A Mathematical Model for Human Intelligence

Curiosity and Intelligence

People have been trying to figure out what intelligence is, and how it differs from person to person, for centuries. Much has been written on the subject, and some of this work has helped people. Unfortunately, much harm has been done as well. Consider, for example, the harm that has been done by those who have had such work tainted by racism, sexism, or some other form of “us and them” thinking. This model is an attempt to eliminate such extraneous factors, and focus on the essence of intelligence. It is necessary to start, therefore, with a clean slate (to the extent possible), and then try to figure out how intelligence works, which must begin with an analysis of what it is.

If two people have the same age — five years old, say — and a battery of tests have been thrown at them to see how much they know (the amount of knowledge at that age), on a wide variety of subjects, person A (represented by the blue curve) may be found to know more, at that age, than person B (represented by the red curve). At that age, one could argue that person A is smarter than person B. Young ages are found on the left side of the graph above, and the two people get older, over their lifespans, as the curves move toward the right side of the graph.

What causes person A to know more than person B, at that age? There can be numerous factors in play, but few will be determined by any conscious choices these two people made over their first five years of life. Person B, for example, might have been affected by toxic substances in utero, while person A had no such disadvantage. On the other hand, person A might simply have been encouraged by his or her parents to learn things, while person B suffered from parental neglect. At age five, schools are not yet likely to have had as much of an impact as other factors.

An important part of this model is the recognition that people change over time. Our circumstances change. Illnesses may come and go. Families move. Wars happen. Suppose that, during the next year, person B is lucky enough to get to enroll in a high-quality school, some distance from the area where these two people live. Person B, simply because he or she is human, does possess curiosity, and curiosity is the key to this model. Despite person B‘s slow start with learning, being in an environment where learning is encouraged works. This person begins to acquire knowledge at a faster rate. On the graph, this is represented by the red curve’s slope increasing. This person is now gaining knowledge at a much faster rate than before.

In the meantime, what is happening with person A? There could be many reasons why the slope of the blue curve decreases, and this decrease simply indicates that knowledge, for this person, is now being gained at a slower rate than before. It is tempting to leap to the assumption that person A is now going to a “bad” school, with teachers who, at best, encourage rote memorization, rather than actual understanding of anything. Could this explain the change in slope? Yes, it could, but so could many other factors. It is undeniable that teachers have an influence on learning, but teacher quality (however it is determined, which is no easy task) is only one factor among many. Encouraging the “blame the teacher” game is not the goal of this model; there are already plenty of others doing that.

Perhaps person A became ill, suffered a high fever, and sustained brain damage as a result. Perhaps he or she is suddenly orphaned, therefore losing a previous, positive influence. There are many other possible factors which could explain this child’s sudden decrease of slope of the blue “learning curve” shown above; our species has shown a talent for inventing horrible things to do to, well, our species. Among the worst of the nightmare scenarios is that, while person B is learning things, at a distant school, the area where person A still lives is plunged into civil war, and/or a genocide-attempt is launched against the ethnic group which person A belongs to, as the result of nothing more than an accident of birth, and the bigotry of others. Later in life, on the graph above, the two curves intersect; beyond that point, person B knows more than person A, despite person B‘s slow start.  To give credit, or blame, to either of these people for this reversal would clearly be, at best, a severely incomplete approach.

At some point, of course, some people take the initiative to begin learning things on their own, becoming autodidacts, with high-slope learning curves. In other words, some people assume personal responsibility for their own learning. Most people do not. Few would be willing to pass such judgment on a child who is five or six years old, but what about a college student? What about a high school senior? What about children who have just turned thirteen years old? For that matter, what about someone my age, which is, as of this writing, 48? It seems that, the older a person is, the more likely we are to apply this “personal responsibility for learning” idea. Especially with adults, the human tendency to apply this idea to individuals may have beneficial results. That does not, however, guarantee that this idea is actually correct.

