Multiplicative Inverse And Gabriel’s Horn

The “Multiplicative inverse” (2024) is also known as the reciprocal of a number $x$, denoted by $\; 1/x \;$ or $\; x^{- 1} \;$ where $x \neq 0$. This can be expressed as the reciprocal function $\; y = 1/x \;$, and a graph of this function is shown in Figure 1.

Multiplying by a number is the same as dividing by its reciprocal, and vice versa. For example, multiplication by 4/5 (or 0.8) will give the same result as division by 5/4 (or 1.25).

The multiplicative inverse property states that a number multiplied by its reciprocal will always equal $1$, since the product of the number and its reciprocal is the identity $\; x (1/x) = 1$. Therefore, the multiplication of a pair of numbers followed by multiplication of the first number’s reciprocal will yield the second number.

Originating from the reciprocal function is a fascinating geometric figure known as a “Gabriel’s horn” (2024). The name refers to the Christian tradition where the archangel Gabriel blows the horn to announce Judgment Day. The properties of this figure were first studied by Italian physicist and mathematician Evangelista Torricelli in the 17th century. So a Gabriel’s Horn is also called “Torricelli’s Trumpet.” It’s also been called a “Tonal Tower” because of the harmonic relationships involved.

Gabriel’s Horn is formed by taking the graph of $\; y = 1/x$, with the domain $x ≥ 1$ and rotating it in three dimensions about the x axis. In Figure 2 below, we created a surface in 3 dimensional space of Gabriel’s Horn from $\; x = 1 \;$ to $\; x = 9 \;$, and we’ve also included a section plane which intersects the figure between $\; x = 1 \;$ and $\; x = 2$.

We’ll be discussing the intersecting plane in Figure 2 further below, but first we need to look at a strange paradox about the volume and surface area of this shape.

1

Gabriel’s Horn Volume And Surface Area

One of the more intriguing things about Gabriel’s Horn is the Painter’s Paradox, which states that it has an infinite surface area but a finite volume! The discovery was made using Cavalieri’s Principle before the invention of calculus, but today, calculus can be used to calculate the volume and surface area of the horn between $\; x = 1 \;$ and $\; x = a \;$, where $\; a > 1 \;$. Using integration, it is possible to find the volume $V$ as follows:

\begin{align}
V = \; \pi \int_1^a\left(\frac{1}{x}\right)^2dx \; = \; \pi \left(1 \; – \; \frac{1}{a}\right)
\end{align}

The value a can be as large as required, but it can be seen from the equation that the volume of the part of the horn between $\; x = 1 \;$ and $\; x = a \;$ will never exceed $\pi$. However, it does gradually draw nearer to $\pi$ as $a$ increases. Mathematically, the volume approaches $\pi$ as $a$ approaches infinity. Using the limit notation of calculus,

\begin{align}
\lim_{a\rightarrow \infty}V = \lim_{a\rightarrow \infty}\pi \left(1 \; – \; \frac{1}{a}\right) \\\\
= \; \pi \lim_{a\rightarrow \infty}\left(1 \; – \; \frac{1}{a}\right) \; = \; \pi \\\\
\end{align}

Now the surface area $A$ on the other hand, can be calculated as follows,

\begin{align}
A = 2\pi \int_1^a\frac{1}{x}\sqrt{1 + \left( – \frac{1}{x^2}\right)^2}dx \\\\
\end{align}

If we can show that something smaller goes to infinity, we also know that something larger will go to infinity as well. So we can make the following statement,

\begin{align}
A \; > \; 2\pi \int_1^a\frac{dx}{x} \; = \; 2\pi \; [ln(x)]_1^a \; = \; 2\pi \; ln(a) \\\\
\end{align}

The surface area formula above gives a lower bound for the area as $2\pi$ times the natural logarithm of $a$. There is no upper bound for the natural logarithm of $a$, as $a$ approaches infinity. This means the horn has an infinite surface area,

\begin{align}
\lim_{a\rightarrow \infty}A \; \geq \; \lim_{a\rightarrow \infty}2\pi \; ln(a) = \infty
\end{align}

In other words, it would take $\pi$ units of paint to fill Gabriel’s Horn, but you couldn’t paint the outside because the surface area is infinite! This is known as the “Painter’s Paradox.”

2

Gabriel’s Horn Intersecting Plane

Above our first section about volume and surface area in Figure 2, we’re showing a three dimensional view of Gabriel’s Horn with an intersecting plane between $x = 1$ and $x = 2$. When we shift the view as shown in Figure 3, we’re looking straight down the $z$ axis at Gabriel’s Horn in the $x,y$ plane. In this two dimensional view, it should be fairly easy to get measurements for the cross section created by the intersecting plane.

