The problem of differential calculus

@1&onlybillyshears

It seems to me that you are confusing the more intuitive explanations of the derivative (gradient of the tangent line, change in function value wrt independent variable etc) with its formal definition. There ARE issues with these intuitive notions. ThatÂ’s why the formal definition was created - specifically to address those issues. The formal definition removes the issues present in any of the intuitive understandings and puts calculus back on a form mathematical footing.

I wonÂ’t belabor this (too much) since IÂ’ve already commented quite a bit on the formal definition, but maybe letÂ’s work through how to use the formal definition to actually calculate a derivative. This is mostly unnecessary as we will see - the intuitive ways give the same result - so long as you realize that the issues with the intuitive method arenÂ’t really present in the formal method.

LetÂ’s go back to our favorite function f(x)=x^2 and calculate its derivative at x=2. We know from previous posts that the differential quotient for any x and any nonzero h is 2x+h. Obviously for x=2 it is 4+h. Let q be the absolute value of the difference between the differential quotient and fÂ’(2). Formally then we say that any number z is defined to be the derivative fÂ’(2) if for any epsilon we can find a delta such that if hepsilon.

LetÂ’s try z=6. Let epsilon=1. q=|4+h-6|=|h-2|. Pick any delta. Let hepsilon. We can in fact try ANY delta we want. If h=0.01, q=1.98>epsilon. This rules out any delta>0.01. If delta>0.01 then h<0.01 and that implies q>1.98>epsilon. This proves that z=6 is not the derivative. We can try any other number we want other than z=4 and we will always be able to find an epsilon and h that makes q>epsilon.

LetÂ’s try z=4 now. We then get that q=h. We pick an epsilon, say 0.001. If the negation holds for this value of epsilon then for any delta we should be able to find a value of hepsilon. That is even if delta=0.00001 we can find h<0.00001 such that q>0.001. But we shows that q=h so if h<0.00001 it cannot be greater than 0.001 so the negation does not hold. We can equally easily show that the original definition does hold. We must show that for every epsilon there is some delta such that for all h

In practice none of this is necessary and we gloss over this formal procedure. We find that the differential quotient is 2x+h then (sloppily) say that “h goes to zero” and pretend that h actually IS zero, concluding that the derivative is 2x. We get the right answer this way, but we create the impression that we are doing something problematic, namely dividing by zero, to get that answer. What we really are doing is the formal procedure shown above. We just donÂ’t normally spell it all out that way. The formal procedure justifies the intuitive result.

I agree with that, I think.

I think it may, yes. Hence my issue with it.

I'm still thinking about what you said about pi. Were you saying that irrational numbers in general are transcendental? That is interesting if so, interesting in any case. I imagine the sequence *the pi computer*, presumably still going as we speak, uses to calculate pi is based on your sequence or similar. An infinite series for an infinite number seems reasonable.

(occultly speaking, pi appears to lack symbolical significance after the first few dps).

It should be "all transendental numbers are irrational", as algebraic irrational numbers such as sqrt(2) are not transendental.

The infinite sequence I chose to express pi converges very slowly. There are much more efficient sequences that converge to a more precise value by several orders of magnitude.

As pi is likely the most studied number, more efficient ways of computing pi have been discovered, such as an iterative algorithm. That is calculating a more precise approximation of pi using a previously calculated approximation. This saves a lot of time because the iterative algorithm does not need to recalculate billions of digits, but instead adds more precise digits very quickly. Infinite series have also advanced that provide extremely precise approximations of pi to billions of digits that rival iterative algorithms in calculation efficiency.

The digits representing pi have been studied for randomness, and it appears that the billions of digits satisfy tests of randomness for multiple digital bases, such as base 2 and 10.

And remember also that infinite series can describe rational, or colloquially "finite" numbers, like the infinite series for 10 I noted above.

How can they be random when they are figured out according to a rule (i.e. whichever program is being used to compute them)?

Edit: And how can a sequence be random in one base but not in another?

