Cosmology

[desiresjab]

All the fancy math in the world is only fancy accounting. The speed of a falling object towards earth owes x to mass, t to time spent falling and r to air resistance; current owes so much to voltage and so much to resistance, and they have a relationship which can be stated in symbols:

I=E/R. Current equals voltage divided by resistance. Accounting.

Sometimes events in the universe are so fancy that it takes highly fancy accounting to account for them. Sometimes specific new accounting tools have to be developed to account for something, but it is always just accounting. Look how fancy and specific some of the tools of mathematicians are. They could not be used to do your bookkeeping. They do the bookkeeping for various aspects of the behavior of objects in the universe which you cannot use everyday arithmetic for.

The awesome thing is, sometimes the accounting tools are discovered and developed before the phenomena themselves are observed!

As for equations, the same old usual suspects apply across every field. An extraordinary number of phenomena around us are described by ordinary first degree equations. The equations that account for other phenomena are more complex and of a higher degree. Quadratic equations bit off another big hunk of what is understandable and can be accounted for.

Even the eight-miles-to-a-side matrices used in the attempts by physicists to simulate or reproduce consciousness, are but attempts to account for consciousness the way -16t² accounts for the position of a falling object.

[desiresjab]

When Newton fully generalized the binomial theorem for the first time, what was he up to? This provides one of the best examples of the accounting tool being developed before the phenomenon it accounts for is even suspected, in this case phenomena many times over. The bionomial theorem is one of the most universally applicable from biology to backgammon to sociology. To Newton it was an exercise in arithmetic. He could have had no idea of the landslide of applications to come, even if he felt certain there would be some applications somewhere or th'other.

I am always comforted that at least a few of our brightest are doing pure mathematics. We are not the first. We could ask ourselves a big question. How did the ancients do it?

How did east Injun mathematicians before the time of Christ come up with at least a way to calculate individual binomial coefficients? We cannot be sure they did not already have the general method Newton developed. An even greater question is: Why? Why would they do it? It is almost certain they had no application. In fact, it looks like humans had to develop the ideas behind the binomial theorem at least three times in our history, forgetting it twice, apparently because it had no use.

The same thing that drove Newton must have driven those second century B.C. east Injuns and twelfth century Moslems who were onto the theorem before. It was about the behavior of numbers. I think they were simply playing at a form of play that requires a lot of concentrated mental focus.

Now, if IQ measures anything it measures the ability to do this kind of abstract accounting. It seems unlikely to me that ancient east Injuns would have measured out at an average (over the entire population) higher than modern Injuns on IQ tests, an average we know is lower than the west, lower than China and lower than Ashkenazi Jews, yet they did get this job done, i.e. the binomial theroem, the coefficients of its expansion.

This tells me human populations with an average as low as 85 will get the job done. They will occasionally produce geniuses enough standard deviations to the right to produce mathematical acheivement. This lower average was perhaps unable produce enough geniuses to sustain mathematics through startup technologies it could be the accounting firm for.

However, we should not kid ourselves that the average European slopping through mud on his way to mass in the fourteenth century would have an IQ of 100 by modern standards, if we had a time machine and could test them. In point of fact, this person is likely right on par with a real dummy these days.

But this European had one mighty advantage that none before him had possessed--Gutenberg. Ideas could be disseminated as never before. Without Gutenberg mankind might have forgotten the binomial theorem a third time.

[YesNo]

Prove that for n ε N, (n!)! is divisible by n!^(n-1)!.

I was thinking of using a mathematical induction argument on this, but I only got as far as the base cases for n = 1, 2 and 3. However, I don't know how to get the inductive step to work algebraically. That is given that the statement is true for n, how can I algebraically manipulate the factorials so that it has to be true for n + 1?

So, I looked up the problem. Here is a combinatorial solution that I am not totally convinced of, but I assume is correct: http://math.stackexchange.com/questi...isible-by-nn-1

What I liked about that solution was the way the person who answered it rewrote the problem as a fraction and then reinterpreted it as a combinatorial problem that must have an integer as a solution. Hence the numerator is divisible by the denominator.

