=\\frac{1}{3}\\left(\\sqrt{1+flatness^{2}} \\pi originpoint^{2} \\frac{1}{\\left(1-flatness^{2}\\right)^{3 / 2}}\\right)\\left(\\frac{|originpoint|}{\\sqrt{1+flatness^{2}}}\\right) \\\\\n&=\\frac{1}{3} \\pi|originpoint|^{3} \\frac{1}{\\left(1-flatness^{2}\\right)^{3 / 2}} .\n\\end{aligned}\\]\n\nThe problem restricts consideration to those planes which cut off a volume \\( \\frac{1}{3} \\pi variable^{3} \\), that is, to planes for which\n\\[|originpoint|=variable \\sqrt{1-flatness^{2}}\\]\nwhere \\( flatness \\) is the tangent of the angle between the plane and the \\( horizontal stationary \\)-plane. We are to find the envelope \\( E \\) of all such planes. Clearly \\( E \\) must share the rotational symmetry of the entire configuration, so it suffices to find the intersection \\( I \\) of \\( E \\) with the \\( orthogonal stationary \\)-plane.\n\nConsider the plane \\( P \\) tangent to \\( E \\) at a point of \\( I \\). Reflection in the \\( orthogonal stationary \\) plane preserves \\( P \\), since it preserves \\( E \\) and fixes each point of \\( I \\). Therefore \\( P \\) is perpendicular to the \\( orthogonal stationary \\)-plane, so it has an equation of the form \\( orthogonal=flatness stationary+originpoint \\) where, as we have seen, \\( |originpoint|=variable \\sqrt{1-flatness^{2}} \\). The line \\( l \\) in which \\( P \\) meets the \\( orthogonal stationary \\)-plane is clearly tangent to \\( I \\). Hence \\( I \\) is the envelope of all lines \\( l \\) in the \\( orthogonal stationary \\)-plane having equations of the form\n(1)\n\\[orthogonal=flatness stationary \\pm variable \\sqrt{1-flatness^{2}} .\\]\n\nThe envelope problem is thus reduced to two dimensions. Differentiating implicitly, we obtain\n(3)\n\\[0=stationary-\\frac{variable flatness}{\\sqrt{1-flatness^{2}}} .\\]\n\nFrom (3) we find\n\\[flatness^{2}=\\frac{stationary^{2}}{stationary^{2}+variable^{2}}, \\quad \\text { so } \\quad 1-flatness^{2}=\\frac{variable^{2}}{stationary^{2}+variable^{2}} .\\]\n\nSubstituting, we finally get\n(4)\n\\[orthogonal^{2}=stationary^{2}+variable^{2}\\]\nas the equation of the envelope \\( I \\).\n\nRestoring the third dimension, the envelope has equation\n\\[orthogonal^{2}=stationary^{2}+horizontal^{2}+variable^{2}\\]\nwhich in the original coordinates becomes\n\\[2 knownvalue certainty=horizontal^{2}+variable^{2} .\\]\n\nGeneralization. Any non-degenerate quadratic cone is affinely equivalent to a right circular cone, and since ratios of volumes are preserved under an affine transformation, the calculations above lead to the conclusion that if all planes of a family cut off solids of the same fixed volume from a quadratic cone, the planes must be tangent to a hyperboloid which is asymptotic to the cone.\n\nRemark. We have seen that all planes tangent to a hyperboloid of two sheets cut off the same volume from the asymptotic cone. This is analogous to the fact that all lines tangent to a hyperbola form with the asymptotes. Consider \\( H \\) the \"hyperboloid of two sheets\" in \\( \\mathbf{R}^{continuous} \\) given by\n\\[knownvalue_{1}{ }^{2}+knownvalue_{2}{ }^{2}+\\cdots+knownvalue_{continuous-1}{ }^{2}=knownvalue_{continuous}{ }^{2}-variable^{2} .\\]\n\nIts asymptotic cone \\( C \\) has equation\n\\[knownvalue_{1}{ }^{2}+knownvalue_{2}{ }^{2}+\\cdots+knownvalue_{continuous-1}{ }^{2}=knownvalue_{continuous}{ }^{2} .\\]\n\nIf \\( P \\) is any hyperplane tangent to \\( H \\), then \\( P \\) and \\( C \\) divide \\( \\mathbf{R}^{continuous} \\) into five regions (seven, if \\( continuous=2 \\) ) of which only one is bounded. The \\( continuous \\)-dimensional volume of this bounded region \\( B_{P} \\) is the same for all choices of \\( P \\). Since the group preserving the quadratic form acts transitively, all the regions \\( B_{P} \\) have the same volume.\n\nAs in the case of three dimensions, all \"hyperboloids of two sheets\" (i.e., non-degenerate, disconnected quadric surfaces) in \\( \\mathbf{R}^{continuous} \\) are equivalent to \\( H \\) under a linear transformation."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "s_n": "hjgrksla",
        "s_n+1": "mxpqolzu",
        "t_n": "vckhpsan",
        "t_n+1": "dzfryuql",
        "t_k+p": "bjezotlw",
        "t_k+i": "ypwnsord",
        "t_k+i+1": "cakrhfue",
        "t_k": "ugrdmvci",
        "p": "wlxehjbn",
        "i": "qemrstuv",
        "k": "ozpvdnla",
        "x": "fltgjwra",
        "y": "nhqksamz",
        "z": "xrspoblu",
        "u": "kevnadzr",
        "v": "swlidfjo",
        "a": "gbocvhyu",
        "m": "jyexuqpa",
        "b": "lhospwzr",
        "L": "cdqitnva",
        "\\\\lambda": "rkespluw",
