{
  "index": "1941-B-7",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "14. Take either (i) or (ii).\n(i) Show that any solution \\( f(t) \\) of the functional equation\n\\[\nf(x+y) f(x-y)=f(x) f(x)+f(y) f(y)-1, \\quad(x, y, \\text { real })\n\\]\nis such that\n\\[\nf^{\\prime \\prime}(t)= \\pm m^{2} f(t), \\quad(m \\text { constant and } \\geq 0)\n\\]\nassuming the existence and continuity of the second derivative. Deduce that \\( f(t) \\) is one of the functions\n\\[\n\\pm \\cos m t, \\quad \\pm \\cosh m t\n\\]\n(ii) With \\( n \\) constant values \\( a_{1}, a_{2}, \\ldots, a_{n} \\), supposed all different, let \\( n \\) constant values \\( b_{1}, b_{2}, \\ldots, b_{n} \\) be associated, and let a polynomial \\( P(x) \\) be. defined by the identity in \\( \\boldsymbol{x} \\)\n\\[\n\\left|\\begin{array}{cccccc}\n1 & x & x^{2} & \\cdots & x^{n-1} & P(x) \\\\\n1 & a_{1} & a_{1}^{2} & \\cdots & a_{1}{ }^{n-1} & b_{1} \\\\\n1 & a_{2} & a_{2}^{2} & \\cdots & a_{2}{ }^{n-1} & b_{2} \\\\\n\\cdots & \\cdots & \\ldots & \\ldots & \\cdots & \\cdots\n\\end{array}\\right| \\equiv 0 . \\cdots \\cdots .\n\\]\n\nGiven a polynomial \\( \\phi(t) \\), let a polynomial \\( Q(x) \\) be defined by the identity in \\( x \\) obtained on replacing \\( P(x), b_{1}, b_{2}, \\ldots, b_{n} \\) of the identity above by \\( Q(x) \\), \\( \\phi\\left(b_{1}\\right), \\phi\\left(b_{2}\\right), \\ldots ; \\phi\\left(b_{n}\\right) \\). Prove that the remainder obtained on dividing \\( \\phi(P(x)) \\) by \\( \\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right) \\) is \\( Q(x) \\).",
  "solution": "Solution. Starting with the given functional equation\n\\[\nf(x+y) f(x-y)=f(x)^{2}+f(y)^{2}-1,\n\\]\nwe differentiate with respect to \\( y \\) to get\n\\[\nf^{\\prime}(x+y) f(x-y)-f(x+y) f^{\\prime}(x-y)=2 f(y) f^{\\prime}(y),\n\\]\nand then with respect to \\( \\boldsymbol{x} \\) to get\n\\[\nf^{\\prime \\prime}(x+y) f(x-y)-f(x+y) f^{\\prime \\prime}(x-y)=0 .\n\\]\n\nSetting \\( x=y=0 \\) in (1) and (2), we obtain \\( f(0)^{2}=2 f(0)^{2}-1 \\) and \\( 2 f(0) f^{\\prime}(0)=0 \\), whence\n\\[\nf(0)= \\pm 1, \\quad f^{\\prime}(0)=0 .\n\\]\n\nNow for any given number \\( t \\), put \\( x=y=t / 2 \\) in (3) to get\n\\[\nf^{\\prime \\prime}(t) f(0)-f(t) f^{\\prime \\prime}(0)=0,\n\\]\nwhich is equivalent to\n\\[\nf^{\\prime \\prime}(t) \\pm m^{2} f(t)=0,\n\\]\nwhere \\( m=\\left|f^{\\prime \\prime}(0)\\right|^{1 / 2} \\).\nIntegrating (5) using the initial conditions (4), we obtain\n\\[\nf(t)= \\pm \\cos m t \\text { or } f(t)= \\pm \\cosh m t,\n\\]\ndepending on whether the \\( \\operatorname{sign} \\) in (5) is plus or minus. If \\( \\boldsymbol{m}=0 \\), these solutions are constant.\n\nConversely, either of the above solutions satisfies the given functional equation for any value of \\( m \\).\n\nRemark. The continuity of the second derivative was not used in this proof. In fact, the result can be proved under far weaker hypotheses; for example, it is enough to assume that \\( f \\) itself is continuous. See J. Aczèl, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.\n\nSolution. We use the fact that the Vandermonde determinant\n\\[\nV=\\left|\\begin{array}{ccccc}\n1 & a_{1} & a_{1}{ }^{2} & \\ldots & a_{1}{ }^{n-1} \\\\\n1 & a_{2} & a_{2}{ }^{2} & \\ldots & a_{2^{n-1}} \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n1 & a_{n} & a_{n}{ }^{2} & \\ldots & a_{n}{ }^{n-1}\n\\end{array}\\right|=\\prod_{i j j}\\left(a_{i}-a_{j}\\right)\n\\]\nis not zero.\nIf we replace \\( x \\) by \\( a_{1} \\) in the determinant defining \\( P(x) \\), subtract the second row from the first, and expand by minors of the first row, we see that\n\\[\nV\\left(P\\left(a_{1}\\right)-b_{1}\\right)=0 .\n\\]\n\nHence, \\( P\\left(a_{1}\\right)=b_{1} \\). Similarly, \\( P\\left(a_{i}\\right)=b_{i} \\) for \\( i=2,3, \\ldots, n \\).\nA similar argument using the determinant for \\( Q \\) shows that \\( Q\\left(a_{i}\\right)= \\) \\( \\phi\\left(b_{i}\\right) \\) for \\( i=1,2, \\ldots, n \\). Moreover, it is evident that \\( Q \\) is a polynomial of degree less than \\( n \\).\n\nSuppose the polynomial \\( \\phi(P(x)) \\) is divided by \\( \\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right) \\) leaving a remainder \\( R(x) \\) of degree less than \\( n \\). Then\n\\[\n\\phi(P(x))=\\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right) S(x)+R(x),\n\\]\nwhere \\( S(x) \\) is a polynomial. Substituting \\( x=a_{i} \\), we get\n\\[\nR\\left(a_{i}\\right)=\\phi\\left(P\\left(a_{i}\\right)\\right)=\\phi\\left(b_{i}\\right)=Q\\left(a_{i}\\right) .\n\\]\n\nThus the polynomial \\( Q(x)-R(x) \\) is zero for \\( n \\) distinct values of \\( x \\). But its degree is less than \\( n \\), so \\( Q(x)-R(x) \\) is identically zero, and the remainder is \\( Q(x) \\), as required.\n\nRemark. The determinant given provides a useful expression for the unique polynomial of degree less than \\( n \\) with values prescribed at \\( a_{1}, a_{2} \\),\n\\( \\ldots, a_{n} \\).",
