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Given two functions $f_1(m) = a^m$ and $f_2(m) = b^m$, how to design another function $f(m)=g(f_1(m), f_2(m))$ such that $f(m)$ is maximized at some finite value $m=m_o$ (with $m_o$ not equal to $0$ or $\infty$). Or prove there is no such function $g(\cdot,\cdot)$.

For instance, $g(x,y) = \frac{x}{y}$ does not meet the requirement, since if $a\le{}b$, $m_o=0$. If $a\ge{}b$, $m_o=\infty$.

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Would $\frac{(a/b)^m}{(c+(a/b)^m)^{2n}}$ where $c > 0$ and $n$ is an integer work? –  J. M. Apr 17 '11 at 18:09
    
Would $g(x,y)=-|\log(kx)\log(ky)|$ work? –  Isaac Apr 17 '11 at 18:12
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Can't you just cheat? Let $m_0$ be the value you wish to maximize at, define $g(f_1(m_0),f_2(m_0))=1$ and for all other values define $g$ to be 0. –  JSchlather Apr 17 '11 at 18:21
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2 Answers

up vote 3 down vote accepted

Just take $g(x,y)$ to be a function of two variables that attains a maximum at $(x,y) = (f_1(m_0), f_2(m_0))$. For example, $g(x,y) = - (x - f_1(m_0))^2 - (y - f_2(m_0))^2$ will work.

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There are many examples. One idea is to note that the maximum of $$H_1(m)=\frac{1}{\frac{a^m}{a}+\frac{a}{a^m}}=\frac{a\cdot a^{m}}{a^2+(a^{m})^2}$$ is reached at $m=1$. So if we let $$f(x,y)=\frac{ax}{a^2+x^2}+\frac{by}{b^2+y^2}$$ then $f(a^m, b^m)$ will reach its maximum at $m=1$.

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