        "A": "zunqkera",
        "\\\\epsilon": "vhypgsod"
      },
      "question": "5. Answer either (i) or (ii).\n(i) Given that the sequence whose \\( qzxwvtnp \\)th term is \\( \\left(hjgrksla+2 mxpqolzu\\right) \\) converges, show that the sequence \\( \\left\\{hjgrksla\\right\\} \\) converges also.\n(page 306)\n(ii) A plane varies so that it includes a cone of constant volume equal to \\( \\pi gbocvhyu^{3} / 3 \\) with the surface the equation of which in rectangular coordinates is \\( 2 fltgjwra nhqksamz=xrspoblu^{2} \\). Find the equation of the envelope of the various positions of this plane.\n\nState the result so that it applies to a general cone (that is, conic surface) of the second order.",
      "solution": "Solution. Suppose \\( \\lim \\left(hjgrksla+2 mxpqolzu\\right)=3 cdqitnva \\). Then\n\\[\n\\lim \\left[\\left(hjgrksla-cdqitnva\\right)+2\\left(mxpqolzu-cdqitnva\\right)\\right]=0\n\\]\n\nPut \\( vckhpsan=hjgrksla-cdqitnva \\); then \\( \\lim \\left(vckhpsan+2 dzfryuql\\right)=0 \\). We shall prove that \\( \\lim vckhpsan=0 \\). This will\nsequence. Given \\( vhypgsod> 0 \\), choose \\( ozpvdnla \\) so that\n\\[\n\\left|vckhpsan+2 dzfryuql\\right|<vhypgsod \\quad \\text { for all } qzxwvtnp \\geq ozpvdnla .\n\\]\n\nBy induction on \\( wlxehjbn \\) we find that\n\\[\nugrdmvci-(-2)^{wlxehjbn} bjezotlw=\\sum_{qemrstuv=0}^{wlxehjbn-1}(-2)^{qemrstuv}\\left(ypwnsord+2 cakrhfue\\right)\n\\]\n\nTherefore\n\\[\n\\left|ugrdmvci-(-2)^{wlxehjbn} bjezotlw\\right| \\leq \\sum_{qemrstuv=0}^{wlxehjbn-1} 2^{qemrstuv}\\left|ypwnsord+2 cakrhfue\\right|<2^{wlxehjbn} vhypgsod,\n\\]\nprovided \\( wlxehjbn \\geq 1 \\). Dividing by \\( 2^{wlxehjbn} \\), we get\nso\n\\[\n\\left|bjezotlw\\right|<vhypgsod+\\frac{1}{2^{j}}\\left|ugrdmvci\\right| .\n\\]\n\nHence\n\\( \\limsup _{wlxehjbn \\rightarrow-\\infty}\\left|bjezotlw\\right| \\leq vhypgsod \\).\nTherefore\n\\[\n\\limsup _{qzxwvtnp \\rightarrow \\infty}\\left|vckhpsan\\right| \\leq vhypgsod .\n\\]\n\nSince \\( vhypgsod \\) was arbitrary, this implies \\( \\lim vckhpsan=0 \\). As we remarked before, this proves that the sequence \\( \\left\\{hjgrksla\\right\\} \\) converges to \\( cdqitnva \\).\n\nRemark. The argument generalizes to prove that if \\( hjgrksla-rkespluw mxpqolzu \\rightarrow zunqkera \\,\\),\nwhere \\( |rkespluw|>1 \\), then \\( hjgrksla \\rightarrow zunqkera /(1-rkespluw) \\).\n\nSolution. Make the change of coordinates\n\\[\n\\begin{array}{l}\nfltgjwra=\\frac{1}{2} \\sqrt{2}(kevnadzr+swlidfjo) \\\\\nnhqksamz=\\frac{1}{2} \\sqrt{2}(kevnadzr-swlidfjo)\n\\end{array}\n\\]\n\nThen the \\( kevnadzr swlidfjo \\)-axes are orthogonal and rotated by \\( \\pi / 4 \\) radians from the \\( fltgjwra nhqksamz \\) axes. The equation of the given surface in the new coordinates is\n\\[\nxrspoblu^{2}+swlidfjo^{2}=kevnadzr^{2}\n\\]\nwhich is a right circular cone with axis the \\( kevnadzr \\)-axis.\nNext we find the volume of the conical region cut off from the solid cone\nby a plane. Because of rotational symmetry we need only consider planes of the form \\( kevnadzr=jyexuqpa swlidfjo+lhospwzr \\). In order that the plane should cut off a bounded\nregion it is necessary that \\( |jyexuqpa|<1 \\). The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base.\nThe altitude is the distance from the origin to the plane, namely\n\\[\n\\frac{|lhospwzr|}{\\sqrt{1+jyexuqpa^{2}}}\n\\]\n\nThe area of the base is \\( \\sqrt{1+jyexuqpa^{2}} \\) times the area of the ellipse obtained by projecting it orthogonally onto the \\( swlidfjo xrspoblu \\)-plane. To find the equation of the\nprojected ellipse we eliminate \\( kevnadzr \\) between the equations of the cone and the plane, getting\n\\[\nxrspoblu^{2}+swlidfjo^{2}=(jyexuqpa swlidfjo+lhospwzr)^{2}\n\\]\n\nCollecting the \\( swlidfjo \\) terms and completing the square we get\n\\[\nxrspoblu^{2}+\\left(1-jyexuqpa^{2}\\right)\\left(swlidfjo-\\frac{jyexuqpa lhospwzr}{1-jyexuqpa^{2}}\\right)^{2}=lhospwzr^{2}\\left(\\frac{1}{1-jyexuqpa^{2}}\\right)\n\\]\n[Note that this is indeed an ellipse because \\( |jyexuqpa|<1 \\) ]. The area of this ellipse is\n\\[\nzunqkera=\\pi lhospwzr^{2} \\frac{1}{\\left(1-jyexuqpa^{2}\\right)^{3 / 2}} .\n\\]\n\nThe volume of the conical region is therefore\n\\[\n\\begin{aligned}\n\\frac{1}{3} \\text { base } \\times \\text { altitude } & =\\frac{1}{3}\\left(\\sqrt{1+jyexuqpa^{2}} \\pi lhospwzr^{2} \\frac{1}{\\left(1-jyexuqpa^{2}\\right)^{3 / 2}}\\right)\\left(\\frac{|lhospwzr|}{\\sqrt{1+jyexuqpa^{2}}}\\right) \\\\\n& =\\frac{1}{3} \\pi|lhospwzr|^{3} \\frac{1}{\\left(1-jyexuqpa^{2}\\right)^{3 / 2}} .