What we’re interested in investigating further is what looks like a proportionally perfect egg shape being traced out by this plane as it intersects with Gabriel’s Horn. And we’ll be analyzing some fascinating details about this shape in section 4 below. Then in section 6, we developed a new app which makes it possible for you to create your own egg shapes using different orientations of the intersecting plane – Interactive Gabriel’s Horn App.

To obtain precise measurements from Figure 3, the goal would be to figure out the exact point where this plane intersects with the $x$ axis, since we can use that as the origin point for other calculations. But right away, we can see a computation for the hypotenuse line $AB$ from the Pythagorean Theorem,

\begin{align}
AB^2 = CA^2 + BC^2 \\\\
AB^2 = 1^2 + \left(\frac{3}{2}\right)^2 = \; 1 + \frac{9}{4} = \frac{13}{4} \\\\
AB = \frac{\sqrt{13}}{2} \\\\
\end{align}

We can also see that right triangles $ \bigtriangleup ODB$ and $ \bigtriangleup OEA$ are similar to $ \bigtriangleup BCA$. We can find scaling factors for the vertical sides of these smaller triangles by noticing the known length of $BC$, and relating that to sides $BD$ and $OE$ (which are also known),

\begin{align}
BC = \frac{3}{2}, \; BD = 1, \; OE = \frac{1}{2} \\\\
\end{align}

By the law of similar triangles, if we find the scaling factor $s$ from $BC$ to $BD$, we can use that to find hypotenuse $OB$ by scaling down hypotenuse $AB$ from the larger triangle.

\begin{align}
BC(s) = BD \\\\
s = \frac{BD}{BC} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \\\\
OB = AB(s) = \frac{\sqrt{13}}{2}\frac{2}{3} = \frac{\sqrt{13}}{3} \\\\
\end{align}

Now we have everything we need to solve for line $OD$, and this will give us the exact origin point $O$ of the intersecting plane with the x axis,

\begin{align}
OB^2 = BD^2 + OD^2 \\\\
\frac{13}{9} = 1 + OD^2 \\\\
OD^2 = \frac{13}{9} \; – \; 1 = \frac{13 \; – \; 9}{9} = \frac{4}{9} \\\\
OD = \sqrt{\frac{4}{9}} = \frac{2}{3} \\\\
\end{align}

Since line $CA$ is one unit in length, line $EA$ is given by,

\begin{align}
EA = CA \; – \; OD = 1 \; – \; \frac{2}{3} \; = \; \frac{1}{3} \\\\
\end{align}

3

Circle at Plane Intersection

In Figure 3, we were showing the origin point $O$ of the intersecting plane with the $x$ axis. We found this location from line $OD$ in the section above to be $2/3$ from the edge of the graph at $\; x = 1$. What we want to look at now is the corresponding circle extending from the origin to the surface of Gabriel’s Horn.

If we take a look at Figure 4, we’ve flipped the 3D view to show the $y, z \;$ plane, so we’re looking straight down the $x$ axis through the horn. The top red circle would then represent a vertical slice of Gabriel’s Horn at the following $x$ value,

\begin{align}
x = 1 + \frac{2}{3} = \frac{5}{3} \\\\
\end{align}

To find the radius $r$ of the top red circle, all we need to do is plug in the $x$ value to find $y$.

\begin{align}
y = \frac{1}{x} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \\\\
r = \frac{3}{5}
\end{align}

4

Egg Shape From Intersecting Plane

One thing you might notice from Figure 4 is an egg shape created by the intersecting plane with Gabriel’s Horn. Since we’re looking straight down the $x$ axis and the section plane is positioned between $x = 1$ and $x = 2$, we’re viewing the egg in perspective when it’s tilted back.

In Figure 5, we extracted a vector slice from the plane’s intersection with the surface of the horn. Then we tilted it up to represent a perpendicular view, and revolved it around the $y$ axis to get our 3D image of the egg.

We know the length of the egg is $\sqrt{13}/2$, and the radius of the top red circle is $3/5$. What’s interesting is when you check $\sqrt{13}/2$ on a calculator, it’s actually very close to $9/5$ with a small remainder. If an egg has a shell, could this small remainder represent the thickness of the shell?

Once we recognize how the length of the egg is just slightly more than $9/5$, we can see that a Vesica Piscis can be constructed by duplicating the top red circle and moving it down by the length of the radius, which is $\; r = 3/5$. The vertical height of these two red circles would then be $3r = 9/5$.

So you can construct a proportionally correct vesica pisces around the yoke in the center of the egg, and it would sit perfectly inside the thickness of a shell!