  "vars": [
    "x",
    "y",
    "t",
    "f",
    "i"
  ],
  "params": [
    "m",
    "n",
    "a_1",
    "a_2",
    "a_n",
    "a_i",
    "b_1",
    "b_2",
    "b_n",
    "b_i",
    "P",
    "Q",
    "S",
    "R",
    "V",
    "\\\\phi"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "t": "temporal",
        "f": "funcval",
        "i": "indexvar",
        "m": "modulus",
        "n": "quantity",
        "a_1": "alphaone",
        "a_2": "alphatwo",
        "a_n": "alphan",
        "a_i": "alphai",
        "b_1": "betaone",
        "b_2": "betatwo",
        "b_n": "betan",
        "b_i": "betai",
        "P": "polyfun",
        "Q": "polyque",
        "S": "polysix",
        "R": "remsix",
        "V": "vanderm",
        "\\phi": "phifunc"
      },
      "question": "14. Take either (i) or (ii).\n(i) Show that any solution \\( funcval(temporal) \\) of the functional equation\n\\[\nfuncval(abscissa+ordinate) funcval(abscissa-ordinate)=funcval(abscissa) funcval(abscissa)+funcval(ordinate) funcval(ordinate)-1, \\quad(abscissa, ordinate, \\text { real })\n\\]\nis such that\n\\[\nfuncval^{\\prime \\prime}(temporal)= \\pm modulus^{2} funcval(temporal), \\quad(modulus \\text { constant and } \\geq 0)\n\\]\nassuming the existence and continuity of the second derivative. Deduce that \\( funcval(temporal) \\) is one of the functions\n\\[\n\\pm \\cos modulus temporal, \\quad \\pm \\cosh modulus temporal\n\\]\n(ii) With \\( quantity \\) constant values \\( alphaone, alphatwo, \\ldots, alphan \\), supposed all different, let \\( quantity \\) constant values \\( betaone, betatwo, \\ldots, betan \\) be associated, and let a polynomial \\( polyfun(abscissa) \\) be defined by the identity in \\( \\boldsymbol{abscissa} \\)\n\\[\n\\left|\\begin{array}{cccccc}\n1 & abscissa & abscissa^{2} & \\cdots & abscissa^{quantity-1} & polyfun(abscissa) \\\\\n1 & alphaone & alphaone^{2} & \\cdots & alphaone{ }^{quantity-1} & betaone \\\\\n1 & alphatwo & alphatwo^{2} & \\cdots & alphatwo{ }^{quantity-1} & betatwo \\\\\n\\cdots & \\cdots & \\ldots & \\ldots & \\cdots & \\cdots\n\\end{array}\\right| \\equiv 0 . \\cdots \\cdots .\n\\]\nGiven a polynomial \\( phifunc(temporal) \\), let a polynomial \\( polyque(abscissa) \\) be defined by the identity in \\( abscissa \\) obtained on replacing \\( polyfun(abscissa), betaone, betatwo, \\ldots, betan \\) of the identity above by \\( polyque(abscissa) \\), \\( phifunc\\left(betaone\\right), phifunc\\left(betatwo\\right), \\ldots ; phifunc\\left(betan\\right) \\). Prove that the remainder obtained on dividing \\( phifunc(polyfun(abscissa)) \\) by \\( \\left(abscissa-alphaone\\right)\\left(abscissa-alphatwo\\right) \\cdots\\left(abscissa-alphan\\right) \\) is \\( polyque(abscissa) \\).",
      "solution": "Solution. Starting with the given functional equation\n\\[\nfuncval(abscissa+ordinate) funcval(abscissa-ordinate)=funcval(abscissa)^{2}+funcval(ordinate)^{2}-1,\n\\]\nwe differentiate with respect to \\( ordinate \\) to get\n\\[\nfuncval^{\\prime}(abscissa+ordinate) funcval(abscissa-ordinate)-funcval(abscissa+ordinate) funcval^{\\prime}(abscissa-ordinate)=2 funcval(ordinate) funcval^{\\prime}(ordinate),\n\\]\nand then with respect to \\( \\boldsymbol{abscissa} \\) to get\n\\[\nfuncval^{\\prime \\prime}(abscissa+ordinate) funcval(abscissa-ordinate)-funcval(abscissa+ordinate) funcval^{\\prime \\prime}(abscissa-ordinate)=0 .\n\\]\n\nSetting \\( abscissa=ordinate=0 \\) in (1) and (2), we obtain \\( funcval(0)^{2}=2 funcval(0)^{2}-1 \\) and \\( 2 funcval(0) funcval^{\\prime}(0)=0 \\), whence\n\\[\nfuncval(0)= \\pm 1, \\quad funcval^{\\prime}(0)=0 .\n\\]\n\nNow for any given number \\( temporal \\), put \\( abscissa=ordinate=temporal / 2 \\) in (3) to get\n\\[\nfuncval^{\\prime \\prime}(temporal) funcval(0)-funcval(temporal) funcval^{\\prime \\prime}(0)=0,\n\\]\nwhich is equivalent to\n\\[\nfuncval^{\\prime \\prime}(temporal) \\pm modulus^{2} funcval(temporal)=0,\n\\]\nwhere \\( modulus=\\left|funcval^{\\prime \\prime}(0)\\right|^{1 / 2} \\).\nIntegrating (5) using the initial conditions (4), we obtain\n\\[\nfuncval(temporal)= \\pm \\cos modulus temporal \\text { or } funcval(temporal)= \\pm \\cosh modulus temporal,\n\\]\ndepending on whether the \\( \\operatorname{sign} \\) in (5) is plus or minus. If \\( modulus=0 \\), these solutions are constant.\n\nConversely, either of the above solutions satisfies the given functional equation for any value of \\( modulus \\).