\n\\end{aligned}\n\\]\n\nThe problem restricts consideration to those planes which cut off a vol-\nme \\( \\frac{1}{3} \\pi gbocvhyu^{3} \\), that is, to planes for which\n\\[\n|lhospwzr|=gbocvhyu \\sqrt{1-jyexuqpa^{2}}\n\\]\nwhere \\( jyexuqpa \\) is the tangent of the angle between the plane and the \\( xrspoblu swlidfjo \\)-plane. We are to find the envelope \\( E \\) of all such planes. Clearly \\( E \\) must share\nthe rotational symmetry of the entire configuration, so it suffices to find\nthe intersection \\( I \\) of \\( E \\) with the \\( kevnadzr swlidfjo \\)-plane.\nConsider the plane \\( P \\) tangent to \\( E \\) at a point of \\( I \\). Reflection in the \\( kevnadzr swlidfjo \\) plane preserves \\( P \\), since it preserves \\( E \\) and fixes each point of \\( I \\). Therefore \\( P \\) is perpendicular to the \\( kevnadzr swlidfjo \\)-plane, so it has an equation of the form \\( kevnadzr=jyexuqpa swlidfjo+lhospwzr \\) where, as we have seen, \\( |lhospwzr|=gbocvhyu \\sqrt{1-jyexuqpa^{2}} \\). The line \\( l \\) in which \\( P \\) meets the \\( kevnadzr swlidfjo \\)-plane is clearly tangent to \\( I \\). Hence \\( I \\) is the envelope of all lines \\( l \\) in the \\( kevnadzr swlidfjo \\)-plane having equations of the form\n(1)\n\\[\nkevnadzr=jyexuqpa swlidfjo \\pm gbocvhyu \\sqrt{1-jyexuqpa^{2}} .\n\\]\n\nThe envelope problem is thus reduced to two dimensions.\nTo find the envelope,\nbetween the equation\n(2)\n\\( \\qquad \\)\nand the equation obtained by differentiating (2) witl\n(3)\n\\[\n0=swlidfjo-\\frac{gbocvhyu jyexuqpa}{\\sqrt{1-jyexuqpa^{2}}} .\n\\]\n\nFrom (3) we find\n\\[\njyexuqpa^{2}=\\frac{swlidfjo^{2}}{swlidfjo^{2}+gbocvhyu^{2}}, \\quad \\text { so } \\quad 1-jyexuqpa^{2}=\\frac{gbocvhyu^{2}}{swlidfjo^{2}+gbocvhyu^{2}} .\n\\]\n\nEquation (2) now becomes\n\\[\nkevnadzr=\\sqrt{1-jyexuqpa^{2}}\\left(\\frac{jyexuqpa}{\\sqrt{1-jyexuqpa^{2}}} swlidfjo+gbocvhyu\\right)=\\sqrt{1-jyexuqpa^{2}}\\left(\\frac{swlidfjo^{2}+gbocvhyu^{2}}{gbocvhyu}\\right),\n\\]\ngiving finally\n(4)\n\\[\nkevnadzr^{2}=swlidfjo^{2}+gbocvhyu^{2}\n\\]\nas the equation of the envelope \\( I \\). The same equation results if we start with the negative sign.\n[Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the\nfamily of tangent lines to the hyperbola (4). This examination reveals that\nthe tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).]\n\\( kevnadzr \\) axis; hence its equation is\n\\[\nkevnadzr^{2}=swlidfjo^{2}+xrspoblu^{2}+gbocvhyu^{2} .\n\\]\n\nTransforming back to the original coordinates its equation is\n\\[\n2 fltgjwra nhqksamz=xrspoblu^{2}+gbocvhyu^{2} .\n\\]\n\nGeneralization. Any non-degenerate quadratic cone is affinely equivalent to a right circular cone, and since ratios of volumes are preserved under an affine transformation, the calculations above lead to the conclusion that if all planes of a family cut off solids of the same fixed\nvolume from a quadratic cone, the planes must be tangent to a hyperboloid which is asymptotic to the cone.\n\nRemark. We have seen that all planes tangent to a hyperboloid of two sheets cut off the same volume from the asymptotic cone. This is analogous to the fact that all lines tangent to a hyperbola form with the asymptotes Consider \\( H \\) the \"hyperboloid of two sheets\" in \\( \\mathbf{R}^{qzxwvtnp} \\) given by\n\\[\nfltgjwra_{1}{ }^{2}+fltgjwra_{2}{ }^{2}+\\cdots+fltgjwra_{qzxwvtnp-1}{ }^{2}=fltgjwra_{qzxwvtnp}{ }^{2}-gbocvhyu^{2} .\n\\]\n\nIts asymptotic cone \\( C \\) has equation\n\\[\nfltgjwra_{1}{ }^{2}+fltgjwra_{2}{ }^{2}+\\cdots+fltgjwra_{qzxwvtnp-1}{ }^{2}=fltgjwra_{qzxwvtnp}{ }^{2} .\n\\]\n\nIf \\( P \\) is any hyperplane tangent to \\( H \\), then \\( P \\) and \\( C \\) divide \\( \\mathbf{R}^{qzxwvtnp} \\) into five regions (seven, if \\( qzxwvtnp=2 \\) ) of which only one is bounded. The \\( qzxwvtnp \\)-dimensional\nvolume of this bounded region \\( B_{P} \\) is the same for all choices of \\( P \\). To or all choices of \\( P \\).\npreserve the quadratic form \\( fltgjwra_{1}{ }^{2}+fltgjwra_{2}{ }^{2}+\\cdots+fltgjwra_{qzxwvtnp-1}{ }^{2}-fltgjwra_{qzxwvtnp}{ }^{2} \\). \\( G \\) clearly preserves \\( H \\) and \\( C \\), and any hyperplane tangent to \\( H \\) is mapped by any element of \\( G \\) to a hyperplane tangent to \\( H \\). Since \\( G \\) acts transitively on \\( H \\), it acts transitively on the set of hyperplanes tangent to \\( H \\), and hence on the are all volume-preserving, and thus all the regions \\( B_{P} \\) have the same\nvolume.\n\nAs in the case of three dimensions, all \"hyperboloids of two sheets\" (i.e.,\nnon-degenerate, disconnected quadric surfaces) in \\( \\mathbf{R}^{qzxwvtnp} \\) are equivalent to \\( H \\)\nunder a linear transformation."
