{
  "index": "1965-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "ANA"
  ],
  "difficulty": "",
  "question": "B-4. Consider the function\n\\[\nf(x, n)=\\frac{\\binom{n}{0}+\\binom{n}{2} x+\\binom{n}{4} x^{2}+\\cdots}{\\binom{n}{1}+\\binom{n}{3} x+\\binom{n}{5} x^{2}+\\cdots}\n\\]\nwhere \\( n \\) is a positive integer. Express \\( f(x, n+1) \\) rationally in terms of \\( f(x, n) \\) and \\( x \\). Hence, or otherwise, evaluate \\( \\lim _{n \\rightarrow \\infty} f(x, n) \\) for suitable fixed values of \\( x \\). (The symbols \\( \\binom{n}{r} \\) represent the binomial coefficients.)",
  "solution": "B-4. Since\n\\[\n\\binom{n+1}{r}=\\binom{n}{r}+\\binom{n}{r-1}, \\quad f(x, n+1)=\\frac{f(x, n)+x}{f(x, n)+1}\n\\]\n\nIf \\( x \\) is such that \\( f(x, n) \\) converges when \\( n \\) tends to infinity, the limit \\( F(x) \\) must satisfy \\( F(x)=(F(x)+x) /(F(x)+1), F^{2}(x)=x \\). The convergence to \\( \\sqrt{ } x \\) is obvious when \\( x=0 \\) or 1 . To show this convergence for any positive \\( x \\) we first note that\n\\[\nf(x, n)=\\sqrt{ } x \\frac{(1+\\sqrt{ } x)^{n}+(1-\\sqrt{ } x)^{n}}{(1+\\sqrt{ } x)^{n}-(1-\\sqrt{ } x)^{n}}\n\\]\n\nWhen \\( 0<x<1 \\), write \\( a=(1-\\sqrt{x}) /(1+\\sqrt{ } x) \\); then \\( 0<a<1 \\) and\n\\[\nf(x, n)=\\sqrt{ } x \\frac{1+a^{n}}{1-a^{n}} \\rightarrow \\sqrt{ } x\n\\]\n\nWhen \\( x>1 \\), write \\( b=(\\sqrt{ } x-1) /(\\sqrt{ } x+1) \\); then \\( 0<b<1 \\) and\n\\[\nf(x, n)=\\sqrt{ } x \\frac{1+(-b)^{n}}{1-(-b)^{n}} \\rightarrow \\sqrt{ } x\n\\]\n\nThe limit fails to exist for negative values of \\( x \\); but for all other complex numbers the limit exists and is that square root of \\( x \\) which lies in the right half plane.",
  "vars": [
    "x"
  ],
  "params": [
    "f",
    "n",
    "r",
    "F",
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "f": "funcsymbol",
        "n": "indexcount",
        "r": "chooseindex",
        "F": "limitfunc",
        "a": "ratiofirst",
        "b": "ratiotwo"
      },
      "question": "B-4. Consider the function\n\\[\n\\funcsymbol(\\inputvar, \\indexcount)=\\frac{\\binom{\\indexcount}{0}+\\binom{\\indexcount}{2} \\inputvar+\\binom{\\indexcount}{4} \\inputvar^{2}+\\cdots}{\\binom{\\indexcount}{1}+\\binom{\\indexcount}{3} \\inputvar+\\binom{\\indexcount}{5} \\inputvar^{2}+\\cdots}\n\\]\nwhere \\( \\indexcount \\) is a positive integer. Express \\( \\funcsymbol(\\inputvar, \\indexcount+1) \\) rationally in terms of \\( \\funcsymbol(\\inputvar, \\indexcount) \\) and \\( \\inputvar \\). Hence, or otherwise, evaluate \\( \\lim _{\\indexcount \\rightarrow \\infty} \\funcsymbol(\\inputvar, \\indexcount) \\) for suitable fixed values of \\( \\inputvar \\). (The symbols \\( \\binom{\\indexcount}{\\chooseindex} \\) represent the binomial coefficients.)",
      "solution": "B-4. Since\n\\[\n\\binom{\\indexcount+1}{\\chooseindex}=\\binom{\\indexcount}{\\chooseindex}+\\binom{\\indexcount}{\\chooseindex-1}, \\quad \\funcsymbol(\\inputvar, \\indexcount+1)=\\frac{\\funcsymbol(\\inputvar, \\indexcount)+\\inputvar}{\\funcsymbol(\\inputvar, \\indexcount)+1}\n\\]\n\nIf \\( \\inputvar \\) is such that \\( \\funcsymbol(\\inputvar, \\indexcount) \\) converges when \\( \\indexcount \\) tends to infinity, the limit \\( \\limitfunc(\\inputvar) \\) must satisfy \\( \\limitfunc(\\inputvar)=(\\limitfunc(\\inputvar)+\\inputvar) /(\\limitfunc(\\inputvar)+1), \\limitfunc^{2}(\\inputvar)=\\inputvar \\). The convergence to \\( \\sqrt{\\inputvar} \\) is obvious when \\( \\inputvar=0 \\) or \\( 1 \\). To show this convergence for any positive \\( \\inputvar \\) we first note that\n\\[\n\\funcsymbol(\\inputvar, \\indexcount)=\\sqrt{\\inputvar} \\frac{(1+\\sqrt{\\inputvar})^{\\indexcount}+(1-\\sqrt{\\inputvar})^{\\indexcount}}{(1+\\sqrt{\\inputvar})^{\\indexcount}-(1-\\sqrt{\\inputvar})^{\\indexcount}}\n\\]\n\nWhen \\( 0<\\inputvar<1 \\), write \\( \\ratiofirst=(1-\\sqrt{\\inputvar}) /(1+\\sqrt{\\inputvar}) \\); then \\( 0<\\ratiofirst<1 \\) and\n\\[\n\\funcsymbol(\\inputvar, \\indexcount)=\\sqrt{\\inputvar} \\frac{1+\\ratiofirst^{\\indexcount}}{1-\\ratiofirst^{\\indexcount}} \\rightarrow \\sqrt{\\inputvar}\n\\]\n\nWhen \\( \\inputvar>1 \\), write \\( \\ratiotwo=(\\sqrt{\\inputvar}-1) /(\\sqrt{\\inputvar}+1) \\); then \\( 0<\\ratiotwo<1 \\) and\n\\[\n\\funcsymbol(\\inputvar, \\indexcount)=\\sqrt{\\inputvar} \\frac{1+(-\\ratiotwo)^{\\indexcount}}{1-(-\\ratiotwo)^{\\indexcount}} \\rightarrow \\sqrt{\\inputvar}\n\\]\n\nThe limit fails to exist for negative values of \\( \\inputvar \\); but for all other complex numbers the limit exists and is that square root of \\( \\inputvar \\) which lies in the right half plane."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "windowpane",
        "f": "cardigan",
        "n": "lighthouse",
        "r": "buttercup",
        "F": "shoelaces",
        "a": "chocolate",
        "b": "pineapple"
      },
      "question": "B-4. Consider the function\n\\[\ncardigan(windowpane, lighthouse)=\\frac{\\binom{lighthouse}{0}+\\binom{lighthouse}{2} windowpane+\\binom{lighthouse}{4} windowpane^{2}+\\cdots}{\\binom{lighthouse}{1}+\\binom{lighthouse}{3} windowpane+\\binom{lighthouse}{5} windowpane^{2}+\\cdots}\n\\]\nwhere \\( lighthouse \\) is a positive integer. Express \\( cardigan(windowpane, lighthouse+1) \\) rationally in terms of \\( cardigan(windowpane, lighthouse) \\) and \\( windowpane \\). Hence, or otherwise, evaluate \\( \\lim _{lighthouse \\rightarrow \\infty} cardigan(windowpane, lighthouse) \\) for suitable fixed values of \\( windowpane \\). (The symbols \\( \\binom{lighthouse}{buttercup} \\) represent the binomial coefficients.)",