Random as in mathematically indistinguishable from digits generated by a random process. As a simple example of this consider generating 10000 random digits (rolling a ten sided die, repeatedly drawing balls numbered 0-9 etc). We want to know if our process was truly random or if it is biased. We can do mathematical tests to determine if our actual values match an expected distribution. We should expect, for instance, to see 1,000 ones in our set. What if we see 999 or 1001? Does that mean our random process is rigged? No, because there is a certain amount of variance inherent to it. We do tests (like the chi -squared test) to determine how likely it is that our data was generated by a truly random process.

Well, if instead of rolling a die or drawing balls, we take a set of 10000 digits from the decimal (or whatever other base) expansion of pi. We subject these digits to the same test we did to see if our random process really is random. We find that indeed the digits of pi pass our tests for randomness.

Unless I misunderstand, the sequence 123456789123456789... would pass that test for randomness.

The decimal expansion of pi is not random WRT Kolmogorov complexity.

https://en.wikipedia.org/wiki/Algorithmi...

You did not misunderstand. I simplified. The distribution of individual digits is only one possible test for randomness. There are others that pi passes and your example fails. A simple example might be to take the first million digits of the sequence and identify 10000 ones. Then look at the next digit after each one. In a random sequence, each digit should appear 1000 times. Your sequence obviously fails this test since every one is followed by a two.

As for Kolmogorov complexity, I am honestly not sure. Yes there are algorithms for calculating pi, but (AFAIK) there are no finite algorithms. An infinite algorithm to calculate an infinite sequence wouldn’t be a compression, would it? There is a difference between say “the decimal expansion of 1/3” and pi. For the decimal expansion of 1/3 there is a simple algorithm to calculate the nth digit of the sequence for any n, namely the nth digit is 3. No such algorithm exists (again AFAIK) to generate the nth digit of pi for arbitrarily large n.

It is an interesting point, however. A

https://en.wikipedia.org/wiki/Computable_number

en

Computable number - Wikipedia

That is different to the apparently stronger condition that there exists a single, non-halting Turing machine that outputs the digits one after another. In some cases, e.g. 0.333..., the two conditions are obviously equivalent. Even if they were equivalent for pi, I wonder if there exist other numbers for which they are not.

I don't know how the rest of the transcendentals behave - I expect similarly - but I wonder if there's value in considering the real number scale as behaving as a readout on a different system of logic. In the same way that we can have essentially anything as a base, the turing approach only renders value after we consider that e and pi can be expressed as a geometric series. We could feasibly create another number system whereby the 'readout' onto that system of such geometric systems behave more similarly than rational and transcendental numbers.

It requires an infinite number of computations to produce a value that is EXACTLY pi, rather than a (really, really close) approximation of pi. I believe the current record for computing digits of pi is somewhere in the trillions, but that is NOT a computation of pi - pi has an infinite number of digits. To actually compute ALL digits of pi, you’d need to use the algorithm to add in an infinite number of terms to the sum. That was my point - can you really call it compression when to represent your infinite sequence of digits, you must specify an infinite number of values to add?

You seemed happy to call it such in the case of 1/3. The algorithm for generating the decimal expansion of pi is merely more complicated; is that in your opinion an essential difference?

[QUOTE=stremba]There is a difference between say “the decimal expansion of 1/3” and pi. For the decimal expansion of 1/3 there is a simple algorithm to calculate the nth digit of the sequence for any n, namely the nth digit is 3. No such algorithm exists (again AFAIK) to generate the nth digit of pi for arbitrarily large n.[/QUOTE]

It’s not just more complicated. I can specify ANY digit of the decimal expansion of 1/3 using a finite “algorithm”. If we let d(n) be the nth digit of whichever sequence we are taking about, then for the decimal expansion of 1/3, d(n)=3 for all n. I have just specified d(n) for ALL n via a finite process.