    },
    "kernel_variant": {
      "question": "\\[\n\\textbf{5.\\; Answer either {\\rm(i)} or {\\rm(ii)}.}\n\\]\n\n\\medskip\n\\textbf{(i)\\; Functional–analytic variant – uniform formulation}\n\nLet $B$ be a complex Banach space, fix an integer $m\\ge 1$, and choose\ncomplex numbers $c_{0},c_{1},\\dots ,c_{m}$, not all zero.  Given a\n$B$–valued sequence $(s_{n})_{n\\ge 0}\\subset B$ define\n\\[\nP(z)=\\sum_{k=0}^{m} c_{k}z^{k},\\qquad \n           R_{n}:=\\sum_{k=0}^{m}c_{k}s_{\\,n+k}\\qquad(n\\ge 0).\n\\]\nAssume  \n\n\\quad$\\bullet$ every zero $\\lambda$ of $P$ satisfies $\\lvert\\lambda\\rvert>1$\n(so in particular $P(1)\\neq 0$);  \n\n\\quad$\\bullet$ the Banach–valued sequence $(R_{n})$ converges in norm to a\nlimit $L\\in B$.\n\n\\smallskip\n(a) Prove that $(s_{n})$ converges in $B$ and that\n\\[\n\\lim_{n\\to\\infty}s_{n}\\;=\\;\\frac{L}{P(1)}.\n\\]\n\n(b) Show that the spectral hypothesis $\\lvert\\lambda\\rvert>1$ is sharp by\nconstructing an explicit pair\n\\[\n(s_{n})_{n\\ge 0}\\subset\\mathbf C,\\qquad P\\in\\mathbf C[z]\n\\]\nwith at least one root on the unit circle for which $(R_{n})$ converges\nwhile $(s_{n})$ does \\emph{not} converge.\n\n\\bigskip\n\\textbf{(ii)\\; Higher–dimensional geometry – fully corrected formulation}\n\nThroughout $\\rho:=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$.\n\n\\medskip\n\\textbf{(a)\\; The four–dimensional model.}\n\nIn $\\mathbf R^{4}$ consider the \\emph{quadric cone surface}\n\\[\nC\\;:\\;x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=x_{4}^{2},\\qquad x_{4}\\ge 0,\n\\]\nand its solid cone\n\\[\n\\mathcal K:=\\{x\\in\\mathbf R^{4}\\;:\\;\\rho\\le x_{4},\\;x_{4}\\ge 0\\}.  \\tag{$\\ast$}\n\\]\nFor a unit vector $u\\in\\mathbf R^{4}$ and $b>0$ let\n\\[\n\\Pi_{u,b}:\\;u\\!\\cdot\\!x=b, \\tag{$\\dagger$}\n\\]\nand define the truncated solid\n\\[\n\\Omega(u,b):=\\mathcal K\\cap\\bigl\\{u\\!\\cdot\\!x\\le b\\bigr\\}.\n\\]\nSuppose that \\emph{every} hyperplane that meets $\\mathcal K$ in a bounded\nregion cuts off the same four–volume\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u,b)\\bigr)=\nV=\\frac{\\pi^{2}a^{4}}{4}\\qquad (a>0). \\tag{$\\ddagger\\ddagger$}\n\\]\n\n(i)   Using the $\\operatorname{O}(3)$ symmetry in the first three coordinates,\n      show that the normal vector may be chosen as\n      \\[\n         u_{\\theta}=(\\sin\\theta,0,0,\\cos\\theta),\\qquad -\\tfrac{\\pi}{4}<\\theta<\\tfrac{\\pi}{4}.\n      \\]\n      For the plane $H_{\\theta,b}:\\sin\\theta\\,x_{1}+\\cos\\theta\\,x_{4}=b$ prove,\n      by an explicit four–volume computation, that $(\\ddagger\\ddagger)$ is\n      equivalent to\n      \\[\n         b^{4}=\\frac{3\\pi a^{4}}{4}\\,\n               \\frac{(1-\\tan^{2}\\theta)^{2}}\n                    {(1+\\tan^{2}\\theta)^{2}}. \\tag{$\\ddagger$}\n      \\]\n\n(ii)  Put $m:=-\\tan\\theta$ ($\\lvert m\\rvert<1$), so that $b=b(m)$ is fixed by\n      $(\\ddagger)$.  Rotating $H_{\\theta,b}$ about the $x_{4}$–axis gives the\n      one–parameter family of hyperplanes  \n      \\[\n        \\sin\\theta\\,(n\\!\\cdot\\!x')+\\cos\\theta\\,x_{4}=b\n        \\quad\\bigl(n\\in S^{2},\\;x'=(x_{1},x_{2},x_{3})\\bigr),\n      \\]\n      whose common envelope is the $C^{1}$ ruled hypersurface\n      \\[\n        \\Sigma_{m}:\\;x_{4}=m\\rho+\\sqrt{1+m^{2}}\\;b(m). \\tag{$\\heartsuit$}\n      \\]\n      Eliminate $m$ from the equations\n      \\[\n          F(m):=x_{4}-m\\rho-\\sqrt{1+m^{2}}\\;b(m)=0,\n          \\qquad\n          F'(m)=0,\n      \\]\n      and prove that the \\emph{collection} of envelopes $\\Sigma_{m}$ has the further\n      envelope\n      \\[\n        x_{4}^{2}-\\bigl(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\bigr)=\n        \\Bigl(\\tfrac{3\\pi}{4}\\Bigr)^{\\!1/2}a^{2}. \\tag{$\\clubsuit$}\n      \\]\n\n\\medskip\n\\textbf{(b)\\; The general quadratic cone.}\n\nLet $n\\ge 2$ and let $Q$ be a symmetric, non–degenerate quadratic\nform on $\\mathbf R^{\\,n+1}$ of \\emph{Lorentzian} signature $(\\,n,1)$.  Define the solid cone\n\\[\n\\mathcal K_{Q}:=\\bigl\\{x\\in\\mathbf R^{\\,n+1}\\;:\\;Q(x)\\le 0,\\;x_{\\,n+1}\\ge 0\\bigr\\}.\n\\]\nFor a unit vector $u\\in\\mathbf R^{\\,n+1}$ and $b>0$ put\n\\[\n\\Pi_{u,b}:\\;u\\!\\cdot\\!x=b,\\qquad\n\\Omega_{Q}(u,b):=\\mathcal K_{Q}\\cap\\{u\\!\\cdot\\!x\\le b\\}.\n\\]\n\nAssume that an affine family of hyperplanes cuts from $\\mathcal K_{Q}$\nbounded regions, all of which have the same $(n+1)$–dimensional volume $V$.\nShow that this family is \\emph{exactly} the set of all hyperplanes that are\ntangent to the unique quadric\n\\[\nQ(x)=-A^{2},\\qquad A>0,\n\\]\nwhere $A$ is determined (and uniquely determined) by the prescribed volume\n$V$.  In particular, after a volume–preserving affine change of variables,\nthe quadric $Q(x)=-A^{2}$ can be written in the canonical form $(\\clubsuit)$.\n\n\n\n\\bigskip",
