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Exercise $5.1$, question 13 of the AQA Further Mathematics for core year 1 and AS has the following question:

$a) ~~$Sketch the curve $~y = \cosh x~$ and the line $~y = x~$ on the same axes. Prove that $~\cosh x > x~,~~ \forall ~x~$.

This exercise is before hyperbolic differentiation, Maclaurin series or Taylor series. I can't see how this proof can be shown algebraically using AS level mathematics. The sketch is given as the answer, would I be correct in assuming that there isn't a simple algebraic proof and the authors have decided the sketch is a proof.

I'm also working on part $b)$ Prove that the point on the curve $~y = \cosh x~$ which is closest to the line $~y = x~$ has coordinates $~\left(\ln\left(1 + \sqrt{2}\right), ~\sqrt{2}\right)~$.

Suggestions would be welcome.

An internet search hasn't helped. Any ideas?

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    $\begingroup$ How on earth is $\cosh$ defined if not by either Taylor series or differential equation? If it is not even defined, then the exercise is meaningless. $\endgroup$
    – user21820
    Commented Feb 8, 2020 at 4:17

4 Answers 4

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Hint: if $x < 1 \implies \cosh x \ge 1 > x $ by AM-GM inequality. If $x \ge 1 \implies e^x -2x > 0$. Few details have been left out for you to fill out.

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    $\begingroup$ AS mathematics is for 16/17 year olds, so AM-GM inequality is beyond their knowledge. $\endgroup$
    – Paul McG
    Commented Jan 25, 2020 at 15:13
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Let $f(x)=\cosh x -x$ and evaluate

$$f'(x)=\sinh x -1 \>\>\bigg\{ \begin{array}{c} \ge 0, \>\>\>x\ge\ln(1+\sqrt2), \\ <0, \>\>\> x<\ln(1+\sqrt2) \ \end{array}$$

where $\sinh^{-1}t=\ln(t+\sqrt{1+t^2})$ is used. So, $f(x)$ strictly decreases for $x<\ln(1+\sqrt2) $ and strictly increases for $x>\ln(1+\sqrt2)$. Then, $$f(x) > f(\ln(1+\sqrt2)) =\cosh(\ln(1+\sqrt2))-\ln(1+\sqrt2)=\sqrt2-\ln(1+\sqrt2)>0$$

Thus, $$\cosh x -x>0$$

Note that the closest point to the line is

$$(\ln(1+\sqrt2),\cosh(\ln(1+\sqrt2))) = (\ln(1+\sqrt2), \sqrt2)$$

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  • $\begingroup$ It's a nice proof only differentiation of hyperbolic functions is not on the syllabus. $\endgroup$
    – Paul McG
    Commented Jan 25, 2020 at 16:32
  • $\begingroup$ @Paul - As a work around, use $\cosh t = \frac{e^t+e^{-t}}2$. So, its derivative is $(\cosh t)' = \frac{e^t-e^{-t}}2=\sinh t$. $\endgroup$
    – Quanto
    Commented Jan 25, 2020 at 16:41
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Consider that

$$\cosh(x) \equiv \frac{e^{x} + e^{-x}}{2}$$

Now just remember that the exponential $e^{x}$ (as well as $e^{-x}$) is always positive. Also you have a SUM between the two.

Argue a bit on it and I believe you are easily done.

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The easiest way to define $\cosh$ is to define it in terms of $\exp$, but that still requires a definition of $\exp$. The problem is, there is no simple way to define $\exp$. Every definition of $\exp$ must pass through a significant amount of real analysis. However, there is a way if the syllabus in question cheats and assumes without justification the following non-trivial facts:

$\exp(x+1) > 2·\exp(x) > 0$ for every real $x$.

$\exp(x) ≥ 1+x$ for every real $x$.

$\exp(-x) = 1 / \exp(x) > 0$ for every real $x$.

$\cosh(x) = (\exp(x)+\exp(-x))/2$ $> (2·\exp(x-1)+0)/2$ $= \exp(x-1) ≥ x$ for every real $x$.

In case it is not clear, there are many other possible ways to derive the same result, but every single way will definitely rely on some non-trivial fact about $\exp$ or a related function.

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