{
  "index": "1975-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { B-5. Let } f_{0}(x)=e^{x} \\text { and } f_{n+1}(x)=x f_{n}^{\\prime}(x) \\text { for } n=0,1,2, \\ldots \\text {. Show that }\\\\\n\\sum_{n=0}^{\\infty} \\frac{f_{n}(1)}{n!}=e^{e}\n\\end{array}",
  "solution": "B-5.\nSince \\( f_{0}(x)=\\sum_{k=0}^{\\infty} x^{k} / k! \\), one easily shows by mathematical induction that \\( f_{n}(x)=\\sum_{k=0}^{\\infty}\\left(k^{n} x^{k} / k!\\right) \\). Then, since all terms are positive, one has\n\\[\n\\sum_{n=0}^{\\infty} \\frac{f_{n}(1)}{n!}=\\sum_{n=0}^{\\infty} \\sum_{k=0}^{\\infty} \\frac{k^{n}}{k!n!}=\\sum_{k=0}^{\\infty} \\frac{1}{k!} \\sum_{n=0}^{\\infty} \\frac{k^{n}}{n!}=\\sum_{k=0}^{\\infty} \\frac{e^{k}}{k!}=e^{e} .\n\\]",
  "vars": [
    "x",
    "n",
    "k",
    "f_0",
    "f_n+1",
    "f_n"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "n": "indexvar",
        "k": "summindex",
        "f_0": "funczero",
        "f_n+1": "funcnext",
        "f_n": "funcgen"
      },
      "question": "\\begin{array}{l}\n\\text { B-5. Let } funczero(variablex)=e^{variablex} \\text { and } funcnext(variablex)=variablex funcgen^{\\prime}(variablex) \\text { for } indexvar=0,1,2, \\ldots \\text {. Show that }\\\\\n\\sum_{indexvar=0}^{\\infty} \\frac{funcgen(1)}{indexvar!}=e^{e}\n\\end{array}",
      "solution": "B-5.\nSince \\( funczero(variablex)=\\sum_{summindex=0}^{\\infty} variablex^{summindex} / summindex! \\), one easily shows by mathematical induction that \\( funcgen(variablex)=\\sum_{summindex=0}^{\\infty}\\left(summindex^{indexvar} variablex^{summindex} / summindex!\\right) \\). Then, since all terms are positive, one has\n\\[\\sum_{indexvar=0}^{\\infty} \\frac{funcgen(1)}{indexvar!}=\\sum_{indexvar=0}^{\\infty} \\sum_{summindex=0}^{\\infty} \\frac{summindex^{indexvar}}{summindex!indexvar!}=\\sum_{summindex=0}^{\\infty} \\frac{1}{summindex!} \\sum_{indexvar=0}^{\\infty} \\frac{summindex^{indexvar}}{indexvar!}=\\sum_{summindex=0}^{\\infty} \\frac{e^{summindex}}{summindex!}=e^{e} .\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "daydream",
        "n": "dandelion",
        "k": "wildberry",
        "f_0": "watermelon",
        "f_n+1": "caterpillar",
        "f_n": "buttercup"
      },
      "question": "\\begin{array}{l}\n\\text { B-5. Let } watermelon(daydream)=e^{daydream} \\text { and } caterpillar(daydream)=daydream buttercup^{\\prime}(daydream) \\text { for } dandelion=0,1,2, \\ldots \\text {. Show that }\\\\\n\\sum_{dandelion=0}^{\\infty} \\frac{buttercup(1)}{dandelion!}=e^{e}\n\\end{array}",
      "solution": "B-5.\nSince \\( watermelon(daydream)=\\sum_{wildberry=0}^{\\infty} daydream^{wildberry} / wildberry! \\), one easily shows by mathematical induction that \\( buttercup(daydream)=\\sum_{wildberry=0}^{\\infty}\\left(wildberry^{dandelion} daydream^{wildberry} / wildberry!\\right) \\). Then, since all terms are positive, one has\n\\[\n\\sum_{dandelion=0}^{\\infty} \\frac{buttercup(1)}{dandelion!}=\\sum_{dandelion=0}^{\\infty} \\sum_{wildberry=0}^{\\infty} \\frac{wildberry^{dandelion}}{wildberry!dandelion!}=\\sum_{wildberry=0}^{\\infty} \\frac{1}{wildberry!} \\sum_{dandelion=0}^{\\infty} \\frac{wildberry^{dandelion}}{dandelion!}=\\sum_{wildberry=0}^{\\infty} \\frac{e^{wildberry}}{wildberry!}=e^{e} .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "n": "limitlesscounter",