      "solution": "Throughout $\\lVert\\cdot\\rVert$ denotes the norm of the underlying real or\ncomplex Banach space, and $\\rho:=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$.\n\n\\bigskip\n\\textbf{(i) Functional–analytic part}\n\n\\smallskip\n\\emph{(a) Convergence of $(s_{n})$.}\nPut $\\delta_{n}:=R_{n}-L$ and\n$t_{n}:=s_{n}-L/P(1)$.  Then\n\\[\n\\sum_{k=0}^{m}c_{k}\\,t_{\\,n+k}=\\delta_{n}\\qquad(n\\ge 0). \\tag{1}\n\\]\nAll zeros of $P$ lie outside the closed unit disc, hence\n$1/P$ has a power–series expansion\n\\[\n\\frac{1}{P(z)}=\\sum_{j=0}^{\\infty}a_{j}z^{j},\\qquad\n|a_{j}|\\le C\\,r^{-j}\\quad\\text{for some }C>0,\\;r>1. \\tag{2}\n\\]\nConvolution of (1) with the sequence $(a_{j})_{j\\ge 0}$ yields\n\\[\nt_{n}=\\sum_{j=0}^{\\infty}a_{j}\\,\\delta_{\\,n+j}\\qquad(n\\ge 0). \\tag{3}\n\\]\nBecause $\\lVert\\delta_{n}\\rVert\\to 0$ and\n$\\sum_{j\\ge 0}r^{-j}<\\infty$, a dominated–tail estimate shows\n$\\lVert t_{n}\\rVert\\to 0$, hence\n\\[\n\\lim_{n\\to\\infty}s_{n}=\\frac{L}{P(1)}.\n\\]\n\n\\smallskip\n\\emph{(b) Sharpness of the spectral condition.}\nTake $P(z)=1-z$ (whose zero lies on the unit circle) and $s_{n}=n$.  Then\n\\[\nR_{n}=s_{n}-s_{n+1}=-1\\qquad(n\\ge 0),\n\\]\nso $(R_{n})$ converges, whereas $(s_{n})$ clearly diverges.  Thus the\nhypothesis $\\lvert\\lambda\\rvert>1$ cannot be weakened.\n\n\\bigskip\n\\textbf{(ii) Geometric part}\n\n\\smallskip\n\\textbf{(a) Exact four–volume of $\\Omega(u_{\\theta},b)$.}\n\nBecause $\\mathcal K$ is invariant under the $\\operatorname{O}(3)$ action on the\nfirst three coordinates, we may rotate the hyperplane so that its unit\nnormal is $u_{\\theta}=(\\sin\\theta,0,0,\\cos\\theta)$ with\n$|\\theta|<\\tfrac{\\pi}{4}$.  Denote $s:=\\sin\\theta$ and $c:=\\cos\\theta\\;(>0)$.\n\n\\medskip\n\\emph{Step 1:  Ray integration.}\nFor $w\\in S^{3}$ in the solid cone one has\n\\(\nt_{\\max}(w)=b/(u_{\\theta}\\!\\cdot\\!w).\n\\)\nWith the Euclidean surface element $d\\sigma$ on $S^{3}$,\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u_{\\theta},b)\\bigr)=\n\\frac{b^{4}}{4}\\!\\int_{D}(u_{\\theta}\\!\\cdot\\!w)^{-4}\\,d\\sigma(w), \\tag{4}\n\\]\nwhere $D:=S^{3}\\cap\\{\\rho<w_{4}\\}$.\n\n\\medskip\n\\emph{Step 2:  Evaluating the integral.}\nA parametrisation of $D$ and two elementary Beta–function evaluations\ngive\n\\[\n\\int_{D}(u_{\\theta}\\!\\cdot\\!w)^{-4}\\,d\\sigma(w)\n      =\\frac{4\\pi}{3}\\,\\cos^{-2}2\\theta. \\tag{5}\n\\]\n\n\\medskip\n\\emph{Step 3:  The volume formula.}\nInserting (5) into (4) yields\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u_{\\theta},b)\\bigr)=\n\\frac{\\pi}{3}\\,b^{4}\\cos^{-2}2\\theta, \\tag{6}\n\\]\nso condition $(\\ddagger\\ddagger)$ is equivalent to $(\\ddagger)$.\n\n\\bigskip\n\\textbf{(b) The first envelope.}\n\nPut $m:=-\\tan\\theta$ ($|m|<1$); then from $(\\ddagger)$\n\\[\nb(m)=\\kappa_{0}\\,\n      \\frac{\\sqrt{1-m^{2}}}{\\sqrt{1+m^{2}}},\n\\qquad\n\\kappa_{0}:=\\Bigl(\\frac{3\\pi}{4}\\Bigr)^{\\!1/4}a. \\tag{7}\n\\]\nRotating $H_{\\theta,b}$ about the $x_{4}$–axis yields the ruled surface\n$(\\heartsuit)$.\n\n\\medskip\nDefine\n\\[\nF(m):=x_{4}-m\\rho-\\sqrt{1+m^{2}}\\;b(m).\n\\]\nA point $x$ belongs to the common envelope of the $\\Sigma_{m}$ iff there\nexists $m$ with $|m|<1$ such that\n\\[\nF(m)=0,\\qquad F'(m)=0. \\tag{8}\n\\]\n\n\\emph{Derivative of $F$.}  \nWith $g(m):=\\sqrt{1+m^{2}}$ one has from (7)\n\\[\nb'(m)=-\\frac{2m\\,b(m)}{1-m^{4}},\\qquad\nF'(m)= -\\rho-\\frac{m}{g(m)}\\,b(m)+\\frac{2m\\,g(m)\\,b(m)}{1-m^{4}}. \\tag{9}\n\\]\n\n\\smallskip\nEquation $F'(m)=0$ gives\n\\[\n\\rho=\\frac{m\\,g(m)\\,b(m)}{1-m^{2}}, \\tag{10}\n\\]\nand inserting (10) in $F(m)=0$ shows\n\\[\nx_{4}=\\frac{g(m)\\,b(m)}{1-m^{2}}. \\tag{11}\n\\]\nHence\n\\[\nx_{4}^{2}-\\rho^{2}=\\frac{g^{2}(m)b^{2}(m)}{1-m^{2}}\n                 =\\kappa_{0}^{2}\n                 =\\Bigl(\\tfrac{3\\pi}{4}\\Bigr)^{1/2}a^{2}, \\tag{12}\n\\]\nso the second envelope is $(\\clubsuit)$.\n\n\\bigskip\n\\textbf{(c) General quadratic cones.}\n\nLet $Q(x)=x\\!\\cdot\\!A x$ with $A=A^{\\top}$, ${\\rm rank}\\,A=n+1$ and\nsignature $(n,1)$.  The Lorentz dual form is\n$Q^{\\!*}(u)=u\\!\\cdot\\!A^{-1}u$.  For the hyperplane\n$\\Pi_{u,b}:u\\!\\cdot\\!x=b$ ($b>0$) define the truncated solid\n\\[\n\\Omega_{Q}(u,b):=\\mathcal K_{Q}\\cap\\{u\\!\\cdot\\!x\\le b\\}.\n\\]\n\n\\medskip\n\\emph{Volume of $\\Omega_{Q}(u,b)$.}\nChoose an affine map sending $\\mathcal K_{Q}$ to the model cone\n$x_{1}^{2}+\\dots+x_{n}^{2}=x_{n+1}^{2}$ and taking $\\Pi_{u,b}$ to a hyperplane\nwith (unit) normal $v$.  Because the map is linear, its Jacobian is a\nconstant.  A direct ray integration in the model cone gives\n\\[\n\\operatorname{vol}_{\\,n+1}\\!\\bigl(\\Omega_{Q}(u,b)\\bigr)=\n\\kappa_{n}\\,\n\\frac{b^{\\,n+1}}{\\lvert Q^{\\!*}(u)\\rvert^{\\frac{n+1}{2}}},\n\\qquad\n\\kappa_{n}:=\\frac{\\operatorname{vol}_{\\,n}(B^{\\,n})}{\\,n+1\\,}, \\tag{13}\n\\]\nwhere $\\operatorname{vol}_{\\,n}(B^{\\,n})=\\dfrac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\frac{n}{2}+1\\bigr)}$ is\nthe volume of the unit $n$–ball.