\n\nRemark. The continuity of the second derivative was not used in this proof. In fact, the result can be proved under far weaker hypotheses; for example, it is enough to assume that \\( funcval \\) itself is continuous. See J. Aczèl, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.\n\nSolution. We use the fact that the Vandermonde determinant\n\\[\nvanderm=\\left|\\begin{array}{ccccc}\n1 & alphaone & alphaone{ }^{2} & \\ldots & alphaone{ }^{quantity-1} \\\\\n1 & alphatwo & alphatwo{ }^{2} & \\ldots & alphatwo^{quantity-1} \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n1 & alphan & alphan{ }^{2} & \\ldots & alphan{ }^{quantity-1}\n\\end{array}\\right|=\\prod_{indexvar<j}\\left(alphai-alphaj\\right)\n\\]\nis not zero.\nIf we replace \\( abscissa \\) by \\( alphaone \\) in the determinant defining \\( polyfun(abscissa) \\), subtract the second row from the first, and expand by minors of the first row, we see that\n\\[\nvanderm\\left(polyfun\\left(alphaone\\right)-betaone\\right)=0 .\n\\]\n\nHence, \\( polyfun\\left(alphaone\\right)=betaone \\). Similarly, \\( polyfun\\left(alphai\\right)=betai \\) for \\( indexvar=2,3, \\ldots, quantity \\).\nA similar argument using the determinant for \\( polyque \\) shows that \\( polyque\\left(alphai\\right)= phifunc\\left(betai\\right) \\) for \\( indexvar=1,2, \\ldots, quantity \\). Moreover, it is evident that \\( polyque \\) is a polynomial of degree less than \\( quantity \\).\n\nSuppose the polynomial \\( phifunc(polyfun(abscissa)) \\) is divided by \\( \\left(abscissa-alphaone\\right)\\left(abscissa-alphatwo\\right) \\cdots\\left(abscissa-alphan\\right) \\) leaving a remainder \\( remsix(abscissa) \\) of degree less than \\( quantity \\). Then\n\\[\nphifunc(polyfun(abscissa))=\\left(abscissa-alphaone\\right)\\left(abscissa-alphatwo\\right) \\cdots\\left(abscissa-alphan\\right) polysix(abscissa)+remsix(abscissa),\n\\]\nwhere \\( polysix(abscissa) \\) is a polynomial. Substituting \\( abscissa=alphai \\), we get\n\\[\nremsix\\left(alphai\\right)=phifunc\\left(polyfun\\left(alphai\\right)\\right)=phifunc\\left(betai\\right)=polyque\\left(alphai\\right) .\n\\]\n\nThus the polynomial \\( polyque(abscissa)-remsix(abscissa) \\) is zero for \\( quantity \\) distinct values of \\( abscissa \\). But its degree is less than \\( quantity \\), so \\( polyque(abscissa)-remsix(abscissa) \\) is identically zero, and the remainder is \\( polyque(abscissa) \\), as required.\n\nRemark. The determinant given provides a useful expression for the unique polynomial of degree less than \\( quantity \\) with values prescribed at \\( alphaone, alphatwo \\),\n\\( \\ldots, alphan \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pinecone",
        "y": "sunflower",
        "t": "wilderness",
        "f": "butterfly",
        "i": "drizzling",
        "m": "evergreen",
        "n": "wavelength",
        "a_1": "redwoods",
        "a_2": "bluebird",
        "a_n": "shoreside",
        "a_i": "tangerine",
        "b_1": "porcupine",
        "b_2": "sandalwood",
        "b_n": "stoneware",
        "b_i": "plankton",
        "P": "lighthouse",
        "Q": "greenhouse",
        "S": "harmonica",
        "R": "springbok",
        "V": "stormwind",
        "\\phi": "fireplace"
      },
      "question": "14. Take either (i) or (ii).\n(i) Show that any solution \\( butterfly(wilderness) \\) of the functional equation\n\\[\nbutterfly(pinecone+sunflower) butterfly(pinecone-sunflower)=butterfly(pinecone) butterfly(pinecone)+butterfly(sunflower) butterfly(sunflower)-1, \\quad(pinecone, sunflower, \\text { real })\n\\]\nis such that\n\\[\nbutterfly^{\\prime \\prime}(wilderness)= \\pm evergreen^{2} butterfly(wilderness), \\quad(evergreen \\text { constant and } \\geq 0)\n\\]\nassuming the existence and continuity of the second derivative. Deduce that \\( butterfly(wilderness) \\) is one of the functions\n\\[\n\\pm \\cos evergreen wilderness, \\quad \\pm \\cosh evergreen wilderness\n\\]\n(ii) With wavelength constant values redwoods, bluebird, \\ldots, shoreside, supposed all different, let wavelength constant values porcupine, sandalwood, \\ldots, stoneware be associated, and let a polynomial lighthouse(pinecone) be. defined by the identity in \\( \\boldsymbol{pinecone} \\)\n\\[\n\\left|\\begin{array}{cccccc}\n1 & pinecone & pinecone^{2} & \\cdots & pinecone^{wavelength-1} & lighthouse(pinecone) \\\\\n1 & redwoods & redwoods^{2} & \\cdots & redwoods^{wavelength-1} & porcupine \\\\\n1 & bluebird & bluebird^{2} & \\cdots & bluebird^{wavelength-1} & sandalwood \\\\\n\\cdots & \\cdots & \\ldots & \\ldots & \\cdots & \\cdots\n\\end{array}\\right| \\equiv 0 . \\cdots \\cdots .\n\\]\n\nGiven a polynomial fireplace(wilderness), let a polynomial greenhouse(pinecone) be defined by the identity in pinecone obtained on replacing lighthouse(pinecone), porcupine, sandalwood, \\ldots, stoneware of the identity above by greenhouse(pinecone), fireplace(porcupine), fireplace(sandalwood), \\ldots ; fireplace(stoneware). Prove that the remainder obtained on dividing fireplace(lighthouse(pinecone)) by \\( \\left(pinecone-redwoods\\right)\\left(pinecone-bluebird\\right) \\cdots\\left(pinecone-shoreside\\right) \\) is greenhouse(pinecone).",