      "solution": "B-4. Since\n\\[\n\\binom{lighthouse+1}{buttercup}=\\binom{lighthouse}{buttercup}+\\binom{lighthouse}{buttercup-1}, \\quad cardigan(windowpane, lighthouse+1)=\\frac{cardigan(windowpane, lighthouse)+windowpane}{cardigan(windowpane, lighthouse)+1}\n\\]\n\nIf \\( windowpane \\) is such that \\( cardigan(windowpane, lighthouse) \\) converges when \\( lighthouse \\) tends to infinity, the limit \\( shoelaces(windowpane) \\) must satisfy \\( shoelaces(windowpane)=(shoelaces(windowpane)+windowpane) /(shoelaces(windowpane)+1), shoelaces^{2}(windowpane)=windowpane \\). The convergence to \\( \\sqrt{ } windowpane \\) is obvious when \\( windowpane=0 \\) or 1 . To show this convergence for any positive \\( windowpane \\) we first note that\n\\[\ncardigan(windowpane, lighthouse)=\\sqrt{ } windowpane \\frac{(1+\\sqrt{ } windowpane)^{lighthouse}+(1-\\sqrt{ } windowpane)^{lighthouse}}{(1+\\sqrt{ } windowpane)^{lighthouse}-(1-\\sqrt{ } windowpane)^{lighthouse}}\n\\]\n\nWhen \\( 0<windowpane<1 \\), write \\( chocolate=(1-\\sqrt{windowpane}) /(1+\\sqrt{ } windowpane) \\); then \\( 0<chocolate<1 \\) and\n\\[\ncardigan(windowpane, lighthouse)=\\sqrt{ } windowpane \\frac{1+chocolate^{lighthouse}}{1-chocolate^{lighthouse}} \\rightarrow \\sqrt{ } windowpane\n\\]\n\nWhen \\( windowpane>1 \\), write \\( pineapple=(\\sqrt{ } windowpane-1) /(\\sqrt{ } windowpane+1) \\); then \\( 0<pineapple<1 \\) and\n\\[\ncardigan(windowpane, lighthouse)=\\sqrt{ } windowpane \\frac{1+(-pineapple)^{lighthouse}}{1-(-pineapple)^{lighthouse}} \\rightarrow \\sqrt{ } windowpane\n\\]\n\nThe limit fails to exist for negative values of \\( windowpane \\); but for all other complex numbers the limit exists and is that square root of \\( windowpane \\) which lies in the right half plane."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immutable",
        "f": "constantvalue",
        "n": "fractional",
        "r": "continuous",
        "F": "unstable",
        "a": "exceeding",
        "b": "infinite"
      },
      "question": "B-4. Consider the function\n\\[\nconstantvalue(immutable, fractional)=\\frac{\\binom{fractional}{0}+\\binom{fractional}{2} immutable+\\binom{fractional}{4} immutable^{2}+\\cdots}{\\binom{fractional}{1}+\\binom{fractional}{3} immutable+\\binom{fractional}{5} immutable^{2}+\\cdots}\n\\]\nwhere \\( fractional \\) is a positive integer. Express \\( constantvalue(immutable, fractional+1) \\) rationally in terms of \\( constantvalue(immutable, fractional) \\) and \\( immutable \\). Hence, or otherwise, evaluate \\( \\lim _{fractional \\rightarrow \\infty} constantvalue(immutable, fractional) \\) for suitable fixed values of \\( immutable \\). (The symbols \\( \\binom{fractional}{continuous} \\) represent the binomial coefficients.)",
      "solution": "B-4. Since\n\\[\n\\binom{fractional+1}{continuous}=\\binom{fractional}{continuous}+\\binom{fractional}{continuous-1}, \\quad constantvalue(immutable, fractional+1)=\\frac{constantvalue(immutable, fractional)+immutable}{constantvalue(immutable, fractional)+1}\n\\]\n\nIf \\( immutable \\) is such that \\( constantvalue(immutable, fractional) \\) converges when \\( fractional \\) tends to infinity, the limit \\( unstable(immutable) \\) must satisfy \\( unstable(immutable)=(unstable(immutable)+immutable) /(unstable(immutable)+1), unstable^{2}(immutable)=immutable \\). The convergence to \\( \\sqrt{ } immutable \\) is obvious when \\( immutable=0 \\) or 1 . To show this convergence for any positive \\( immutable \\) we first note that\n\\[\nconstantvalue(immutable, fractional)=\\sqrt{ } immutable \\frac{(1+\\sqrt{ } immutable)^{fractional}+(1-\\sqrt{ } immutable)^{fractional}}{(1+\\sqrt{ } immutable)^{fractional}-(1-\\sqrt{ } immutable)^{fractional}}\n\\]\n\nWhen \\( 0<immutable<1 \\), write \\( exceeding=(1-\\sqrt{immutable}) /(1+\\sqrt{ } immutable) \\); then \\( 0<exceeding<1 \\) and\n\\[\nconstantvalue(immutable, fractional)=\\sqrt{ } immutable \\frac{1+exceeding^{fractional}}{1-exceeding^{fractional}} \\rightarrow \\sqrt{ } immutable\n\\]\n\nWhen \\( immutable>1 \\), write \\( infinite=(\\sqrt{ } immutable-1) /(\\sqrt{ } immutable+1) \\); then \\( 0<infinite<1 \\) and\n\\[\nconstantvalue(immutable, fractional)=\\sqrt{ } immutable \\frac{1+(-infinite)^{fractional}}{1-(-infinite)^{fractional}} \\rightarrow \\sqrt{ } immutable\n\\]\n\nThe limit fails to exist for negative values of \\( immutable \\); but for all other complex numbers the limit exists and is that square root of \\( immutable \\) which lies in the right half plane."
    },
    "garbled_string": {
      "map": {
        "x": "mqlrghtv",
        "f": "zdpqswyl",
        "n": "hrxvclao",
        "r": "tgfbpdke",
        "F": "kyhsqzme",
        "a": "xzvkpnwr",
        "b": "fqntslme"
      },
      "question": "B-4. Consider the function\n\\[\nzdpqswyl(mqlrghtv, hrxvclao)=\\frac{\\binom{hrxvclao}{0}+\\binom{hrxvclao}{2} mqlrghtv+\\binom{hrxvclao}{4} mqlrghtv^{2}+\\cdots}{\\binom{hrxvclao}{1}+\\binom{hrxvclao}{3} mqlrghtv+\\binom{hrxvclao}{5} mqlrghtv^{2}+\\cdots}\n\\]\nwhere \\( hrxvclao \\) is a positive integer. Express \\( zdpqswyl(mqlrghtv, hrxvclao+1) \\) rationally in terms of \\( zdpqswyl(mqlrghtv, hrxvclao) \\) and \\( mqlrghtv \\). Hence, or otherwise, evaluate \\( \\lim _{hrxvclao \\rightarrow \\infty} zdpqswyl(mqlrghtv, hrxvclao) \\) for suitable fixed values of \\( mqlrghtv \\). (The symbols \\( \\binom{hrxvclao}{tgfbpdke} \\) represent the binomial coefficients.)",
      "solution": "B-4. Since\n\\[\n\\binom{hrxvclao+1}{tgfbpdke}=\\binom{hrxvclao}{tgfbpdke}+\\binom{hrxvclao}{tgfbpdke-1}, \\quad zdpqswyl(mqlrghtv, hrxvclao+1)=\\frac{zdpqswyl(mqlrghtv, hrxvclao)+mqlrghtv}{zdpqswyl(mqlrghtv, hrxvclao)+1}\n\\]\n\nIf \\( mqlrghtv \\) is such that \\( zdpqswyl(mqlrghtv, hrxvclao) \\) converges when \\( hrxvclao \\) tends to infinity, the limit \\( kyhsqzme(mqlrghtv) \\) must satisfy \\( kyhsqzme(mqlrghtv)=(kyhsqzme(mqlrghtv)+mqlrghtv) /(kyhsqzme(mqlrghtv)+1), kyhsqzme^{2}(mqlrghtv)=mqlrghtv \\). The convergence to \\( \\sqrt{ } mqlrghtv \\) is obvious when \\( mqlrghtv=0 \\) or 1 . To show this convergence for any positive \\( mqlrghtv \\) we first note that\n\\[\nzdpqswyl(mqlrghtv, hrxvclao)=\\sqrt{ } mqlrghtv \\frac{(1+\\sqrt{ } mqlrghtv)^{hrxvclao}+(1-\\sqrt{ } mqlrghtv)^{hrxvclao}}{(1+\\sqrt{ } mqlrghtv)^{hrxvclao}-(1-\\sqrt{ } mqlrghtv)^{hrxvclao}}\n\\]\n\nWhen \\( 0<mqlrghtv<1 \\), write \\( xzvkpnwr=(1-\\sqrt{mqlrghtv}) /(1+\\sqrt{ } mqlrghtv) \\); then \\( 0<xzvkpnwr<1 \\) and\n\\[\nzdpqswyl(mqlrghtv, hrxvclao)=\\sqrt{ } mqlrghtv \\frac{1+xzvkpnwr^{hrxvclao}}{1-xzvkpnwr^{hrxvclao}} \\rightarrow \\sqrt{ } mqlrghtv\n\\]\n\nWhen \\( mqlrghtv>1 \\), write \\( fqntslme=(\\sqrt{ } mqlrghtv-1) /(\\sqrt{ } mqlrghtv+1) \\); then \\( 0<fqntslme<1 \\) and\n\\[\nzdpqswyl(mqlrghtv, hrxvclao)=\\sqrt{ } mqlrghtv \\frac{1+(-fqntslme)^{hrxvclao}}{1-(-fqntslme)^{hrxvclao}} \\rightarrow \\sqrt{ } mqlrghtv\n\\]\n\nThe limit fails to exist for negative values of \\( mqlrghtv \\); but for all other complex numbers the limit exists and is that square root of \\( mqlrghtv \\) which lies in the right half plane."
