{
  "index": "1970-A-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-1. Show that the power series for the function\n\\[\ne^{a x} \\cos b x \\quad(a>0, b>0)\n\\]\nin powers of \\( x \\) has either no zero coefficients or infinitely many zero coefficients.",
  "solution": "A-1 Note that \\( e^{a x} \\cos b x \\) is the real part of \\( e^{(a+i b) x} \\). Thus the power series is\n\\[\ne^{a x} \\cos b x=\\sum_{n=0}^{\\infty} \\operatorname{Re}\\left\\{(a+i b)^{n}\\right\\} \\frac{x^{n}}{n!} .\n\\]\n\nIn this form, it is easily seen that if \\( x^{n} \\) has a zero coefficient, then \\( x^{k n} \\) has a zero coefficient for every odd value of \\( k \\).",
  "vars": [
    "x",
    "n",
    "k"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "n": "counter",
        "k": "integer",
        "a": "positive",
        "b": "parameter"
      },
      "question": "A-1. Show that the power series for the function\n\\[\ne^{positive variable} \\cos parameter variable \\quad(positive>0, parameter>0)\n\\]\nin powers of \\( variable \\) has either no zero coefficients or infinitely many zero coefficients.",
      "solution": "A-1 Note that \\( e^{positive variable} \\cos parameter variable \\) is the real part of \\( e^{(positive+i parameter) variable} \\). Thus the power series is\n\\[\ne^{positive variable} \\cos parameter variable=\\sum_{counter=0}^{\\infty} \\operatorname{Re}\\left\\{(positive+i parameter)^{counter}\\right\\} \\frac{variable^{counter}}{counter!} .\n\\]\n\nIn this form, it is easily seen that if \\( variable^{counter} \\) has a zero coefficient, then \\( variable^{integer counter} \\) has a zero coefficient for every odd value of \\( integer \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "companion",
        "n": "architecture",
        "k": "backpack",
        "a": "waterfall",
        "b": "sunflower"
      },
      "question": "A-1. Show that the power series for the function\n\\[\ne^{waterfall companion} \\cos sunflower companion \\quad(waterfall>0, sunflower>0)\n\\]\nin powers of \\( companion \\) has either no zero coefficients or infinitely many zero coefficients.",
      "solution": "A-1 Note that \\( e^{waterfall companion} \\cos sunflower companion \\) is the real part of \\( e^{(waterfall+i sunflower) companion} \\). Thus the power series is\n\\[\ne^{waterfall companion} \\cos sunflower companion=\\sum_{architecture=0}^{\\infty} \\operatorname{Re}\\left\\{(waterfall+i sunflower)^{architecture}\\right\\} \\frac{companion^{architecture}}{architecture!} .\n\\]\n\nIn this form, it is easily seen that if \\( companion^{architecture} \\) has a zero coefficient, then \\( companion^{backpack architecture} \\) has a zero coefficient for every odd value of \\( backpack \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "n": "continuous",
        "k": "fractional",
        "a": "negativeval",
        "b": "stationary"
      },
      "question": "A-1. Show that the power series for the function\n\\[\ne^{negativeval constantval} \\cos stationary constantval \\quad(negativeval>0, stationary>0)\n\\]\nin powers of \\( constantval \\) has either no zero coefficients or infinitely many zero coefficients.",
      "solution": "A-1 Note that \\( e^{negativeval constantval} \\cos stationary constantval \\) is the real part of \\( e^{(negativeval+i stationary) constantval} \\). Thus the power series is\n\\[\ne^{negativeval constantval} \\cos stationary constantval=\\sum_{continuous=0}^{\\infty} \\operatorname{Re}\\left\\{(negativeval+i stationary)^{continuous}\\right\\} \\frac{constantval^{continuous}}{continuous!} .\n\\]\n\nIn this form, it is easily seen that if \\( constantval^{continuous} \\) has a zero coefficient, then \\( constantval^{fractional continuous} \\) has a zero coefficient for every odd value of \\( fractional \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "k": "bvlpsezm",
        "a": "rpqdgnfz",
        "b": "slhmgxtr"
      },
      "question": "A-1. Show that the power series for the function\n\\[\ne^{rpqdgnfz qzxwvtnp} \\cos slhmgxtr qzxwvtnp \\quad(rpqdgnfz>0, slhmgxtr>0)\n\\]\nin powers of \\( qzxwvtnp \\) has either no zero coefficients or infinitely many zero coefficients.",
      "solution": "A-1 Note that \\( e^{rpqdgnfz qzxwvtnp} \\cos slhmgxtr qzxwvtnp \\) is the real part of \\( e^{(rpqdgnfz+i slhmgxtr) qzxwvtnp} \\). Thus the power series is\n\\[\ne^{rpqdgnfz qzxwvtnp} \\cos slhmgxtr qzxwvtnp=\\sum_{hjgrksla=0}^{\\infty} \\operatorname{Re}\\left\\{(rpqdgnfz+i slhmgxtr)^{hjgrksla}\\right\\} \\frac{qzxwvtnp^{hjgrksla}}{hjgrksla!} .\n\\]\n\nIn this form, it is easily seen that if \\( qzxwvtnp^{hjgrksla} \\) has a zero coefficient, then \\( qzxwvtnp^{bvlpsezm hjgrksla} \\) has a zero coefficient for every odd value of \\( bvlpsezm \\)."