I must stop analyzing the graph above for now, because the best person for me to examine, at this point, in detail, is not on the graph above. He is, however the person I know better than anyone else: myself. I’ve been me now for over 48 years, and have been “doing math problems for fun” (as my blog’s header-cartoon puts it) for as long as I can remember. This is unusual, but, if I’m honest, I have to admit that there are inescapable and severe limits on the degree to which I can make a valid claim that I deserve credit for any of this. I did not select my parents, nor did I ask either of them to give me stacks of books about mathematics, as well as the mathematical sciences. They simply noticed that, when still young, I was curious about certain things, and provided me with resources I could use to start learning, early, at a rapid rate . . . and then I made this a habit, for, to me, learning is fun, if (and only if) the learning is in a field I find interesting. I had absolutely nothing to do with creating this situation. My parents had the money to buy those math books; not all children are as fortunate in this respect. Later still, I had the opportunity to attend an excellent high school, with an award-winning teacher of both chemistry and physics. To put it bluntly, I lucked out. As Sam Harris, the neuroscientist, has written, “You cannot make your own luck.”

At no point in my life have I managed to learn how to create my own luck, although I have certainly tried, so I have now reached the point where I must admit that, in this respect, Sam Harris is correct. For example, I am in college, again, working on a second master’s degree, but this would not be the case without many key factors simply falling into place. I didn’t create the Internet, and my coursework is being done on-line. I did not choose to be born in a nation with federal student loan programs, and such student loans are paying my tuition. I did not create the university I am attending, nor did I place professors there whose knowledge exceeds my own, regarding many things, thus creating a situation where I can learn from them. I did not choose to have Asperger’s Syndrome, especially not in a form which has given me many advantages, given that my “special interests” lie in mathematics and the mathematical sciences, which are the primary subjects I have taught, throughout my career as a high school teacher. The fact that I wish to be honest compels me to admit that I cannot take credit for any of this — not even the fact that I wish to be honest. I simply observed that lies create bad situations, especially when they are discovered, and so I began to try to avoid the negative consequences of lying, by breaking myself of that unhelpful habit. 

The best we can do, in my opinion, is try to figure out what is really going on in various situations, and discern which factors help people learn at a faster rate, then try to increase the number of people influenced by these helpful factors, rather than harmful ones. To return to the graph above, we will improve the quality of life, for everyone, if we can figure out ways to increase the slope of people’s learning-curves. That slope could be called the learning coefficient, and it is simply the degree to which a person’s knowledge is changing over time, at any given point along that person’s learning-curve. This learning coefficient can change for anyone, at any age, for numerous reasons, a few of which were already described above. Learning coefficients therefore vary from person to person, and also within each person, at different times in an individual’s lifetime. This frequently-heard term “lifelong learning” translates, on such graphs, to keeping learning coefficients high throughout our lives. The blue and red curves on the graph above change slope only early in life, but such changes can, of course, occur at other ages, as well.

It is helpful to understand what factors can affect learning coefficients. Such factors include people’s families, health, schools and teachers, curiosity, opportunities (or lack thereof), wealth and income, government laws and policies, war and/or peace, and, of course, luck, often in the form of accidents of birth. Genetic factors, also, will be placed on this list by many people. I am not comfortable with such DNA-based arguments, and am not including them on this list, for that reason, but I am also willing to admit that this may be an error on my part. This is, of course, a partial list; anyone reading this is welcome to suggest other possible factors, as comments on this post. 

On Pertrigonometric Functions

Pertrigonometric functions are modifications of the three primary trigonometric functions. Unlike the familiar sine, cosine, and tangent functions, the “pertrig” functions include triangle perimeter in their right-triangle-based definitions, which are given in the bulleted list below. The longer form of “pertrigonometric functions” is “perimeter-based trigonometric functions,” and the shorter, informal version is “pertrig functions.”

  • The persine of an acute angle (abbreviated “pers”) equals the length of the side opposite that angle, in a right triangle, divided by the triangle’s perimeter.
  • The percosine of an acute angle (abbreviated “perc”) equals the length of the leg adjacent to that angle, in a right triangle, divided by the triangle’s perimeter.
  • The pertangent of an acute angle (abbreviated “pert”) equals the length of the hypotenuse of a right triangle containing this acute angle, divided by the triangle’s perimeter.

