{
  "index": "1989-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $f$ be a function on $[0,\\infty)$, differentiable and satisfying\n\\[\nf'(x)=-3f(x)+6f(2x)\n\\]\nfor $x>0$. Assume that $|f(x)|\\le e^{-\\sqrt{x}}$ for $x\\ge 0$ (so that\n$f(x)$ tends rapidly to $0$ as $x$ increases).\nFor $n$ a non-negative integer, define\n\\[\n\\mu_n=\\int_0^\\infty x^n f(x)\\,dx\n\\]\n(sometimes called the $n$th moment of $f$).\n\\begin{enumerate}\n\\item[a)] Express $\\mu_n$ in terms of $\\mu_0$.\n\\item[b)] Prove that the sequence $\\{\\mu_n \\frac{3^n}{n!}\\}$ always converges,\nand that the limit is $0$ only if $\\mu_0=0$.\n\\end{enumerate}",
  "solution": "Solution. a. As \\( x \\rightarrow \\infty, f(x) \\) tends to 0 faster than any negative power of \\( x \\), so the integral defining \\( \\mu_{n} \\) converges. Multiply the functional equation by \\( x^{n} \\) for some \\( n \\geq 1 \\) and integrate from 0 to some finite \\( B>0 \\) :\n\\[\n\\int_{0}^{B} x^{n} f^{\\prime}(x) d x=-3 \\int_{0}^{B} x^{n} f(x) d x+6 \\int_{0}^{B} x^{n} f(2 x) d x\n\\]\n\nUsing integration by parts [ \\( \\mathrm{Spv}, \\mathrm{Ch} .18 \\), Theorem 1] on the left, and the substitution \\( u=2 x \\) on the last term converts this into\n\\[\n\\left.x^{n} f(x)\\right|_{0} ^{B}-n \\int_{0}^{B} f(x) x^{n-1} d x=-3 \\int_{0}^{B} x^{n} f(x) d x+\\frac{6}{2^{n+1}} \\int_{0}^{B / 2} u^{n} f(u) d u\n\\]\n\nTaking limits as \\( B \\rightarrow \\infty \\) yields\n\\[\n-n \\mu_{n-1}=-3 \\mu_{n}+\\frac{6}{2^{n+1}} \\mu_{n}\n\\]\nso\n\\[\n\\mu_{n}=\\frac{n}{3}\\left(1-2^{-n}\\right)^{-1} \\mu_{n-1}\n\\]\nfor all \\( n \\geq 1 \\). By induction, we obtain\n\\[\n\\mu_{n}=\\frac{n!}{3^{n}}\\left(\\prod_{m=1}^{n}\\left(1-2^{-m}\\right)\\right)^{-1} \\mu_{0}\n\\]\nb. Since \\( \\sum_{m=1}^{\\infty} 2^{-m} \\) converges, \\( \\prod_{m=1}^{\\infty}\\left(1-2^{-m}\\right) \\) converges to some nonzero limit \\( L \\), and\n\\[\n\\mu_{n} \\frac{3^{n}}{n!}=\\left(\\prod_{m=1}^{n}\\left(1-2^{-m}\\right)\\right)^{-1} \\mu_{0} \\rightarrow L^{-1} \\mu_{0}\n\\]\nas \\( n \\rightarrow \\infty \\). This limit \\( L^{-1} \\mu_{0} \\) is finite, and equals zero if and only if \\( \\mu_{0}=0 \\).\nRemark. We show that nonzero functions \\( f(x) \\) satisfying the conditions of the problem do exist. Given that \\( f(x) \\) tends to zero rapidly as \\( x \\rightarrow \\infty \\), one expects \\( f(2 x) \\) to be negligible compared to \\( f(x) \\) for large \\( x \\), and hence one can guess that \\( f(x) \\) can be approximated by a solution to \\( y^{\\prime}=-3 y \\), for example \\( e^{-3 x} \\). But then the error in the differential equation \\( f^{\\prime}(x)=-3 f(x)+6 f(2 x) \\) is of order \\( e^{-6 x} \\) as \\( x \\rightarrow \\infty \\), leading one to guess that \\( e^{-3 x}+a_{1} e^{-6 x} \\) will be a better approximation, and so on, finally leading one to guess\n\\[\nf(x)=e^{-3 x}+a_{1} e^{-6 x}+a_{2} e^{-12 x}+a_{3} e^{-24 x}+\\cdots,\n\\]\nwhere the coefficients \\( a_{i} \\) are to be solved for. Let \\( a_{0}=1 \\). If we differentiate (1) term by term, temporarily disregarding convergence issues, then for \\( k \\geq 1 \\), the coefficient of \\( e^{-3 \\cdot 2^{k} x} \\) in the expression \\( f^{\\prime}(x)+3 f(x)-6 f(2 x) \\) is \\( -3 \\cdot 2^{k} a_{k}+3 a_{k}-6 a_{k-1} \\). If this is to be zero, then \\( a_{k}=\\frac{-2}{2^{k}-1} a_{k-1} \\).\n\nAll this so far has been motivation. Now we define the sequence \\( a_{0}, a_{1}, \\ldots \\) by \\( a_{0}=1 \\) and \\( a_{k}=\\frac{-2}{2^{k}-1} a_{k-1} \\) for \\( k \\geq 1 \\), so\n\\[\na_{k}=\\frac{(-2)^{n}}{(2-1)\\left(2^{2}-1\\right) \\cdots\\left(2^{k}-1\\right)}\n\\]\nand define \\( f(x) \\) by (1). The series \\( \\sum_{k=0}^{\\infty}\\left|a_{k}\\right| \\) converges by the Ratio Test [Spv, Ch. 22, Theorem 3], so (1) converges to a continuous function on \\( [0, \\infty) \\). Moreover, \\( \\left|a_{k} e^{-3 \\cdot 2^{k} x}\\right| \\leq\\left|a_{k}\\right| \\) for all complex \\( x \\) with \\( \\operatorname{Re}(x)>0 \\), so by the Weierstrass \\( M \\)-test and Weierstrass's Theorem [Ah, pp. 37, 176], (1) converges to a holomorphic function in this region. In particular, \\( f(x) \\) is differentiable on \\( (0, \\infty) \\), and can be differentiated term by term, so \\( f^{\\prime}(x)=-3 f(x)+6 f(2 x) \\) holds.\n\nFinally, we will show that if \\( \\epsilon>0 \\) is sufficiently small, then \\( \\epsilon f(x) \\), which still satisfies the differential equation, now also satisfies \\( |\\epsilon f(x)| \\leq e^{-\\sqrt{x}} \\). Since\n\\[\nf(x)=O\\left(e^{-3 x}+e^{-6 x}+e^{-12 x}+\\cdots\\right)=O\\left(e^{-3 x}\\right)=o\\left(e^{-\\sqrt{x}}\\right)\n\\]\nas \\( x \\rightarrow \\infty \\), we have \\( |f(x)| \\leq e^{-\\sqrt{x}} \\) for all \\( x \\) greater than some \\( x_{0} \\). Let \\( M= \\) \\( \\sup _{x \\in\\left[0, x_{0}\\right]} \\frac{|f(x)|}{e^{-\\sqrt{x}}} \\). If \\( \\epsilon \\in(0,1) \\) is chosen so that \\( \\epsilon M \\leq 1 \\), then \\( |\\epsilon f(x)| \\leq e^{-\\sqrt{x}} \\) both for \\( x \\in\\left[0, x_{0}\\right] \\) and for \\( x \\in\\left(x_{0}, \\infty\\right) \\), as desired.",
  "vars": [
    "x",
    "n",
    "u",
    "k",
    "m",
    "i"
  ],
  "params": [
    "f",
    "B",
    "x_0",
    "\\\\mu_0",
    "\\\\mu_n",
    "\\\\mu_n-1",
    "a_k",
    "a_0",
    "a_1",
    "a_k-1",
    "L",
    "M",
    "\\\\epsilon"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablevalue",
        "n": "indexcount",
        "u": "substitution",
        "k": "summandindex",
        "m": "productindex",
        "i": "seriesindex",
        "f": "givenfunction",
        "B": "upperlimit",
        "x_0": "initialpoint",
        "\\mu_0": "momentzero",
        "\\mu_n": "momentindex",
        "\\mu_{n-1}": "momentprev",
        "a_k": "coefficientk",
        "a_0": "coefficientzero",
        "a_1": "coefficientone",
        "a_{k-1}": "coefficientprev",
        "L": "productlimit",
        "M": "supremumvalue",
        "\\epsilon": "smallpositive"
      },