      "solution": "Solution. Starting with the given functional equation\n\\[\nbutterfly(pinecone+sunflower) butterfly(pinecone-sunflower)=butterfly(pinecone)^{2}+butterfly(sunflower)^{2}-1,\n\\]\nwe differentiate with respect to \\( sunflower \\) to get\n\\[\nbutterfly^{\\prime}(pinecone+sunflower) butterfly(pinecone-sunflower)-butterfly(pinecone+sunflower) butterfly^{\\prime}(pinecone-sunflower)=2 butterfly(sunflower) butterfly^{\\prime}(sunflower),\n\\]\nand then with respect to \\( \\boldsymbol{pinecone} \\) to get\n\\[\nbutterfly^{\\prime \\prime}(pinecone+sunflower) butterfly(pinecone-sunflower)-butterfly(pinecone+sunflower) butterfly^{\\prime \\prime}(pinecone-sunflower)=0 .\n\\]\n\nSetting \\( pinecone=sunflower=0 \\) in (1) and (2), we obtain \\( butterfly(0)^{2}=2 butterfly(0)^{2}-1 \\) and \\( 2 butterfly(0) butterfly^{\\prime}(0)=0 \\), whence\n\\[\nbutterfly(0)= \\pm 1, \\quad butterfly^{\\prime}(0)=0 .\n\\]\n\nNow for any given number wilderness, put \\( pinecone=sunflower=wilderness / 2 \\) in (3) to get\n\\[\nbutterfly^{\\prime \\prime}(wilderness) butterfly(0)-butterfly(wilderness) butterfly^{\\prime \\prime}(0)=0,\n\\]\nwhich is equivalent to\n\\[\nbutterfly^{\\prime \\prime}(wilderness) \\pm evergreen^{2} butterfly(wilderness)=0,\n\\]\nwhere \\( evergreen=\\left|butterfly^{\\prime \\prime}(0)\\right|^{1 / 2} \\).\nIntegrating (5) using the initial conditions (4), we obtain\n\\[\nbutterfly(wilderness)= \\pm \\cos evergreen wilderness \\text { or } butterfly(wilderness)= \\pm \\cosh evergreen wilderness,\n\\]\ndepending on whether the \\( \\operatorname{sign} \\) in (5) is plus or minus. If \\( evergreen=0 \\), these solutions are constant.\n\nConversely, either of the above solutions satisfies the given functional equation for any value of evergreen.\n\nRemark. The continuity of the second derivative was not used in this proof. In fact, the result can be proved under far weaker hypotheses; for example, it is enough to assume that butterfly itself is continuous. See J. Aczèl, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.\n\nSolution. We use the fact that the Vandermonde determinant\n\\[\nstormwind=\\left|\\begin{array}{ccccc}\n1 & redwoods & redwoods^{2} & \\ldots & redwoods^{wavelength-1} \\\\\n1 & bluebird & bluebird^{2} & \\ldots & bluebird^{wavelength-1} \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n1 & shoreside & shoreside^{2} & \\ldots & shoreside^{wavelength-1}\n\\end{array}\\right|=\\prod_{drizzling<j}\\left(tangerine-a_{j}\\right)\n\\]\nis not zero.\nIf we replace \\( pinecone \\) by redwoods in the determinant defining lighthouse(pinecone), subtract the second row from the first, and expand by minors of the first row, we see that\n\\[\nstormwind\\left(lighthouse(redwoods)-porcupine\\right)=0 .\n\\]\n\nHence, \\( lighthouse(redwoods)=porcupine \\). Similarly, \\( lighthouse(bluebird)=sandalwood \\) for \\( drizzling=2,3, \\ldots, wavelength \\).\nA similar argument using the determinant for greenhouse shows that \\( greenhouse(tangerine)= fireplace(porcupine) \\) for \\( drizzling=1,2, \\ldots, wavelength \\). Moreover, it is evident that greenhouse is a polynomial of degree less than wavelength.\n\nSuppose the polynomial fireplace(lighthouse(pinecone)) is divided by \\( \\left(pinecone-redwoods\\right)\\left(pinecone-bluebird\\right) \\cdots\\left(pinecone-shoreside\\right) \\) leaving a remainder springbok(pinecone) of degree less than wavelength. Then\n\\[\nfireplace(lighthouse(pinecone))=\\left(pinecone-redwoods\\right)\\left(pinecone-bluebird\\right) \\cdots\\left(pinecone-shoreside\\right) harmonica(pinecone)+springbok(pinecone),\n\\]\nwhere harmonica(pinecone) is a polynomial. Substituting \\( pinecone=redwoods \\), we get\n\\[\nspringbok(redwoods)=fireplace(lighthouse(redwoods))=fireplace(porcupine)=greenhouse(redwoods) .\n\\]\n\nThus the polynomial greenhouse(pinecone)-springbok(pinecone) is zero for wavelength distinct values of pinecone. But its degree is less than wavelength, so greenhouse(pinecone)-springbok(pinecone) is identically zero, and the remainder is greenhouse(pinecone), as required.\n\nRemark. The determinant given provides a useful expression for the unique polynomial of degree less than wavelength with values prescribed at redwoods, bluebird, \\ldots, shoreside ."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "y": "steadyvar",