After defining these terms, I used Geometer’s Sketchpad to construct a right triangle containing a 10º angle, and then used the “measure” and “calculate” functions to find the values of pers(10º), perc(10º), and pert(10º). Since these are ratios, they would have the same values shown for larger or smaller right triangles which contain 10º angles.

pertrig functions

An observation: the pertangents of complementary angles are equal. Why? Because complementary angles appear in all right triangles, as pairs of acute angles in the same triangle. For each such complementary angle pair, therefore, the same triangle is used to define pertangent. The hypotenuse/perimeter ratio (which is pertangent) would, it follows, remain unchanged — because both its numerator and denominator remain unchanged.  This relationship does not hold for the tangent function; instead, the tangents of complementary acute angles are reciprocals of each other.

Of course, I wanted to know more than just the pers, perc, and pert values for 10º, but I had no desire to repeat the same calculations, many more times, to form a table. Instead, I simply graphed the functions, again using Geometer’s Sketchpad. The units on the x-axis are degrees, not radians.

pertrig functions B

In the graph above, the dark blue curve is the persine function, with the sine function in light blue, for comparison. Similarly, percosine is shown in red, with cosine shown in pink. Finally, pertangent is shown with a heavy, dark green curve, while tangent is shown as a thinner, light green curve.

Entering the equations for these curves was a little tricky, due to the fact that I wanted this graph to venture beyond 0 and 90 degrees, in both directions, on the x-axis. When that is done, the unit circle must be used (in place of right-triangle based definitions), simply because no right triangles contain angles outside this range. The radius of the unit circle is 1, by definition, and that is the hypotenuse of the right triangle which exists in the zero-to-ninety degree part of the domain of the graph above. As a consequence of setting the length of the hypotenuse of each right triangle at 1, the side opposite the angle in question (used for persine) becomes, simply, the sine of that angle, while the adjacent leg’s length is the angle’s cosine. It then follows that the perimeter (the denominator of the pers, perc, and pert ratios) is equal to sin(x) + cos(x) + 1.

Calculations are shown on the graph above, and you can click on the graph to enlarge it, to make them more readable. In these calculations, one more adjustment had to be made, and that was to the perimeter portion of each pertrigonometric ratio. Using sin(x) + cos(x) + 1 works fine for perimeter, for the zero-to-ninety degree portion of the domain, but, outside that, negative numbers intrude, for values of sin(x) and/or cos(x). It is my contention that triangle perimeter only makes sense as a sum of absolute values of a triangle’s three side lengths. To obtain absolute values for both sin(x) and cos(x) in the perimeter-part of each calculation, then, I simply squared each of these two functions, and then took the square roots of those squares. The result of this can be seen on the graph, in the curve for the pertangent function, which resembles a child’s drawing of waves in the ocean. On the y-axis, it never reaches as low as 0.4, and its maximum value is clearly exactly 0.5 — at the sharp “wave peaks.” At the (smooth) troughs, the actual minimum is equal to the square root of two, minus one, or ~0.414, although I have not yet figured out exactly why that is the case — I simply noticed it on the graph — but, to investigate it further, I know where to look: the 45-45-90 triangle, since these minima are hit when x = (45 ± 90n) degrees, where n is any integer. The pertangent function has the shortest period of all the functions shown above, at a mere 90º. For tangent, by contrast, the period is 180º. All four of the other functions shown have periods of a full 360º.

It is striking that the pertangent and tangent curves bear little resemblance to each other, while marked resemblances do exist between the persine and sine curves, as well as between the percosine and cosine curves. In informal terms, the persine curve is a shorter and spikier (but still recognizable) version of the sine curve (vertically, with the amplitude exactly one-half as great for the shorter persine curve, relative to the sine curve), but, horizontally, the two curves are synchronized. The same relationship holds for the percosine and cosine curves. Also, it is well-known that the cosine curve is simply the sine curve, phase-shifted one-quarter cycle (or 90º, or π/2 radians) to the left. This phase-shift relationship between the cosine and sine curves holds, precisely, for the percosine and persine curves.

