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)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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