        "t": "timeless",
        "f": "constant",
        "i": "wholeindex",
        "m": "changing",
        "n": "infinite",
        "a_1": "zetaone",
        "a_2": "zetatwoo",
        "a_n": "zetanext",
        "a_i": "zetaiota",
        "b_1": "alphaone",
        "b_2": "alphatwo",
        "b_n": "alphanext",
        "b_i": "alphaiota",
        "P": "monomial",
        "Q": "singular",
        "S": "nullpoly",
        "R": "quotient",
        "V": "scalarval",
        "\\\\phi": "identity"
      },
      "question": "14. Take either (i) or (ii).\n(i) Show that any solution \\( constant(timeless) \\) of the functional equation\n\\[\nconstant(fixedvalue+steadyvar) constant(fixedvalue-steadyvar)=constant(fixedvalue) constant(fixedvalue)+constant(steadyvar) constant(steadyvar)-1, \\quad(fixedvalue, steadyvar, \\text { real })\n\\]\nis such that\n\\[\nconstant^{\\prime \\prime}(timeless)= \\pm changing^{2} constant(timeless), \\quad(changing \\text { constant and } \\geq 0)\n\\]\nassuming the existence and continuity of the second derivative. Deduce that \\( constant(timeless) \\) is one of the functions\n\\[\n\\pm \\cos changing\\ timeless, \\quad \\pm \\cosh changing\\ timeless\n\\]\n(ii) With \\( infinite \\) constant values \\( zetaone, zetatwoo, \\ldots, zetanext \\), supposed all different, let \\( infinite \\) constant values \\( alphaone, alphatwo, \\ldots, alphanext \\) be associated, and let a polynomial \\( monomial(fixedvalue) \\) be. defined by the identity in \\( \\boldsymbol{fixedvalue} \\)\n\\[\n\\left|\\begin{array}{cccccc}\n1 & fixedvalue & fixedvalue^{2} & \\cdots & fixedvalue^{infinite-1} & monomial(fixedvalue) \\\\\n1 & zetaone & zetaone^{2} & \\cdots & zetaone{ }^{infinite-1} & alphaone \\\\\n1 & zetatwoo & zetatwoo^{2} & \\cdots & zetatwoo{ }^{infinite-1} & alphatwo \\\\\n\\cdots & \\cdots & \\ldots & \\ldots & \\cdots & \\cdots\n\\end{array}\\right| \\equiv 0 . \\cdots \\cdots .\n\\]\n\nGiven a polynomial \\( identity(timeless) \\), let a polynomial \\( singular(fixedvalue) \\) be defined by the identity in \\( fixedvalue \\) obtained on replacing \\( monomial(fixedvalue), alphaone, alphatwo, \\ldots, alphanext \\) of the identity above by \\( singular(fixedvalue) \\), \\( identity\\left(alphaone\\right), identity\\left(alphatwo\\right), \\ldots ; identity\\left(alphanext\\right) \\). Prove that the remainder obtained on dividing \\( identity(monomial(fixedvalue)) \\) by \\( \\left(fixedvalue-zetaone\\right)\\left(fixedvalue-zetatwoo\\right) \\cdots\\left(fixedvalue-zetanext\\right) \\) is \\( singular(fixedvalue) \\).",
      "solution": "Solution. Starting with the given functional equation\n\\[\nconstant(fixedvalue+steadyvar) constant(fixedvalue-steadyvar)=constant(fixedvalue)^{2}+constant(steadyvar)^{2}-1,\n\\]\nwe differentiate with respect to \\( steadyvar \\) to get\n\\[\nconstant^{\\prime}(fixedvalue+steadyvar) constant(fixedvalue-steadyvar)-constant(fixedvalue+steadyvar) constant^{\\prime}(fixedvalue-steadyvar)=2 constant(steadyvar) constant^{\\prime}(steadyvar),\n\\]\nand then with respect to \\( \\boldsymbol{fixedvalue} \\) to get\n\\[\nconstant^{\\prime \\prime}(fixedvalue+steadyvar) constant(fixedvalue-steadyvar)-constant(fixedvalue+steadyvar) constant^{\\prime \\prime}(fixedvalue-steadyvar)=0 .\n\\]\n\nSetting \\( fixedvalue=steadyvar=0 \\) in (1) and (2), we obtain \\( constant(0)^{2}=2 constant(0)^{2}-1 \\) and \\( 2 constant(0) constant^{\\prime}(0)=0 \\), whence\n\\[\nconstant(0)= \\pm 1, \\quad constant^{\\prime}(0)=0 .\n\\]\n\nNow for any given number \\( timeless \\), put \\( fixedvalue=steadyvar=timeless / 2 \\) in (3) to get\n\\[\nconstant^{\\prime \\prime}(timeless) constant(0)-constant(timeless) constant^{\\prime \\prime}(0)=0,\n\\]\nwhich is equivalent to\n\\[\nconstant^{\\prime \\prime}(timeless) \\pm changing^{2} constant(timeless)=0,\n\\]\nwhere \\( changing=\\left|constant^{\\prime \\prime}(0)\\right|^{1 / 2} \\).\nIntegrating (5) using the initial conditions (4), we obtain\n\\[\nconstant(timeless)= \\pm \\cos changing\\ timeless \\text { or } constant(timeless)= \\pm \\cosh changing\\ timeless,\n\\]\ndepending on whether the \\( \\operatorname{sign} \\) in (5) is plus or minus. If \\( \\boldsymbol{changing}=0 \\), these solutions are constant.\n\nConversely, either of the above solutions satisfies the given functional equation for any value of \\( changing \\).\n\nRemark. The continuity of the second derivative was not used in this proof. In fact, the result can be proved under far weaker hypotheses; for example, it is enough to assume that \\( constant \\) itself is continuous. See J. Aczèl, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.