1
$\begingroup$

Use the Mean Value Theorem to prove $\cosh(x) \ge 1 + \frac{x^2}{2}$ in the interval $[0,x]$, given $\sinh(x) \ge x$ for all $x \gt 0$.

I tried using $f(x) = \cosh(x)$, but to no avail. All help appreciated, thanks!

$\endgroup$
  • $\begingroup$ Are you sure you copied this correctly. It seems that the variable $x$ is used in two different meanings in: prove $\cosh(x) \ge 1 + \frac{x^2}{2}$ in the interval $[0,x]$. (This makes the question unclear. Of course, I might have misunderstood something.) $\endgroup$ – Martin Sleziak Jun 13 '13 at 12:57
  • $\begingroup$ @MartinSleziak puu.sh/3eJMs.png $\endgroup$ – Clinton Jun 13 '13 at 12:59
1
$\begingroup$

What about the function $f(x)=\cosh (x)-1-\frac{x^2}2$.

You clearly have $f(0)=0$. Can you show $f'(x)\ge0$. If you use these facts and Mean value theorem, what do you get?

$\endgroup$
0
$\begingroup$

Let's assume $\cosh (x) \leq 1 + \frac{x^2}{2}$, within that interval and let $f(x) = \cosh (x)$. Then, the mean value theorem tells us that there exists an $x_0 \in [0,x]$ such that

$$f'(x_0) = \frac{f(x) - f(0)}{x - 0}.$$

If $f(x) = \cosh (x) \implies f'(x) = \sinh (x)$. From the constraint given, we know that $\sinh (x_0) \geq x_0$. Now, let's work out the RHS of the MVT equality:

$$1) \,\,\,\,\,\,\,\,f(x) = \cosh (x) = \cosh (x)$$ $$2) \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f(0) = \cosh (0) = 1$$ $$3) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, x - 0 = x.$$

Subbing this all in, we get:

$$\sinh (x_0) = \frac{\cosh (x) - 1}{x}.$$

As $\sinh (x_0) \geq x_0$ we get

$$x_0 \leq \frac{\cosh (x) - 1}{x}$$ $$x \cdot x_0 \leq \cosh (x) - 1$$ $$\implies \cosh(x) \geq 1 + x \cdot x_0.$$

We know that $x_o \leq x$ and so, the biggest value this $x_0$ can possible take is $x$, putting this in gives us

$$\cosh (x) \leq 1 + x \cdot x = 1 +x^2$$

which is a contradiction as we said that $\cosh (x) \leq 1 + \frac{x^2}{2} < 1 + x^2$, as $x > 0$.

$\endgroup$
  • $\begingroup$ $\sinh(x)\geq x$ implies $\frac{\cosh(x)-1}{x}\geq x$ $\endgroup$ – Amr Jun 13 '13 at 12:36
  • $\begingroup$ @Amr Oh yeah, sorry, typo. $\endgroup$ – Kaish Jun 13 '13 at 12:38
  • $\begingroup$ It seems to me that your argument can be used to show that $\cosh(x)\geq 1+x^2$ (a better lower bound) $\endgroup$ – Amr Jun 13 '13 at 12:39
  • $\begingroup$ @Amr As opposed to $\geq 1 + \frac{x^2}{2}$? So it doesn't answer this question properly? $\endgroup$ – Kaish Jun 13 '13 at 12:40
  • $\begingroup$ OK . There is a mistake $\endgroup$ – Amr Jun 13 '13 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.