There is a simple reason why persine, percosine, and pertangent all peak at exactly y = ½. All three functions generalize, for acute angles, to this ratio — (some side of a right triangle)/(perimeter of that same triangle) — and no side of any triangle can ever exceed, nor even reach, half that same triangle’s perimeter. In all three cases, the maximum y-value is only reached, even in the zero-to-ninety degree portion of the domain, for “degenerate cases” — angles of 0º or 90º, which are, of course, not acute angles at all. Interpreted as triangles, these are cases where either a triangle becomes so short that it collapses to a single segment, or the opposite degenerate situation: two parallel lines, connected by a single segment. If you try to make either (or both) of the acute angles in a right triangle into an additional right angle, after all, that’s what you get.

To my knowledge, no one has described these pertrigonometric functions before, by this or any other name, although I could be wrong. (If I am wrong on this point, please let me know in a comment.) Regardless of whether this is their first appearance, or not, I did not invent them. The reason for this is simple: nothing in mathematics is ever “invented” — only discovered — for mathematics existed long before human beings existed, let alone started writing things down. How do I know this? Simple: there was a universe here before there were people, and all evidence indicates that it operated under the same laws of physics we observe today — and all evidence to date also indicates that those laws are mathematical in nature. Therefore, with the “pertrig” functions, I either discovered them, or, if they have been found before, then I independently rediscovered them.

Finally, I’ll address that question so often asked, about numerous things, in mathematics classes: what are these pertrigonometric functions used for? As far as I know, the answer in this case, so far, is absolutely nothing, other than delighting me by their very existence. It is possible that this may change, for someone might find a way to make a profitable application of these functions — and I won’t get any money if they do, either, for I am not copyrighting any of this. Nothing in mathematics is subject to ownership.

Honestly, though, I hope no one ever finds any practical, “real-world” use, at all, for pers, perc, or pert. Right now, they are pure mathematical ideas, unsullied by tawdry, real-world applications, and, well, I like that. I am far from the only person who ever had such an attitude about a mathematical idea, either — such views are actually fairly common in the mathematical community. Most of those who try to discover previously-unseen things in mathematics do so solely, or primarily, for one reason: the joy of discovery, in its purest form.

Nested Trigonometric Functions: Sine and Cosine

The sine and cosine curves themselves are shown for reference, and all four possible combinations of sine and/or cosine which nest one function inside another (with values for the outer function to evaluate ranging from -2π to 2π) are also shown.

six functions

Working Towards a Solution of the Hemisphere Problem

The hemisphere problem referred to here was described in the previous post. To reword it somewhat, consider this hemisphere, half of a sphere of radius r. The orange cross-section is a circle parallel to the hemisphere’s yellow, circular base.

hemisphere

We are to find the height of the yellow section with the orange circular top (which I shall call x), as a fraction of r, such that the yellow and red sections above have equal volumes.

Since the volume of a hemisphere is (2/3)πr³ (that’s half a sphere’s volume), each of these two sections must have half the hemisphere’s volume, or (1/3)πr³.

Moreover, the top (red) portion is a “spherical cap,” described here on Wikipedia, as was pointed out to me, by a friend, on Facebook. On that Wikipedia page, you can find this diagram, as well as the formula shown below for the volume of the purple spherical cap in the diagram.

spherical cap

spherical cap volume

Now, as our goal is to find x, as described at the top of this post, it important to remember that r = x + h, where r is the radius of the original sphere (and height of the hemisphere), h is the height of the spherical cap, and x is the height of the hemisphere, after the spherical cap is removed. We now have two expressions for the volume of the spherical cap: (1/3)πr³ (because it is a fourth of the volume of the original sphere), and (1/6)πh(3a² + h²) from the Wikipedia article on the spherical cap (so all of this assumes, then, accuracy in that Wikipedia article). Setting them equal to each other,

(1/3)πr³ = (1/6)πh(3a² + h²)

Next, I’ll clean this up by multiplying left and right by 6/π, to cancel fractions and π from both sides.