\n\n\\medskip\n\\emph{Equal–volume condition.}\nIf all truncations have the same volume $V$, then\n\\[\nb^{\\,n+1}=\\lambda\\,\\lvert Q^{\\!*}(u)\\rvert^{\\frac{n+1}{2}},\\qquad\n\\lambda:=\\frac{V}{\\kappa_{n}}. \\tag{14}\n\\]\n\n\\medskip\n\\emph{Tangent hyperplanes.}\nDefine\n\\[\np(u):=\\lambda^{\\frac{1}{n+1}}\n       \\frac{A^{-1}u}{\\lvert Q^{\\!*}(u)\\rvert^{1/2}}. \\tag{15}\n\\]\nThen\n\\[\nQ\\bigl(p(u)\\bigr)=-\\lambda^{\\frac{2}{n+1}},\\qquad\nu\\!\\cdot\\!p(u)=b. \\tag{16}\n\\]\nHence $\\Pi_{u,b}$ is tangent to the quadric\n\\[\nQ(x)=-\\lambda^{\\frac{2}{n+1}}=:-A^{2}. \\tag{17}\n\\]\nConversely, every hyperplane tangent to $Q(x)=-A^{2}$ satisfies\n(14), hence belongs to the given family.  Therefore the equal–volume\nhyperplanes are \\emph{exactly} the tangent hyperplanes to (17).  An\naffine unimodular change of coordinates transforms (17) into the\nhyperboloid $(\\clubsuit)$, completing the proof.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.434357",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher algebraic sophistication:  Part (i) requires knowledge of the spectrum of a linear operator, power-series inversion in Banach algebras, and interchange of limits—none of which appears in the original problem.\n\n2. More variables & higher dimension:  The geometric section moves from ℝ³ to ℝ⁴ (and then to arbitrary n), demanding the use of n-dimensional volume formulas and a careful treatment of rotational symmetry in higher codimension.\n\n3. Additional constraints:  Instead of a single linear relation s_n+ks_{n+1}, we deal with an arbitrary finite‐order linear recurrence with arbitrary coefficients; root-location conditions must be analysed and shown to be sharp.\n\n4. Deeper theoretical content:  The analytic proof employs functional calculus for bounded operators and analyticity of 1/P on the closed unit disk; the geometric proof uses group actions (O(3), the orthogonal group of Q) and affine equivalence of quadrics.\n\n5. Multiple interacting concepts:  Each part mixes classical analysis (infinite series, convergence), algebra (polynomial factorisation, spectral theory), geometry (envelopes, hyperquadrics), and higher-dimensional volume calculations.\n\nTogether these elements render the enhanced variant substantially more intricate and demanding than both the original textbook exercise and the previous kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "\\[\n\\textbf{5.\\; Answer either {\\rm(i)} or {\\rm(ii)}.}\n\\]\n\n\\medskip\n\\textbf{(i)\\; Functional–analytic variant – uniform formulation}\n\nLet $B$ be a complex Banach space, fix an integer $m\\ge 1$, and choose\ncomplex numbers $c_{0},c_{1},\\dots ,c_{m}$, not all zero.  Given a\n$B$–valued sequence $(s_{n})_{n\\ge 0}\\subset B$ define\n\\[\nP(z)=\\sum_{k=0}^{m} c_{k}z^{k},\\qquad \n           R_{n}:=\\sum_{k=0}^{m}c_{k}s_{\\,n+k}\\qquad(n\\ge 0).\n\\]\nAssume  \n\n\\quad$\\bullet$ every zero $\\lambda$ of $P$ satisfies $\\lvert\\lambda\\rvert>1$\n(so in particular $P(1)\\neq 0$);  \n\n\\quad$\\bullet$ the Banach–valued sequence $(R_{n})$ converges in norm to a\nlimit $L\\in B$.\n\n\\smallskip\n(a) Prove that $(s_{n})$ converges in $B$ and that\n\\[\n\\lim_{n\\to\\infty}s_{n}\\;=\\;\\frac{L}{P(1)}.\n\\]\n\n(b) Show that the spectral hypothesis $\\lvert\\lambda\\rvert>1$ is sharp by\nconstructing an explicit pair\n\\[\n(s_{n})_{n\\ge 0}\\subset\\mathbf C,\\qquad P\\in\\mathbf C[z]\n\\]\nwith at least one root on the unit circle for which $(R_{n})$ converges\nwhile $(s_{n})$ does \\emph{not} converge.\n\n\\bigskip\n\\textbf{(ii)\\; Higher–dimensional geometry – fully corrected formulation}\n\nThroughout $\\rho:=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$.\n\n\\medskip\n\\textbf{(a)\\; The four–dimensional model.}\n\nIn $\\mathbf R^{4}$ consider the \\emph{quadric cone surface}\n\\[\nC\\;:\\;x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=x_{4}^{2},\\qquad x_{4}\\ge 0,\n\\]\nand its solid cone\n\\[\n\\mathcal K:=\\{x\\in\\mathbf R^{4}\\;:\\;\\rho\\le x_{4},\\;x_{4}\\ge 0\\}.  \\tag{$\\ast$}\n\\]\nFor a unit vector $u\\in\\mathbf R^{4}$ and $b>0$ let\n\\[\n\\Pi_{u,b}:\\;u\\!\\cdot\\!x=b, \\tag{$\\dagger$}\n\\]\nand define the truncated solid\n\\[\n\\Omega(u,b):=\\mathcal K\\cap\\bigl\\{u\\!\\cdot\\!x\\le b\\bigr\\}.\n\\]\nSuppose that \\emph{every} hyperplane that meets $\\mathcal K$ in a bounded\nregion cuts off the same four–volume\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u,b)\\bigr)=\nV=\\frac{\\pi^{2}a^{4}}{4}\\qquad (a>0). \\tag{$\\ddagger\\ddagger$}\n\\]\n\n(i)   Using the $\\operatorname{O}(3)$ symmetry in the first three coordinates,\n      show that the normal vector may be chosen as\n      \\[\n         u_{\\theta}=(\\sin\\theta,0,0,\\cos\\theta),\\qquad -\\tfrac{\\pi}{4}<\\theta<\\tfrac{\\pi}{4}.