      "question": "Let $givenfunction$ be a function on $[0,\\infty)$, differentiable and satisfying\n\\[\ngivenfunction'(variablevalue)=-3givenfunction(variablevalue)+6givenfunction(2variablevalue)\n\\]\nfor $variablevalue>0$. Assume that $|givenfunction(variablevalue)|\\le e^{-\\sqrt{variablevalue}}$ for $variablevalue\\ge 0$ (so that\ngivenfunction(variablevalue) tends rapidly to $0$ as variablevalue increases).\nFor $indexcount$ a non-negative integer, define\n\\[\nmomentindex=\\int_0^\\infty variablevalue^{indexcount} givenfunction(variablevalue)\\,dvariablevalue\n\\]\n(sometimes called the $indexcount$th moment of givenfunction).\n\\begin{enumerate}\n\\item[a)] Express $momentindex$ in terms of $momentzero$.\n\\item[b)] Prove that the sequence $\\{momentindex \\frac{3^{indexcount}}{indexcount!}\\}$ always converges,\nand that the limit is $0$ only if $momentzero=0$.\n\\end{enumerate}",
      "solution": "Solution. a. As \\( variablevalue \\rightarrow \\infty, givenfunction(variablevalue) \\) tends to 0 faster than any negative power of \\( variablevalue \\), so the integral defining \\( momentindex \\) converges. Multiply the functional equation by \\( variablevalue^{indexcount} \\) for some \\( indexcount \\geq 1 \\) and integrate from 0 to some finite \\( upperlimit>0 \\) :\n\\[\n\\int_{0}^{upperlimit} variablevalue^{indexcount} givenfunction^{\\prime}(variablevalue)\\,d variablevalue=-3 \\int_{0}^{upperlimit} variablevalue^{indexcount} givenfunction(variablevalue)\\,d variablevalue+6 \\int_{0}^{upperlimit} variablevalue^{indexcount} givenfunction(2 variablevalue)\\,d variablevalue\n\\]\n\nUsing integration by parts on the left, and the substitution \\( substitution=2 variablevalue \\) on the last term converts this into\n\\[\n\\left. variablevalue^{indexcount} givenfunction(variablevalue)\\right|_{0}^{upperlimit}-indexcount \\int_{0}^{upperlimit} givenfunction(variablevalue) variablevalue^{indexcount-1}\\,d variablevalue=-3 \\int_{0}^{upperlimit} variablevalue^{indexcount} givenfunction(variablevalue)\\,d variablevalue+\\frac{6}{2^{indexcount+1}} \\int_{0}^{upperlimit / 2} substitution^{indexcount} givenfunction(substitution)\\,d substitution\n\\]\n\nTaking limits as \\( upperlimit \\rightarrow \\infty \\) yields\n\\[\n-indexcount \\, momentprev=-3 \\, momentindex+\\frac{6}{2^{indexcount+1}} \\, momentindex\n\\]\nso\n\\[\nmomentindex=\\frac{indexcount}{3}\\left(1-2^{-indexcount}\\right)^{-1} \\, momentprev\n\\]\nfor all \\( indexcount \\geq 1 \\). By induction, we obtain\n\\[\nmomentindex=\\frac{indexcount!}{3^{indexcount}}\\left(\\prod_{productindex=1}^{indexcount}\\left(1-2^{-productindex}\\right)\\right)^{-1} \\, momentzero\n\\]\n\nb. Since \\( \\sum_{productindex=1}^{\\infty} 2^{-productindex} \\) converges, \\( \\prod_{productindex=1}^{\\infty}\\left(1-2^{-productindex}\\right) \\) converges to some nonzero limit \\( productlimit \\), and\n\\[\nmomentindex \\frac{3^{indexcount}}{indexcount!}=\\left(\\prod_{productindex=1}^{indexcount}\\left(1-2^{-productindex}\\right)\\right)^{-1} \\, momentzero \\rightarrow productlimit^{-1} momentzero\n\\]\nas \\( indexcount \\rightarrow \\infty \\). This limit \\( productlimit^{-1} momentzero \\) is finite, and equals zero if and only if \\( momentzero=0 \\).\n\nRemark. We show that nonzero functions \\( givenfunction(variablevalue) \\) satisfying the conditions of the problem do exist. Given that \\( givenfunction(variablevalue) \\) tends to zero rapidly as \\( variablevalue \\rightarrow \\infty \\), one expects \\( givenfunction(2 variablevalue) \\) to be negligible compared to \\( givenfunction(variablevalue) \\) for large \\( variablevalue \\), and hence one can guess that \\( givenfunction(variablevalue) \\) can be approximated by a solution to \\( y^{\\prime}=-3 y \\), for example \\( e^{-3 variablevalue} \\). But then the error in the differential equation \\( givenfunction^{\\prime}(variablevalue)=-3 givenfunction(variablevalue)+6 givenfunction(2 variablevalue) \\) is of order \\( e^{-6 variablevalue} \\) as \\( variablevalue \\rightarrow \\infty \\), leading one to guess that \\( e^{-3 variablevalue}+coefficientone e^{-6 variablevalue}+a_{2} e^{-12 variablevalue}+a_{3} e^{-24 variablevalue}+\\cdots \\) will be a better approximation, and so on, finally leading one to guess\n\\[\ngivenfunction(variablevalue)=e^{-3 variablevalue}+coefficientone e^{-6 variablevalue}+a_{2} e^{-12 variablevalue}+a_{3} e^{-24 variablevalue}+\\cdots,\n\\]\nwhere the coefficients \\( a_{seriesindex} \\) are to be solved for. Let \\( coefficientzero=1 \\). If we differentiate (1) term by term, temporarily disregarding convergence issues, then for \\( summandindex \\geq 1 \\), the coefficient of \\( e^{-3 \\cdot 2^{summandindex} variablevalue} \\) in the expression \\( givenfunction^{\\prime}(variablevalue)+3 givenfunction(variablevalue)-6 givenfunction(2 variablevalue) \\) is \\( -3 \\cdot 2^{summandindex} coefficientk+3 coefficientk-6 coefficientprev \\). If this is to be zero, then \\( coefficientk=\\frac{-2}{2^{summandindex}-1} coefficientprev \\).\n\nAll this so far has been motivation. Now we define the sequence \\( coefficientzero, coefficientone, \\ldots \\) by \\( coefficientzero=1 \\) and \\( coefficientk=\\frac{-2}{2^{summandindex}-1} coefficientprev \\) for \\( summandindex \\geq 1 \\), so\n\\[\ncoefficientk=\\frac{(-2)^{indexcount}}{(2-1)\\left(2^{2}-1\\right) \\cdots\\left(2^{summandindex}-1\\right)}\n\\]\nand define \\( givenfunction(variablevalue) \\) by (1). The series \\( \\sum_{summandindex=0}^{\\infty}\\left|coefficientk\\right| \\) converges by the Ratio Test, so (1) converges to a continuous function on \\( [0, \\infty) \\). Moreover, \\( \\left|coefficientk e^{-3 \\cdot 2^{summandindex} variablevalue}\\right| \\leq\\left|coefficientk\\right| \\) for all complex \\( variablevalue \\) with \\( \\operatorname{Re}(variablevalue)>0 \\), so by the Weierstrass supremumvalue-test and Weierstrass's Theorem, (1) converges to a holomorphic function in this region. In particular, \\( givenfunction(variablevalue) \\) is differentiable on \\( (0, \\infty) \\), and can be differentiated term by term, so \\( givenfunction^{\\prime}(variablevalue)=-3 givenfunction(variablevalue)+6 givenfunction(2 variablevalue) \\) holds.\n\nFinally, we will show that if \\( smallpositive>0 \\) is sufficiently small, then \\( smallpositive\\,givenfunction(variablevalue) \\), which still satisfies the differential equation, now also satisfies \\( |smallpositive\\,givenfunction(variablevalue)| \\leq e^{-\\sqrt{variablevalue}} \\). Since\n\\[\ngivenfunction(variablevalue)=O\\left(e^{-3 variablevalue}+e^{-6 variablevalue}+e^{-12 variablevalue}+\\cdots\\right)=O\\left(e^{-3 variablevalue}\\right)=o\\left(e^{-\\sqrt{variablevalue}}\\right)\n\\]\nas \\( variablevalue \\rightarrow \\infty \\), we have \\( |givenfunction(variablevalue)| \\leq e^{-\\sqrt{variablevalue}} \\) for all \\( variablevalue \\) greater than some \\( initialpoint \\). Let \\( supremumvalue=\\sup_{variablevalue\\in[0,initialpoint]} \\frac{|givenfunction(variablevalue)|}{e^{-\\sqrt{variablevalue}}} \\). If \\( smallpositive\\in(0,1) \\) is chosen so that \\( smallpositive\\,supremumvalue \\leq 1 \\), then \\( |smallpositive\\,givenfunction(variablevalue)| \\leq e^{-\\sqrt{variablevalue}} \\) both for \\( variablevalue\\in[0,initialpoint] \\) and for \\( variablevalue\\in(initialpoint,\\infty) \\), as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "n": "raspberry",
