A visual demonstration does not require
algebraic manipulations or calculations complex, just images that
constitute in themselves the proof. These images are usually in 2D
because the pages of our books are. Some use pieces or
three-dimensional figures, but... can we visualize
higher dimensions? Can we demonstrate theorems visually in the fourth dimension?
In this video, we will learn to see in 4D and convince ourselves that this is possible!
For this, we will go back to one of the most famous episodes in the History
of Mathematics. An episode that concerns Carl Frederich Gauss himself... rather
Gauss when he was a 5-year-old schoolboy, in his arithmetic class the teacher
proposed that they sum the numbers from 1 to 100 to keep them entertained, to which young
Gauss responded almost instantly with 5050! How could Gauss perform
this calculation so quickly? One way to see it is through a
visual demonstration. Summing the numbers from 1 to 100 is equivalent to counting how many squares
there are in a staircase with a base and height of 100. Counting the number of squares in this triangle
can be a complicated and tedious task, but paradoxically, counting how many squares there are
in TWO identical triangles is much easier. How is this possible? The reason is
that we can organize the information more efficiently. We can rotate one of
these triangles and move it so that it fits with the previous one forming a rectangle.
Counting squares in a rectangle is an easy task because we only need to multiply the number
of squares in the base 101 by the number of squares in the height 100. In this way, we have
that twice the sum from 1 to 100 is 101 times 100, which gives 10100, and dividing this product
by 2, we have that the sought sum is 5050. If instead of summing the numbers from 1 to 100
we want to sum the first n natural numbers, the resulting formula would be obtained by the
same argument and would be n times (n+1) divided by 2. This visual demonstration is very surprising
for its efficiency: changing 99 sums for just one multiplication and one division
by 2. But also for its simplicity. We wonder if this idea has analogs
in higher dimensions. Let's see: If we want to sum, this time, the
first n square numbers, we would have to calculate 1 squared plus 2 squared,
plus 3 squared and so on up to n squared. And this is equivalent to counting how many little cubes there are
in a pyramid with a square base of side n. Is there a simple formula for
the number of little cubes in this pyramid? Surely you are not surprised if I tell you that,
although this is difficult, it is easier to count how many cubes are in three pyramids like this.
Indeed, we can rotate, turn, and move these three pyramids in three-dimensional space to form a block.
The idea is that counting the number of cubes in a block is reduced to multiplying its dimensions.
But, in doing this, we see that we do not get a perfect block because the last layer of this
block has two levels. This is a small inconvenience that is resolved with a bit of
masonry. We must take out our radial and cut the cubes of this last floor so that we can
fit them and get our perfect block, which in mathematics is known as a parallelepiped.
Now we would have that the width of the block is n, the length is n+1 and the height is n + one half. Therefore,
3 times the sum from 1 squared to n squared is n times n+1 times n + one half. And by isolating the three that multiplies
we have a very simple formula. The sum of the first n squares is equal to n
times n+1 times n + one half divided by 3. The next case, the sum of the first
n cubes, also admits a very simple and surprising formula known as the THEOREM OF NICOMACHUS in honor of the Neopythagorean mathematician Nicomachus of
Gerasa to whom it is referenced in his works. Let's see the visual demonstration
of this theorem because it is very beautiful! This time we have n cubes with edges of 1, 2, 3, up to n. And we want to count how many
little cubes there are in total. To be able to sum these little cubes efficiently we will
decompose each of these cubes into their square floors and if the number of floors is
even we will split the last one in half. These pieces can be attached to the previous pieces
forming a square. For example, the cube with an edge of three will be separated
into three square floors. And these three squares can be attached to the previous square
forming a larger square. The next cube has an even number of floors so
the last floor must also be split into two pieces as well. These two pieces are
attached to the sides of the previous square and the remaining square floors fit perfectly
forming a larger square. In this way, to count the number of little cubes
of the n cubes with edges 1, 2, 3, up to n, it is enough to count the little cubes of this square and for
that we only need to do side by side. Considering that the side of this square is the sum
of 1+2+3 up to n, we obtain that the sum of the first n cubes is precisely the square of
the sum of the first n numbers, AMAZING! If we analyze the three previous visual demonstrations, we observe that, while the first two are completely analogous, the third, that is, the Nicomachus Theorem, is quite different.
The formula does have some analogy with the two previous cases because, if we consider that
the sum of the first n natural numbers could be written as the quotient n times n+1 divided
by 2, raising this fraction to the square we obtain n squared times n+1 squared divided by 4.