\n      \\]\n      For the plane $H_{\\theta,b}:\\sin\\theta\\,x_{1}+\\cos\\theta\\,x_{4}=b$ prove,\n      by an explicit four–volume computation, that $(\\ddagger\\ddagger)$ is\n      equivalent to\n      \\[\n         b^{4}=\\frac{3\\pi a^{4}}{4}\\,\n               \\frac{(1-\\tan^{2}\\theta)^{2}}\n                    {(1+\\tan^{2}\\theta)^{2}}. \\tag{$\\ddagger$}\n      \\]\n\n(ii)  Put $m:=-\\tan\\theta$ ($\\lvert m\\rvert<1$), so that $b=b(m)$ is fixed by\n      $(\\ddagger)$.  Rotating $H_{\\theta,b}$ about the $x_{4}$–axis gives the\n      one–parameter family of hyperplanes  \n      \\[\n        \\sin\\theta\\,(n\\!\\cdot\\!x')+\\cos\\theta\\,x_{4}=b\n        \\quad\\bigl(n\\in S^{2},\\;x'=(x_{1},x_{2},x_{3})\\bigr),\n      \\]\n      whose common envelope is the $C^{1}$ ruled hypersurface\n      \\[\n        \\Sigma_{m}:\\;x_{4}=m\\rho+\\sqrt{1+m^{2}}\\;b(m). \\tag{$\\heartsuit$}\n      \\]\n      Eliminate $m$ from the equations\n      \\[\n          F(m):=x_{4}-m\\rho-\\sqrt{1+m^{2}}\\;b(m)=0,\n          \\qquad\n          F'(m)=0,\n      \\]\n      and prove that the \\emph{collection} of envelopes $\\Sigma_{m}$ has the further\n      envelope\n      \\[\n        x_{4}^{2}-\\bigl(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\bigr)=\n        \\Bigl(\\tfrac{3\\pi}{4}\\Bigr)^{\\!1/2}a^{2}. \\tag{$\\clubsuit$}\n      \\]\n\n\\medskip\n\\textbf{(b)\\; The general quadratic cone.}\n\nLet $n\\ge 2$ and let $Q$ be a symmetric, non–degenerate quadratic\nform on $\\mathbf R^{\\,n+1}$ of \\emph{Lorentzian} signature $(\\,n,1)$.  Define the solid cone\n\\[\n\\mathcal K_{Q}:=\\bigl\\{x\\in\\mathbf R^{\\,n+1}\\;:\\;Q(x)\\le 0,\\;x_{\\,n+1}\\ge 0\\bigr\\}.\n\\]\nFor a unit vector $u\\in\\mathbf R^{\\,n+1}$ and $b>0$ put\n\\[\n\\Pi_{u,b}:\\;u\\!\\cdot\\!x=b,\\qquad\n\\Omega_{Q}(u,b):=\\mathcal K_{Q}\\cap\\{u\\!\\cdot\\!x\\le b\\}.\n\\]\n\nAssume that an affine family of hyperplanes cuts from $\\mathcal K_{Q}$\nbounded regions, all of which have the same $(n+1)$–dimensional volume $V$.\nShow that this family is \\emph{exactly} the set of all hyperplanes that are\ntangent to the unique quadric\n\\[\nQ(x)=-A^{2},\\qquad A>0,\n\\]\nwhere $A$ is determined (and uniquely determined) by the prescribed volume\n$V$.  In particular, after a volume–preserving affine change of variables,\nthe quadric $Q(x)=-A^{2}$ can be written in the canonical form $(\\clubsuit)$.\n\n\n\n\\bigskip",
      "solution": "Throughout $\\lVert\\cdot\\rVert$ denotes the norm of the underlying real or\ncomplex Banach space, and $\\rho:=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$.\n\n\\bigskip\n\\textbf{(i) Functional–analytic part}\n\n\\smallskip\n\\emph{(a) Convergence of $(s_{n})$.}\nPut $\\delta_{n}:=R_{n}-L$ and\n$t_{n}:=s_{n}-L/P(1)$.  Then\n\\[\n\\sum_{k=0}^{m}c_{k}\\,t_{\\,n+k}=\\delta_{n}\\qquad(n\\ge 0). \\tag{1}\n\\]\nAll zeros of $P$ lie outside the closed unit disc, hence\n$1/P$ has a power–series expansion\n\\[\n\\frac{1}{P(z)}=\\sum_{j=0}^{\\infty}a_{j}z^{j},\\qquad\n|a_{j}|\\le C\\,r^{-j}\\quad\\text{for some }C>0,\\;r>1. \\tag{2}\n\\]\nConvolution of (1) with the sequence $(a_{j})_{j\\ge 0}$ yields\n\\[\nt_{n}=\\sum_{j=0}^{\\infty}a_{j}\\,\\delta_{\\,n+j}\\qquad(n\\ge 0). \\tag{3}\n\\]\nBecause $\\lVert\\delta_{n}\\rVert\\to 0$ and\n$\\sum_{j\\ge 0}r^{-j}<\\infty$, a dominated–tail estimate shows\n$\\lVert t_{n}\\rVert\\to 0$, hence\n\\[\n\\lim_{n\\to\\infty}s_{n}=\\frac{L}{P(1)}.\n\\]\n\n\\smallskip\n\\emph{(b) Sharpness of the spectral condition.}\nTake $P(z)=1-z$ (whose zero lies on the unit circle) and $s_{n}=n$.  Then\n\\[\nR_{n}=s_{n}-s_{n+1}=-1\\qquad(n\\ge 0),\n\\]\nso $(R_{n})$ converges, whereas $(s_{n})$ clearly diverges.  Thus the\nhypothesis $\\lvert\\lambda\\rvert>1$ cannot be weakened.\n\n\\bigskip\n\\textbf{(ii) Geometric part}\n\n\\smallskip\n\\textbf{(a) Exact four–volume of $\\Omega(u_{\\theta},b)$.}\n\nBecause $\\mathcal K$ is invariant under the $\\operatorname{O}(3)$ action on the\nfirst three coordinates, we may rotate the hyperplane so that its unit\nnormal is $u_{\\theta}=(\\sin\\theta,0,0,\\cos\\theta)$ with\n$|\\theta|<\\tfrac{\\pi}{4}$.  Denote $s:=\\sin\\theta$ and $c:=\\cos\\theta\\;(>0)$.\n\n\\medskip\n\\emph{Step 1:  Ray integration.}\nFor $w\\in S^{3}$ in the solid cone one has\n\\(\nt_{\\max}(w)=b/(u_{\\theta}\\!\\cdot\\!w).\n\\)\nWith the Euclidean surface element $d\\sigma$ on $S^{3}$,\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u_{\\theta},b)\\bigr)=\n\\frac{b^{4}}{4}\\!\\int_{D}(u_{\\theta}\\!\\cdot\\!w)^{-4}\\,d\\sigma(w), \\tag{4}\n\\]\nwhere $D:=S^{3}\\cap\\{\\rho<w_{4}\\}$.\n\n\\medskip\n\\emph{Step 2:  Evaluating the integral.}\nA parametrisation of $D$ and two elementary Beta–function evaluations\ngive\n\\[\n\\int_{D}(u_{\\theta}\\!\\cdot\\!w)^{-4}\\,d\\sigma(w)\n      =\\frac{4\\pi}{3}\\,\\cos^{-2}2\\theta. \\tag{5}\n\\]\n\n\\medskip\n\\emph{Step 3:  The volume formula.}\nInserting (5) into (4) yields\n\\[\n\\operatorname{vol}_{4}\\!