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Let $f\left(x\right):=e^{x}+e^{-x}+2$ and $g_{\beta}\left(x\right):=4e^{\beta x^{2}}$. Do there exist $a>0$ and $\beta>0$, such that $f\left(x\right)\le g\left(x\right)$ for all $x$, $0\le x\le a$?

Thanks for any helpful answers!

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1 Answer 1

Hint: The Taylor expansion of $e^x+e^{-x}+2$ about $0$ is $$4+\frac{2}{2!}x^2+\frac{2}{4!}x^4+\frac{2}{6!}x^6+\frac{2}{8!}x^8+\cdots.$$ (Just expand $e^x$ and $e^{-x}$ as usual and observe the partial cancellations.)

The Taylor expansion of $4e^{\beta x^2}$ about $0$ is $$4+\frac{4\beta}{1!}x^2+\frac{4\beta^2}{2!}x^4+\frac{4\beta^3}{3!} x^6+\frac{4\beta^4}{4!}x^8+\cdots.$$ (Write down the usual expansion of $e^t$, and replace $t$ by $\beta x^2$.)

Now can you find $\beta$ so that the coefficient of $x^{2k}$ in the second series is always $\ge$ the corresponding coefficient of $x^{2k}$ in the first series?

Remark: The problem becomes more challenging if we ask instead about $g(x)\le f(x)$. My guess is that that is what may actually be intended, since then we need to worry about the size of $x$.

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