In this way, the sum of the first natural numbers is a quotient with denominator 2
and two factors in the numerator, the sum of the first n squares is a quotient
with denominator 3 and three factors in the numerator and the sum of the first n cubes is a quotient
with denominator 4 and 4 factors in the numerator. But the first demonstration was in 2D and had
denominator 2, the second visual demonstration was in 3D like the denominator. However,
the demonstration we saw of the Nicomachus Theorem does not take place
in the same dimension as the denominator. Will we be able to give a demonstration
of the NICOMACHUS THEOREM in 4D? Is it possible to give a demonstration in
dimension 4 by taking 4 pyramids of 4-dimensional hypercubes that generalize
the previous two demonstrations?? Do we have any way to see the
fourth dimension that allows us to continue with our project?
The answer is YES, but to understand it we have to think about how a three-dimensional object
could be seen by a being from a flat universe. Let's understand the three-dimensional pyramid
as if we were beings from Flatland, who can only visualize 2 dimensions.
Instead of observing the complete pyramid, we can take a 2D X-ray of the pyramid so that each cube is
scanned only once. In this way, instead of counting how many
little cubes there are in a 3-dimensional pyramid a Flatland being can count how many
squares there are in its flat X-rays and the total number is the same.
If we start with the highest floor we would have a first photograph
formed by a solitary square. We continue with the lower floor and we would have
4 squares, that is 2 squared. The immediately lower floor would provide us
with a photograph of 3 squared, that is, 9 squares and so on until
we reach n squared squares. It is no surprise that there are as many cubes in
the pyramid as there are squares in these sections because precisely the number of cubes in the
pyramid was the sum of the first n squares. We will call this collection of squares the
main sections which is nothing more than what is known in technical drawing
as the plan of our pyramid. The advantage of our scanning technique
is that there is not only one way to make a 2-dimensional X-ray of
the 3-dimensional pyramid. We can make a side scan of the pyramid
What sections will we get in this case? The first X-ray in this case is
precisely a triangle with a base and height of n which coincided with the sum of the first n numbers.
The next X-ray is the same triangle, but truncated, so
it has lost the upper floor. The following X-rays give us
all the truncated triangles. We will call these sections formed by truncated triangles
the secondary sections which correspond to the profile of the pyramid.
But that is indeed somewhat surprising. Since the number of squares in these
sections must coincide with the number of little cubes in the pyramid, both must be
equal to each other. We are affirming that the sum of the first n squares is equal to
the sum of the first truncated natural numbers. But, although this video is about visual demonstrations,
allow me to see this theorem also algebraically.
We want to see that the sum of the truncated triangular numbers coincides with the sum
of the first squares. But if we make these sums instead of orderly horizontally,
vertically, we have that, 1 is the same as 1 squared. 2 times 2
is the same as 2 squared, 3 times 3 is 3 squared and so on until n times n is n
squared. And the theorem is proved. The advantage of this algebraic version
is that it allows it to be extended to higher dimensions and see that the sum of the
little cubes in the truncated pyramids, that is, the sum of the first n squares and the
successive sums losing one term each time is equal to the sum of the first cubes.
Indeed, if we sum vertically 1 squared is the same as 1 cubed.
Two times 2 squared is 2 cubed, 3 times 3 squared is 3 cubed and so on
until n times n squared is n cubed. This will be of great importance to prove
our Nicomachus's Theorem in 4D. but before we get to that, let's train with
our section technique and see the demonstration of the sum of the first n squares
using 3 three-dimensional pyramids as if we were beings from Flatland.
Remember that for this we rotated our pyramids and fit them forming a block to
which we had to do some DIY. This block was a figure in dimension 3 to
which we are going to do our 2D scan to obtain the same information, but in dimension 2.
Let's get to it! Let's start visualizing just one of these pyramids,
the blue one. If we take an X-ray choosing one direction we will obtain secondary
sections in all cases except one. In this way, we obtain the
secondary sections of the blue pyramid, which as we already know are truncated triangles.
If we add a second pyramid and scan the figure again. We will have the same
truncated triangles as before to which we will add some new red truncated triangles.
A detail of great importance is that the first of these sections coincides with the visual
demonstration of the previous dimension, that is, with the sum of the first n
numbers using two triangles. The following sections will be
identical to the first, but truncating the triangles one floor more in each section.
As we see this truncation leaves free a space that has an “almost square” shape.
Finally, if we add the third pyramid it will be oriented so that the sections
are now main and will therefore give rise to squares. These squares occupy
the free space left by the truncated triangles and not being entirely square
we have to adjust these protruding squares. Precisely this was the reason why
it was necessary to do some masonry work to adjust these protruding squares.
But in our sections, which we can see as 2-dimensional puzzles, this part is
even clearer, as what we have to do is cut the pieces of the first
row and exchange pieces between the first and last section, second and penultimate, etc.