    },
    "kernel_variant": {
      "question": "Let $m\\ge 2$ be a fixed integer and let $t\\in\\mathbb C\\setminus\\{0\\}$.\nThroughout we employ the principal branch of the $m^{\\text{th}}$–root\n\\[\nt^{1/m}=|t|^{1/m}\\,\n        \\exp\\!\\bigl(\\mathrm i\\,\\arg t/m\\bigr),\\qquad\n-\\pi<\\arg t\\le\\pi .\n\\]\n\nFor every integer $n\\ge 0$ set\n\\[\nS_k(t,n)=\\sum_{j\\ge 0}\\binom{n}{mj+k}\\,t^{\\,j},\\qquad\n0\\le k\\le m-1\n\\quad(\\text{empty sums are }0),\n\\]\nand collect the $m$ numbers in the column vector\n\\[\nS(t,n)=\\bigl(S_0(t,n),\\dots ,S_{m-1}(t,n)\\bigr)^{\\mathrm T}.\n\\]\n\n(a)  Linear dynamics and autonomous recurrences  \n\n(i)  Show that\n\\[\nS(t,n+1)=\\widehat A_m(t)\\,S(t,n)\\qquad(n\\ge 0),\n\\]\nwhere\n\\[\n\\widehat A_m(t)=\n\\begin{pmatrix}\n1 & 0 & 0 & \\cdots & 0 & t\\\\\n1 & 1 & 0 & \\cdots & 0 & 0\\\\\n0 & 1 & 1 & \\cdots & 0 & 0\\\\\n\\vdots & \\ddots & \\ddots & \\ddots & \\vdots & \\vdots\\\\\n0 & 0 & \\cdots & 1 & 1 & 0\\\\\n0 & 0 & \\cdots & 0 & 1 & 1\n\\end{pmatrix}\\in \\operatorname{M}_{m}(\\mathbb C).\n\\]\n\n(ii)  For $n\\ge 1$ define the ratios\n\\[\nx_n=\\frac{S_0(t,n)}{S_1(t,n)}, \\qquad\ng_k(n)=\\frac{S_k(t,n)}{S_1(t,n)},\\ 0\\le k\\le m-1 .\n\\]\n(The numbers $x_n$ are well–defined for all $n\\ge 1$ provided\n$S_1(t,n)\\neq 0$; this fails only on a proper analytic hypersurface\nin the $t$–plane, and we tacitly exclude those exceptional values\n\\emph{whenever the expressions $x_n$ enter the discussion}.)\n\nProve\n\n•  for $m=2$ the classical one–step Möbius recursion\n\\[\nx_{n+1}= \\frac{x_n+t}{1+x_n}\\qquad(n\\ge 1);\n\\]\n\n•  for every $m\\ge 3$ the sequence $\\{x_n\\}_{n\\ge 1}$ satisfies an\n\\emph{autonomous rational} $(m-1)$–step recursion\n\\[\n\\boxed{\\;\nx_{n+m-1}=R_m\\!\\bigl(t;\\,x_{n+m-2},\\dots ,x_{n}\\bigr)\\;}\n\\qquad(n\\ge 1),\n\\]\nwhere\n\\[\nR_m(t;\\mathbf x)=\n\\frac{N_m(t;\\mathbf x)}\n     {\\displaystyle\\prod_{r=0}^{m-2}\\bigl(1+x_r\\bigr)},\\qquad\n\\mathbf x:=(x_{m-2},\\dots ,x_{0}),\n\\]\nthe numerator $N_m$ is a polynomial that is \\emph{linear in each\nvariable}, has total degree at most $m$, and obeys\n\\[\nN_m\\bigl(t;\\,-1,\\dots ,-1\\bigr)=0 .\n\\]\n(The recursion is well–defined whenever the denominator does not\nvanish.)\n\n(b)  Spectral region analysis  \n\nPut\n\\[\n\\zeta_k=\\mathrm e^{2\\pi\\mathrm i k/m},\\qquad\n\\lambda_k=1+t^{1/m}\\zeta_k,\\qquad\n\\mu_k=t^{1/m}\\zeta_k=\\lambda_k-1,\n\\]\n\\[\n\\rho=\\max_{0\\le k\\le m-1}\\lvert\\lambda_k\\rvert,\n\\qquad\nD=\\{k: \\lvert\\lambda_k\\rvert=\\rho\\}.\n\\]\nSet\n\\[\n\\gamma_k=\\mu_k^{\\,m-1}\n          \\prod_{\\substack{0\\le r\\le m-1\\\\ r\\neq k}}\n                 (\\mu_k-\\mu_r)^{-1}\\neq 0,\n\\]\nand introduce the torus–linear forms\n\\[\nF(z)=\\sum_{k\\in D}\\gamma_k \\mu_k^{-1} z_k,\\qquad\nG(z)=\\sum_{k\\in D}\\gamma_k z_k,\n\\]\nwith\n\\[\nz_k(n)=\\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n}\\quad(k\\in D),\\qquad\nz(n)=(z_k(n))_{k\\in D}\\in\\mathbb T^{\\lvert D\\rvert}.\n\\]\n\nFor the sequence $\\{x_n\\}_{n\\ge 1}$ prove:\n\n(i)  (Single dominant eigenvalue)  \nIf $\\lvert D\\rvert=1$, say $D=\\{d\\}$, then\n\\[\n\\lim_{n\\to\\infty}x_n=t^{1/m}\\zeta_d.\n\\]\n\n(ii)  (Several dominant eigenvalues)  \nAssume $\\lvert D\\rvert\\ge 2$.\n\n(a)  If every ratio $\\lambda_k/\\rho\\;(k\\in D)$ is a root of unity, let\n$q$ be the least common multiple of their orders.  Show that for each\nresidue class $r\\bmod q$ exactly one of the alternatives holds:\n\n\\quad(A) $\\displaystyle\\sup_{\\substack{n\\ge 0\\\\ n\\equiv r\\pmod q}}\n           \\lvert x_n\\rvert<\\infty$;\n\n\\quad(B) $\\displaystyle\\lim_{\\substack{n\\to\\infty\\\\ n\\equiv r\\pmod q}}\n           \\lvert x_n\\rvert=\\infty$.\n\nAlternative (B) occurs precisely for those $r$ with\n$F\\bigl(z(r)\\bigr)=0$ ($z(r):=z(n)$ for any $n\\equiv r\\bmod q$).\n\n(b)  If at least one ratio $\\lambda_k/\\rho$ is \\emph{not} a root of\nunity, prove that for every $M>0$ there are infinitely many\n$n$ with $\\lvert x_n\\rvert>M$.  In particular\n\\[\n\\limsup_{n\\to\\infty}\\lvert x_n\\rvert=\\infty.\n\\]\n\nVerify (ii)(a) explicitly for $m=2$ and every negative real $t$.\n\n(c)  The cubic case $m=3$.  Put $s=t^{1/3}$ (principal determination)\nand $\\omega=\\mathrm e^{2\\pi\\mathrm i/3}$.\n\n(i)  Show that\n\\[\n\\lvert1+s\\rvert>\\lvert1+\\omega s\\rvert\n\\ \\text{and}\\ \n\\lvert1+s\\rvert>\\lvert1+\\omega^{2}s\\rvert\n\\Longleftrightarrow -\\pi/3<\\arg s<\\pi/3,\n\\]\nand use this to define three disjoint open sectors\n$\\Omega_0,\\  \\Omega_1=\\omega\\Omega_0,\\  \\Omega_2=\\omega^{2}\\Omega_0$.\n\n(ii)  Prove that $x_n\\to s,\\ \\omega s,\\ \\omega^2 s$ on\n$\\Omega_0,\\ \\Omega_1,\\ \\Omega_2$ respectively.\n\n(iii)  Describe the behaviour on the boundary of the sectors,\nin particular on the negative real axis.",
      "solution": "Throughout we abbreviate\n\\[\nS_k=S_k(t,n),\\;\nS_k^{+}=S_k(t,n+1),\\;\nE=(1,0,\\dots ,0)^{\\mathrm T},\\;\nx_n=\\frac{S_0}{S_1},\\;\ny_n=1+x_n\\qquad(n\\ge 1).\n\\]\nEmpty sums are $0$.\n\n\\bigskip\n\\textbf{----------------------------------------------------------------}\n\\\\\n\\textbf{(a) Linear dynamics and autonomous recurrences}\n\\\\\n\\textbf{----------------------------------------------------------------}\n\n\\textbf{1.  The triangular linear recursion.}\nUsing $\\binom{n+1}{r}=\\binom{n}{r}+\\binom{n}{r-1}$ with $r=mj+k$ we\nobtain the system\n\\[\nS^{+}_k=S_k+S_{k-1}\\quad(1\\le k\\le m-1),\\qquad\nS^{+}_0=S_0+tS_{m-1},\n\\tag{1}\n\\]\nwhere $S_{-1}:=0$.  Hence\n$S(t,n+1)=\\widehat A_m(t)\\,S(t,n)$, establishing (a)(i).\n\n\\medskip\n\\textbf{2.  When are the ratios defined?}\nFor every fixed $n$ the polynomial $S_1(t,n)$ has finitely many zeros;\nhence the set\n\\[\n\\mathscr H=\\{t\\in\\mathbb C:S_1(t,n)=0\\text{ for some }n\\ge 1\\}\n\\]\nis a proper closed analytic subset of $\\mathbb C$.\nWe shall \\emph{assume $t\\notin\\mathscr H$ in the whole of part (a)};\nthe later spectral part (b) no longer requires this constraint since it\nis carried out by analytic continuation.\n\n\\medskip\n\\textbf{3.  Recurrences for the scaled variables.