        "k": "continuousindex",
        "f_0": "terminalfunction",
        "f_n+1": "stagnantfunction",
        "f_n": "staticfunction"
      },
      "question": "\\begin{array}{l}\n\\text { B-5. Let } terminalfunction(fixedvalue)=e^{fixedvalue} \\text { and } stagnantfunction(fixedvalue)=fixedvalue staticfunction^{\\prime}(fixedvalue) \\text { for } limitlesscounter=0,1,2, \\ldots \\text {. Show that }\\\\\n\\sum_{limitlesscounter=0}^{\\infty} \\frac{staticfunction(1)}{limitlesscounter!}=e^{e}\n\\end{array}",
      "solution": "B-5.\nSince \\( terminalfunction(fixedvalue)=\\sum_{continuousindex=0}^{\\infty} fixedvalue^{continuousindex} / continuousindex! \\), one easily shows by mathematical induction that \\( staticfunction(fixedvalue)=\\sum_{continuousindex=0}^{\\infty}\\left(continuousindex^{limitlesscounter} fixedvalue^{continuousindex} / continuousindex!\\right) \\). Then, since all terms are positive, one has\n\\[\n\\sum_{limitlesscounter=0}^{\\infty} \\frac{staticfunction(1)}{limitlesscounter!}=\\sum_{limitlesscounter=0}^{\\infty} \\sum_{continuousindex=0}^{\\infty} \\frac{continuousindex^{limitlesscounter}}{continuousindex!limitlesscounter!}=\\sum_{continuousindex=0}^{\\infty} \\frac{1}{continuousindex!} \\sum_{limitlesscounter=0}^{\\infty} \\frac{continuousindex^{limitlesscounter}}{limitlesscounter!}=\\sum_{continuousindex=0}^{\\infty} \\frac{e^{continuousindex}}{continuousindex!}=e^{e} .\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qwertyui",
        "n": "asdfghjk",
        "k": "zxcvbnml",
        "f_0": "plmoknji",
        "f_n+1": "bvghytre",
        "f_n": "ujmikolp"
      },
      "question": "\\begin{array}{l}\n\\text { B-5. Let } plmoknji(qwertyui)=e^{qwertyui} \\text { and } bvghytre(qwertyui)=qwertyui ujmikolp^{\\prime}(qwertyui) \\text { for } asdfghjk=0,1,2, \\ldots \\text {. Show that }\\\\\n\\sum_{asdfghjk=0}^{\\infty} \\frac{ujmikolp(1)}{asdfghjk!}=e^{e}\n\\end{array}",
      "solution": "B-5.\nSince \\( plmoknji(qwertyui)=\\sum_{zxcvbnml=0}^{\\infty} qwertyui^{zxcvbnml} / zxcvbnml! \\), one easily shows by mathematical induction that \\( ujmikolp(qwertyui)=\\sum_{zxcvbnml=0}^{\\infty}\\left(zxcvbnml^{asdfghjk} qwertyui^{zxcvbnml} / zxcvbnml!\\right) \\). Then, since all terms are positive, one has\n\\[\n\\sum_{asdfghjk=0}^{\\infty} \\frac{ujmikolp(1)}{asdfghjk!}=\\sum_{asdfghjk=0}^{\\infty} \\sum_{zxcvbnml=0}^{\\infty} \\frac{zxcvbnml^{asdfghjk}}{zxcvbnml!asdfghjk!}=\\sum_{zxcvbnml=0}^{\\infty} \\frac{1}{zxcvbnml!} \\sum_{asdfghjk=0}^{\\infty} \\frac{zxcvbnml^{asdfghjk}}{asdfghjk!}=\\sum_{zxcvbnml=0}^{\\infty} \\frac{e^{zxcvbnml}}{zxcvbnml!}=e^{e} .\n\\]"
    },
    "kernel_variant": {
      "question": "Let \n  H₀(x,y,z)=exp ( z·e^{\\,y·e^{\\,x}} ).\n\nFor n = 0,1,2,… define recursively  \n  H_{\\,n+1}(x,y,z)=\\bigl(x\\,∂/∂x + y\\,∂/∂y + z\\,∂/∂z\\bigr)\\,H_{\\,n}(x,y,z).\n\nEvaluate the infinite series  \n  S = Σ_{n=0}^{∞} H_{\\,n}(1,2,3) / n!,\n\nand prove that the series is absolutely convergent.\n\n(Here the differential operator acts component-wise on the entire function H₀.)\n\n───────────────────────────────────────────────",
