MHB Range of k for Real x in $\dfrac{e^x+e^{-x}}{2}\le e^{kx^2}$

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The discussion focuses on determining the range of \( k \) for which the inequality \( \frac{e^x + e^{-x}}{2} \le e^{kx^2} \) holds for all real \( x \). It is established that using Taylor expansions for \( \cosh x \) and \( e^{kx^2} \) leads to the conclusion that the condition \( \cosh x \le e^{kx^2} \) is satisfied if \( k \ge \frac{1}{2} \). This means the minimum value of \( k \) required for the inequality to hold universally is \( \frac{1}{2} \). The discussion emphasizes the importance of Taylor series in deriving this result. Overall, the range of \( k \) is confirmed to be \( k \ge \frac{1}{2} \).
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Find the range of $k$ for which $\dfrac{e^x+e^{-x}}{2}\le e^{kx^2}$ holds for all real $x$.
 
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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 $x$.

[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

$\chi$ $\sigma$
 
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

$\chi$ $\sigma$

Well done, chisigma! And thanks for participating!(Yes)
 
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