The discussion focuses on determining the range of the parameter \( k \) for which the inequality \( \frac{e^x + e^{-x}}{2} \le e^{kx^2} \) holds for all real values of \( x \). Through the analysis of Taylor expansions, it is established that the condition \( \cosh x \le e^{kx^2} \) is satisfied if and only if \( k \ge \frac{1}{2} \). This conclusion is drawn from the comparison of the series expansions of \( \cosh x \) and \( e^{kx^2} \).
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anemone said:Find the range of $k$ for which $\dfrac{e^x+e^{-x}}{2}\le e^{kx^2}$ holds for all real $x$.
chisigma said:[sp]Considering the Taylor expansions...
$\displaystyle \cosh x = 1 + \frac{x^{2}}{2} + ...\ (1)$
$\displaystyle e^{k\ x^{2}} = 1 + k\ x^{2} + ...\ (2)$
... the condition $\displaystyle \cosh x \le e^{k\ x^{2}}, \forall x$ is equivalent to say $\displaystyle k \ge \frac{1}{2}$...[/sp]
Kind regards
$\chi$ $\sigma$