        "u": "watermelon",
        "k": "blackberry",
        "m": "strawberry",
        "i": "butterscotch",
        "f": "caterpillar",
        "B": "sunflower",
        "x_0": "rainforest",
        "\\mu_0": "peanutbutter",
        "\\mu_n": "cheesecake",
        "\\mu_n-1": "macadamia",
        "a_k": "blueberry",
        "a_0": "dragonfruit",
        "a_1": "passionfruit",
        "a_k-1": "coconutty",
        "L": "waterlily",
        "M": "tangerine",
        "\\epsilon": "planktonic"
      },
      "question": "Let $caterpillar$ be a function on $[0,\\infty)$, differentiable and satisfying\n\\[\ncaterpillar'(pineapple)=-3caterpillar(pineapple)+6caterpillar(2pineapple)\n\\]\nfor $pineapple>0$. Assume that $|caterpillar(pineapple)|\\le e^{-\\sqrt{pineapple}}$ for $pineapple\\ge 0$ (so that\n$caterpillar(pineapple)$ tends rapidly to $0$ as $pineapple$ increases).\nFor $raspberry$ a non-negative integer, define\n\\[\ncheesecake=\\int_0^\\infty pineapple^{raspberry}\\,caterpillar(pineapple)\\,d pineapple\n\\]\n(sometimes called the $raspberry$th moment of $caterpillar$).\n\\begin{enumerate}\n\\item[a)] Express $cheesecake$ in terms of $peanutbutter$.\n\\item[b)] Prove that the sequence $\\{cheesecake \\frac{3^{raspberry}}{raspberry!}\\}$ always converges,\nand that the limit is $0$ only if $peanutbutter=0$.\n\\end{enumerate}",
      "solution": "Solution. a. As $pineapple \\rightarrow \\infty$, $caterpillar(pineapple)$ tends to $0$ faster than any negative power of $pineapple$, so the integral defining $cheesecake$ converges. Multiply the functional equation by $pineapple^{raspberry}$ for some $raspberry \\ge 1$ and integrate from $0$ to some finite $sunflower>0$:\n\\[\n\\int_{0}^{sunflower} pineapple^{raspberry} caterpillar'(pineapple)\\,d pineapple=-3\\int_{0}^{sunflower} pineapple^{raspberry} caterpillar(pineapple)\\,d pineapple+6\\int_{0}^{sunflower} pineapple^{raspberry} caterpillar(2pineapple)\\,d pineapple\n\\]\nUsing integration by parts and the substitution $watermelon=2pineapple$ on the last term converts this into\n\\[\n\\left. pineapple^{raspberry}caterpillar(pineapple)\\right|_{0}^{sunflower}-raspberry \\int_{0}^{sunflower} caterpillar(pineapple)\\,pineapple^{raspberry-1}\\,d pineapple=-3\\int_{0}^{sunflower} pineapple^{raspberry}caterpillar(pineapple)\\,d pineapple+\\frac{6}{2^{raspberry+1}}\\int_{0}^{sunflower/2} watermelon^{raspberry}caterpillar(watermelon)\\,d watermelon\n\\]\nTaking limits as $sunflower \\rightarrow \\infty$ yields\n\\[\n-raspberry\\,macadamia=-3\\,cheesecake+\\frac{6}{2^{raspberry+1}}\\,cheesecake,\n\\]\nso\n\\[\ncheesecake=\\frac{raspberry}{3}\\left(1-2^{-raspberry}\\right)^{-1} macadamia\n\\]\nfor all $raspberry \\ge 1$. By induction,\n\\[\ncheesecake=\\frac{raspberry!}{3^{raspberry}}\\left(\\prod_{strawberry=1}^{raspberry}\\left(1-2^{-strawberry}\\right)\\right)^{-1} peanutbutter.\n\\]\n\nb. Since $\\sum_{strawberry=1}^{\\infty}2^{-strawberry}$ converges, $\\prod_{strawberry=1}^{\\infty}(1-2^{-strawberry})$ converges to some non-zero limit $waterlily$, and\n\\[\ncheesecake\\,\\frac{3^{raspberry}}{raspberry!}=\\left(\\prod_{strawberry=1}^{raspberry}(1-2^{-strawberry})\\right)^{-1}peanutbutter\\;\\longrightarrow\\;waterlily^{-1}peanutbutter\n\\]\nas $raspberry \\rightarrow \\infty$. This limit is finite, and is $0$ iff $peanutbutter=0$.\n\nRemark. Non-zero functions satisfying the hypotheses do exist. Because $caterpillar(pineapple)$ decays rapidly, one expects $caterpillar(2pineapple)$ to be negligible compared to $caterpillar(pineapple)$ for large $pineapple$, so $caterpillar(pineapple)$ can be approximated by a solution of $y'= -3y$, e.g.\n$e^{-3pineapple}$. Correcting successively suggests an expansion\n\\[\ncaterpillar(pineapple)=e^{-3pineapple}+passionfruit\\,e^{-6pineapple}+a_{2}e^{-12pineapple}+a_{3}e^{-24pineapple}+\\cdots,\n\\]\nwhere the coefficients $a_{butterscotch}$ are chosen so that $caterpillar'(pineapple)+3caterpillar(pineapple)-6caterpillar(2pineapple)=0$. Writing $dragonfruit=1$ and defining recursively\n$blueberry=\\dfrac{-2}{2^{blackberry}-1}coconutty$ for $blackberry\\ge1$ gives\n\\[\nblueberry=\\frac{(-2)^{raspberry}}{(2-1)(2^{2}-1)\\cdots(2^{blackberry}-1)}.\n\\]\nThe series $\\sum_{blackberry=0}^{\\infty}|blueberry|$ converges by the ratio test, so the expansion converges absolutely and uniformly on compact subsets of $[0,\\infty)$, yielding a continuous, even holomorphic, solution. Finally, choose $planktonic>0$ so small that $|planktonic\\,caterpillar(pineapple)|\\le e^{-\\sqrt{pineapple}}$ for all $pineapple\\ge0$. Set\n\\[\ntangerine=\\sup_{pineapple\\in[0,rainforest]}\\frac{|caterpillar(pineapple)|}{e^{-\\sqrt{pineapple}}},\n\\]\nand pick $planktonic<1/tangerine$; then the required bound holds on $[0,\\infty)$, completing the construction."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "n": "fractional",
        "u": "unchanged",
        "k": "infinite",
        "m": "continuous",
        "i": "realvalue",
        "f": "constantfn",
        "B": "lowerside",
        "x_0": "endpoint",
        "\\mu_0": "stillness",
        "\\mu_n": "momentless",
        "\\mu_n-1": "momentvoid",
        "a_k": "fixedcoef",
        "a_0": "basecoef",
        "a_1": "firstcoef",
        "a_k-1": "prevcoef",
        "L": "limitless",
        "M": "minimumval",
        "\\epsilon": "bigdelta"
      },