Inverse Relationship of Remainder to a Circle?

If we take a look at that small remainder described above, we get the following result,

\begin{align}
\frac{\sqrt{13}}{2} \; – \; \frac{9}{5} = 0.002775637732 \\\\
\end{align}

What we find interesting is what happens when you take the inverse of that remainder,

\begin{align}
\frac{1}{\frac{\sqrt{13}}{2} \; – \; \frac{9}{5}} = 360.2775638 \\\\
\end{align}

So this leftover number appears to be inversely related to the number of degrees in a circle, and it could also have a cyclical property since the decimal remainder is repeated in the inverse calculation above.

A Closer Approximation to $\pi$

We should keep in mind that our actual shell thickness $S_t$ would be half of our small remainder, since that remainder is measuring both sides of the egg shape.

\begin{align}
S_t = \frac{1}{2}\left(\frac{\sqrt{13}}{2} \; – \; \frac{9}{5}\right) = \frac{5\sqrt{13} \; – \; 18}{20} \\\\
\end{align}

In our Squaring the Circle Insights article, we derived the exact same value of $\pi$ that was found by Archimedes in his calculations. We found this result by using a completely different method (deriving it from a squaring the circle diagram), so we think this could be adding more validity to his original definition for the upper bound of pi.

If we let $\pi_A$ be the Archimedean upper bound for $\pi$, this can be expressed by the following inequality,

\begin{align}
\pi_A \leq \frac{22}{7} \\\\
\end{align}

So if we reduce this upper bound from Archimedes $ \; \pi_A \; $ by our theoretical shell thickness $S_t$, we can find a closer approximation, $ \; \pi_s$, to the accepted value of pi.

\begin{align}
\pi_s = {\mathrm{\pi}}_{{\mathrm{A}}} \; – \; {\mathrm{S}}_{{\mathrm{t}}} \\\\
\pi_s = \frac{22}{7} \; – \; \frac{\left(5\sqrt{13} \; – \; 18\right)}{20} \\\\
\pi_s = \frac{566 \; – \; 35\sqrt{13}}{140} \\\\
\pi_s = 3.141469324 \\\\
\end{align}

To show the difference between $\pi_s$ and the accepted value of $\pi$, we’re showing both results together below,

\begin{align}
\pi_s = 3.141469324 \\
\pi = 3.141592654 \\\\
\end{align}

Evaluating a Tiny Triangle?

Since this small remainder representing our theoretical shell thickness $S_t$ is the result of cutting through the graph at an angle between $x = 1$ and $x = 2$, would it be possible to examine a tiny right triangle at point $B$ in Figure 3 and calculate its $x$ and $y$ values? This could be worth looking into, since we might be able to use those measurements to tighten up and improve Archimedes’ upper and lower bounds for $\pi$.

5

Improving Archimedes Approximation

Archimedes claimed the value of $\pi$ was between $22/7$ (approximately 3.142857) as an upper bound, and $223/71$ (approximately 3.140845) as a lower bound, essentially proving the following inequality,

\begin{align}
\frac{223}{71} \leq \; \pi \; \leq \frac{22}{7} \\\\
3.140845 \leq \; \pi \; \leq 3.142857 \\\\
\end{align}

So an interesting place to look for $\pi$ would be right in the middle of these upper and lower bounds, and we can do that by simply taking the average of those numbers when added together,

\begin{align}
\frac{3.140845 + 3.142857}{2} = 3.141851 \\\\
\end{align}

This is getting pretty close to the accepted value of $\pi = 3.141592654$, but is there a way to tighten up Archimedes’ upper and lower bounds with the shell thickness from our egg shape?

Constructing The Tiny Triangle

In Figure 6, we’re taking a closer look at triangle $ \bigtriangleup ODB$. If we can find angle $\theta$, then by using the law of similar triangles we can evaluate a tiny triangle at point $B$ with a hypotenuse length equal to our theoretical shell thickness $S_t$.

