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Can someone help me with this analysis question:

I have to show that $\sinh(x)$ is strictly increasing on all of $\Bbb R$, and that $\cosh(x)$ is strictly increasing on the interval: $[0,\infty)$ and strictly decreasing on the interval: $(-\infty, 0]$.

I'm not allowed to differentiate or anything like that. I think I have to do it with inequalities.

I've been giving a hint to use the trigonometric addition formulas.

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  • $\begingroup$ thank you GNU supporter! :) $\endgroup$ – tonytouch Feb 17 '17 at 15:23
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It's easy to see that $\sinh(-x)=-\sinh(x)$ and $\cosh(-x)=\cosh(x)$, so it suffices to show that each function is increasing on $[0,\infty)$. Their increasing/decreasing behaviors on $(-\infty,0]$ follow from their odd/even symmetries.

For $\sinh(x)$, this is easy, provided you know that $e^x$ is an increasing function: That makes $e^{-x}$ a decreasing function, which makes $-e^{-x}$ an increasing function, and that makes their sum increasing. (This argument actually works for all $x$, not just for $x\ge0$. It's the next paragraph that uses the restriction.)

For $\cosh(x)$, it's a little trickier. since you are summing a decreasing function to an increasing function. But here we're aided by that fact that $\sinh(x)\ge0$ for all $x\in[0,\infty)$, so that $\sinh^2(x)$, being the product of non-negative increasing functions, is also increasing on $[0,\infty)$, which means $\cosh(x)=\sqrt{\sinh^2(x)+1}$ in increasing (being a composition with the increasing square root function).

Added later: Here is a second approach that may be more along the lines of what the OP is looking for. Again, we'll restrict to $x\ge0$ (using odd/even symmetry for negative $x$), and we'll take as given that $\sinh(x)$ and $\cosh(x)$ are both non-negative for $x\ge0$, from which the hyperbolic identity $\cosh^2(x)-\sinh^2(x)=1$ implies $\cosh(x)=\sqrt{1+\sinh^2(x)}\ge1$.

From the addition formulae

$$\sinh(x+u)=\sinh(x)\cosh(u)+\cosh(x)\sinh(u)$$

and

$$\cosh(x+u)=\cosh(x)\cosh(u)+\sinh(x)\sinh(u)$$

we have, for $x,u\ge0$,

$$\sinh(x+u)\ge\sinh(x)\cdot1+0=\sinh(x)$$

and

$$\cosh(x+u)\ge\cosh(x)\cdot1+0=\cosh(x)$$

so that $\sinh(x)$ and $\cosh(x)$ are non-decreasing on $[0,\infty)$. If we know (or prove) that $\sinh(u)$ is positive for $u\gt0$, then $\cosh(u)\gt1$, and we get $\sinh(x+u)\gt\sinh(x)$ and $\cosh(x+u)\gt\cosh(x)$, which means $\sinh(x)$ and $\cosh(x)$ are strictly increasing on $[0,\infty)$.

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  • $\begingroup$ Hello thank you for your answer Barry Cipra. I have tried showing it this way: Sinh(x) > 0 for all x and that sinh(x) < 0, for x < 0. Then I tried: for y > 0, then sinh(x+y) = a* sin(x)+b where a=>1 and b > 0. Is that correct? $\endgroup$ – tonytouch Feb 17 '17 at 15:48
  • $\begingroup$ @tonytouch, even setting aside what I assume are typos in your comment, I don't think that's the way to go. $\endgroup$ – Barry Cipra Feb 17 '17 at 15:51
  • $\begingroup$ Oh damn. But I just know that I have to use the additions formulas, and right now in class, we are only using inequalites and using the definitions. So I think I can't use that -exp(-x) is an increasing function. $\endgroup$ – tonytouch Feb 17 '17 at 15:53
  • $\begingroup$ @tonytouch, since exponential functions are always strictly positive, you can safely multiply both sides of any inequality by $e^{-x}$ or $e^{-y}$, so $$x\lt y\implies e^x\lt e^y\implies 1\lt e^{-x}e^y\implies e^{-y}\lt e^{-x}\implies -e^{-x}\lt-e^{-y}$$ $\endgroup$ – Barry Cipra Feb 17 '17 at 15:58
  • $\begingroup$ Thank you Barry. It's just because in our class we are only allowed to use what's what we have read so far, (limits and continuity) and the appendix, so sometimes I get confused about what I'm allowed to do, but i'll try what you said. thank you a lot! $\endgroup$ – tonytouch Feb 17 '17 at 16:02
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exp(x) is increasing.

exp(-x) is decreasing, so -exp(-x) is increasing.

The sum of two increasing functions is increasing.

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  • $\begingroup$ thank you for your answer! $\endgroup$ – tonytouch Feb 17 '17 at 15:57
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if $f(x)=\frac{e^x-e^{-x}}{2}$ then we have $$f'(x)=\frac{e^x+e^{-x}}{2}>0$$ and our function i strictly monotonously increasing.

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    $\begingroup$ Hello Dr. Sonnhard. I'm not allowed to use differentiation. $\endgroup$ – tonytouch Feb 17 '17 at 15:25
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$f(x)=\exp(x)$ and $g(x)=- \exp(-x)$ are both increasing functions.

Therefore, so is $\sinh(x) = \frac12(f(x)+g(x))$.

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  • $\begingroup$ Beat me by less than a minute. $\endgroup$ – marty cohen Feb 17 '17 at 15:26
  • $\begingroup$ hello Ihf, thank you for your answer. But I think I have to show it with inequalities, that's why they told me to use the addtions formulas. $\endgroup$ – tonytouch Feb 17 '17 at 15:30
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$$\sinh x=\sum_{k\ge 0}\frac{x^{2k+1}}{(2k+1)!},$$ and each $x^{2k+1}$ is an increasing function.

Similar argument for $$\cosh x=\sum_{k\ge 0}\frac{x^{2k}}{(2k)!}.$$

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  • $\begingroup$ Thanks for your answer Bernard! $\endgroup$ – tonytouch Feb 17 '17 at 15:58

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