      "question": "Let $constantfn$ be a function on $[0,\\infty)$, differentiable and satisfying\n\\[\nconstantfn'(constantval)=-3constantfn(constantval)+6constantfn(2constantval)\n\\]\nfor $constantval>0$. Assume that $|constantfn(constantval)|\\le e^{-\\sqrt{constantval}}$ for $constantval\\ge 0$ (so that\n$constantfn(constantval)$ tends rapidly to $0$ as $constantval$ increases).\nFor $fractional$ a non-negative integer, define\n\\[\nmomentless=\\int_0^\\infty constantval^{fractional} constantfn(constantval)\\,dconstantval\n\\]\n(sometimes called the $fractional$th moment of $constantfn$).\n\\begin{enumerate}\n\\item[a)] Express $momentless$ in terms of $stillness$.\n\\item[b)] Prove that the sequence $\\{momentless \\frac{3^{fractional}}{fractional!}\\}$ always converges,\nand that the limit is $0$ only if $stillness=0$.\n\\end{enumerate}",
      "solution": "Solution. a. As \\( constantval \\rightarrow \\infty, constantfn(constantval) \\) tends to 0 faster than any negative power of \\( constantval \\), so the integral defining \\( momentless \\) converges. Multiply the functional equation by \\( constantval^{fractional} \\) for some \\( fractional \\geq 1 \\) and integrate from 0 to some finite \\( lowerside>0 \\) :\n\\[\n\\int_{0}^{lowerside} constantval^{fractional} constantfn^{\\prime}(constantval) d constantval=-3 \\int_{0}^{lowerside} constantval^{fractional} constantfn(constantval) d constantval+6 \\int_{0}^{lowerside} constantval^{fractional} constantfn(2 constantval) d constantval\n\\]\n\nUsing integration by parts [ \\( \\mathrm{Spv}, \\mathrm{Ch} .18 \\), Theorem 1] on the left, and the substitution \\( unchanged=2 constantval \\) on the last term converts this into\n\\[\n\\left.constantval^{fractional} constantfn(constantval)\\right|_{0} ^{lowerside}-fractional \\int_{0}^{lowerside} constantfn(constantval) constantval^{fractional-1} d constantval=-3 \\int_{0}^{lowerside} constantval^{fractional} constantfn(constantval) d constantval+\\frac{6}{2^{fractional+1}} \\int_{0}^{lowerside / 2} unchanged^{fractional} constantfn(unchanged) d unchanged\n\\]\n\nTaking limits as \\( lowerside \\rightarrow \\infty \\) yields\n\\[\n-fractional momentvoid=-3 momentless+\\frac{6}{2^{fractional+1}} momentless\n\\]\nso\n\\[\nmomentless=\\frac{fractional}{3}\\left(1-2^{-fractional}\\right)^{-1} momentvoid\n\\]\nfor all \\( fractional \\geq 1 \\). By induction, we obtain\n\\[\nmomentless=\\frac{fractional!}{3^{fractional}}\\left(\\prod_{continuous=1}^{fractional}\\left(1-2^{-continuous}\\right)\\right)^{-1} stillness\n\\]\n\nb. Since \\( \\sum_{continuous=1}^{\\infty} 2^{-continuous} \\) converges, \\( \\prod_{continuous=1}^{\\infty}\\left(1-2^{-continuous}\\right) \\) converges to some nonzero limit \\( limitless \\), and\n\\[\nmomentless \\frac{3^{fractional}}{fractional!}=\\left(\\prod_{continuous=1}^{fractional}\\left(1-2^{-continuous}\\right)\\right)^{-1} stillness \\rightarrow limitless^{-1} stillness\n\\]\nas \\( fractional \\rightarrow \\infty \\). This limit \\( limitless^{-1} stillness \\) is finite, and equals zero if and only if \\( stillness=0 \\).\n\nRemark. We show that nonzero functions \\( constantfn(constantval) \\) satisfying the conditions of the problem do exist. Given that \\( constantfn(constantval) \\) tends to zero rapidly as \\( constantval \\rightarrow \\infty \\), one expects \\( constantfn(2 constantval) \\) to be negligible compared to \\( constantfn(constantval) \\) for large \\( constantval \\), and hence one can guess that \\( constantfn(constantval) \\) can be approximated by a solution to \\( y^{\\prime}=-3 y \\), for example \\( e^{-3 constantval} \\). But then the error in the differential equation \\( constantfn^{\\prime}(constantval)=-3 constantfn(constantval)+6 constantfn(2 constantval) \\) is of order \\( e^{-6 constantval} \\) as \\( constantval \\rightarrow \\infty \\), leading one to guess that \\( e^{-3 constantval}+firstcoef e^{-6 constantval} \\) will be a better approximation, and so on, finally leading one to guess\n\\[\nconstantfn(constantval)=e^{-3 constantval}+firstcoef e^{-6 constantval}+fixedcoef e^{-12 constantval}+\\cdots,\n\\]\nwhere the coefficients \\( fixedcoef \\) are to be solved for. Let \\( basecoef=1 \\). If we differentiate (1) term by term, temporarily disregarding convergence issues, then for \\( infinite \\geq 1 \\), the coefficient of \\( e^{-3 \\cdot 2^{infinite} constantval} \\) in the expression \\( constantfn^{\\prime}(constantval)+3 constantfn(constantval)-6 constantfn(2 constantval) \\) is \\( -3 \\cdot 2^{infinite} fixedcoef+3 fixedcoef-6 prevcoef \\). If this is to be zero, then \\( fixedcoef=\\frac{-2}{2^{infinite}-1} prevcoef \\).\n\nAll this so far has been motivation. Now we define the sequence \\( basecoef, firstcoef, \\ldots \\) by \\( basecoef=1 \\) and \\( fixedcoef=\\frac{-2}{2^{infinite}-1} prevcoef \\) for \\( infinite \\geq 1 \\), so\n\\[\nfixedcoef=\\frac{(-2)^{fractional}}{(2-1)\\left(2^{2}-1\\right) \\cdots\\left(2^{fractional}-1\\right)}\n\\]\nand define \\( constantfn(constantval) \\) by (1). The series \\( \\sum_{infinite=0}^{\\infty}\\left|fixedcoef\\right| \\) converges by the Ratio Test [Spv, Ch. 22, Theorem 3], so (1) converges to a continuous function on \\( [0, \\infty) \\). Moreover, \\( \\left|fixedcoef e^{-3 \\cdot 2^{infinite} constantval}\\right| \\leq\\left|fixedcoef\\right| \\) for all complex \\( constantval \\) with \\( \\operatorname{Re}(constantval)>0 \\), so by the Weierstrass \\( M \\)-test and Weierstrass's Theorem [Ah, pp. 37, 176], (1) converges to a holomorphic function in this region. In particular, \\( constantfn(constantval) \\) is differentiable on \\( (0, \\infty) \\), and can be differentiated term by term, so \\( constantfn^{\\prime}(constantval)=-3 constantfn(constantval)+6 constantfn(2 constantval) \\) holds.\n\nFinally, we will show that if \\( bigdelta>0 \\) is sufficiently small, then \\( bigdelta constantfn(constantval) \\), which still satisfies the differential equation, now also satisfies \\( |bigdelta constantfn(constantval)| \\leq e^{-\\sqrt{constantval}} \\). Since\n\\[\nconstantfn(constantval)=O\\left(e^{-3 constantval}+e^{-6 constantval}+e^{-12 constantval}+\\cdots\\right)=O\\left(e^{-3 constantval}\\right)=o\\left(e^{-\\sqrt{constantval}}\\right)\n\\]\nas \\( constantval \\rightarrow \\infty \\), we have \\( |constantfn(constantval)| \\leq e^{-\\sqrt{constantval}} \\) for all \\( constantval \\) greater than some \\( endpoint \\). Let \\( minimumval= \\sup _{constantval \\in\\left[0, endpoint\\right]} \\frac{|constantfn(constantval)|}{e^{-\\sqrt{constantval}}} \\). If \\( bigdelta \\in(0,1) \\) is chosen so that \\( bigdelta\\,minimumval \\leq 1 \\), then \\( |bigdelta constantfn(constantval)| \\leq e^{-\\sqrt{constantval}} \\) both for \\( constantval \\in\\left[0, endpoint\\right] \\) and for \\( constantval \\in\\left(endpoint, \\infty\\right) \\), as desired."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "u": "rmbvcxye",