We can’t do this for pi. We have a variety of choices for algorithms for calculating pi to any desired accuracy, limited only by processing power and the amount of time we want to spend calculating it. We have found d(n) for pi, but only up to a certain limit (currently for n= something like 10 trillion). We do this by adding successive, ever decreasing terms of an infinite sum to get increasingly accurate partial sums. What we cannot do, is specify pi EXACTLY by a finite process. There is no finite formula allowing us to calculate d(n) in the expansion of pi for ALL n. For n>the 10 trillion (or whatever the exact number is now) we have no formula at all. Even theoretically there is no way to calculate d(n) for any nnin a finite number of steps.

This is indeed fundamentally different than the case of expansion of any rational number (any rational number has a decimal expansion with a repeating pattern that can be specified in a finite instruction, possibly preceded by a finite sequence of digits not following that pattern - e.g. 1/6=0.166666…). For any rational we can give.a finite method for determining any arbitrary d(n) no matter how big n may be. Thus we can specify the infinite sequence EXACTLY in a finite manner - that is algorithmic compression.

Your definition of exact specification of a number is that the digits in its decimal expansion be regarded as elements of a countably infinite, linearly ordered set, all existing together at the same time, and it is known simultaneously what each of these elements is?

Anything known will never be truly random. I could write 5 numbers on a piece of paper and show everyone except one, and it will appear random to that one person, yet not random to everyone else.

Maybe, but we are talking about mathematical randomness. If you looked at say digits 1 billion to 1.1 billion of pi, you’d be looking at one million digits. If we rolled a ten-sided die one million times and recorded the results we could compare the two sequences of digits. Imagine doing this another 998 times, giving us 1000 lists of a million digits each.

Saying that the digits of pi are random means that there is no way (other than actually knowing what the 1 billionth through 1.1 billionth digits are) to point to one out of the 1000 lists and say “this one is the digits of pi and all the others were generated via dice rolling”. The million digits from pi have exactly the same mathematical properties as the lists of dice rolls. In math when we have two entities with exactly the same properties we regard them as equivalent. Nobody doubts that the list of rolls is random, so we regard the list of digits from pi, which pass all randomness testing, as random as well.

Think of it this way: the lists generated via dice rolls are not truly random either. If we knew very precisely the initial position of the die, how fast it rotated, how hard it was tossed, etc., we could predict what the result of a given roll will be. We don’t have such information though so the rolls appear random. Similarly, I don’t happen to know that a particular sequence of digits is a sequence appearing deep in the expansion of pi. It appears random. It can equally well be said that both the die rolling sequence and the digits of pi sequence only appear random. So what makes one “really” random and the other not random?

Yes. The question was about algorithmic compressibility of sequences. A sequence is algorithmically compressible precisely if there is a way to completely specify all terms of that sequence that is shorter than the sequence itself (obviously for any sequence there is a way to specify the sequence that is of equal length to it - simply enumerating its terms). For countably infinite sequences to be algorithmically compressible, it must be possible to specify all the terms using a finite algorithm. A algorithm in which an infinite number of steps is needed to generate all the terms is not shorter than just listing all the terms.

That is possible for the expansion of any rational number, such as 1/3, where we can simply say for any n, the nth term equals 3. We have algorithmically compressed this sequence by providing a finite way to calculate any term in the sequence. (A particularly simple way in this case). I was questioning whether such an algorithmic compression is possible for pi. We certainly can approximate the sequence of digits with a finite algorithm, but we cannot, to my knowledge, specify the exact sequence using a finite algorithm.

I agree. The point I was trying to make is that the only things that are random are things we don't understand or have knowledge of. Once we have gained an understanding of how that thing behaves, then it is no longer random. One could take this to extremes, such as external inputs to a RNG, such as noise, which is not truly random.

I guess quantum behavior is about the closest thing to true randomness.

And if pi is a normal number, then one trying to match a pattern with lookup sequences to predict the next digit will likely fail, because those sequences can appear infinite times.

I would say a PRNG with random seed using pi should perform just as well as other PRNGs, albeit much slower.

So the sequence 01234567891011121314... obtained by concatenating successive natural numbers is not algorithmically compressible?

When we talk about randomness we have to be careful whether we define it as unknown, unknowable, or unknowable given our current knowledge.