If your first reaction to mentions of mathematics is, 'Ugh, math is stupid and boring,' then I invite you to read A Mathematician's Lament by Paul Lockhart, who agrees with your assessment and shows why math is actually fun.
Everything from this point on is very basic, involving very little computation, and what there is is only trivial integration, which you need not understand to get the proof. It, like most proofs in basic mathematics, can be shown with natural language and diagrams. Now, on with the show.
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Question: What ratio of the volume of a 4-D cube is taken up by the largest 4-D pyramid that fits inside it? If possible, what is the general expression for the same problem on all dimensions n?
Guess: Without working out a proof, I posit that the answer is likely to be ¼. This is extrapolated from the n=2 and n=3 versions of the question, for which the answers are ½ and ⅓ respectively. These facts are embodied in the formulas for area and volume of a triangle and pyramid: lh/2 and lwh/3, respectively.
Reasoning
In his open letter A Mathematician's Lament, Paul Lockhart gives an elegant visual proof of the n=2 answer. Drawing a line through the top vertex of the triangle gives us two new rectangles, each of which is clearly half filled by two new triangles. Thus, the whole triangle is half the whole rectangle.
Let us assume from this point that all squares, cubes and hypercubes in our examples are in their unit form; all sides equal 1.
While the 3-D version is subject to a similar visual proof, by splitting it into 4 segments along the z-axis centered around the top vertex, it is much less useful, as these objects are much less obviously one third of their respective cuboids. If the 3-D visual proof is not self-evident, we may assume that the 4-D version will be less so.
Back to the drawing board.
Idea: We can reach ‘up’ a dimension without straining human intelligence by using time as the 4th dimension.
Proof of concept in 2-D: We can adequately represent the 2-D version of this problem using only a single spatial dimension, and one temporal dimension. Consider that the ‘base’ of our 2-D pyramid is a line, which coincides exactly with the bottom side of the square.
We may draw horizontal lines across the figure, and in the process take a cross-section of the triangle. This cross-section takes the form of points on a line. Consider the cross sections taken at the heights h=0, h=½ and h=1
The first has the triangle points coinciding exactly with the bottom corners of the square. The second places them at ¼ and ¾ on the line [0, 1], leaving exactly half the line between them. Finally, the third has a single point the center of the line; the top vertex of the triangle.
But what use are these? These segments, when laid out, are precisely the 1-D representations of the 2-D figure in the temporal dimension, at t=0, t=½, t=1.
Proof of concept in 3-D: We may expand to the 3-D version, replacing the height dimension once again with time. The cross-sections are now taken by horizontal xy planes at z=t, and the pyramid is adequately represented by a shrinking square set within the static square of the cube.
Proof of concept in 4-D: The method holds, and so we export to the fourth dimension. The cross sections now take the form of growing cubes, set within static cubes. These, of course, represent cross-sections taken of the hyper-pyramid by R^3 spaces in R^4 space (my non-specialist vocabulary breaks down here, but the metaphor yet functions) at א=t.
(I use the א though it has other meanings in higher mathematics for two reasons. First, it allows me to flex my usage of a Hebrew keyboard, a rare event indeed. Second, on such a keyboard, א is in fact tied to the t key, which I consider kabbalistically appropriate).
Calculation
Now that we have a robust model for pyramids in cubes in higher dimensions, how can we calculate their exact volume? My calculus professor, Shiva Chidambaram, encouraged me to try using integration to solve the problem, since we are currently learning multivariable integration. However, he also informed me that there is a simpler proof that requires only single variable integration.
Going back to the 2-D example, we can backfit a simple solution.
Our slope for the pyramid, as shown here, is y=x. Replacing height with time, and redrawing our triangle for simplicity, we can show simply that f(t)=x.
The general integral is trivial: F(t)=(x^2)/2. We can integrate on our bound of [0, 1] with similar ease, and the result is ½, as expected.
Scaling up to 3-D, we find ourselves now calculating the volume of the pyramid from square cross-sections. By redrawing the examples shown previously, it is clear that the area is given by f(t)=x^2.
Our integral, similarly trivial, is (x^3)/3, and on [0, 1], integrates to ⅓.
Having shown the above, I am confident in scaling this approach up to 4-D. The size of the cross section cubes is given by f(t)=x^3, the integral by F(t)=(x^4)/4, and on [0, 1] is equal to ¼. This confirms my original guess. Score one for intuition.
Quod Erat Demonstrandum
Further, I am confident in scaling this approach up to all dimensions n. For the volume of an n-dimensional pyramid set in an n-dimensional cube, the volume may be given as (s^n)/n, where s equals the length of the cube’s sides, and the ratio of volume is simply 1/n.
So for a 3-D cube of side length 4, the pyramid’s volume is 64/3. For a 5-D cube of side length 3, the pyramid’s volume is 243/5, and so on.
QED
Remaining question
All this begs the question. Is there a simpler proof that the one outlined above, preferably one that does not require even trivial integration? That is yet to be seen, and I refuse to look up the answer.
Musical observation
Mozart’s Don Giovanni is perfectly adequate listening material for composing mathematics. Madamina (tin tin tin) il catalogo e questo...
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