        "k": "ldfqpnso",
        "m": "vjsoknhe",
        "i": "tpyalwcz",
        "f": "sbnqzmlr",
        "B": "mxldarvo",
        "x_0": "uwaeflyo",
        "\\mu_0": "zptgkram",
        "\\mu_n": "gcfomrwh",
        "\\mu_n-1": "wdcqzneb",
        "a_k": "nyxgofsl",
        "a_0": "flirpsad",
        "a_1": "hkgamqre",
        "a_k-1": "pqmsdnei",
        "L": "owyfrqzt",
        "M": "bcjthvra",
        "\\epsilon": "dsqplnvo"
      },
      "question": "Let $sbnqzmlr$ be a function on $[0,\\infty)$, differentiable and satisfying\n\\[\nsbnqzmlr'(qzxwvtnp)=-3sbnqzmlr(qzxwvtnp)+6sbnqzmlr(2qzxwvtnp)\n\\]\nfor $qzxwvtnp>0$. Assume that $|sbnqzmlr(qzxwvtnp)|\\le e^{-\\sqrt{qzxwvtnp}}$ for $qzxwvtnp\\ge 0$ (so that\n$sbnqzmlr(qzxwvtnp)$ tends rapidly to $0$ as $qzxwvtnp$ increases).\nFor $hjgrksla$ a non-negative integer, define\n\\[\ngcfomrwh=\\int_0^\\infty qzxwvtnp^{hjgrksla} sbnqzmlr(qzxwvtnp)\\,dqzxwvtnp\n\\]\n(sometimes called the $hjgrksla$th moment of $sbnqzmlr$).\n\\begin{enumerate}\n\\item[a)] Express $gcfomrwh$ in terms of $zptgkram$.\n\\item[b)] Prove that the sequence $\\{gcfomrwh \\frac{3^{hjgrksla}}{hjgrksla!}\\}$ always converges,\nand that the limit is $0$ only if $zptgkram=0$.\n\\end{enumerate}",
      "solution": "Solution. a. As \\( qzxwvtnp \\rightarrow \\infty, sbnqzmlr(qzxwvtnp) \\) tends to 0 faster than any negative power of \\( qzxwvtnp \\), so the integral defining \\( gcfomrwh \\) converges. Multiply the functional equation by \\( qzxwvtnp^{hjgrksla} \\) for some \\( hjgrksla \\geq 1 \\) and integrate from 0 to some finite \\( mxldarvo>0 \\) :\n\\[\n\\int_{0}^{mxldarvo} qzxwvtnp^{hjgrksla} sbnqzmlr^{\\prime}(qzxwvtnp) d qzxwvtnp=-3 \\int_{0}^{mxldarvo} qzxwvtnp^{hjgrksla} sbnqzmlr(qzxwvtnp) d qzxwvtnp+6 \\int_{0}^{mxldarvo} qzxwvtnp^{hjgrksla} sbnqzmlr(2 qzxwvtnp) d qzxwvtnp\n\\]\n\nUsing integration by parts [ \\( \\mathrm{Spv}, \\mathrm{Ch} .18 \\), Theorem 1] on the left, and the substitution \\( rmbvcxye=2 qzxwvtnp \\) on the last term converts this into\n\\[\n\\left.qzxwvtnp^{hjgrksla} sbnqzmlr(qzxwvtnp)\\right|_{0} ^{mxldarvo}-hjgrksla \\int_{0}^{mxldarvo} sbnqzmlr(qzxwvtnp) qzxwvtnp^{hjgrksla-1} d qzxwvtnp=-3 \\int_{0}^{mxldarvo} qzxwvtnp^{hjgrksla} sbnqzmlr(qzxwvtnp) d qzxwvtnp+\\frac{6}{2^{hjgrksla+1}} \\int_{0}^{mxldarvo / 2} rmbvcxye^{hjgrksla} sbnqzmlr(rmbvcxye) d rmbvcxye\n\\]\n\nTaking limits as \\( mxldarvo \\rightarrow \\infty \\) yields\n\\[\n-hjgrksla wdcqzneb=-3 gcfomrwh+\\frac{6}{2^{hjgrksla+1}} gcfomrwh\n\\]\nso\n\\[\ngcfomrwh=\\frac{hjgrksla}{3}\\left(1-2^{-hjgrksla}\\right)^{-1} wdcqzneb\n\\]\nfor all \\( hjgrksla \\geq 1 \\). By induction, we obtain\n\\[\ngcfomrwh=\\frac{hjgrksla!}{3^{hjgrksla}}\\left(\\prod_{vjsoknhe=1}^{hjgrksla}\\left(1-2^{-vjsoknhe}\\right)\\right)^{-1} zptgkram\n\\]\nb. Since \\( \\sum_{vjsoknhe=1}^{\\infty} 2^{-vjsoknhe} \\) converges, \\( \\prod_{vjsoknhe=1}^{\\infty}\\left(1-2^{-vjsoknhe}\\right) \\) converges to some nonzero limit \\( owyfrqzt \\), and\n\\[\ngcfomrwh \\frac{3^{hjgrksla}}{hjgrksla!}=\\left(\\prod_{vjsoknhe=1}^{hjgrksla}\\left(1-2^{-vjsoknhe}\\right)\\right)^{-1} zptgkram \\rightarrow owyfrqzt^{-1} zptgkram\n\\]\nas \\( hjgrksla \\rightarrow \\infty \\). This limit \\( owyfrqzt^{-1} zptgkram \\) is finite, and equals zero if and only if \\( zptgkram=0 \\).\nRemark. We show that nonzero functions \\( sbnqzmlr(qzxwvtnp) \\) satisfying the conditions of the problem do exist. Given that \\( sbnqzmlr(qzxwvtnp) \\) tends to zero rapidly as \\( qzxwvtnp \\rightarrow \\infty \\), one expects \\( sbnqzmlr(2 qzxwvtnp) \\) to be negligible compared to \\( sbnqzmlr(qzxwvtnp) \\) for large \\( qzxwvtnp \\), and hence one can guess that \\( sbnqzmlr(qzxwvtnp) \\) can be approximated by a solution to \\( y^{\\prime}=-3 y \\), for example \\( e^{-3 qzxwvtnp} \\). But then the error in the differential equation \\( sbnqzmlr^{\\prime}(qzxwvtnp)=-3 sbnqzmlr(qzxwvtnp)+6 sbnqzmlr(2 qzxwvtnp) \\) is of order \\( e^{-6 qzxwvtnp} \\) as \\( qzxwvtnp \\rightarrow \\infty \\), leading one to guess that \\( e^{-3 qzxwvtnp}+hkgamqre e^{-6 qzxwvtnp} \\) will be a better approximation, and so on, finally leading one to guess\n\\[\nsbnqzmlr(qzxwvtnp)=e^{-3 qzxwvtnp}+hkgamqre e^{-6 qzxwvtnp}+nyxgofsl e^{-12 qzxwvtnp}+\\cdots,\n\\]\nwhere the coefficients \\( nyxgofsl \\) are to be solved for. Let \\( flirpsad=1 \\). If we differentiate (1) term by term, temporarily disregarding convergence issues, then for \\( ldfqpnso \\geq 1 \\), the coefficient of \\( e^{-3 \\cdot 2^{ldfqpnso} qzxwvtnp} \\) in the expression \\( sbnqzmlr^{\\prime}(qzxwvtnp)+3 sbnqzmlr(qzxwvtnp)-6 sbnqzmlr(2 qzxwvtnp) \\) is \\( -3 \\cdot 2^{ldfqpnso} nyxgofsl+3 nyxgofsl-6 pqmsdnei \\). If this is to be zero, then \\( nyxgofsl=\\frac{-2}{2^{ldfqpnso}-1} pqmsdnei \\).\n\nAll this so far has been motivation. Now we define the sequence \\( flirpsad, hkgamqre, \\ldots \\) by \\( flirpsad=1 \\) and \\( nyxgofsl=\\frac{-2}{2^{ldfqpnso}-1} pqmsdnei \\) for \\( ldfqpnso \\geq 1 \\), so\n\\[\nnyxgofsl=\\frac{(-2)^{hjgrksla}}{(2-1)\\left(2^{2}-1\\right) \\cdots\\left(2^{ldfqpnso}-1\\right)}\n\\]\nand define \\( sbnqzmlr(qzxwvtnp) \\) by (1). The series \\( \\sum_{ldfqpnso=0}^{\\infty}\\left|nyxgofsl\\right| \\) converges by the Ratio Test [Spv, Ch. 22, Theorem 3], so (1) converges to a continuous function on \\( [0, \\infty) \\). Moreover, \\( \\left|nyxgofsl e^{-3 \\cdot 2^{ldfqpnso} qzxwvtnp}\\right| \\leq\\left|nyxgofsl\\right| \\) for all complex \\( qzxwvtnp \\) with \\( \\operatorname{Re}(qzxwvtnp)>0 \\), so by the Weierstrass \\( M \\)-test and Weierstrass's Theorem [Ah, pp. 37, 176], (1) converges to a holomorphic function in this region. In particular, \\( sbnqzmlr(qzxwvtnp) \\) is differentiable on \\( (0, \\infty) \\), and can be differentiated term by term, so \\( sbnqzmlr^{\\prime}(qzxwvtnp)=-3 sbnqzmlr(qzxwvtnp)+6 sbnqzmlr(2 qzxwvtnp) \\) holds.