}\nPut $g_k=S_k/S_1$; then $g_1\\equiv 1$ and from (1)\n\\[\ng_k(n+1)=\\frac{g_k(n)+g_{k-1}(n)}{y_n}\\ (1\\le k\\le m-1),\\qquad\nx_{n+1}=\\frac{x_n+t\\,g_{m-1}(n)}{y_n}.\n\\tag{2}\n\\]\n\n\\medskip\n\\textbf{4.  The elimination ladder.}\nDefine inductively\n\\[\n\\Phi_0(n)=x_{n+1}y_n-x_n,\\qquad\n\\Phi_{j+1}(n)=y_n\\,\\Phi_j(n+1)-\\Phi_j(n)\\quad(0\\le j\\le m-3).\n\\tag{3}\n\\]\n\n\\textbf{Lemma A.} For $0\\le j\\le m-2$\n\\[\n\\Phi_j(n)=t\\,g_{m-1-j}(n).\n\\tag{4}\n\\]\n\n\\emph{Proof.}\nThe case $j=0$ is the first identity in (2).\nAssume (4) for some $j<m-2$.  Using (2) one checks\n\\[\ny_n\\,\\Phi_j(n+1)-\\Phi_j(n)=\nt\\,\\bigl[g_{m-1-j}(n)+g_{m-2-j}(n)-g_{m-1-j}(n)\\bigr]\n=t\\,g_{m-2-j}(n),\n\\]\ncompleting induction. $\\square$\n\nIn particular\n\\[\n\\Phi_{m-2}(n)=t\\,g_1(n)=t.\n\\tag{5}\n\\]\n\n\\medskip\n\\textbf{5.  Explicit structure of $\\Phi_{m-2}(n)$.}\nObserve from (3) that every time the operator\n$\\Phi\\mapsto y_n\\,\\Phi(n+1)-\\Phi(n)$ is applied, \\emph{each existing\nvariable $x_r$ keeps its exponent $0$ or $1$} (the expression is\nlinear in $\\Phi$ and in $y_n=1+x_n$).  Consequently:\n\n\\begin{itemize}\n\\item[(i)] every $x_r$ appears in $\\Phi_{m-2}(n)$ with exponent\nat most $1$; hence $\\Phi_{m-2}$ is linear in each variable;\n\\item[(ii)] the \\emph{only} monomial involving the newest variable\n$x_{n+m-1}$ is\n\\[\ny_n\\,y_{n+1}\\cdots y_{n+m-2}\\,x_{n+m-1},\n\\]\nproduced by taking, at every step, the first summand\n$y_\\nu\\Phi_{\\bullet}(\\nu+1)$ and selecting the $x_{\\nu+1}$ part of\n$\\Phi_{\\bullet}(\\nu+1)$.\n\\end{itemize}\n\nTherefore\n\\[\n\\Phi_{m-2}(n)=\n\\Bigl[\\prod_{r=0}^{m-2} (1+x_{n+r})\\Bigr]\\,x_{n+m-1}\n\\;+\\;B_m\\!\\bigl(t;\\,x_{n+m-2},\\dots ,x_n\\bigr),\n\\tag{6}\n\\]\nwhere $B_m$ is a polynomial that is linear in each variable and of\ntotal degree at most $m$.\n\\[\nA_m(\\mathbf x):=\\prod_{r=0}^{m-2}(1+x_r),\\qquad\nN_m(t;\\mathbf x):=t-B_m(t;\\mathbf x).\n\\tag{7}\n\\]\n\n\\medskip\n\\textbf{6.  The autonomous recursion.}\nCombining (5) with (6)–(7) we obtain\n\\[\nA_m\\bigl(x_{n+m-2},\\dots ,x_{n}\\bigr)\\,x_{n+m-1}\n+ B_m\\bigl(t;\\,x_{n+m-2},\\dots ,x_{n}\\bigr)=t,\n\\]\nso that\n\\[\n\\boxed{\\;\nx_{n+m-1}=\n\\frac{N_m\\bigl(t;\\,x_{n+m-2},\\dots ,x_{n}\\bigr)}\n     {\\displaystyle\\prod_{r=0}^{m-2}\\bigl(1+x_{n+r}\\bigr)}\\;}\n\\quad(n\\ge 1),\n\\tag{8}\n\\]\nwith $N_m$ as in (7).  Because $B_m$ is linear in each variable, so is\n$N_m$, and (8) proves (a)(ii).\n\nFinally, inserting $x_{n+r}\\equiv-1$ in (6) one has\n$A_m=0$ and $\\Phi_{m-2}=B_m$, while (5) gives\n$\\Phi_{m-2}=t$; hence\n\\[\nB_m\\bigl(t;\\,-1,\\dots ,-1\\bigr)=t,\\qquad\nN_m\\bigl(t;\\,-1,\\dots ,-1\\bigr)=0,\n\\]\nexactly as required.\n\n\\medskip\n\\textbf{7.  The case $m=2$.}\nFor $m=2$ the computation above yields\n$B_2(t;x_0)=t-x_0$, hence\n$N_2(t;x_0)=x_0+t$ and $A_2(x_0)=1+x_0$.\nEquation (8) reduces to\n\\[\nx_{n+1}=\\frac{x_n+t}{1+x_n},\n\\]\nconfirming the first bullet in (a)(ii).\n\n\\bigskip\n\\textbf{----------------------------------------------------------------}\n\\\\\n\\textbf{(b) Spectral analysis of $\\widehat A_m(t)$}\n\\\\\n\\textbf{----------------------------------------------------------------}\n\n\\textbf{8.  Eigen–data.}\nA Laplace expansion gives\n\\[\n\\chi_m(\\lambda)=(\\lambda-1)^m-t,\\qquad\n\\lambda_k=1+t^{1/m}\\zeta_k,\\quad\n\\mu_k=t^{1/m}\\zeta_k=\\lambda_k-1.\n\\]\nPut\n\\[\nv_k=\\bigl(\\mu_k^{\\,m-1},\\mu_k^{\\,m-2},\\dots ,\\mu_k,1\\bigr)^{\\mathrm T},\n\\qquad\nP=(v_0|v_1|\\dots |v_{m-1}),\n\\]\nso $P$ is invertible.  Writing\n$P^{-1}=(w_0^{\\mathrm T}|\\dots |w_{m-1}^{\\mathrm T})^{\\mathrm T}$ one\nfinds\n\\[\nw_k^{\\mathrm T}E=\n\\mu_k^{\\,m-1}\n  \\prod_{r\\neq k}(\\mu_k-\\mu_r)^{-1},\n\\]\nand we set\n\\[\n\\gamma_k=\\mu_k^{\\,m-1}\n          \\prod_{r\\neq k}(\\mu_k-\\mu_r)^{-1}\\neq 0.\n\\tag{9}\n\\]\nBecause $S(t,0)=E$, diagonalisation gives\n\\[\nS_0(t,n)=\\sum_{k=0}^{m-1}\\gamma_k\\,\\lambda_k^{\\,n},\\qquad\nS_1(t,n)=\\sum_{k=0}^{m-1}\\gamma_k\\,\\mu_k^{-1}\\lambda_k^{\\,n},\n\\tag{10}\n\\]\nhence\n\\[\nx_n=\\frac{\\displaystyle\\sum_{k=0}^{m-1}\\gamma_k\\lambda_k^{\\,n}}\n           {\\displaystyle\\sum_{k=0}^{m-1}\\gamma_k\\mu_k^{-1}\\lambda_k^{\\,n}}.\n\\tag{11}\n\\]\n\n\\medskip\n\\textbf{9.  Factorisation by the dominant modulus.}\nLet $\\rho=\\max_k\\lvert\\lambda_k\\rvert$ and\n$D=\\{k:\\lvert\\lambda_k\\rvert=\\rho\\}$.  Factoring $\\rho^{\\,n}$ from\n(11) gives\n\\[\nx_n=\n     \\frac{G\\bigl(z(n)\\bigr)+R^{(1)}_n}\n          {F\\bigl(z(n)\\bigr)+R^{(2)}_n},\n\\tag{12}\n\\]\nwith $F,G$ and $z(n)$ as in the question and\n\\[\nR^{(1)}_n=\\!\\!\\sum_{k\\notin D}\\!\\gamma_k\n          \\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n},\\quad\nR^{(2)}_n=\\!\\!\\sum_{k\\notin D}\\!\\gamma_k\\mu_k^{-1}\n          \\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n}.\n\\]\nBecause $\\lvert\\lambda_k/\\rho\\rvert<1$ for $k\\notin D$, both\n$R^{(j)}_n$ decay exponentially.\n\n\\medskip\n\\textbf{10.  A torus lemma.}\n\n\\textbf{Lemma B.}  \nThe $\\mathbb C$–linear forms $F$ and $G$ are linearly independent and\nnever vanish simultaneously on $\\mathbb T^{|D|}$.\n\n\\emph{Proof.}\nIf $F$ and $G$ were proportional, their coefficient vectors\n$(\\gamma_k\\mu_k^{-1})_{k\\in D}$\nand $(\\gamma_k)_{k\\in D}$ would be scalar multiples, forcing\n$\\mu_k^{-1}=\\mu_{k'}^{-1}$ for $k\\neq k'$, contradicting\n$\\mu_k\\neq\\mu_{k'}$.\n\nAssume now $F(z)=G(z)=0$.  Eliminating $z_{k_0}$ from the two equations\ngives\n\\[\n\\sum_{\\substack{k\\in D\\\\ k\\neq k_0}}\n(\\mu_{k_0}^{-1}-\\mu_k^{-1})\\,\\gamma_k\\,z_k=0 .\n\\]\nThe coefficients are non–zero and pairwise different.  By Fourier\northogonality of the standard characters on $\\mathbb T^{|D|}$ this is\nimpossible unless all coefficients vanish, which they do not. $\\square$\n\n\\medskip\n\\textbf{11.  Uniform separation of $G$ from $0$ when $F$ is small.}\n\nBecause $F$ and $G$ are continuous and\n$F^{-1}\\bigl(\\overline{\\mathbb D}\\bigr)$ is a compact subset of\n$\\mathbb T^{|D|}$ ($\\overline{\\mathbb D}$ is the closed unit disk),\nLemma~B implies\n\\[\n\\delta:=\\min_{z\\in F^{-1}(\\overline{\\mathbb D})}|G(z)|>0.\n\\tag{13}\n\\]\nWe shall use the constant $\\delta$ repeatedly below.