We calculated values earlier for lines $OD$ and $OB$, so from these can get $\theta$,

\begin{align}
\cos \left(\theta \right) = \frac{OD}{OB} \\\\
\theta = \cos^{- 1}\left(\frac{2}{\sqrt{13}}\right) = 56.30993247^{\circ} \\\\
\end{align}

We also want to evaluate a line perpendicular to the hypotenuse, since this might be able to approximate a normal vector to the curve at point $B$. So we should also find angle $\alpha$ from right triangle $ \bigtriangleup OFD$ in Figure 6.

\begin{align}
\alpha = 180^{\circ} \; – \; 90^{\circ} \; – \; 56.30993247^{\circ} \\\\
\alpha = 33.69006753^{\circ}
\end{align}

Decreasing Archimedes’ Upper Bound

Now we can start finding measurements for our tiny triangle at point $B$. Since the $y$ value is the longest length of the triangle we have running across the shell thickness $S_t$, it might be reasonable to use that number for reducing our upper bound. We can define this measurement as line $DB’$ and calculate that as follows,

\begin{align}
\sin \left(\theta \right) = \frac{DB’}{S_t} \\\\
DB’ = S_t\sin \left(\theta \right) = 0.001154735096 \\\\
\end{align}

Now we can subtract $DB’$ from Archimedes’ upper bound for pi, and name our improved upper bound $\pi_u$,

\begin{align}
\pi_u = \frac{22}{7} \; – \; S_t\sin \left(\theta \right) = 3.141702408 \\\\
\end{align}

Increasing Archimedes’ Lower Bound

Next, we can evaluate the perpendicular line $DF’$ for the tiny triangle. This is the shortest measurement of our new triangle and it approximates a normal vector at point $B$, so it would be a good value to use for increasing Archimedes’ lower bound. First we need to find $OD’$ since this will be the hypotenuse of our tiny triangle $ \bigtriangleup OFD’$,

\begin{align}
\cos \left(\theta \right) = \frac{OD’}{S_t} \\\\
OD’ = S_t\cos \left(\theta \right) = 0.0007698233974 \\\\
\end{align}

Then we can find $DF’$ using the angle $\alpha$ shown in Figure 6,

\begin{align}
\cos \left(\alpha \right) = \frac{DF’}{OD’} \\\\
DF’ = OD’\cos \left(\alpha \right) = 0.000640531784 \\\\
\end{align}

Now we can increase Archimedes’ lower bound for pi, and name our improved lower bound $\pi_l$,

\begin{align}
\pi_l = \frac{223}{71} + OD’\cos \left(\alpha \right) = 3.141485602 \\\\
\end{align}

Calculating a New Average

Taking the new average $\pi_{lu}$ of our improved lower and upper bounds from Archimedes, we can see how close we’re getting to the accepted version of $\pi$ as shown below.

\begin{align}
\pi_{lu} = \frac{\pi_l + \pi_u}{2} \\\\
\pi_{lu} = 3.141594005 \\\\
\pi = 3.141592654 \\
\end{align}

6

Interactive Gabriel’s Horn

The following interactive app allows you to set two points to control the orientation of the intersecting plane, with X1 and X2 representing where the points intersect with the top and the bottom surface of the horn. Note the different tools you can use when you hover your mouse over the app window (pan, zoom, rotate, etc.). Your mouse scroll wheel is also functional, so scrolling up or down will zoom in or out for example. You can download your current image view (far left camera tool icon), and you can download the outline of the oval or circle resulting from your plane’s intersection. We also created an extended version of this app on a separate full width page — which includes a dashboard with more metrics from each selection of X1 and X2.

> Gabriel’s Horn Intersecting Plane App

Conclusion

Viktor Schauberger (1885 – 1958) studied Gabriel’s Horn extensively for its ability to create structured water, and there are many other reasons why we should pay closer attention to the vortex formed by this graph.

We also wrote an article recently about the newly discovered Chestahedron Seven Sided Geometry, where we noticed a close relationship between that shape, the $\; y = 1/x \;$ graph, and Gabriel’s Horn. We encountered several calculations that contained small and unusual remainders in that article as well, so it’s an interesting read!

Gabriel’s Horn sits right on the boundary between the finite and the infinite, and this makes us curious about that small remainder we found between $\sqrt{13}/2$ and $9/5$. We’ll be looking into this further, so stay tuned!

Follow Us On Social Media

Also be sure to follow our social networks where you can stay up to date on our latest research! And feel free to share your thoughts in the Comments section below.

Cite This Page As:

The Math Zone. “Multiplicative Inverse And Gabriel’s Horn.” From MathZone.io — A Modern Exploration of Ancient Mathematics. https://mathzone.io/multiplicative-inverse-and-gabriels-horn/

Work Cited

Wikipedia contributors. (2024, June 29). Multiplicative inverse. In Wikipedia, The Free Encyclopedia. Retrieved 17:41, October 12, 2024, from https://en.wikipedia.org/w/index.php?title=Multiplicative_inverse&oldid=1231720726

Wikipedia contributors. (2024, October 11). Gabriel’s horn. In Wikipedia, The Free Encyclopedia. Retrieved 19:38, October 12, 2024, from https://en.wikipedia.org/w/index.php?title=Gabriel%27s_horn&oldid=1250545676