\n\nFinally, we will show that if \\( dsqplnvo>0 \\) is sufficiently small, then \\( dsqplnvo sbnqzmlr(qzxwvtnp) \\), which still satisfies the differential equation, now also satisfies \\( |dsqplnvo sbnqzmlr(qzxwvtnp)| \\leq e^{-\\sqrt{qzxwvtnp}} \\). Since\n\\[\nsbnqzmlr(qzxwvtnp)=O\\left(e^{-3 qzxwvtnp}+e^{-6 qzxwvtnp}+e^{-12 qzxwvtnp}+\\cdots\\right)=O\\left(e^{-3 qzxwvtnp}\\right)=o\\left(e^{-\\sqrt{qzxwvtnp}}\\right)\n\\]\nas \\( qzxwvtnp \\rightarrow \\infty \\), we have \\( |sbnqzmlr(qzxwvtnp)| \\leq e^{-\\sqrt{qzxwvtnp}} \\) for all \\( qzxwvtnp \\) greater than some \\( uwaeflyo \\). Let \\( bcjthvra= \\) \\( \\sup _{qzxwvtnp \\in\\left[0, uwaeflyo\\right]} \\frac{|sbnqzmlr(qzxwvtnp)|}{e^{-\\sqrt{qzxwvtnp}}} \\). If \\( dsqplnvo \\in(0,1) \\) is chosen so that \\( dsqplnvo bcjthvra \\leq 1 \\), then \\( |dsqplnvo sbnqzmlr(qzxwvtnp)| \\leq e^{-\\sqrt{qzxwvtnp}} \\) both for \\( qzxwvtnp \\in\\left[0, uwaeflyo\\right] \\) and for \\( qzxwvtnp \\in\\left(uwaeflyo, \\infty\\right) \\), as desired."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf:[0,\\infty)\\longrightarrow\\mathbb R\n\\]  \nbe a continuously differentiable function that, for every $x>0$, satisfies the delay–differential equation  \n\\[\nf'(x)=-6\\,f(x)+15\\,f(2x)-10\\,f(3x).\\tag{1}\n\\]\n\nAssume the rapid–decay bound  \n\\[\n\\lvert f(x)\\rvert\\le e^{-x^{4/5}}\\qquad(x\\ge 0).\\tag{2}\n\\]\n\nFor every non–negative integer $n$ define the $n$-th (ordinary) moment of $f$ by  \n\\[\n\\mu_n:=\\int_{0}^{\\infty}x^{n}f(x)\\,dx.\\tag{3}\n\\]\n\n\\begin{enumerate}\n\\item[(a)]  Prove that each $\\mu_n$ is finite and that the moments satisfy the first–order recurrence  \n\\[\n\\mu_n=\\frac{n}{6}\\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}\\mu_{n-1}\\qquad(n\\ge 1).\\tag{4}\n\\]\n\n\\item[(b)]  Deduce the closed formula  \n\\[\n\\mu_n=\\frac{n!}{6^{\\,n}}\n        \\Bigl(\\prod_{m=1}^{n}\\bigl(1-\\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}}\\bigr)\\Bigr)^{-1}\n        \\mu_0\\qquad(n\\ge 0).\\tag{5}\n\\]\n\n\\item[(c)]  Define the scaled moments $\\rho_n:=\\mu_n\\,6^{\\,n}/n!$.  Prove that the limit  \n\\[\n\\rho_\\infty:=\\lim_{n\\to\\infty}\\rho_n\n\\]\nexists and equals  \n\\[\n\\rho_\\infty=L^{-1}\\mu_0,\n\\qquad\nL:=\\prod_{m=1}^{\\infty}\\Bigl(1-\\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}}\\Bigr).\\tag{6}\n\\]\nShow that $L$ is finite and non–zero, and that $\\rho_\\infty=0$ if and only if $\\mu_0=0$.\n\n\\item[(d)]  Let  \n\\[\nF(s):=\\int_{0}^{\\infty}e^{-s x}\\,f(x)\\,dx\\qquad(\\operatorname{Re}s>0)\\tag{7}\n\\]\nbe the Laplace transform of $f$.\n\n\\begin{enumerate}\n\\item[(i)]  Show that $F$ extends analytically to the half–plane $\\operatorname{Re}s>-6$ and that, there, it satisfies  \n\\[\n(s+6)F(s)=f(0)+\\frac{15}{2}\\,F\\!\\Bigl(\\frac{s}{2}\\Bigr)-\\frac{10}{3}\\,F\\!\\Bigl(\\frac{s}{3}\\Bigr).\\tag{8}\n\\]\n\n\\item[(ii)]  Prove that $F$ is analytic at $s=0$ and that  \n\\[\n\\mu_0=\\frac{6}{11}\\,f(0)\\qquad\n\\bigl(\\text{equivalently }F(0)=\\tfrac{6}{11}f(0)\\bigr).\\tag{9}\n\\]\n\n\\item[(iii)]  Show further that $F$ possesses at most a simple pole at $s=-6$ and that this pole is present unless the exceptional cancellation identity  \n\\[\nf(0)+\\frac{15}{2}\\,F(-3)-\\frac{10}{3}\\,F(-2)=0\\tag{10}\n\\]\nholds.\n\\end{enumerate}\n\\end{enumerate}\n\n\\bigskip",
      "solution": "Throughout we keep the notation introduced in the statement.\n\n\\paragraph{Step 0.  Convergence of the moments.}\nBecause $\\lvert f(x)\\rvert\\le e^{-x^{4/5}}$, for every fixed $n$ we have\n\\[\nx^{n}\\lvert f(x)\\rvert\\le x^{n}e^{-x^{4/5}}\n          =\\exp\\bigl(n\\log x-x^{4/5}\\bigr)\n          \\le \\exp\\!\\Bigl(-\\tfrac12 x^{4/5}\\Bigr)\n\\]\nfor $x$ sufficiently large; the right–hand side is integrable on $(0,\\infty)$,\nhence each $\\mu_n$ converges absolutely.\n\n\\paragraph{Step 1.  Derivation of the recurrence (4).}\nMultiply (1) by $x^{n}$ and integrate over $(0,\\infty)$.  \nBecause $x^{n}f(x)\\rightarrow 0$ as $x\\to\\infty$ by (2) and $x^{n}f(x)=\\mathrm O(x^{n})$ near $0$, the boundary term in integration by parts vanishes:\n\\[\n\\int_{0}^{\\infty}x^{n}f'(x)\\,dx\n   =[x^{n}f(x)]_{0}^{\\infty}-n\\int_{0}^{\\infty}x^{n-1}f(x)\\,dx\n   =-n\\mu_{n-1}.\n\\]\nFor the dilated terms use the substitutions $u=2x$ and $u=3x$,\n\\[\n\\int_{0}^{\\infty}x^{n}f(2x)\\,dx=2^{-(n+1)}\\mu_{n},\n\\qquad\n\\int_{0}^{\\infty}x^{n}f(3x)\\,dx=3^{-(n+1)}\\mu_{n}.\n\\]\nInsert these three integrals into (1) and rearrange to obtain (4).\n\n\\paragraph{Step 2.  Closed formula (5).}\nWrite (4) as $\\mu_n=A_{n}\\mu_{n-1}$ with\n\\[\nA_{n}:=\\frac{n}{6}\\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}.\n\\]\nIterating gives (5).\n\n\\paragraph{Step 3.  Convergence of the scaled moments (6).}\nSet $\\rho_n:=\\mu_n6^{\\,n}/n!$.  \nThen  \n\\[\n\\rho_n=\\rho_{n-1}\n        \\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}.\n\\tag{11}\n\\]\nDefine $B_m:=1-\\dfrac{5}{2^{\\,m+2}}+\\dfrac{5}{3^{\\,m+2}}$.\nEach $B_m>0$ because the two positive summands are smaller than $1$.  \nMoreover\n\\[\n\\lvert B_m-1\\rvert\\le \\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}},\n\\qquad\n\\sum_{m=1}^{\\infty}\\Bigl(\\frac{1}{2^{\\,m}}+\\frac{1}{3^{\\,m}}\\Bigr)<\\infty,\n\\]\nso $\\sum_{m=1}^{\\infty}\\lvert B_m-1\\rvert<\\infty$ and the infinite product\n$L:=\\prod_{m=1}^{\\infty}B_m$ converges absolutely to a finite, non–zero\nlimit.  From (11) we deduce\n\\[\n\\rho_n=\\rho_0\\prod_{m=1}^{n}B_m^{-1}\\xrightarrow[n\\to\\infty]{} L^{-1}\\rho_0,\n\\]\nwhich is precisely (6).  Clearly $\\rho_\\infty=0$ iff $\\rho_0=0$, i.e.\\ iff\n$\\mu_0=0$.\n\n\\paragraph{Step 4.  Laplace transform and analytic continuation.}\n\n\\subparagraph{4.1  Laplace transform of the differential equation.}\nFor $\\operatorname{Re}s>0$ the rapid decay (2) justifies termwise Laplace transformation of (1):\n\\[\nsF(s)-f(0)=-6F(s)+\\frac{15}{2}\\,F\\!\\Bigl(\\frac{s}{2}\\Bigr)\n                       -\\frac{10}{3}\\,F\\!\\Bigl(\\frac{s}{3}\\Bigr).\n\\]\nRearranging gives (8).\n\n\\subparagraph{4.2  Analytic continuation to $\\operatorname{Re}s>-6$.