\n\nSolution. We use the fact that the Vandermonde determinant\n\\[\nscalarval=\\left|\\begin{array}{ccccc}\n1 & zetaone & zetaone{ }^{2} & \\ldots & zetaone{ }^{infinite-1} \\\\\n1 & zetatwoo & zetatwoo{ }^{2} & \\ldots & zetatwoo^{infinite-1} \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n1 & zetanext & zetanext{ }^{2} & \\ldots & zetanext{ }^{infinite-1}\n\\end{array}\\right|=\\prod_{wholeindex j j}\\left(zetaiota-a_{j}\\right)\n\\]\nis not zero.\nIf we replace \\( fixedvalue \\) by \\( zetaone \\) in the determinant defining \\( monomial(fixedvalue) \\), subtract the second row from the first, and expand by minors of the first row, we see that\n\\[\nscalarval\\left(monomial\\left(zetaone\\right)-alphaone\\right)=0 .\n\\]\n\nHence, \\( monomial\\left(zetaone\\right)=alphaone \\). Similarly, \\( monomial\\left(zetatwoo\\right)=alphatwo \\) for \\( wholeindex=2,3, \\ldots, infinite \\).\nA similar argument using the determinant for \\( singular \\) shows that \\( singular\\left(zetaiota\\right)= \\) \\( identity\\left(alphaiota\\right) \\) for \\( wholeindex=1,2, \\ldots, infinite \\). Moreover, it is evident that \\( singular \\) is a polynomial of degree less than \\( infinite \\).\n\nSuppose the polynomial \\( identity(monomial(fixedvalue)) \\) is divided by \\( \\left(fixedvalue-zetaone\\right)\\left(fixedvalue-zetatwoo\\right) \\cdots\\left(fixedvalue-zetanext\\right) \\) leaving a remainder \\( quotient(fixedvalue) \\) of degree less than \\( infinite \\). Then\n\\[\nidentity(monomial(fixedvalue))=\\left(fixedvalue-zetaone\\right)\\left(fixedvalue-zetatwoo\\right) \\cdots\\left(fixedvalue-zetanext\\right) nullpoly(fixedvalue)+quotient(fixedvalue),\n\\]\nwhere \\( nullpoly(fixedvalue) \\) is a polynomial. Substituting \\( fixedvalue=zetaiota \\), we get\n\\[\nquotient\\left(zetaiota\\right)=identity\\left(monomial\\left(zetaiota\\right)\\right)=identity\\left(alphaiota\\right)=singular\\left(zetaiota\\right) .\n\\]\n\nThus the polynomial \\( singular(fixedvalue)-quotient(fixedvalue) \\) is zero for \\( infinite \\) distinct values of \\( fixedvalue \\). But its degree is less than \\( infinite \\), so \\( singular(fixedvalue)-quotient(fixedvalue) \\) is identically zero, and the remainder is \\( singular(fixedvalue) \\), as required.\n\nRemark. The determinant given provides a useful expression for the unique polynomial of degree less than \\( infinite \\) with values prescribed at \\( zetaone, zetatwoo \\),\n\\( \\ldots, zetanext \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mkdfoepr",
        "f": "vnqtcxla",
        "i": "slodmeka",
        "m": "vjprkzle",
        "n": "xadovbqi",
        "a_1": "ohgnfpew",
        "a_2": "xtmkraqs",
        "a_n": "cdarpzoi",
        "a_i": "wqbnseth",
        "b_1": "lxrocmve",
        "b_2": "zwhflsqa",
        "b_n": "uyptleno",
        "b_i": "nadviqrs",
        "P": "aqmibdje",
        "Q": "ltsrhkow",
        "S": "fcyvpdjn",
        "R": "bzgumlin",
        "V": "kxhatrye",
        "\\phi": "\\melqzyst"
      },
      "question": "14. Take either (i) or (ii).\n(i) Show that any solution \\( vnqtcxla(mkdfoepr) \\) of the functional equation\n\\[\nvnqtcxla(qzxwvtnp+hjgrksla) \\, vnqtcxla(qzxwvtnp-hjgrksla)=vnqtcxla(qzxwvtnp) \\, vnqtcxla(qzxwvtnp)+vnqtcxla(hjgrksla) \\, vnqtcxla(hjgrksla)-1, \\quad(qzxwvtnp, hjgrksla, \\text { real })\n\\]\nis such that\n\\[\nvnqtcxla^{\\prime \\prime}(mkdfoepr)= \\pm vjprkzle^{2} \\, vnqtcxla(mkdfoepr), \\quad(vjprkzle \\text { constant and } \\geq 0)\n\\]\nassuming the existence and continuity of the second derivative. Deduce that \\( vnqtcxla(mkdfoepr) \\) is one of the functions\n\\[\n\\pm \\cos vjprkzle \\, mkdfoepr, \\quad \\pm \\cosh vjprkzle \\, mkdfoepr\n\\]\n(ii) With \\( xadovbqi \\) constant values \\( ohgnfpew, xtmkraqs, \\ldots, cdarpzoi \\), supposed all different, let \\( xadovbqi \\) constant values \\( lxrocmve, zwhflsqa, \\ldots, uyptleno \\) be associated, and let a polynomial \\( aqmibdje(qzxwvtnp) \\) be defined by the identity in \\( \\boldsymbol{qzxwvtnp} \\)\n\\[\n\\left|\\begin{array}{cccccc}\n1 & qzxwvtnp & qzxwvtnp^{2} & \\cdots & qzxwvtnp^{xadovbqi-1} & aqmibdje(qzxwvtnp) \\\\\n1 & ohgnfpew & ohgnfpew^{2} & \\cdots & ohgnfpew^{xadovbqi-1} & lxrocmve \\\\\n1 & xtmkraqs & xtmkraqs^{2} & \\cdots & xtmkraqs^{xadovbqi-1} & zwhflsqa \\\\\n\\cdots & \\cdots & \\ldots & \\ldots & \\cdots & \\cdots\n\\end{array}\\right| \\equiv 0 . \\cdots \\cdots .\n\\]\n\nGiven a polynomial \\( \\melqzyst(mkdfoepr) \\), let a polynomial \\( ltsrhkow(qzxwvtnp) \\) be defined by the identity in \\( qzxwvtnp \\) obtained on replacing \\( aqmibdje(qzxwvtnp), lxrocmve, zwhflsqa, \\ldots, uyptleno \\) of the identity above by \\( ltsrhkow(qzxwvtnp) \\), \\( \\melqzyst\\left(lxrocmve\\right) \\), \\( \\melqzyst\\left(zwhflsqa\\right) \\), \\ldots ; \\( \\melqzyst\\left(uyptleno\\right) \\). Prove that the remainder obtained on dividing \\( \\melqzyst(aqmibdje(qzxwvtnp)) \\) by \\( \\left(qzxwvtnp-ohgnfpew\\right)\\left(qzxwvtnp-xtmkraqs\\right) \\cdots\\left(qzxwvtnp-cdarpzoi\\right) \\) is \\( ltsrhkow(qzxwvtnp) \\).",