In this way we obtain identical rectangles for which the length of the block coincides
with the base of the rectangles. The height of the block coincides with the height of the rectangles.
But... how is the width of the block interpreted in our 2D sections? Indeed! The width
of the block coincides with the total number of rectangles because there are as many sections, and
therefore as many rectangles, as the length of the block in the direction we were taking the
X-rays, that is, the width of the block. In short, the number of little cubes in
three three-dimensional pyramids is obtained by multiplying n, the number of rectangles,
by their bases and heights, that is: n+1 and n+1/2. And from here our
formula was obtained by isolating the 3. Perfect! We are now trained to understand the
visual demonstration of Nicomachus's Theorem in 4D. Let's remember that this theorem gives us a formula
for the sum of the first n cubes, that is, we wanted to count how many little cubes there are
in a collection formed by a little cube a cube with an edge of 2, another with an edge
of 3 and so on up to a cube with an edge of n. But why do you think we colored these
cubes the same color? In the classical demonstration of Nicomachus's Theorem we colored the
cubes differently for a reason: they were different objects that
were decomposed into smaller pieces to form a kind of square tapestry.
In our Nicomachus's Theorem 4D the different cubes are of a single color because they are
actually the same object! It is altogether a 4-dimensional pyramid seen through its 3D main sections. But besides the main sections, we could
see a pyramid by secondary sections. The first of these 3D sections of the 4D pyramid
is precisely a 3D pyramid. And the following sections were truncated pyramids.
Indeed! We had seen algebraically that the sum of cubes, which are the main
sections of the 4D pyramid, coincided with the truncated sums of squares, which are
the secondary sections of the 4D pyramid. Counting the number of hypercubes in a
4D pyramid can be a difficult task, but this can be simplified if instead
of a 4D pyramid we take 4! But as we were working with sections,
we start by taking 3 pyramids seen in their 3D secondary sections. The first of these
sections is exactly the visual demonstration of the sum of powers in the lower dimension,
just as it happened in the 2D demonstration of the sum of squares we already saw.
Next, the same movement of pieces is done in all sections with
truncated pyramids and what we get are blocks in which an
“almost cubic” gap is left, getting larger each time. In these gaps, we fit our
fourth 4D pyramid seen in its main sections. These main sections
were cubes and therefore fit almost perfectly in the different gaps.
As they were left a bit out of the piece, now is when we have to do the
masonry work to fix these blocks. In the last section we separate
the nxn square slab that protrudes. In our image n is 4, so we have a
4x4 square that we separate into bars of length 4. And now we see that in each section there is a
space of length 4 free that appeared because the upper floor of the blocks was not complete.
It is in each of these spaces where we will fit the bars of length 4. The result
of this fitting operation is that the last section is already a perfect block and we have to
keep working with the remaining sections. In the penultimate section we have a
3x3 square slab that protrudes from the gap. We separate it from the block and divide it
into three bars of length 3. In the remaining sections we see a space of length 3 empty
in which we will fit our three bars. In this way, we see that the penultimate section
is already a perfect block and we will have to proceed similarly with each of the sections.
In the second section we have a 2x2 square slab that is divided into two bars of length 2
that fit into the first and second section and finally in the first section we have a
cube that protrudes and a gap to fill. Once this last cube is fitted we have
n 3D blocks of identical dimensions. Length n+1, width n, height n+1. And since
there are as many blocks as sections, that is, n, to count the hypercubes in
4 4D pyramids, or what is the same, calculate 4 times the sum of the first cubes, it is enough to
multiply n squared by (n+1) squared. And to get to the known formula,
it is enough to isolate the 4. You may wonder what happens in
higher dimensions. This problem, of summing the powers of the first
natural numbers for any exponent has a formula known as Faulhaber's formula in honor of the German mathematician Johann Faulhaber who
studied the cases up to exponent 17. However, the general formula is due to
Jakob Bernoulli and was published in 1713 in his posthumous work ‘Ars Conjectandi’. In fact, the
numbers used in this formula are known as Bernoulli numbers and appear
in countless parts of mathematics. So much so that the pioneer of
computing ADA LOVELACE dedicated nothing less than the first program in history
precisely to the calculation of these numbers… Faulhaber's formula is demonstrated purely algebraically,
but can we demonstrate more cases in higher dimensions
visually using our technique? In dimension 5, Faulhaber's formula can be written
as follows and a mysterious irreducible factor appears in the numerator in the rationals and its visual explanation
has resisted the mathematical community. If you are interested in knowing the visual explanation of this mysterious irreducible factor,
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