\\bigl(\\Omega(u_{\\theta},b)\\bigr)=\n\\frac{\\pi}{3}\\,b^{4}\\cos^{-2}2\\theta, \\tag{6}\n\\]\nso condition $(\\ddagger\\ddagger)$ is equivalent to $(\\ddagger)$.\n\n\\bigskip\n\\textbf{(b) The first envelope.}\n\nPut $m:=-\\tan\\theta$ ($|m|<1$); then from $(\\ddagger)$\n\\[\nb(m)=\\kappa_{0}\\,\n      \\frac{\\sqrt{1-m^{2}}}{\\sqrt{1+m^{2}}},\n\\qquad\n\\kappa_{0}:=\\Bigl(\\frac{3\\pi}{4}\\Bigr)^{\\!1/4}a. \\tag{7}\n\\]\nRotating $H_{\\theta,b}$ about the $x_{4}$–axis yields the ruled surface\n$(\\heartsuit)$.\n\n\\medskip\nDefine\n\\[\nF(m):=x_{4}-m\\rho-\\sqrt{1+m^{2}}\\;b(m).\n\\]\nA point $x$ belongs to the common envelope of the $\\Sigma_{m}$ iff there\nexists $m$ with $|m|<1$ such that\n\\[\nF(m)=0,\\qquad F'(m)=0. \\tag{8}\n\\]\n\n\\emph{Derivative of $F$.}  \nWith $g(m):=\\sqrt{1+m^{2}}$ one has from (7)\n\\[\nb'(m)=-\\frac{2m\\,b(m)}{1-m^{4}},\\qquad\nF'(m)= -\\rho-\\frac{m}{g(m)}\\,b(m)+\\frac{2m\\,g(m)\\,b(m)}{1-m^{4}}. \\tag{9}\n\\]\n\n\\smallskip\nEquation $F'(m)=0$ gives\n\\[\n\\rho=\\frac{m\\,g(m)\\,b(m)}{1-m^{2}}, \\tag{10}\n\\]\nand inserting (10) in $F(m)=0$ shows\n\\[\nx_{4}=\\frac{g(m)\\,b(m)}{1-m^{2}}. \\tag{11}\n\\]\nHence\n\\[\nx_{4}^{2}-\\rho^{2}=\\frac{g^{2}(m)b^{2}(m)}{1-m^{2}}\n                 =\\kappa_{0}^{2}\n                 =\\Bigl(\\tfrac{3\\pi}{4}\\Bigr)^{1/2}a^{2}, \\tag{12}\n\\]\nso the second envelope is $(\\clubsuit)$.\n\n\\bigskip\n\\textbf{(c) General quadratic cones.}\n\nLet $Q(x)=x\\!\\cdot\\!A x$ with $A=A^{\\top}$, ${\\rm rank}\\,A=n+1$ and\nsignature $(n,1)$.  The Lorentz dual form is\n$Q^{\\!*}(u)=u\\!\\cdot\\!A^{-1}u$.  For the hyperplane\n$\\Pi_{u,b}:u\\!\\cdot\\!x=b$ ($b>0$) define the truncated solid\n\\[\n\\Omega_{Q}(u,b):=\\mathcal K_{Q}\\cap\\{u\\!\\cdot\\!x\\le b\\}.\n\\]\n\n\\medskip\n\\emph{Volume of $\\Omega_{Q}(u,b)$.}\nChoose an affine map sending $\\mathcal K_{Q}$ to the model cone\n$x_{1}^{2}+\\dots+x_{n}^{2}=x_{n+1}^{2}$ and taking $\\Pi_{u,b}$ to a hyperplane\nwith (unit) normal $v$.  Because the map is linear, its Jacobian is a\nconstant.  A direct ray integration in the model cone gives\n\\[\n\\operatorname{vol}_{\\,n+1}\\!\\bigl(\\Omega_{Q}(u,b)\\bigr)=\n\\kappa_{n}\\,\n\\frac{b^{\\,n+1}}{\\lvert Q^{\\!*}(u)\\rvert^{\\frac{n+1}{2}}},\n\\qquad\n\\kappa_{n}:=\\frac{\\operatorname{vol}_{\\,n}(B^{\\,n})}{\\,n+1\\,}, \\tag{13}\n\\]\nwhere $\\operatorname{vol}_{\\,n}(B^{\\,n})=\\dfrac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\frac{n}{2}+1\\bigr)}$ is\nthe volume of the unit $n$–ball.\n\n\\medskip\n\\emph{Equal–volume condition.}\nIf all truncations have the same volume $V$, then\n\\[\nb^{\\,n+1}=\\lambda\\,\\lvert Q^{\\!*}(u)\\rvert^{\\frac{n+1}{2}},\\qquad\n\\lambda:=\\frac{V}{\\kappa_{n}}. \\tag{14}\n\\]\n\n\\medskip\n\\emph{Tangent hyperplanes.}\nDefine\n\\[\np(u):=\\lambda^{\\frac{1}{n+1}}\n       \\frac{A^{-1}u}{\\lvert Q^{\\!*}(u)\\rvert^{1/2}}. \\tag{15}\n\\]\nThen\n\\[\nQ\\bigl(p(u)\\bigr)=-\\lambda^{\\frac{2}{n+1}},\\qquad\nu\\!\\cdot\\!p(u)=b. \\tag{16}\n\\]\nHence $\\Pi_{u,b}$ is tangent to the quadric\n\\[\nQ(x)=-\\lambda^{\\frac{2}{n+1}}=:-A^{2}. \\tag{17}\n\\]\nConversely, every hyperplane tangent to $Q(x)=-A^{2}$ satisfies\n(14), hence belongs to the given family.  Therefore the equal–volume\nhyperplanes are \\emph{exactly} the tangent hyperplanes to (17).  An\naffine unimodular change of coordinates transforms (17) into the\nhyperboloid $(\\clubsuit)$, completing the proof.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.375700",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher algebraic sophistication:  Part (i) requires knowledge of the spectrum of a linear operator, power-series inversion in Banach algebras, and interchange of limits—none of which appears in the original problem.\n\n2. More variables & higher dimension:  The geometric section moves from ℝ³ to ℝ⁴ (and then to arbitrary n), demanding the use of n-dimensional volume formulas and a careful treatment of rotational symmetry in higher codimension.\n\n3. Additional constraints:  Instead of a single linear relation s_n+ks_{n+1}, we deal with an arbitrary finite‐order linear recurrence with arbitrary coefficients; root-location conditions must be analysed and shown to be sharp.\n\n4. Deeper theoretical content:  The analytic proof employs functional calculus for bounded operators and analyticity of 1/P on the closed unit disk; the geometric proof uses group actions (O(3), the orthogonal group of Q) and affine equivalence of quadrics.\n\n5. Multiple interacting concepts:  Each part mixes classical analysis (infinite series, convergence), algebra (polynomial factorisation, spectral theory), geometry (envelopes, hyperquadrics), and higher-dimensional volume calculations.\n\nTogether these elements render the enhanced variant substantially more intricate and demanding than both the original textbook exercise and the previous kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}