One might also be able to use the gamma function which would allow one to rewrite the problem as integrals. But I didn't see how that would make the inductive step any easier to calculate.

[desiresjab]

I was thinking of using a mathematical induction argument on this, but I only got as far as the base cases for n = 1, 2 and 3. However, I don't know how to get the inductive step to work algebraically. That is given that the statement is true for n, how can I algebraically manipulate the factorials so that it has to be true for n + 1?

So, I looked up the problem. Here is a combinatorial solution that I am not totally convinced of, but I assume is correct: http://math.stackexchange.com/questi...isible-by-nn-1

What I liked about that solution was the way the person who answered it rewrote the problem as a fraction and then reinterpreted it as a combinatorial problem that must have an integer as a solution. Hence the numerator is divisible by the denominator.

One might also be able to use the gamma function which would allow one to rewrite the problem as integrals. But I didn't see how that would make the inductive step any easier to calculate.

I cannot ever remember using induction successfully with a problem other than homework assignments long ago. It is not one of the techniques. The notation is difficult and requires brain-wracking precision. It is not one of my techniques, which I will eventually regret.

Some of these problems come from textbooks, and the student is expected to use the same techniques studied in each section to solve the problems. You might have noticed I seldom do this. I use my own bag of preferred tricks on almost everything, picking up additional information as I go.

One thing I find torturous is long, algebraic manipulations looking for derivations. This was Euler's style. I avoid that when I can. Induction would be a good example. Several times in my solutions, however, I do point to induction as the final step of a process without actually doing it, if I am pretty sure of myself.

* * * * *

Both are merely factorials, it should suffice to show the denominator is smaller than or equal to the numerator, since any smaller factorial divides a larger one.

n!^(n-1)!=₁n! ₂n! ₃n!...(_n-1)n!

(n!)!=n!_(n!)!Π⁽ⁿ⁺¹⁾ i)

n!(_(n!)!Π⁽ⁿ⁺¹⁾i)
₁n! ₂n! ₃n!..._(n-1)n!

Obviously, a factor of n! could be canceled in this fraction. That does not yet show the top is at least equal to the bottom.

The concept is the one we used before with this notation. The numerator consists of factors which are all higher multiples of all the factors of the denominator.

Look at the fraction again. The numerator is making itself _(n!)!Π⁽ⁿ⁺¹⁾ i) copies of n!, leaving at least one copy after cancellation of every factor in the denominator, since .

Note: The subscript and superscript on the multiplication operator capitol Pi are not powers and only show where to begin and end the multiplication sequence. I am unable to put both of them on the same side of the operator, or I would.

[YesNo]

Even the eight-miles-to-a-side matrices used in the attempts by physicists to simulate or reproduce consciousness, are but attempts to account for consciousness the way -16t² accounts for the position of a falling object.

I was thinking about this while we were walking around Oak Park looking at early Frank Lloyd Wright houses. I don't think it is possible to reproduce consciousness with a mathematical structure.

One might be able to simulate some aspects of consciousness or find correlates of consciousness in the brain, but at some point this fails to reproduce consciousness. The reason is because the mathematical structure, based on determinism and randomness, cannot make a choice. However, consciousness could be characterized as having an ability to make a choice, no matter how limited. That implies that the property of making a choice cannot be mapped to a deterministic-random structure.

This property of making a choice is not the "hard" problem of consciousness. That would have to do with subjective experience and "qualia". That so-called hard problem limits consciousness to sentient animals (and perhaps plants) who can be expected to have subjective experiences.

This is a harder problem of consciousness and allows anything, including indeterministic quantum reality, to be conscious in its own way with or without qualia.

Of course the counter argument is that nothing is able to make a choice which is why the indeterminism of quantum reality is shocking. If one defines choice as being outside a deterministic/uniformly random structure, then that quantum indeterminism can be interpreted as making a choice. But we don't have to deal with reality directly at that level. We see it as larger clumps where it behaves more predictably. We can make tables, chairs and computers out of it and think it is all mathematically predictable because those tables, chairs and computers as tables, chairs and computers don't make choices. They are not conscious as such.

