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Show that a given function satisfies a functional equation

by RoRi

Define a function for a > 0 by

    \[ f(x) = \frac{1}{2} (a^x + a^{-x}). \]

Prove that

    \[ f(x+y) + f(x-y) = 2 f(x) f(y). \]

Proof. First, we compute f(x+y) and f(x-y) using the definition of f above, and the properties of the exponential a^x established in the previous exercise (Section 6.17, Exercise #36), we have

    \[ f(x+y) = \frac{a^{2(x+y)+1}}{2a^xa^y}, \qquad f(x-y) = \frac{a^{2x} + a^{2y}}{2a^x a^y}. \]

Therefore,

    \begin{align*} f(x+y) + f(x-y) &= \frac{a^{2(x+y)} + 1 + a^{2x} + a^{2y}}{2a^x a^y} \\ &= \frac{1}{2} \left( a^{2x+2y} + a^{2x} + a^{2y} + 1 \right) \left(a^{-x}a^{-y} \right) \\ &= \frac{1}{2} \left( a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x}a^{-y} \right) \\ &= \frac{1}{2} \left( a^x + a^{-x} \right) \left(a^y + a^{-y} \right) \\ &= 2 f(x) f(y). \qquad \blacksquare \end{align*}

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