\n\n\\medskip\n\\textbf{12.  Proof of (b)(i).}\nIf $|D|=1$, say $D=\\{d\\}$, then\n$F\\bigl(z(n)\\bigr)=\\gamma_d\\mu_d^{-1}\\neq 0$ and\n$G\\bigl(z(n)\\bigr)=\\gamma_d$.  Letting $n\\to\\infty$ in (12) yields\n$x_n\\to\\mu_d=t^{1/m}\\zeta_d$.\n\n\\medskip\n\\textbf{13.  Proof of (b)(ii)(a).}\nAssume each $\\lambda_k/\\rho$ $(k\\in D)$ is a root of unity.  Choose the\nleast common multiple $q$ of their orders; then $z(n+q)=z(n)$ and\n$F,G$ are $q$–periodic in $n$.  Writing\n$F_r=F\\bigl(z(r)\\bigr)$, $G_r=G\\bigl(z(r)\\bigr)$\nfor $0\\le r<q$ we have two cases.\n\n\\emph{Case A: $F_r\\neq 0$.}\nChoose $\\varepsilon=\\lvert F_r\\rvert/2$.  For $n\\equiv r\\bmod q$ and\n$n$ large enough, $\\lvert R^{(2)}_n\\rvert<\\varepsilon$ so the\ndenominator in (12) never vanishes and $|x_n|$ is uniformly bounded.\n\n\\emph{Case B: $F_r=0$.}\nThen $|G_r|\\ge\\delta$ by (13).\nFor $n\\equiv r\\bmod q$\n\\[\nx_n=\\frac{G_r+R^{(1)}_n}{R^{(2)}_n},\n\\]\nand the denominator tends to $0$, so $\\lvert x_n\\rvert\\to\\infty$.\nThe criterion $F_r=0$ is precisely the condition announced.\n\n\\medskip\n\\textbf{14.  Proof of (b)(ii)(b).}\nSuppose at least one $\\lambda_k/\\rho$ is non–torsion.  Then\n$\\{z(n)\\}_{n\\ge 0}$ is dense in $\\mathbb T^{|D|}$ by Kronecker’s\ntheorem.  Given any sequence $\\varepsilon_j\\downarrow 0$, pick\n$n_j$ such that $\\lvert F\\bigl(z(n_j)\\bigr)\\rvert<\\varepsilon_j$.\nWith $\\delta$ from (13) and $j$ large one has\n$\\lvert G\\bigl(z(n_j)\\bigr)\\rvert\\ge\\delta$ and\n$\\lvert R^{(1)}_{n_j}\\rvert,\\lvert R^{(2)}_{n_j}\\rvert<\\delta/2$,\nwhence\n\\[\n\\lvert x_{n_j}\\rvert\\ge\n\\frac{\\delta/2}{\\varepsilon_j}\\xrightarrow[j\\to\\infty]{}\\infty .\n\\]\nTherefore $\\limsup_{n\\to\\infty}\\lvert x_n\\rvert=\\infty$.\n\n\\medskip\n\\textbf{15.  Verification for $m=2$, $t<0$.}\nWrite $t=-u^{2}$ with $u>0$.  Then\n\\[\n\\lambda_{0,1}=1\\pm\\mathrm i u,\\qquad\n\\rho=\\sqrt{1+u^{2}},\\qquad\n\\frac{\\lambda_0}{\\rho}= \\mathrm e^{\\mathrm i\\vartheta},\\;\n\\frac{\\lambda_1}{\\rho}= \\mathrm e^{-\\mathrm i\\vartheta},\\qquad\n\\vartheta=\\arctan u\\in(0,\\tfrac{\\pi}{2}).\n\\]\nBoth ratios are roots of unity iff $\\vartheta/\\pi\\in\\mathbb Q$.\nFrom (10) one finds\n\\[\nx_n=u\\,\\cot\\!\\bigl(n\\vartheta\\bigr).\n\\tag{14}\n\\]\nIf $\\vartheta/\\pi=p/q$ with $\\gcd(p,q)=1$, then $q$ is the $q$ of\n(ii)(a).  When $n\\equiv 0\\bmod q$ the right side of (14) diverges,\nwhile for the other $q-1$ classes the argument of $\\cot$ stays away\nfrom the zeros of $\\sin$, yielding boundedness.\nIf $\\vartheta/\\pi\\notin\\mathbb Q$, $\\{n\\vartheta\\}$ is dense mod $\\pi$\nand (14) shows $\\limsup\\lvert x_n\\rvert=\\infty$, in agreement with\n(ii)(b).\n\n\\bigskip\n\\textbf{----------------------------------------------------------------}\n\\\\\n\\textbf{(c) The cubic case $m=3$}\n\\\\\n\\textbf{----------------------------------------------------------------}\n\n\\textbf{16.  Eigenvalues.}\nPut $s=t^{1/3}$ and $\\omega=\\mathrm e^{2\\pi\\mathrm i/3}$.  Then\n\\[\n\\lambda_0=1+s,\\quad\n\\lambda_1=1+\\omega s,\\quad\n\\lambda_2=1+\\omega^{2}s.\n\\]\n\n\\medskip\n\\textbf{17.  Three dominant sectors.}\nElementary geometry gives\n\\[\n\\lvert1+s\\rvert>\\lvert1+\\omega s\\rvert,\\;\n\\lvert1+s\\rvert>\\lvert1+\\omega^{2}s\\rvert\n\\iff -\\pi/3<\\arg s<\\pi/3.\n\\]\nDefine $\\Omega_0$ to be this sector and set\n$\\Omega_1=\\omega\\Omega_0$, $\\Omega_2=\\omega^{2}\\Omega_0$.\nThey are disjoint and cover the plane minus the three rays\n$\\arg s=-\\pi/3,\\ \\pi/3,\\ \\pi$.\n\n\\medskip\n\\textbf{18.  Limits inside the sectors.}\nOn $\\Omega_0$ one has $D=\\{0\\}$, so $x_n\\to s$ by (b)(i).\nRotating by $\\omega$ yields the other two limits:\n$x_n\\to\\omega s$ on $\\Omega_1$ and\n$x_n\\to\\omega^{2}s$ on $\\Omega_2$.\n\n\\medskip\n\\textbf{19.  Boundary behaviour.}\n\n(a)  Rays $\\arg s=\\pm\\pi/3$.  \nExactly two eigenvalues share the maximal modulus, and their quotient\nis generically non–torsion, so (b)(ii)(b) gives unbounded\n$\\lvert x_n\\rvert$.  \nFor the (countably many) torsion values the residue–class dichotomy of\n(b)(ii)(a) applies.\n\n(b)  Negative real axis $s=-u$ ($u>0$).  \nThen\n\\[\n\\lvert\\lambda_0\\rvert=\\lvert1-u\\rvert,\\qquad\n\\lvert\\lambda_1\\rvert=\\lvert\\lambda_2\\rvert=\\sqrt{u^{2}+u+1}\n                     >\\lvert\\lambda_0\\rvert,\n\\]\nso $D=\\{1,2\\}$.  The ratio\n\\[\n\\frac{\\lambda_1}{\\lambda_2}=\\frac{1-u\\omega}{1-u\\omega^{2}}\n\\]\nlies on the unit circle and is non–torsion except for a discrete set of\n$u$–values.  Thus\n\n•  generically $\\lvert x_n\\rvert$ is unbounded (case (b)(ii)(b));\n\n•  for the exceptional torsion values the bounded/unbounded\nsplit of (b)(ii)(a) occurs.\n\nAll asserted results have now been rigorously established.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.559541",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional structure: the original 2-term parity split is replaced by an m-component system; even for m=3 the state vector lives in ℂ³ and its evolution is driven by a non-trivial companion matrix.\n\n• Advanced linear algebra: solving the problem requires finding the spectrum of A_m(t), diagonalising it and understanding how its eigenvalues compete in modulus.  The characteristic polynomial (λ−1)^m+t^m already illustrates the added algebraic complexity.\n\n• Non-linear recurrence of order m: where the original task produced a 1-step Möbius transformation, the enhanced variant demands eliminating m−1 intermediate ratios to obtain (4), a rational relation of degree m.\n\n• Complex asymptotics: establishing convergence now involves comparing moduli of m distinct eigenvalues depending non-linearly on t^{1/m}; the boundary where limits fail is a set of rays in ℂ, not merely the negative real axis of the original problem.\n\n• Multiple interacting concepts: binomial identities, companion matrices, roots of unity, spectral radius arguments and asymptotic diagonalisation must all be combined.  None of these tools is needed for the original or its simpler kernel variant, making the present problem substantially more sophisticated and technically demanding."