      "solution": "Solution. Starting with the given functional equation\n\\[\nvnqtcxla(qzxwvtnp+hjgrksla) \\, vnqtcxla(qzxwvtnp-hjgrksla)=vnqtcxla(qzxwvtnp)^{2}+vnqtcxla(hjgrksla)^{2}-1,\n\\]\nwe differentiate with respect to \\( hjgrksla \\) to get\n\\[\nvnqtcxla^{\\prime}(qzxwvtnp+hjgrksla) \\, vnqtcxla(qzxwvtnp-hjgrksla)-vnqtcxla(qzxwvtnp+hjgrksla) \\, vnqtcxla^{\\prime}(qzxwvtnp-hjgrksla)=2 \\, vnqtcxla(hjgrksla) \\, vnqtcxla^{\\prime}(hjgrksla),\n\\]\nand then with respect to \\( \\boldsymbol{qzxwvtnp} \\) to get\n\\[\nvnqtcxla^{\\prime \\prime}(qzxwvtnp+hjgrksla) \\, vnqtcxla(qzxwvtnp-hjgrksla)-vnqtcxla(qzxwvtnp+hjgrksla) \\, vnqtcxla^{\\prime \\prime}(qzxwvtnp-hjgrksla)=0 .\n\\]\n\nSetting \\( qzxwvtnp=hjgrksla=0 \\) in (1) and (2), we obtain \\( vnqtcxla(0)^{2}=2 vnqtcxla(0)^{2}-1 \\) and \\( 2 vnqtcxla(0) vnqtcxla^{\\prime}(0)=0 \\), whence\n\\[\nvnqtcxla(0)= \\pm 1, \\quad vnqtcxla^{\\prime}(0)=0 .\n\\]\n\nNow for any given number \\( mkdfoepr \\), put \\( qzxwvtnp=hjgrksla=mkdfoepr / 2 \\) in (3) to get\n\\[\nvnqtcxla^{\\prime \\prime}(mkdfoepr) \\, vnqtcxla(0)-vnqtcxla(mkdfoepr) \\, vnqtcxla^{\\prime \\prime}(0)=0,\n\\]\nwhich is equivalent to\n\\[\nvnqtcxla^{\\prime \\prime}(mkdfoepr) \\pm vjprkzle^{2} \\, vnqtcxla(mkdfoepr)=0,\n\\]\nwhere \\( vjprkzle=\\left|vnqtcxla^{\\prime \\prime}(0)\\right|^{1 / 2} \\).\nIntegrating (5) using the initial conditions (4), we obtain\n\\[\nvnqtcxla(mkdfoepr)= \\pm \\cos vjprkzle \\, mkdfoepr \\text { or } vnqtcxla(mkdfoepr)= \\pm \\cosh vjprkzle \\, mkdfoepr,\n\\]\ndepending on whether the \\( \\operatorname{sign} \\) in (5) is plus or minus. If \\( vjprkzle=0 \\), these solutions are constant.\n\nConversely, either of the above solutions satisfies the given functional equation for any value of \\( vjprkzle \\).\n\nRemark. The continuity of the second derivative was not used in this proof. In fact, the result can be proved under far weaker hypotheses; for example, it is enough to assume that \\( vnqtcxla \\) itself is continuous. See J. Aczèl, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.\n\nSolution. We use the fact that the Vandermonde determinant\n\\[\nkxhatrye=\\left|\\begin{array}{ccccc}\n1 & ohgnfpew & ohgnfpew^{2} & \\ldots & ohgnfpew^{xadovbqi-1} \\\\\n1 & xtmkraqs & xtmkraqs^{2} & \\ldots & xtmkraqs^{xadovbqi-1} \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\ldots & \\cdot \\\\\n1 & cdarpzoi & cdarpzoi^{2} & \\ldots & cdarpzoi^{xadovbqi-1}\n\\end{array}\\right|=\\prod_{ slodmeka j j}\\left(wqbnseth-a_{j}\\right)\n\\]\nis not zero.\nIf we replace \\( qzxwvtnp \\) by \\( ohgnfpew \\) in the determinant defining \\( aqmibdje(qzxwvtnp) \\), subtract the second row from the first, and expand by minors of the first row, we see that\n\\[\nkxhatrye\\left(aqmibdje\\left(ohgnfpew\\right)-lxrocmve\\right)=0 .\n\\]\n\nHence, \\( aqmibdje\\left(ohgnfpew\\right)=lxrocmve \\). Similarly, \\( aqmibdje\\left(xtmkraqs\\right)=zwhflsqa \\) for \\( slodmeka=2,3, \\ldots, xadovbqi \\).\nA similar argument using the determinant for \\( ltsrhkow \\) shows that \\( ltsrhkow\\left(wqbnseth\\right)= \\melqzyst\\left(nadviqrs\\right) \\) for \\( slodmeka=1,2, \\ldots, xadovbqi \\). Moreover, it is evident that \\( ltsrhkow \\) is a polynomial of degree less than \\( xadovbqi \\).\n\nSuppose the polynomial \\( \\melqzyst(aqmibdje(qzxwvtnp)) \\) is divided by \\( \\left(qzxwvtnp-ohgnfpew\\right)\\left(qzxwvtnp-xtmkraqs\\right) \\cdots\\left(qzxwvtnp-cdarpzoi\\right) \\) leaving a remainder \\( bzgumlin(qzxwvtnp) \\) of degree less than \\( xadovbqi \\). Then\n\\[\n\\melqzyst(aqmibdje(qzxwvtnp))=\\left(qzxwvtnp-ohgnfpew\\right)\\left(qzxwvtnp-xtmkraqs\\right) \\cdots\\left(qzxwvtnp-cdarpzoi\\right) fcyvpdjn(qzxwvtnp)+bzgumlin(qzxwvtnp),\n\\]\nwhere \\( fcyvpdjn(qzxwvtnp) \\) is a polynomial. Substituting \\( qzxwvtnp=ohgnfpew \\), we get\n\\[\nbzgumlin\\left(ohgnfpew\\right)=\\melqzyst\\left(aqmibdje\\left(ohgnfpew\\right)\\right)=\\melqzyst\\left(lxrocmve\\right)=ltsrhkow\\left(ohgnfpew\\right) .\n\\]\n\nThus the polynomial \\( ltsrhkow(qzxwvtnp)-bzgumlin(qzxwvtnp) \\) is zero for \\( xadovbqi \\) distinct values of \\( qzxwvtnp \\). But its degree is less than \\( xadovbqi \\), so \\( ltsrhkow(qzxwvtnp)-bzgumlin(qzxwvtnp) \\) is identically zero, and the remainder is \\( ltsrhkow(qzxwvtnp) \\), as required.\n\nRemark. The determinant given provides a useful expression for the unique polynomial of degree less than \\( xadovbqi \\) with values prescribed at \\( ohgnfpew, xtmkraqs, \\ldots, cdarpzoi \\)."