      "solution": "Step 1.  Exponential generating function.  \nSet  \n  F(x,y,z,t) = Σ_{n=0}^{∞} H_{\\,n}(x,y,z)\\,t^{n}/n!.         (0)\n\nDifferentiating term wise yields  \n  ∂F/∂t = (x ∂F/∂x + y ∂F/∂y + z ∂F/∂z), F(x,y,z,0)=H₀(x,y,z).     (1)\n\nEquation (1) is a linear first-order PDE in the four variables (x,y,z,t).\n\nStep 2.  Method of characteristics.  \nAlong a curve s ↦ (x(s),y(s),z(s),t(s)) we impose\n\n  t′(s)=1, x′(s)=−x(s), y′(s)=−y(s), z′(s)=−z(s).     (2)\n\nWith this choice, using (1) we have  \n\n  dF/ds = F_t t′ − (xF_x + yF_y + zF_z)=0,\n\nso F stays constant on such curves.  Solving (2) with initial data\n(x₀,y₀,z₀,0) at s=0 gives  \n\n  x(s)=x₀e^{−s}, y(s)=y₀e^{−s}, z(s)=z₀e^{−s}, t(s)=s.    (3)\n\nEliminating s by putting s=t leads to (x₀,y₀,z₀)=(x e^{t},y e^{t},z e^{t}), and therefore\n\n  F(x,y,z,t)=H₀\\!\\bigl(xe^{t},\\,ye^{t},\\,ze^{t}\\bigr).     (4)\n\nStep 3.  Insert H₀.  \nSince H₀(u,v,w)=exp ( w·e^{\\,v·e^{\\,u}} ),\n\n  F(x,y,z,t)=exp [\\, ze^{t}·exp( ye^{t}·exp( xe^{t} ) )\\,].     (5)\n\nStep 4.  Evaluate at (1,2,3,t=1).  \nFrom (0) we have S = F(1,2,3,1); using (5):\n\n  S = exp [\\,3e·exp( 2e·exp(e) )\\,].                      (6)\n\nStep 5.  Absolute convergence.  \nThe map t ↦ F(1,2,3,t) is an entire function.  Fix an arbitrary radius R>1 (for instance R=2) and denote  \n\n  M_R := max_{|t|=R} |F(1,2,3,t)| < ∞                   (7)\n\n(Cauchy’s estimate applies because the circle |t|=R is compact and F is continuous).  \nFor n≥0, Cauchy’s integral formula for the n-th derivative at 0 gives  \n\n  |H_{\\,n}(1,2,3)| = |F^{(n)}_t(1,2,3,0)|\n            ≤ M_R·n!/R^{\\,n}.                       (8)\n\nHence  \n\n  |H_{\\,n}(1,2,3)|/n! ≤ M_R / R^{\\,n}.                  (9)\n\nBecause R>1, the majorant series Σ_{n≥0} M_R/R^{\\,n} is geometric and convergent; thus Σ_{n≥0} H_{\\,n}(1,2,3)/n! converges absolutely.  The sum is precisely the value computed in (6).\n\nTherefore  \n\n  S = exp [\\,3e·exp( 2e·exp(e) )\\,]. □\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.622276",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensions: the problem now involves three independent variables instead of one.  \n2. Sophisticated structure: the initial function contains iterated (nested) exponentials, and the differential operator is the weighted radial operator in ℝ³.  \n3. Deeper theory: solving the problem requires setting up and solving a first–order linear PDE via the method of characteristics rather than elementary series manipulation.  \n4. Multiple interacting concepts: the solution blends power-series generating functions, partial differential equations, characteristic curves, and properties of exponential functions.  \n5. More steps and insight: one must (i) recognise that the series is the exponential of an operator, (ii) translate the recursion into a PDE, (iii) solve the PDE on ℝ³, and (iv) trace characteristics back to evaluate the generating function—far more elaborate than the single-variable induction/evaluation needed for the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let \n  H₀(x,y,z)=exp ( z·e^{\\,y·e^{\\,x}} ).\n\nFor n = 0,1,2,… define recursively  \n  H_{\\,n+1}(x,y,z)=\\bigl(x\\,∂/∂x + y\\,∂/∂y + z\\,∂/∂z\\bigr)\\,H_{\\,n}(x,y,z).\n\nEvaluate the infinite series  \n  S = Σ_{n=0}^{∞} H_{\\,n}(1,2,3) / n!,\n\nand prove that the series is absolutely convergent.\n\n(Here the differential operator acts component-wise on the entire function H₀.)\n\n───────────────────────────────────────────────",