}\n\nThe obstacle encountered in the original proof was to deal with the possible\nblow–up of the factor $\\lvert s+6\\rvert^{-1}$ when $s$ approaches the line\n$\\operatorname{Re}s=-6$.  We bypass this difficulty by working in a \\emph{weighted}\nBanach space and by invoking the Banach fixed–point theorem.\n\nFix $\\sigma\\in(-6,0]$.  Write  \n\\[\n\\Pi_\\sigma:=\\{s\\in\\mathbb C\\mid\\operatorname{Re}s>\\sigma\\}.\n\\]\nGiven an integer $\\alpha\\ge 4$, set  \n\\[\n\\lVert G\\rVert_{\\sigma,\\alpha}:=\\sup_{s\\in\\Pi_\\sigma}\n        \\frac{|G(s)|}{(1+|s|)^{\\alpha}}\\qquad\n        \\bigl(G\\text{ holomorphic on }\\Pi_\\sigma\\bigr).\n\\]\nDenote the corresponding Banach space by $\\mathcal A_{\\sigma,\\alpha}$.\n\nDefine the operator  \n\\[\n(\\mathcal T G)(s):=\\frac{f(0)+\\dfrac{15}{2}G(s/2)-\\dfrac{10}{3}G(s/3)}{s+6},\n\\qquad s\\in\\Pi_\\sigma.\n\\]\nFor $G_1,G_2\\in\\mathcal A_{\\sigma,\\alpha}$ we estimate, using\n$\\lvert s+6\\rvert\\ge 6+\\sigma$ and  \n$(1+|s/2|)^{\\alpha}\\le 2^{-\\alpha}(1+|s|)^{\\alpha}$,  \n$(1+|s/3|)^{\\alpha}\\le 3^{-\\alpha}(1+|s|)^{\\alpha}$,\n\\[\n\\begin{aligned}\n\\lVert\\mathcal T G_1-\\mathcal T G_2\\rVert_{\\sigma,\\alpha}\n&\\le\\Bigl(\\frac{15}{2(6+\\sigma)}\\,2^{-\\alpha}\n          +\\frac{10}{3(6+\\sigma)}\\,3^{-\\alpha}\\Bigr)\n   \\lVert G_1-G_2\\rVert_{\\sigma,\\alpha}\\\\[4pt]\n&=:C_{\\sigma,\\alpha}\\,\\lVert G_1-G_2\\rVert_{\\sigma,\\alpha}.\n\\end{aligned}\n\\]\nBecause $6+\\sigma>0$, one can pick $\\alpha$ so large that $C_{\\sigma,\\alpha}<1$:\nfor instance $\\alpha=8$ works uniformly for every $\\sigma\\in(-6,-5]$, then\n$\\alpha=9$ works on $(-5,-4]$, and so on.  Hence $\\mathcal T$ is a\n\\emph{contraction} on $\\mathcal A_{\\sigma,\\alpha}$.\n\nThe Banach fixed–point theorem yields a unique element $F_{\\sigma,\\alpha}\\in\n\\mathcal A_{\\sigma,\\alpha}$ satisfying $\\mathcal T F_{\\sigma,\\alpha}=F_{\\sigma,\\alpha}$,\ni.e.\\ Equation (8) on $\\Pi_\\sigma$.  \nIf $\\sigma'<\\sigma$, the restrictions of $F_{\\sigma',\\alpha'}$ and\n$F_{\\sigma,\\alpha}$ to $\\Pi_\\sigma$ coincide by uniqueness; therefore the\nfunctions $F_{\\sigma,\\alpha}$ glue together to a single holomorphic function\non the half–plane $\\operatorname{Re}s>-6$.  But for $\\sigma=0$ we know that\n$F_{0,\\alpha}$ equals the original Laplace transform (7); hence the glued\nfunction is the \\emph{analytic continuation} of $F$ wanted in part (i).\nThis proves the first assertion of (i) rigorously.\n\n\\subparagraph{4.3  Regularity at $s=0$ and identity (9).}\nBy dominated convergence,\n\\[\nF(0)=\\int_{0}^{\\infty}f(x)\\,dx=\\mu_0. \\tag{13}\n\\]\nPut $s=0$ in (8):\n\\[\n6F(0)=f(0)+\\frac{15}{2}F(0)-\\frac{10}{3}F(0)\n      =f(0)+\\frac{25}{6}F(0),\n\\]\nhence $(11/6)F(0)=f(0)$, i.e.\\ (9).  \nEquation (8) shows that, near $s=0$,\n\\[\nF(s)=\\frac{f(0)+\\mathrm O(s)}{6+s},\n\\]\nso $F$ is analytic at $s=0$.\n\n\\subparagraph{4.4  Behaviour at $s=-6$.}\nWrite $N(s):=f(0)+\\dfrac{15}{2}F(s/2)-\\dfrac{10}{3}F(s/3)$, analytic near\n$s=-6$.  Then (8) gives\n\\[\nF(s)=\\frac{N(-6)}{s+6}+O(1)\\qquad(s\\to-6),\n\\]\nso $F$ has at most a simple pole at $s=-6$, and the pole is absent\nprecisely when $N(-6)=0$, i.e.\\ when (10) holds.\n\n\\hfill$\\square$\n\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.709696",
        "was_fixed": false,
        "difficulty_analysis": "•  Two independent delays (2x and 3x) replace the single delay of the original, leading to the more intricate recurrence (4) where the correction factor involves a linear combination of 2^{−(n+2)} and 3^{−(n+2)}.  \n•  The explicit solution (5) now contains a double Euler-type infinite product rather than the single product (1−2^{−m})^{−1} that appeared previously.  Proving convergence requires handling two geometric series simultaneously.  \n•  Part (d) introduces the Laplace transform and a functional equation mixing F(s), F(s/2) and F(s/3).  Analysing this equation and its analytic implications forces the solver to combine differential‐equation techniques with complex analysis.  \n•  Altogether, extra delays, an extra analytic task (meromorphic continuation of F) and the need to control a more complicated infinite product make this variant substantially harder and longer than both the original problem and the first kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf:[0,\\infty)\\longrightarrow\\mathbb R\n\\]  \nbe a continuously differentiable function that, for every $x>0$, satisfies the delay–differential equation  \n\\[\nf'(x)=-6\\,f(x)+15\\,f(2x)-10\\,f(3x).\\tag{1}\n\\]\n\nAssume the rapid–decay bound  \n\\[\n\\lvert f(x)\\rvert\\le e^{-x^{4/5}}\\qquad(x\\ge 0).\\tag{2}\n\\]\n\nFor every non–negative integer $n$ define the $n$-th (ordinary) moment of $f$ by  \n\\[\n\\mu_n:=\\int_{0}^{\\infty}x^{n}f(x)\\,dx.\\tag{3}\n\\]\n\n\\begin{enumerate}\n\\item[(a)]  Prove that each $\\mu_n$ is finite and that the moments satisfy the first–order recurrence  \n\\[\n\\mu_n=\\frac{n}{6}\\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}\\mu_{n-1}\\qquad(n\\ge 1).\\tag{4}\n\\]\n\n\\item[(b)]  Deduce the closed formula  \n\\[\n\\mu_n=\\frac{n!}{6^{\\,n}}\n        \\Bigl(\\prod_{m=1}^{n}\\bigl(1-\\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}}\\bigr)\\Bigr)^{-1}\n        \\mu_0\\qquad(n\\ge 0).\\tag{5}\n\\]\n\n\\item[(c)]  Define the scaled moments $\\rho_n:=\\mu_n\\,6^{\\,n}/n!\\,.$  Prove that the limit  \n\\[\n\\rho_\\infty:=\\lim_{n\\to\\infty}\\rho_n\n\\]\nexists and equals  \n\\[\n\\rho_\\infty=L^{-1}\\mu_0,\n\\qquad\nL:=\\prod_{m=1}^{\\infty}\\Bigl(1-\\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}}\\Bigr).\\tag{6}\n\\]\nShow that $L$ is finite and non–zero, and that $\\rho_\\infty=0$ if and only if $\\mu_0=0$.\n\n\\item[(d)]  Let  \n\\[\nF(s):=\\int_{0}^{\\infty}e^{-s x}\\,f(x)\\,dx\\qquad(\\operatorname{Re}s>0)\\tag{7}\n\\]\nbe the Laplace transform of $f$.\n\n\\begin{enumerate}\n\\item[(i)]  Show that $F$ extends analytically to the half–plane $\\operatorname{Re}s>-6$ and that, there, it satisfies  \n\\[\n(s+6)F(s)=f(0)+\\frac{15}{2}\\,F\\!\\Bigl(\\frac{s}{2}\\Bigr)-\\frac{10}{3}\\,F\\!\\Bigl(\\frac{s}{3}\\Bigr).\\tag{8}\n\\]\n\n\\item[(ii)]  Prove that $F$ is analytic at $s=0$ and that  \n\\[\n\\mu_0=\\frac{6}{11}\\,f(0)\\qquad\n\\bigl(\\text{equivalently }F(0)=\\tfrac{6}{11}f(0)\\bigr).\\tag{9}\n\\]\n\n\\item[(iii)]  Show further that $F$ possesses at most a simple pole at $s=-6$ and that this pole is present unless the exceptional cancellation identity  \n\\[\nf(0)+\\frac{15}{2}\\,F(-3)-\\frac{10}{3}\\,F(-2)=0\\tag{10}\n\\]\nholds.\n\\end{enumerate}\n\\end{enumerate}\n\n\\bigskip",