    },
    "kernel_variant": {
      "question": "Let a and b be real numbers with b ≠ 0 and consider the Maclaurin expansion\n\ne^{ax}\\,\\sin (bx)=\\sum_{n=0}^{\\infty}c_n x^{n}.\n\n(1)  Show that c₀ = 0.\n\n(2)  Prove that, apart from this constant term, the sequence of coefficients either contains no further zeros or contains infinitely many of them.  Equivalently,\n\na)  either c_n ≠ 0 for every n ≥ 1, or\n\nb)  c_n = 0 for infinitely many indices n ≥ 1.\n\n(No hypothesis beyond b ≠ 0 is needed; in particular the signs of a and b are irrelevant.)",
      "solution": "Step 1.  An explicit formula for the coefficients.\n\nWrite z = a + i b ( b ≠ 0, so z is not real).  Because\n\ne^{ax}\\sin(bx)= \\operatorname{Im}\\{e^{(a+ib)x}\\}= \\operatorname{Im}\\Bigl\\{\\sum_{n=0}^{\\infty}\\frac{z^{\\,n}}{n!}x^{n}\\Bigr\\}\n              = \\sum_{n=0}^{\\infty}\\frac{\\operatorname{Im}(z^{\\,n})}{n!}\\;x^{n},\n\nthe Maclaurin coefficient is\n\nc_n = \\dfrac{\\operatorname{Im}(z^{\\,n})}{n!}.            (1)\n\nStep 2.  The constant term.\n\nSince z^{0}=1 is real, (1) gives c₀ = Im(1)/0! = 0.\n\nStep 3.  When does a further coefficient vanish?\n\nPut θ = arg z, chosen in (−π, π)\\{0} (b ≠ 0 guarantees θ ≠ 0,±π).  Then z = |z|e^{iθ} and\n\nz^{\\,n}=|z|^{n}e^{i nθ},   so   Im(z^{\\,n}) = 0 ⇔ \\sin(nθ)=0 ⇔ nθ ∈ π\\mathbb Z.            (2)\n\nThus, for n ≥ 1,\n\nc_n = 0  ⇔  nθ/π ∈ \\mathbb Z.                                    (3)\n\nStep 4.  Two cases depending on θ/π.\n\n(i)  θ/π is irrational.\n\nIf θ/π ∉ \\mathbb Q, equality (3) cannot hold for any positive integer n, so c_n ≠ 0 for every n ≥ 1.  The series then contains exactly one zero coefficient, namely c₀.\n\n(ii)  θ/π is rational.\n\nWrite θ/π = p/q in lowest terms, where q ≥ 1.  Condition (3) becomes n·p/q ∈ \\mathbb Z, i.e. q | n.  All positive multiples n = kq (k = 1,2,3, …) satisfy this, so c_{kq} = 0 for every k ≥ 1.  There are therefore infinitely many vanishing coefficients.\n\nStep 5.  Conclusion.\n\nApart from the constant term c₀ = 0, either no further coefficient vanishes (case (i)) or infinitely many do (case (ii)).  This proves the required dichotomy.",
      "_meta": {
        "core_steps": [
          "Rewrite e^{ax} cos bx as Re e^{(a+ib)x}.",
          "Use Maclaurin expansion: coef(x^n)=Re[(a+ib)^n]/n!.",
          "If this real part vanishes, (a+ib)^n is purely imaginary.",
          "Purely imaginary numbers raised to any odd power stay purely imaginary, so Re[(a+ib)^{kn}]=0 for all odd k.",
          "Hence either no coefficient ever vanishes or infinitely many do."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Sign restriction on the real parameters",
            "original": "a>0, b>0"
          },
          "slot2": {
            "description": "Trigonometric factor could be sine instead of cosine (then use Im instead of Re)",
            "original": "cos"
          },
          "slot3": {
            "description": "Taking the real part; could equivalently take imaginary part if the trig factor is changed",
            "original": "Re{…}"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}