[desiresjab]

I was thinking about this while we were walking around Oak Park looking at early Frank Lloyd Wright houses. I don't think it is possible to reproduce consciousness with a mathematical structure.

One might be able to simulate some aspects of consciousness or find correlates of consciousness in the brain, but at some point this fails to reproduce consciousness. The reason is because the mathematical structure, based on determinism and randomness, cannot make a choice. However, consciousness could be characterized as having an ability to make a choice, no matter how limited. That implies that the property of making a choice cannot be mapped to a deterministic-random structure.

This property of making a choice is not the "hard" problem of consciousness. That would have to do with subjective experience and "qualia". That so-called hard problem limits consciousness to sentient animals (and perhaps plants) who can be expected to have subjective experiences.

This is a harder problem of consciousness and allows anything, including indeterministic quantum reality, to be conscious in its own way with or without qualia.

Of course the counter argument is that nothing is able to make a choice which is why the indeterminism of quantum reality is shocking. If one defines choice as being outside a deterministic/uniformly random structure, then that quantum indeterminism can be interpreted as making a choice. But we don't have to deal with reality directly at that level. We see it as larger clumps where it behaves more predictably. We can make tables, chairs and computers out of it and think it is all mathematically predictable because those tables, chairs and computers as tables, chairs and computers don't make choices. They are not conscious as such.

Good post. But then, I did not say I thought it was possible or impossible, just that that is what the boys and girls with the eight-milies-on-a-side matrices were trying to do, some very fancy accounting.

Because we are unable to define consciousness adequately it would be harder to reproduce and know if we had. Movies are full of rebellious machines.

Personally, I would not be surprised if some connection between consciousness and prime numbers exists. That is why, of all the unsolved problems in elementary mathematics, I consider the Goldbach conjecture the most important, for it is unlikely that it would be proven without shedding much light on the additive properties of primes, something almost nothing is known of.

I will post a problem now my bag of tricks seems insufficient for. I will have to pick up some new tricks, unless I notice a way I have missed. The problem looks simple enough.

Let n=n₁+n₂...+n_k where n are nonnegative integers. Prove that the quantity

n! ............................... Is an integer.
n₁! +n₂!...n_k!

[desiresjab]

Something is wrong! I must have stated the problem wrong. Maybe it asks for under what conditions, because 5+2=7, but 5!+2! does not divide 7!

I will have to recheck the wording of this problem, since a counter example is so available.

[YesNo]

I checked it doesn't work with 5, 2 and 7 using a Jupyter notebook as you noted. However, 3, 4 and 7 works.

I think the following is true: The sum of the factorials of two consecutive positive integers divides the factorial of their sum.

It doesn't work for 0, 1 and the sum 1 so the integers have to be positive.

A more general statement might work: The sum of the factorials of consecutive positive integers divides the factorial of their sum. I don't have a proof for this. It is just a conjecture. It looks like a nice pattern.

I wonder if the following works? Does the sum of the factorials of consecutive primes divides the factorial of their sum? That would be another nice pattern. If it is true.

[desiresjab]

I have to find the question and read it again. I think it is related to De Polignac's formula, because that is the section I found it under. It was also a solved problem.

Those other patterns are interesting. The answer I have to give to them is: I don't know. It is the kind of stuff you might find in the sources I am consulting.

In the meantime, here is a problem that is a killer. It might not look like it, but it is.

Prove that 7 divides (2222⁵⁵⁵⁵+5555²²²²).

It cannot be done what seems the intuitive way of reducing everything in sight (mod 7).

That is the scary part. I always thought you could reduce with impunity any time you felt like it or it was handy. Apparently that is not the case, otherwise 3⁴+43 would be divisible by 7, but it is not. This simple looking problem shows the traps involved with congruence theory. I wish I knew why 3⁴+4³ does not work, but the truth has not sprung on my brain yet.

I did not solve this problem, I looked at the answer. It took a while before their answer made sense, because I have a stubborn brain. I see why their answer is correct, but I do not see why the simpler version is incorrect. It is another problem in itself. I will not be able to let go of this until I understand it. I foresee torture.

I will let you look at this one for a spell while I go look up the other one.

