4
$\begingroup$

Question: In a hyperbolic right angled triangle, the two legs have hyperbolic lengths of $3$ and $4$. What is the hyperbolic length of the hypotenuse? Is this larger or smaller than $5$?

I'm having trouble with this problem mainly because I am not allowed to use a calculator. So I have to do everything manually.

I used the formula:

$$ cosh(c)=cosh(a)cosh(b) $$

where $a=3$ and $b=4$

I know $cosh(x)=\frac{e^x+e^{-x}}{2}$

With that being said, by plugging the values of $a$ and $b$

$$ cosh(c)=cosh(3)cosh(4) $$

and substituting into $cosh(x)$

$$ cosh(c)=\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}$$

I know:

$$ cosh^{-1}(x)=ln(x+\sqrt{x^2-1})$$

In this case, my $x$ would be $\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}$

Plugging into the inverse, I would obtain:

$$ c=cosh^{-1}(x)=ln(\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}+\sqrt{(\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2})^2-1})$$

Maybe my algebra is a bit rusty but how would I know if $c$ is less than or greater than $5$. The only way I could think is to simplify above but I'm not sure how to simplify above. It became way too messy. Any help?

If there is an alternate way, I'm open to hear that as well.

$\endgroup$
  • $\begingroup$ Just calculate(get a advanced calculator) : cosh(3) is a bit more than 10 , cosh(4) is a bit more than 27 so the product cosh(3)cosh(4) is more than 270, arcosh(270) is more than 6.2 so definitly longer than 5 (notice all values above are rounded down, btw cosh(5) is less than 75 so the hypothenusea is definitly longer in the hyperbolic case. $\endgroup$ – Willemien Apr 29 '15 at 7:40
  • $\begingroup$ Start by noticing that in the Euclidean case, $3^2+4^2=5^2$. So if you only have to decide whether the length is smallr or greater than $5$, you only have to see whether it's smaller or larger than in the Euclidean case. You might find ways to decide this without performing any actual computation. $\endgroup$ – MvG Apr 29 '15 at 9:39
  • $\begingroup$ alternative calculation forget the $e^{-x} $ so you get $\cosh (c) > e^7 / 4 $ then for arcosh forget everything after $ \ln(x) $ and you get $\ln(e^7 /4 ) = 7- \ln(4) = 7 - 1.4 = 5.6 $ still much above 5, ps a hyperbolic length of 3 is really long, hyperbolic geometry has an absolute measure $\endgroup$ – Willemien Apr 30 '15 at 7:43
3
$\begingroup$

\begin{align*} \cosh\sqrt{a^2+b^2} &= \sum_{m=0}^\infty \frac{(a^2+b^2)^m}{(2m)!} \\ &= \sum_{m=0}^\infty \sum_{n=0}^m \binom mn \frac{a^{2(m-n)}b^{2n}}{(2m)!} \\ &= \sum_{n=0}^\infty \sum_{m=n}^\infty \binom mn \frac{a^{2(m-n)}b^{2n}}{(2m)!} \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{m+n}n \frac{a^{2m}b^{2n}}{(2(m+n))!} \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{\binom{m+n}n}{\binom{2(m+n)}{2n}} \cdot \frac{a^{2m}b^{2n}}{(2m)!\,(2n)!} \\ &\le \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{a^{2m}b^{2n}}{(2m)!\,(2n)!} \\ &= \sum_{m=0}^\infty \frac{a^{2m}}{(2m)!} \sum_{n=0}^\infty \frac{b^{2n}}{(2n)!} \\ &= \cosh a \cosh b \\ &= \cosh c \end{align*} which since $\cosh$ is increasing on $[0,\infty)$ yields $\sqrt{a^2+b^2}\le c$. We have equality iff $a=0$ or $b=0$.

$\endgroup$

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.