\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2017.3.4 Question 4

1. Notice that \(a = e^{\ln a}\) and hence \(a^x = e^{x \ln a}\), \(a^{\frac {x}{\ln a}} = e^x\)we have \begin {align*} F(y) & = \exp \left (\frac {1}{y} \int _0^y \ln f(x) \Diff x\right ) \\ & = a^{\frac {1}{y \ln a} \cdot \int _0^y \ln f(x) \Diff x} \\ & = a^{\frac {1}{y} \cdot \int _0^y \frac {\ln f(x)}{\ln a} \Diff x} \\ & = a^{\frac {1}{y} \cdot \int _0^y \log _a f(x) \Diff x} \end {align*}

   as desired.

2. We have \begin {align*} H(y) & = \exp \left (\frac {1}{y} \int _0^y \ln f(x) g(x) \Diff x\right ) \\ & = \exp \left [\frac {1}{y} \int _0^y \left (\ln f(x) + \ln g(x)\right ) \Diff x\right ] \\ & = \exp \left [\frac {1}{y} \left (\int _0^y \ln f(x) \Diff x + \int _0^y \ln g(x) \Diff x\right )\right ] \\ & = \exp \left (\frac {1}{y} \int _0^y \ln f(x) \Diff x\right ) \cdot \exp \left (\frac {1}{y} \int _0^y \ln g(x) \Diff x\right ) \\ & = F(y) \cdot G(y). \end {align*}
3. Let \(f(x) = b^x\). \begin {align*} F(y) & = \exp \left (\frac {1}{y} \int _0^y \ln f(x) \Diff x\right ) \\ & = b^{\frac {1}{y} \int _0^y \log _b f(x) \Diff x} \\ & = b^{\frac {1}{y} \int _0^y \log _b b^x \Diff x} \\ & = b^{\frac {1}{y} \int _0^y x \Diff x} \\ & = b^{\frac {1}{y} \cdot \frac {y^2}{2}} \\ & = b^{\frac {y}{2}} \\ & = \sqrt {b^y}. \end {align*}

4. Since \(F(y) = \sqrt {f(y)}\), we notice that \(f(y) = F(y)^2 = \exp \left (\frac {2}{y} \int _0^y \ln f(x) \Diff x\right )\), and therefore \(\ln f(y) = \frac {2}{y} \int _0^y \ln f(x) \Diff x\).

   We substitute \(g(y) = \ln f(y)\), and therefore \[ y g(y) = 2 \int _0^y g(y) \Diff x. \]

   Therefore, differentiating both sides with respect to \(y\) gives us \[ yg'(y) + g(y) = 2 g(y), \] and therefore \[ -g(y) + y g'(y) = 0. \]

   Multiplying \(y^{-2}\) on both sides gives us \[ -y^{-2}g(y) + y^{-1} g'(y) = 0, \] and therefore \[ \frac {\Diff }{\Diff y} \frac {g(y)}{y} = 0, \] and therefore \[ \frac {g(y)}{y} = C \implies g(y) = Cy. \]

   Therefore, we have \begin {align*} f(y) & = \exp g(y) \\ & = \exp (Cy) \\ & = b^y \end {align*}

   if we substitute \(b = \exp (C) > 0\), and therefore \(f(x) = b^y\) as desired.