[desiresjab]

I made a horrible oversight. No wonder the proposition was not true as stated. All the separate n_k's sum to n all right. but then he asks if the product of those numbers (not the sum as I printed) divides n! and is an integer. It now looks easy, but the Polignac Formula and method they are using is complicated.

Now that I have righted the mistake in the question itself, I am wondering if some of my old tricks might solve this one easier than all of De Polignac's torturous Eulerian algebra. Will get back.

[desiresjab]

Prove that if n₁+n₂.... +n_k=n, then

......n!.........
n₁! n₂!... n_k!

is an integer.

That is the problem stated correctly. I does not seem like a monster now. Maybe it is, though. Later.

[YesNo]

Not having the plus sign in the denominator makes this look easier. The first n₁ easily cancels.

For my last conjecture about consecutive primes, a counterexample would be 5 and 7 showing the conjecture is false.

Edit: What you are asking is if the coefficients of the terms in the "multinomial theorem" are integers. Here is some discussion of it: https://en.wikipedia.org/wiki/Multinomial_theorem

They would be expected to be integers since they are involved with counting elements in sets. That would mean that the fraction above is an integer.

[desiresjab]

I have been away for a few days.

What you say about the coefficients is quite likely true. While I was away I only thought about the problem in my head, no paper or computers. My goal on many such problems is to reduce the solution to a simple diagram and a little algebra. Just how elementary can this grade of a problem be made to look? Look how far Eisenstein reduced QR with his simple diagram. This is a much simpler problem. The diagram will be a lot simpler. I already know the diagram. What I have left to do is the little bit of algebra. I am rather tired after my drive. It might be tomorrow.

[desiresjab]

x+y=n...n-y......n-x

*
*
*
*............*
*............*..........*
*............*..........*
*............*..........*

n............x..........y


x+y has as many units left over at the top as n-x has total units. Those three large numbers as a product will be greater than what the three smaller numbers in y! produce. This proves the numerator is always greater than or equal to the denominator, a condition for an integer.

n will always have y units left over at the top after it matches every unit of x. In the case above, n has three units left over, i.e. three consecutive numbers. Being of equal length as y, that stretch of consecutive integers is forced to contain every factor of puny y factorial, always getting the first one of "1" for free. Getting that first factor of "1" for free guarantees that it will not miss any factors of y factorial.

Now, if we imagine placing y directly on top of x, we see that 5 is a multiple of 1, 6 is a multiple of 2 and 3, and 7 is a prime. 7 had to be a prime, because there were no more factors, which had just been used and could of course not be used again the very next number.

This proves they will always divide evenly.

This "proof" is intuitively strong, I believe, but of course it is algebraically weak. I could use fewer words, but I prefer to be as clear as possible for anyone making an effort to follow. A modern algebraist would not use many words at all. Wordy proofs are sort of antique in nature. But the fact is, if you can show it to yourself verbally, you will truly see it.

Suppose I had chosen 6 and 1 for the additives, instead of 4 and 3. That is the easiest case of all, because 1! affects nothing, and we can plainly see that 6! is a factor of 7!.

I am experimentally certain one could go a step further and show that (4!)(3!) is a factor of (5!)(2!), (5!)(2!) is a factor of 6!... right on up the line as far as it goes. I can see that but I have not demonstrated it.

[YesNo]

I trust algebraic proofs as much as I trust the results of a computer program. There could be something wrong in either that has little to do with the original problem. Being convinced is the intuitive part of the proof.

One way that I think about factorials is to view them as the number of ways to order n objects. In the first position one could have n choices. After that choice there are n - 1 choices for the next position and so on all the way down to the last piece and there is only one way to order it. Multiply all those together and you get n! For the multinomial formula coefficient the numerator is the way to order n objects. The denominator is the way to order subsets of those objects separately.

I was looking at Carmichael's "Theory of Numbers" and he approaches the problem by considering the highest power of a prime p in n! which I am not familiar with but sounds interesting. Then if the highest power of a prime in the denominator is less than or equal to the highest power of the prime in the numerator the fraction would be an integer. This might be more useful than the multinomial coefficient approach since it could answer more questions.

I agree that it is good to find multiple ways to solve something. The easier, the better.