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Prove by using Taylor formula that in a neighbourhood of zero this inequality holds:

$\cosh x \le 12 \frac{x^2+2}{24-x^4}$

Please explicitate everything you do :)

Thanks in advance

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    We can explain what *we* do, but what about what *you* have done so far? – 2017-01-02
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    I tried to look after the taylor expansion of cosh(x). My thought was to manipulate the first three terms and use the fact that any remainder will tend to 0 in the limit. But I am pretty sure to be wrong – 2017-01-02

2 Answers 2

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The hint pretty much tells you everything you need to do: take the Taylor expansion of the difference of the two sides, until you see a non-zero term. Then check the sign of that term near $0$.

In particular, the theorem that we will use is the following:

Theorem. Let $f,g$ be two functions defined on a (pointed) neighborhood of zero, and such that $g$ does not cancel there. Then, if $f(x) = g(x) + o(g(x))$ when $x\to 0$, $f$ and $g$ keep the same sign on a neighborhood of $0$.

Proof. We have $\frac{f(x)}{g(x)} = \frac{g(x)+o(g(x))}{g(x)} = 1+o(1) \xrightarrow[x\to 0]{} 1$, and thus there exists $\delta > 0$ such that $\frac{f(x)}{g(x)} \in \left(\frac{1}{2},\frac{3}{2}\right)$ for all $x\in(\delta,\delta)$. $\qquad\square$

Now, we will invoke this theorem at the end: we will prove that $12\cdot \frac{2+x^2}{24-x^4} - \cosh x = \alpha x^c + o(x^c)$ for some $\alpha\neq 0,c>0$ (by performing a Taylor expansion of both sides of the inequality until the point where they differ) and be quite happy to notice that $c$ is even, and $\alpha > 0$. Then, we will "set" $g(x) = \alpha x^c$ and conclude.

When $x\to 0$, we have (this one is standard) $$ \cosh x = 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{720}+o(x^6) $$ and $$\begin{align} 12\cdot \frac{2+x^2}{24-x^4} &= \frac{1+\frac{x^2}{2}}{1-\frac{x^4}{24}} = \left(1+\frac{x^2}{2}\right)\left(1+\frac{x^4}{24}+O(x^8)\right)\\ &= 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{48}+O(x^8)\\ &= 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{48}+o(x^6) \end{align}$$ and thus $$ 12\cdot \frac{2+x^2}{24-x^4} - \cosh x = \frac{7}{360}x^6 + o(x^6). $$ Since $\frac{7}{360}>0$, the difference is positive on a neighborhood of $0$, as it is equivalent to $\frac{7}{360}x^6$. This shows that on that very same neighborhood of $0$, the inequality $$ 12\cdot \frac{2+x^2}{24-x^4} > \cosh x$$ holds by the above theorem.

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    I don't understand the second equality on the first line after "and". Who does a minus become a plus for 1 - x^4/24 ? – 2017-01-02
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    Use the following classic Taylor expansion: when $u\to 0$, $\frac{1}{1-u} = 1+u+u^2+u^3+\dots+u^k + o(u^k)$. Here, $u\stackrel{\rm def}{=}\frac{x^4}{24}$. [This is a specific case of the Taylor expansion of $(1+u)^\alpha$, where we have here $\alpha=-1$.] – 2017-01-02
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    Oh I see, thanks – 2017-01-02
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$$12 \frac{x^2+2}{24-x^4}=\frac{1+\frac{1}{2}x^2}{1-\frac{x^4}{24}}=\left(1+\frac{1}{2}x^2\right)\sum_{n\ge0}\frac{x^{4n}}{24^n}\ge\sum_{n\ge0}\frac{x^{2n}}{(2n)!}=\cosh x$$ with the inequality coming by comparing coefficients of $x^{2n}$.

  • $(4n)!\ge 24^{n}$: check $n=0,1$ explicitly, note that lhs grows by a factor $\ge5\cdot6\cdot7\cdot8>24$ for larger $n$, induction finishes the job.
  • $(4n+2)!\ge2\cdot 24^{n}$:check $n=0$ explicitly, note that lhs grows by a factor $\ge 3\cdot4\cdot5\cdot6>24$ for larger $n$, induction finishes the job.