      "solution": "Throughout we keep the notation introduced in the statement.\n\n\\paragraph{Step 0.  Convergence of the moments.}\nBecause $\\lvert f(x)\\rvert\\le e^{-x^{4/5}}$, for every fixed $n$ we have\n\\[\nx^{n}\\lvert f(x)\\rvert\\le x^{n}e^{-x^{4/5}}\n          =\\exp\\bigl(n\\log x-x^{4/5}\\bigr)\n          \\le \\exp\\!\\Bigl(-\\tfrac12 x^{4/5}\\Bigr)\n\\]\nfor $x$ large enough; the right–hand side is integrable on $(0,\\infty)$,\nhence each $\\mu_n$ converges absolutely.\n\n\\paragraph{Step 1.  Derivation of the recurrence (4).}\nMultiply (1) by $x^{n}$ and integrate over $(0,\\infty)$.  \nBecause $x^{n}f(x)\\rightarrow 0$ as $x\\to\\infty$ by (2) and $x^{n}f(x)=\\mathrm O(x^{n})$ near $0$, the boundary term in integration by parts vanishes:\n\\[\n\\int_{0}^{\\infty}x^{n}f'(x)\\,dx\n   =[x^{n}f(x)]_{0}^{\\infty}-n\\int_{0}^{\\infty}x^{n-1}f(x)\\,dx\n   =-n\\mu_{n-1}.\n\\]\nFor the dilated terms use the substitutions $u=2x$ and $u=3x$,\n\\[\n\\int_{0}^{\\infty}x^{n}f(2x)\\,dx=2^{-(n+1)}\\mu_{n},\n\\qquad\n\\int_{0}^{\\infty}x^{n}f(3x)\\,dx=3^{-(n+1)}\\mu_{n}.\n\\]\nInsert these three integrals into (1) and rearrange to obtain (4).\n\n\\paragraph{Step 2.  Closed formula (5).}\nWrite (4) as $\\mu_n=A_{n}\\mu_{n-1}$ with\n\\[\nA_{n}:=\\frac{n}{6}\\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}.\n\\]\nIterating gives (5).\n\n\\paragraph{Step 3.  Convergence of the scaled moments (6).}\nSet $\\rho_n:=\\mu_n6^{\\,n}/n!$.  \nThen  \n\\[\n\\rho_n=\\rho_{n-1}\n        \\Bigl(1-\\frac{5}{2^{\\,n+2}}+\\frac{5}{3^{\\,n+2}}\\Bigr)^{-1}.\n\\tag{11}\n\\]\nDefine $B_m:=1-\\dfrac{5}{2^{\\,m+2}}+\\dfrac{5}{3^{\\,m+2}}$.\nEach $B_m>0$ because the two positive summands are smaller than $1$.  \nMoreover\n\\[\n\\lvert B_m-1\\rvert\\le \\frac{5}{2^{\\,m+2}}+\\frac{5}{3^{\\,m+2}},\n\\qquad\n\\sum_{m=1}^{\\infty}\\Bigl(\\frac{1}{2^{\\,m}}+\\frac{1}{3^{\\,m}}\\Bigr)<\\infty,\n\\]\nso $\\sum_{m=1}^{\\infty}\\lvert B_m-1\\rvert<\\infty$ and the infinite product\n$L:=\\prod_{m=1}^{\\infty}B_m$ converges absolutely to a finite, non–zero\nlimit.  From (11) we deduce\n\\[\n\\rho_n=\\rho_0\\prod_{m=1}^{n}B_m^{-1}\\xrightarrow[n\\to\\infty]{} L^{-1}\\rho_0,\n\\]\nwhich is precisely (6).  Clearly $\\rho_\\infty=0$ iff $\\rho_0=0$, i.e.\\ iff\n$\\mu_0=0$.\n\n\\paragraph{Step 4.  Laplace transform and analytic continuation.}\n\n\\subparagraph{4.1  Laplace transform of the differential equation.}\nFor $\\operatorname{Re}s>0$ the rapid decay (2) justifies termwise Laplace transformation of (1):\n\\[\nsF(s)-f(0)=-6F(s)+\\frac{15}{2}\\,F\\!\\Bigl(\\frac{s}{2}\\Bigr)\n                       -\\frac{10}{3}\\,F\\!\\Bigl(\\frac{s}{3}\\Bigr).\n\\]\nRearranging gives (8).\n\n\\subparagraph{4.2  Analytic continuation to $\\operatorname{Re}s>-6$.}\nRewrite (8) as  \n\\[\nF(s)=\\frac{f(0)+\\dfrac{15}{2}F(s/2)-\\dfrac{10}{3}F(s/3)}{s+6}.\n\\tag{12}\n\\]\n\nFix $s_0$ with $-6<\\operatorname{Re}s_0\\le 0$.  \nChoose $k\\ge 0$ minimal such that $\\operatorname{Re}(s_0/2^{\\,k})>0$.\nIterate (12) $k$ times to obtain\n\\[\nF(s_0)=\\frac{P_k\\bigl(F(s_0/2^{\\,k}),f(0)\\bigr)}{Q_k(s_0)},\n\\]\nwhere $P_k$ is an explicit linear polynomial in $F(s_0/2^{\\,k})$ and $f(0)$\nand $Q_k$ a product of $k$ factors of the form $(s_0/2^{\\,j}+6)$.\nSince $F(s_0/2^{\\,k})$ is known (its real part is positive), the right–hand\nside defines $F(s_0)$.  \nBecause the construction is algebraic in $s$ and uses finitely many\napplications of the analytic map $s\\mapsto F(s/2)$ or $s\\mapsto F(s/3)$,\nthe resulting $F$ is analytic at $s_0$.  \nDoing this for every $s_0$ with $\\operatorname{Re}s_0>-6$ gives a unique\nanalytic continuation of $F$ to that half–plane (except possibly at\n$s=-6$).\n\nAlternatively, one can solve (12) by the Neumann series   \n\\[\nF(s)=\\frac{f(0)}{s+6}\n      +\\sum_{n=0}^{\\infty}\\Bigl(\\frac{15}{2(s+6)}T_2\n                                 -\\frac{10}{3(s+6)}T_3\\Bigr)^{\\!n}\n        \\!\\!\\Bigl(\\frac{f(0)}{s+6}\\Bigr),\n\\]\nwhere $T_a$ denotes the dilation operator $(T_a G)(s):=G(s/a)$.  \nFor $\\operatorname{Re}s>-6$ the operator norm of\n$\\tfrac{15}{2(s+6)}T_2-\\tfrac{10}{3(s+6)}T_3$ is $<1$, so the series\nconverges uniformly on compacta, again yielding an analytic extension.\n\n\\subparagraph{4.3  Regularity at $s=0$ and identity (9).}\nBy dominated convergence,\n\\[\nF(0)=\\int_{0}^{\\infty}f(x)\\,dx=\\mu_0. \\tag{13}\n\\]\nPut $s=0$ in (8):\n\\[\n6F(0)=f(0)+\\frac{15}{2}F(0)-\\frac{10}{3}F(0)\n      =f(0)+\\frac{25}{6}F(0),\n\\]\nhence $(11/6)F(0)=f(0)$, i.e.\\ (9).  \nEquation (12) shows that near $s=0$\n\\[\nF(s)=\\frac{f(0)+\\mathrm O(s)}{6+s},\n\\]\nso $F$ is analytic at $s=0$.\n\n\\subparagraph{4.4  Behaviour at $s=-6$.}\nWrite $N(s):=f(0)+\\dfrac{15}{2}F(s/2)-\\dfrac{10}{3}F(s/3)$, analytic near\n$s=-6$.  Then (12) gives\n\\[\nF(s)=\\frac{N(-6)}{s+6}+O(1)\\qquad(s\\to-6),\n\\]\nso $F$ has at most a simple pole at $s=-6$, and the pole is absent\nprecisely when $N(-6)=0$, i.e.\\ when (10) holds.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.553353",
        "was_fixed": false,
        "difficulty_analysis": "•  Two independent delays (2x and 3x) replace the single delay of the original, leading to the more intricate recurrence (4) where the correction factor involves a linear combination of 2^{−(n+2)} and 3^{−(n+2)}.  \n•  The explicit solution (5) now contains a double Euler-type infinite product rather than the single product (1−2^{−m})^{−1} that appeared previously.  Proving convergence requires handling two geometric series simultaneously.  \n•  Part (d) introduces the Laplace transform and a functional equation mixing F(s), F(s/2) and F(s/3).  Analysing this equation and its analytic implications forces the solver to combine differential‐equation techniques with complex analysis.  \n•  Altogether, extra delays, an extra analytic task (meromorphic continuation of F) and the need to control a more complicated infinite product make this variant substantially harder and longer than both the original problem and the first kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}