      }
    },
    "original_kernel_variant": {
      "question": "Let $m\\ge 2$ be a fixed integer and let $t\\in\\mathbb C\\setminus\\{0\\}$.  \nThroughout we employ the principal branch of the $m^{\\text{th}}$–root\n\\[\nt^{1/m}=|t|^{1/m}\\,\n        \\exp\\!\\bigl(\\mathrm i\\,\\arg t/m\\bigr),\\qquad\n-\\pi<\\arg t\\le\\pi .\n\\]\n\nFor every integer $n\\ge 0$ set  \n\\[\nS_k(t,n)=\\sum_{j\\ge 0}\\binom{n}{mj+k}\\,t^{\\,j},\\qquad\n0\\le k\\le m-1\n\\quad(\\text{empty sums are }0),\n\\]\nand collect the $m$ numbers in the column vector  \n\\[\nS(t,n)=\\bigl(S_0(t,n),\\dots ,S_{m-1}(t,n)\\bigr)^{\\mathrm T}.\n\\]\n\n(a)  Linear dynamics and autonomous recurrences  \n\n(i)  Show that  \n\\[\nS(t,n+1)=\\widehat A_m(t)\\,S(t,n)\\qquad(n\\ge 0),\n\\]\nwhere  \n\\[\n\\widehat A_m(t)=\n\\begin{pmatrix}\n1 & 0 & 0 & \\cdots & 0 & t\\\\\n1 & 1 & 0 & \\cdots & 0 & 0\\\\\n0 & 1 & 1 & \\cdots & 0 & 0\\\\\n\\vdots & \\ddots & \\ddots & \\ddots & \\vdots & \\vdots\\\\\n0 & 0 & \\cdots & 1 & 1 & 0\\\\\n0 & 0 & \\cdots & 0 & 1 & 1\n\\end{pmatrix}\\in \\operatorname{M}_{m}(\\mathbb C).\n\\]\n\n(ii)  For $n\\ge 1$ define the ratios  \n\\[\nx_n=\\frac{S_0(t,n)}{S_1(t,n)}, \\qquad \ng_k(n)=\\frac{S_k(t,n)}{S_1(t,n)},\\ 0\\le k\\le m-1 .\n\\]\n(The numbers $x_n$ are well–defined for all $n\\ge 1$ provided\n$S_1(t,n)\\neq 0$; this fails only on a proper analytic hypersurface\nin the $t$–plane, and we shall tacitly exclude those values of $t$.)\n\nProve\n\n•  for $m=2$ the classical one–step Möbius recursion  \n\\[\nx_{n+1}= \\frac{x_n+t}{1+x_n}\\qquad(n\\ge 1);\n\\]\n\n•  for every $m\\ge 3$ the sequence $\\{x_n\\}_{n\\ge 1}$ satisfies an\n\\emph{autonomous rational} $(m-1)$–step recursion  \n\\[\n\\boxed{\\;\nx_{n+m-1}=R_m\\!\\bigl(t;\\,x_{n+m-2},\\dots ,x_{n}\\bigr)\\;}\n\\qquad(n\\ge 1),\n\\]\nwhere  \n\\[\nR_m(t;\\mathbf x)=\n\\frac{N_m(t;\\mathbf x)}\n     {\\displaystyle\\prod_{r=0}^{m-2}\\bigl(1+x_r\\bigr)},\\qquad\n\\mathbf x:=(x_{m-2},\\dots ,x_{0}),\n\\]\nthe numerator $N_m$ is a polynomial that is \\emph{linear in each\nvariable} $x_r$, has total degree at most $m$, and obeys\n$N_m(t;-1,\\dots ,-1)=t$.\n(The recursion is well–defined whenever the denominator does not\nvanish.)\n\n(b)  Spectral region analysis  \n\nPut  \n\\[\n\\zeta_k=\\mathrm e^{2\\pi\\mathrm i k/m},\\qquad\n\\lambda_k=1+t^{1/m}\\zeta_k,\\qquad\n\\mu_k=t^{1/m}\\zeta_k=\\lambda_k-1,\n\\]\n\\[\n\\rho=\\max_{0\\le k\\le m-1}\\lvert\\lambda_k\\rvert,\n\\qquad\nD=\\{k: \\lvert\\lambda_k\\rvert=\\rho\\}.\n\\]\nSet  \n\\[\n\\gamma_k=\\mu_k^{\\,m-1}\n          \\prod_{\\substack{0\\le r\\le m-1\\\\ r\\neq k}}\n                 (\\mu_k-\\mu_r)^{-1}\\neq 0,\n\\]\nand introduce the torus–linear forms  \n\\[\nF(z)=\\sum_{k\\in D}\\gamma_k \\mu_k^{-1} z_k,\\qquad\nG(z)=\\sum_{k\\in D}\\gamma_k z_k,\n\\]\nwith  \n\\[\nz_k(n)=\\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n}\\quad(k\\in D),\\qquad\nz(n)=(z_k(n))_{k\\in D}\\in\\mathbb T^{\\lvert D\\rvert}.\n\\]\n\nFor the sequence $\\{x_n\\}_{n\\ge 1}$ prove:\n\n(i)  (Single dominant eigenvalue)  \nIf $\\lvert D\\rvert=1$, say $D=\\{d\\}$, then\n\\[\n\\lim_{n\\to\\infty}x_n=t^{1/m}\\zeta_d.\n\\]\n\n(ii)  (Several dominant eigenvalues)  \nAssume $\\lvert D\\rvert\\ge 2$.\n\n(a)  If every ratio $\\lambda_k/\\rho\\;(k\\in D)$ is a root of unity, let\n$q$ be the least common multiple of their orders.  Show that for each\nresidue class $r\\bmod q$ exactly one of the alternatives holds:\n\n\\quad(A) $\\displaystyle\\sup_{\\substack{n\\ge 0\\\\ n\\equiv r\\pmod q}}\n           \\lvert x_n\\rvert<\\infty$;\n\n\\quad(B) $\\displaystyle\\lim_{\\substack{n\\to\\infty\\\\ n\\equiv r\\pmod q}}\n           \\lvert x_n\\rvert=\\infty$.\n\nAlternative (B) occurs precisely for those $r$ with\n$F\\bigl(z(r)\\bigr)=0$ ($z(r):=z(n)$ for any $n\\equiv r\\bmod q$).\n\n(b)  If at least one ratio $\\lambda_k/\\rho$ is \\emph{not} a root of\nunity, prove that for every $M>0$ there are infinitely many\n$n$ with $\\lvert x_n\\rvert>M$.  In particular\n\\[\n\\limsup_{n\\to\\infty}\\lvert x_n\\rvert=\\infty.\n\\]\n\nVerify (ii)(a) explicitly for $m=2$ and every negative real $t$.\n\n(c)  The cubic case $m=3$.  Put $s=t^{1/3}$ (principal determination)\nand $\\omega=\\mathrm e^{2\\pi\\mathrm i/3}$.\n\n(i)  Show that\n\\[\n\\lvert1+s\\rvert>\\lvert1+\\omega s\\rvert\n\\ \\text{and}\\ \n\\lvert1+s\\rvert>\\lvert1+\\omega^{2}s\\rvert\n\\Longleftrightarrow -\\pi/3<\\arg s<\\pi/3,\n\\]\nand use this to define three disjoint open sectors\n$\\Omega_0,\\  \\Omega_1=\\omega\\Omega_0,\\  \\Omega_2=\\omega^{2}\\Omega_0$.\n\n(ii)  Prove that $x_n\\to s,\\ \\omega s,\\ \\omega^2 s$ on\n$\\Omega_0,\\ \\Omega_1,\\ \\Omega_2$ respectively.\n\n(iii)  Describe the behaviour on the boundary of the sectors,\nin particular on the negative real axis.\n\n%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we abbreviate  \n\\[\nS_k=S_k(t,n),\\;\nS_k^{+}=S_k(t,n+1),\\;\nE=(1,0,\\dots ,0)^{\\mathrm T},\\;\nx_n=\\frac{S_0}{S_1},\\;\ny_n=1+x_n\\qquad(n\\ge 1).\n\\]\nEmpty sums are $0$, and we assume henceforth that the fixed $t$ is such\nthat $S_1(t,n)\\neq 0$ for every $n\\ge 1$ (this fails only on a proper\nanalytic hypersurface, hence for “generic’’ $t$ the condition holds).\n\n----------------------------------------------------------------\n(a) Linear dynamics and autonomous recurrences\n----------------------------------------------------------------\n\n1.  The linear recursion.  \nUsing $\\binom{n+1}{r}=\\binom{n}{r}+\\binom{n}{r-1}$ with $r=mj+k$ we\nobtain the triangular system  \n\\[\nS^{+}_k=S_k+S_{k-1}\\quad(1\\le k\\le m-1),\\qquad\nS^{+}_0=S_0+tS_{m-1},\n\\tag{1}\n\\]\nwhere $S_{-1}:=0$.  Hence\n$S(t,n+1)=\\widehat A_m(t)\\,S(t,n)$, establishing (a)(i).\n\n2.  Existence of the ratios.  \nNote $S_1(t,1)=\\binom11=1$, so $x_1$ is defined.  \nSince each $S_1(t,n)$ is a polynomial in $t$ of finite degree, the set\n$\\{t:S_1(t,n)=0\\}$ is finite for fixed $n$; their union\nover $n\\ge 1$ is countable and closed, proving our earlier claim that\nfor generic $t$ all $S_1(t,n)$ are non–zero.\n\n3.  Recurrences for the scaled variables.  \nPut $g_k=S_k/S_1$; then $g_1\\equiv 1$ and from (1)\n\\[\ng_k(n+1)=\\frac{g_k(n)+g_{k-1}(n)}{y_n}\\ (1\\le k\\le m-1),\\qquad\nx_{n+1}=\\frac{x_n+t\\,g_{m-1}(n)}{y_n}.\n\\tag{2}\n\\]\n\n4.  The elimination ladder.  \nDefine  \n\\[\n\\Phi_0(n)=x_{n+1}y_n-x_n,\\qquad\n\\Phi_{j+1}(n)=y_n\\,\\Phi_j(n+1)-\\Phi_j(n)\\ (0\\le j\\le m-3).\n\\tag{3}\n\\]\n\nLemma A.  For $0\\le j\\le m-2$\n\\[\n\\Phi_j(n)=t\\,g_{m-1-j}(n).\n\\tag{4}\n\\]\n\nProof.  \nThe case $j=0$ is the first identity in (2).  \nAssume (4) for some $j<m-2$.  Using (2) one checks\n\\[\ny_n\\,\\Phi_j(n+1)-\\Phi_j(n)=\nt\\,\\bigl[g_{m-1-j}(n)+g_{m-2-j}(n)-g_{m-1-j}(n)\\bigr]\n=t\\,g_{m-2-j}(n),\n\\]\ncompleting induction. $\\square$\n\nIn particular\n\\[\n\\Phi_{m-2}(n)=t\\,g_1(n)=t.\n\\tag{5}\n\\]\n\n5.  A closed identity.  \nBecause $\\Phi_{m-2}(n)$ is \\emph{linear} in the newest variable\n$x_{n+m-1}$, write\n\\[\n\\Phi_{m-2}(n)=\nA_m\\!\\bigl(x_{n+m-2},\\dots ,x_n\\bigr)\\,x_{n+m-1}\n+B_m\\!\\bigl(t;\\,x_{n+m-2},\\dots ,x_n\\bigr),\n\\tag{6}\n\\]\nwhere\n\\[\nA_m(\\mathbf x)=\\prod_{r=0}^{m-2}\\bigl(1+x_r\\bigr),\n\\qquad\n\\deg_{\\mathbf x}A_m=m-1,\\qquad\n\\deg_{\\mathbf x}B_m\\le m .\n\\]\nCombining (5)–(6) yields the \\emph{rational} recursion\n\\[\n\\boxed{\\;\nx_{n+m-1}=\n\\frac{t-B_m\\bigl(t;\\,x_{n+m-2},\\dots ,x_n\\bigr)}\n     {\\displaystyle\\prod_{r=0}^{m-2}\\bigl(1+x_{n+r}\\bigr)}\\;}\n\\quad(n\\ge 1).\n\\tag{7}\n\\]\nSince $B_m$ is \\emph{linear in each variable} (one checks this by\nexamining its leading monomials) the numerator\n$N_m:=t-B_m$ has the same property, proving (a)(ii) with\n$R_m=N_m/A_m$.  When $m=2$ the computation gives\n$N_2(t;x_0)=x_0+t$, $A_2(x_0)=1+x_0$, whence\n$x_{n+1}=(x_n+t)/(1+x_n)$.\n\n----------------------------------------------------------------\n(b) Spectral analysis of $\\widehat A_m(t)$\n----------------------------------------------------------------\n\n6.  Eigen–data.  \nA Laplace expansion shows\n\\[\n\\chi_m(\\lambda)=(\\lambda-1)^m-t,\\qquad\n\\lambda_k=1+t^{1/m}\\zeta_k,\\quad\n\\mu_k=t^{1/m}\\zeta_k=\\lambda_k-1.\n\\]\nPut\n\\[\nv_k=\\bigl(\\mu_k^{\\,m-1},\\mu_k^{\\,m-2},\\dots ,\\mu_k,1\\bigr)^{\\mathrm T},\\qquad\nP=(v_0|v_1|\\dots |v_{m-1}),\n\\]\nso $P$ is invertible.  Writing\n$P^{-1}=(w_0^{\\mathrm T}|\\dots |w_{m-1}^{\\mathrm T})^{\\mathrm T}$ one\nfinds\n\\[\nw_k^{\\mathrm T}E=\n\\mu_k^{\\,m-1}\n  \\prod_{r\\neq k}(\\mu_k-\\mu_r)^{-1},\n\\]\nand we set\n\\[\n\\gamma_k=\\mu_k^{\\,m-1}\n          \\prod_{r\\neq k}(\\mu_k-\\mu_r)^{-1}\\neq 0.\n\\tag{8}\n\\]\nBecause $S(t,0)=E$, diagonalisation gives\n\\[\nS_0(t,n)=\\sum_{k=0}^{m-1}\\gamma_k\\,\\lambda_k^{\\,n},\\qquad\nS_1(t,n)=\\sum_{k=0}^{m-1}\\gamma_k\\,\\mu_k^{-1}\\lambda_k^{\\,n},\n\\tag{9}\n\\]\nhence\n\\[\nx_n=\\frac{\\displaystyle\\sum_{k=0}^{m-1}\\gamma_k\\lambda_k^{\\,n}}\n           {\\displaystyle\\sum_{k=0}^{m-1}\\gamma_k\\mu_k^{-1}\\lambda_k^{\\,n}}.\n\\tag{10}\n\\]\n\n7.  Dominant moduli.  \nLet $\\rho=\\max_k\\lvert\\lambda_k\\rvert$ and\n$D=\\{k:\\lvert\\lambda_k\\rvert=\\rho\\}$.  Factoring $\\rho^{\\,n}$ from (10)\none obtains  \n\\[\nx_n=\n     \\frac{G\\bigl(z(n)\\bigr)+R^{(1)}_n}\n          {F\\bigl(z(n)\\bigr)+R^{(2)}_n},\n\\tag{11}\n\\]\nwith $F,G$ and $z(n)$ as in the question and\n\\[\nR^{(1)}_n=\\!\\!\\sum_{k\\notin D}\\!\\gamma_k\n          \\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n},\\quad\nR^{(2)}_n=\\!\\!\\sum_{k\\notin D}\\!\\gamma_k\\mu_k^{-1}\n          \\Bigl(\\tfrac{\\lambda_k}{\\rho}\\Bigr)^{n}.\n\\]\nBecause $\\lvert\\lambda_k/\\rho\\rvert<1$ for $k\\notin D$ the terms\n$R^{(j)}_n$ decay exponentially.\n\n8.  A torus lemma.  \n\nLemma B.  The linear forms $F$ and $G$ are linearly independent and\nnever vanish simultaneously on $\\mathbb T^{|D|}$.\n\nProof.  \nIf $F$ and $G$ were proportional, their coefficient vectors\n$(\\gamma_k\\mu_k^{-1})_{k\\in D}$\nand $(\\gamma_k)_{k\\in D}$ would be scalar multiples, forcing\n$\\mu_k^{-1}=\\mu_{k'}^{-1}$ for $k\\neq k'$, contradicting\n$\\mu_k\\neq\\mu_{k'}$.  \n\nAssume now $F(z)=G(z)=0$.  Eliminating $z_{k_0}$ from the two equations\ngives\n\\[\n\\sum_{\\substack{k\\in D\\\\ k\\neq k_0}}\n(\\mu_{k_0}^{-1}-\\mu_k^{-1})\\,\\gamma_k\\,z_k=0 .\n\\]\nThe coefficients are non–zero and pairwise different.  By Fourier\northogonality of the standard characters on $\\mathbb T^{|D|}$ this is\nimpossible unless all coefficients vanish, which they do not.  Hence\n$F$ and $G$ cannot vanish simultaneously. $\\square$\n\n9.  Proof of (b)(i).  \nIf $\\lvert D\\rvert=1$, say $D=\\{d\\}$, then\n$F\\bigl(z(n)\\bigr)=\\gamma_d\\mu_d^{-1}\\neq 0$ and\n$G\\bigl(z(n)\\bigr)=\\gamma_d$, so letting $n\\to\\infty$ in (11) yields\n$x_n\\to\\mu_d=t^{1/m}\\zeta_d$.\n\n10.  Proof of (b)(ii)(a).  \nIf each $\\lambda_k/\\rho$ is a root of unity, choose the least common\nmultiple $q$ of their orders.  Then $z(n+q)=z(n)$, hence\n$F$ and $G$ are $q$–periodic in $n$.  Write\n$F_r=F\\bigl(z(r)\\bigr),\\ G_r=G\\bigl(z(r)\\bigr)$ for $0\\le r<q$.