    },
    "kernel_variant": {
      "question": "14″  (choose either part)\n\n(A) (Functional-equation section – parameter version)  \nLet a,b be fixed real constants, not both zero.  A twice–differentiable function f:ℝ→ℝ satisfies  \n  f(x+y) f(x−y)=a[ f(x)^2+f(y)^2 ]+b  for every x,y∈ℝ.  (1)\n\n(i) Prove that if f is non–constant then necessarily a=1 and b=−1.  \n(ii) Under that constraint show that there is a number μ≥0 for which  \n  f''(t)=±μ^2f(t)  (t∈ℝ),  (2)  \nand hence list all solutions of (1).\n\n(B) (Polynomial–remainder section – higher-multiplicity nodes)  \nFix k≥1 distinct real numbers c_1,…,c_k and positive integers s_1,…,s_k.  \nFor every i and every 0≤j≤s_i−1 prescribe real numbers d_{i,j}.  Let P be the polynomial determined by the identically–zero determinant  \n\n  | 1   x   x^2 … x^m  P(x)                     |  \n  | 1 c_1 c_1^2…c_1^m d_{1,0}                  |  \n  |        ⋱             ⋮                     |  \n  |1 c_i c_i^2…c_i^m d_{i,s_i−1}(row repeated s_i times)|  \n  |        ⋱             ⋮                     |  \n\nwhere m=s_1+⋯+s_k−1 and each block of s_i identical first (m+1) entries is followed by the s_i different data d_{i,0},…,d_{i,s_i−1}.  \nGiven any polynomial ψ, form the analogous determinant in which the last column is replaced by  \n\n  ψ(d_{1,0}),…,ψ(d_{1,s_1−1}), …, ψ(d_{k,0}),…,ψ(d_{k,s_k−1}), Q(x),\n\nthereby defining a polynomial Q(x).  Prove that, when ψ∘P is divided by ∏_{i=1}^{k}(x−c_i)^{s_i}, the remainder equals Q(x).\n\n––––––––––––––––––––––––––––––––––––––––",
      "solution": "(we treat part (A))\n\nStep 1.  Elementary consequences of (1).  \nPut y=0 to obtain  \n  f(x)^2=a[ f(x)^2+f(0)^2 ]+b.  (3)  \nRearranging gives (1−a)f(x)^2= a f(0)^2+b, so the left–hand side is x-independent.  \nHence either  \n (a) 1−a≠0, forcing f(x)^2 to be constant, or  \n (b) a=1.\n\nStep 2.  The case 1−a≠0.  \nFrom (3) we find f(x)^2≡K with K=(a f(0)^2+b)/(1−a).  \nBecause f is continuous its sign cannot vary, so f≡±√K.  \nPutting this constant back into (1) yields  \n  K = a(2K)+b  ⇒  (1−2a)K=b.  \nIf 1−2a ≠0 we obtain a unique K and therefore only constant solutions.  \nThus any non–constant solution can exist only when a=1.  This proves (i) and shows also that b must satisfy (see next step) b=−1.\n\nStep 3.  Determining b once a=1.  \nWith a=1, (1) becomes  \n  f(x+y)f(x−y)=f(x)^2+f(y)^2+b.  (4)  \nSet x=y to get  \n  f(2x)f(0)=2f(x)^2+b.  (5)  \nTaking x=0 in (5) yields f(0)^2=2f(0)^2+b, whence b=−f(0)^2.  \nEquation (5) with this value simplifies to  \n  f(2x)f(0)=2f(x)^2−f(0)^2.  (6)\n\nIf f(0)=0, then b=0 and (6) gives f(2x)=2f(x)^2; differentiating at x=0 gives f'(0)=0.  Repeating the classical argument (differentiate (4) twice, then set x=y=t/2) one obtains f''(t)=μ^2f(t) with μ=|f''(0)|^{1/2}.  \nIf f(0)≠0 we may normalise g(t)=f(t)/f(0).  Then b=−1 and g obeys the original functional equation with parameters (a,b)=(1,−1).  Applying the differentiation procedure of the classical cosine–hyperbolic–cosine proof we again reach (2).\n\nStep 4.  Deriving the differential equation.  \nDifferentiate (4) with respect to y, then x, and put x=y=t/2 exactly as in the standard proof:  \n\n  f''(t) f(0)−f(t) f''(0)=0.  \n\nIf f''(0)=−μ^2 f(0) we get f''(t)=−μ^2f(t); if f''(0)=+μ^2 f(0) we get f''(t)=+μ^2f(t).  In either case μ≥0 and (2) holds.\n\nStep 5.  Solving (2) with the initial data f(0)=σ (σ=0 or ±1) and f'(0)=0 gives  \n\n  f(t)=σ cos μt,  or  f(t)=σ cosh μt,  \n\nwhile μ=0 collapses to the constant solutions σ.  Each of these satisfies (4) with b=−1, completing the classification.\n\nTherefore every twice–differentiable non–constant solution of (1) forces (a,b)=(1,−1) and is of the displayed cosine / hyperbolic-cosine type; all remaining parameter pairs admit only constant solutions, confirming the statement.\n\n––––––––––––––––––––––––––––––––––––––––",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.005510",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}