      "solution": "Step 1.  Exponential generating function.  \nSet  \n  F(x,y,z,t) = Σ_{n=0}^{∞} H_{\\,n}(x,y,z)\\,t^{n}/n!.         (0)\n\nDifferentiating term wise yields  \n  ∂F/∂t = (x ∂F/∂x + y ∂F/∂y + z ∂F/∂z), F(x,y,z,0)=H₀(x,y,z).     (1)\n\nEquation (1) is a linear first-order PDE in the four variables (x,y,z,t).\n\nStep 2.  Method of characteristics.  \nAlong a curve s ↦ (x(s),y(s),z(s),t(s)) we impose\n\n  t′(s)=1, x′(s)=−x(s), y′(s)=−y(s), z′(s)=−z(s).     (2)\n\nWith this choice, using (1) we have  \n\n  dF/ds = F_t t′ − (xF_x + yF_y + zF_z)=0,\n\nso F stays constant on such curves.  Solving (2) with initial data\n(x₀,y₀,z₀,0) at s=0 gives  \n\n  x(s)=x₀e^{−s}, y(s)=y₀e^{−s}, z(s)=z₀e^{−s}, t(s)=s.    (3)\n\nEliminating s by putting s=t leads to (x₀,y₀,z₀)=(x e^{t},y e^{t},z e^{t}), and therefore\n\n  F(x,y,z,t)=H₀\\!\\bigl(xe^{t},\\,ye^{t},\\,ze^{t}\\bigr).     (4)\n\nStep 3.  Insert H₀.  \nSince H₀(u,v,w)=exp ( w·e^{\\,v·e^{\\,u}} ),\n\n  F(x,y,z,t)=exp [\\, ze^{t}·exp( ye^{t}·exp( xe^{t} ) )\\,].     (5)\n\nStep 4.  Evaluate at (1,2,3,t=1).  \nFrom (0) we have S = F(1,2,3,1); using (5):\n\n  S = exp [\\,3e·exp( 2e·exp(e) )\\,].                      (6)\n\nStep 5.  Absolute convergence.  \nThe map t ↦ F(1,2,3,t) is an entire function.  Fix an arbitrary radius R>1 (for instance R=2) and denote  \n\n  M_R := max_{|t|=R} |F(1,2,3,t)| < ∞                   (7)\n\n(Cauchy’s estimate applies because the circle |t|=R is compact and F is continuous).  \nFor n≥0, Cauchy’s integral formula for the n-th derivative at 0 gives  \n\n  |H_{\\,n}(1,2,3)| = |F^{(n)}_t(1,2,3,0)|\n            ≤ M_R·n!/R^{\\,n}.                       (8)\n\nHence  \n\n  |H_{\\,n}(1,2,3)|/n! ≤ M_R / R^{\\,n}.                  (9)\n\nBecause R>1, the majorant series Σ_{n≥0} M_R/R^{\\,n} is geometric and convergent; thus Σ_{n≥0} H_{\\,n}(1,2,3)/n! converges absolutely.  The sum is precisely the value computed in (6).\n\nTherefore  \n\n  S = exp [\\,3e·exp( 2e·exp(e) )\\,]. □\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.497278",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensions: the problem now involves three independent variables instead of one.  \n2. Sophisticated structure: the initial function contains iterated (nested) exponentials, and the differential operator is the weighted radial operator in ℝ³.  \n3. Deeper theory: solving the problem requires setting up and solving a first–order linear PDE via the method of characteristics rather than elementary series manipulation.  \n4. Multiple interacting concepts: the solution blends power-series generating functions, partial differential equations, characteristic curves, and properties of exponential functions.  \n5. More steps and insight: one must (i) recognise that the series is the exponential of an operator, (ii) translate the recursion into a PDE, (iii) solve the PDE on ℝ³, and (iv) trace characteristics back to evaluate the generating function—far more elaborate than the single-variable induction/evaluation needed for the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}