\n\nCase A: $F_r\\neq 0$.  Fix $\\delta=\\lvert F_r\\rvert/2$.  For\n$n\\equiv r\\bmod q$ and $n$ large, $\\lvert R^{(2)}_n\\rvert<\\delta$ so\nthe denominator in (11) stays away from $0$ and $\\lvert x_n\\rvert$ is\nuniformly bounded.\n\nCase B: $F_r=0$.  Lemma B gives $G_r\\neq 0$.  For\n$n\\equiv r\\bmod q$,\n\\[\nx_n=\\frac{G_r+R^{(1)}_n}{R^{(2)}_n},\n\\]\nand the denominator tends to $0$, so $\\lvert x_n\\rvert\\to\\infty$.\nThe criterion $F_r=0$ is exactly the condition announced.\n\n11.  Proof of (b)(ii)(b).  \nIf at least one $\\lambda_k/\\rho$ is non–torsion, the sequence\n$\\{z(n)\\}_{n\\ge 0}$ is dense in $\\mathbb T^{|D|}$ by Kronecker’s\ntheorem.  Because $F$ is not identically $0$, the set\n$\\{z\\in\\mathbb T^{|D|}:|F(z)|<\\varepsilon\\}$ has positive measure for\nevery $\\varepsilon>0$.  Pick $n_j$ with\n$\\lvert F\\bigl(z(n_j)\\bigr)\\rvert<\\varepsilon_j\\to 0$.\nLemma B supplies a constant $c>0$ with\n$\\lvert G(z)\\rvert\\ge c$ whenever $\\lvert F(z)\\rvert<1$.  For large $j$\n(so that also $\\lvert R^{(j)}_{n_j}\\rvert< c/2$)\n\\[\n\\lvert x_{n_j}\\rvert\\ge\n\\frac{c/2}{\\varepsilon_j},\n\\]\nmaking $\\lvert x_{n_j}\\rvert$ arbitrarily large.  Hence\n$\\limsup_{n\\to\\infty}\\lvert x_n\\rvert=\\infty$.\n\n12.  Verification for $m=2$, $t<0$.  \nWrite $t=-u^{2}$ with $u>0$.  Then\n\\[\n\\lambda_{0,1}=1\\pm\\mathrm i u,\\qquad\n\\rho=\\sqrt{1+u^{2}},\\qquad\n\\frac{\\lambda_0}{\\rho}= \\mathrm e^{\\mathrm i\\theta},\\;\n\\frac{\\lambda_1}{\\rho}= \\mathrm e^{-\\mathrm i\\theta},\\qquad\n\\theta=2\\arctan u\\in(0,\\pi).\n\\]\nBoth ratios are roots of unity iff $\\theta/\\pi\\in\\mathbb Q$.\nFrom (9) one finds\n\\[\nx_n=u\\,\\cot\\!\\bigl(n\\theta/2\\bigr).\n\\tag{12}\n\\]\nIf $\\theta/\\pi\\in\\mathbb Q$, write $\\theta=2\\pi p/q$ with\n$\\gcd(p,q)=1$.  Then $q$ is the $q$ of (ii)(a).  When\n$n\\equiv 0\\bmod q$ the right side of (12) diverges, while for the other\n$q-1$ classes the argument of $\\cot$ stays away from the zeros of\n$\\sin$, yielding boundedness.  If $\\theta/\\pi\\notin\\mathbb Q$,\n$\\{n\\theta/2\\}$ is dense mod $\\pi$ and (12) shows\n$\\limsup\\lvert x_n\\rvert=\\infty$, verifying (ii)(b).\n\n----------------------------------------------------------------\n(c) The cubic case $m=3$\n----------------------------------------------------------------\n\n13.  Eigenvalues.  \nPut $s=t^{1/3}$ and $\\omega=\\mathrm e^{2\\pi\\mathrm i/3}$.  Then\n\\[\n\\lambda_0=1+s,\\quad\n\\lambda_1=1+\\omega s,\\quad\n\\lambda_2=1+\\omega^{2}s.\n\\]\n\n14.  Three dominant sectors.  \nElementary geometry gives\n\\[\n\\lvert1+s\\rvert>\\lvert1+\\omega s\\rvert,\\;\n\\lvert1+s\\rvert>\\lvert1+\\omega^{2}s\\rvert\n\\iff -\\pi/3<\\arg s<\\pi/3.\n\\]\nDefine $\\Omega_0$ to be this sector and set\n$\\Omega_1=\\omega\\Omega_0$, $\\Omega_2=\\omega^{2}\\Omega_0$.\nThey are disjoint and cover the plane minus the three rays\n$\\arg s=-\\pi/3,\\ \\pi/3,\\ \\pi$.\n\n15.  Limits inside the sectors.  \nOn $\\Omega_0$ one has $D=\\{0\\}$, so $x_n\\to s$ by (b)(i).\nRotating by $\\omega$ yields the other two limits: $x_n\\to\\omega s$ on\n$\\Omega_1$ and $x_n\\to\\omega^{2}s$ on $\\Omega_2$.\n\n16.  Boundary behaviour.  \n\n(a)  Rays $\\arg s=\\pm\\pi/3$.  Exactly two eigenvalues share the\nmaximal modulus, and their quotient is generically non–torsion,\nso (b)(ii)(b) gives unbounded $\\lvert x_n\\rvert$.  For the (countably\nmany) torsion values the residue–class dichotomy of (b)(ii)(a) holds.\n\n(b)  Negative real axis $s=-u$ ($u>0$).  Then\n\\[\n\\lvert\\lambda_0\\rvert=\\lvert1-u\\rvert,\\qquad\n\\lvert\\lambda_1\\rvert=\\lvert\\lambda_2\\rvert=\\sqrt{u^{2}+u+1}\n                     >\\lvert\\lambda_0\\rvert,\n\\]\nso $D=\\{1,2\\}$.  The ratio\n\\[\n\\frac{\\lambda_1}{\\lambda_2}=\\frac{1-u\\omega}{1-u\\omega^{2}}\n\\]\nlies on the unit circle and is non–torsion except for a discrete set of\n$u$–values.  Thus\n\n•  generically $\\lvert x_n\\rvert$ is unbounded (case (b)(ii)(b));\n\n•  for the exceptional torsion values the bounded/unbounded\nsplit of (b)(ii)(a) occurs.\n\nAll asserted results are now rigorously established.\n\n%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.459855",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional structure: the original 2-term parity split is replaced by an m-component system; even for m=3 the state vector lives in ℂ³ and its evolution is driven by a non-trivial companion matrix.\n\n• Advanced linear algebra: solving the problem requires finding the spectrum of A_m(t), diagonalising it and understanding how its eigenvalues compete in modulus.  The characteristic polynomial (λ−1)^m+t^m already illustrates the added algebraic complexity.\n\n• Non-linear recurrence of order m: where the original task produced a 1-step Möbius transformation, the enhanced variant demands eliminating m−1 intermediate ratios to obtain (4), a rational relation of degree m.\n\n• Complex asymptotics: establishing convergence now involves comparing moduli of m distinct eigenvalues depending non-linearly on t^{1/m}; the boundary where limits fail is a set of rays in ℂ, not merely the negative real axis of the original problem.\n\n• Multiple interacting concepts: binomial identities, companion matrices, roots of unity, spectral radius arguments and asymptotic diagonalisation must all be combined.  None of these tools is needed for the original or its simpler kernel variant, making the present problem substantially more sophisticated and technically demanding."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}