2r³ = h(3a² + h²)

A right triangle exists in the blue-and-purple figure above, and the unlabeled leg is x, the problem’s original goal. I’ll add an “x” to this diagram.

spherical cap with xUsing this right triangle, and the Pythagorean Theorem, it can be seen that a² = r² – x². Also, since r = x + h, it follows that h = r – x. By substitution for a and h, then,

2r³ = h(3a² + h²)

becomes

2r³ = (r – x)[3(r² – x²) + (r – x)²], which then distributes to

2r³ = (r – x)(3r² – 3x² + r² – 2rx + x²), which expands as

2r³ = 3r³ – 3rx² + r³ – 2r²x + rx² -3r²x  + 3x³ – r²x + 2rx² – x³, which simplifies to

0 = 2r³ – 6r²x + 2x³, which becomes, by division:

0 = r³ – 3r²x + x³.

I’m trying to find x, as a fraction of r, meaning that x = kr, and I want k. On that basis, I’ll now substitute kr for each x in the last equation above.

0 = r³ – 3r²(kr) + (kr)³, which then becomes

0 = r³ – 3kr³ + k³r³, or

3kr³ = r³ + k³r³, and then dividing by r³ yields

3k = 1 + k³, which can be rearranged to

3k – k³ = 1, which factors,on the left side, to yield

(k)(3 – k²) = 1.

For k and (3 – k²) to have a product of one, they must be reciprocals. Therefore, 1/k = 3 – k². I can then graph y = 1/k, as well as y = 3 – k², and find the solution by seeing where the graphed functions intersect above the k-axis, with a “k” value between zero and one, since no value of k less than zero or greater than one would make sense, as a solution to the original problem. (I’ll be using k coordinates, and a k-axis, in place of the usual x coordinates and x-axis.) Here’s the initial graph:

graph1

The only intersection in the specified range of zero to one is between point A and point B, so I brought them together, as closely as I could get Geometer’s Sketchpad to let me.

graph2

With points A and B almost on top of each other, k = 0.34958 by one equation, and k = 0.34820 by the other. To two significant figures, then, I can conclude that the horizontal cut in the original problem should be made 35% of the way from the base of the hemisphere to the hemisphere’s top.

Using a graphing calculator, a more precise answer of 0.34729636 was obtained. I’d still like to have an exact answer, but this will do for now.

—-

Later addition: a helpful reader led me to a Wolfram Alpha site where I could get an exact answer, as well as a decimal approximation with a greater degree of precision. In the pic below, I have omitted the two solutions of the third-order polynomial which are not the one solution of interest. Here’s the one which is:

wolfram

Now, however, I have another mystery: how can an exact answer, with all those imaginary units in it, have a real-only approximation? To this question, at least for now, I don’t even have the beginnings of an answer.

—-

Even later post-script: I have been assured by friends on Facebook that the imaginary units in the above exact solution somehow cancel, although I must concede that I still do not see how, myself. I’ve also been shown another way to express the solution, for 2sin(π/18) is also ≈ 0.3472963553338606977034333. This surprised me, due the the lack of any explicit appearance by a π/18 (= 10°) angle in the original problem, and the fact that no trigonometric functions were used to solve it.

I Now Have Empirical Evidence for the Existence of My Own Brain!

Pic-03202015-001

A doctor needed to look at my brainwaves (and a bunch of other MSLs, also known as “medical squiggly lines”), as recorded during a sleep study, so of course I asked him if I could see them myself. Who wouldn’t want to see their own brainwaves?

The WM Function

As if the world needed another trigonometric function, I give it the WM function, named after the letters apparent in its graph. It is the sine of the sine of x, times 2π, if x is measured in radians, as in the graph below.

WM function

Trigonometric Stained-Glass Windows

trigart

To make the virtual “painting” above, I plotted simple and moderately-complex trigonometric functions on a single coordinate plane, as shown below, using Geometer’s Sketchpad. I then erased all the text, etc., copied-and-pasted a screenshot into MS-Paint, and used that program to make the finished image above.

trigartexpl