The Derivative of a Number

http://rjlipton.wordpress.com/2014/08/19/the-derivative-of-a-number/

by j2kun • 12 years ago

101 47 12 years ago

rjlipton.wordpress.com

nilsimsa• 12 years ago

Shouldn't derivatives by respect to change of another variable. For example d/dx of f(x).

pfortuny2• 12 years ago

No, it can be an operator on any kind of (usually) ring or more general algebraic structure. What you refer to is the "usual" derivative of functions of one variable, just one of many derivatives one can define.

sp332• 12 years ago

Where would it be used? It seems like declaring that the derivative of every prime number = 1 is entirely arbitrary.

quotemstr• 12 years ago

You can even take the derivative of a grammar to get a parser!

http://matt.might.net/papers/might2011derivatives.pdf

nilkn• 12 years ago

The standard way of generalizing the derivative is to require it to be linear and to satisfy a version of the product rule. See, for instance, derivations:

http://en.wikipedia.org/wiki/Derivation_(differential_algebr...

In the case of this article, the proposed definition is not linear, so it is indeed a bizarre candidate for a derivative.

tel7• 12 years ago

That's the intuition used to develop the concept, but it becomes increasingly difficult to apply that intuition to in more exotic locales.

Thus, it's important to eventually seek out more abstract ways of characterizing the derivative (and integral). In more advanced mathematics, you usually state that the derivative is any operation which follows two rules

    1. Linearity, d(ax + by) = a d(x) + b d(y)
    2. The Product Rule, d(xy) = x d(y) + d(x) y

and then try to squeeze things until that operation is defined uniquely.[0]

Likewise, it's often valuable to define integration as nothing more than the relationship such that

      I(region, derivative(quantity)) 
    = I(boundary(region), quantity)

which is known as the Generalized Stokes Rule. It basically is the "Fundamental Theorem of Calculus" on steroids and it gives a characterization of integration in terms of nothing more than it's algebraic/topological relationship with derivation... which is itself abstracted as mentioned above.

---

Why do all this? Because you can squeeze most of Calculus so that it depends only upon this "abstract interface" and then apply things you learned from calculus all over the place.

---

Finally, note that this is more like a "proposed" derivative than "the" derivative on natural numbers. The author notes that linearity fails, for instance. Thus, some intuition might "port over" but we shouldn't expect too much of it to do so.

Which echoes back to your original question—there's not really a notion of instantaneous change for us to be talking about... so how much sense does it make to talk about a derivative here?

Apparently, more than no sense at all, but less than you might want.

[0] Note that all we need to state this property of the derivative is a notion of multiplication and addition. This structure is, at its most abstract usually called a ring (but can be made even weaker if needed). An example "exotic" ring might be concurrent processes. If P and Q are two processes then P*Q is P "followed by" Q and P + Q is P and Q "together". Can we write a derivative here? Who knows? (As another comment in this thread suggests, this kind of formulation can be used to consider the "derivative of a grammar" to be a parser! It's also well-known that the derivative of an algebraic data type is its "zipper"!)

AnimalMuppet3• 12 years ago

That's the intuition used to develop the concept, but it becomes increasingly difficult to apply that intuition to in more exotic locales.

Then maybe it's better to use a different term for things that are different. Maybe it's better to keep the term "derivative" for the rate of change of one thing with respect to another thing, and to let the generalizations that aren't that be called something else.

gus_massa• 12 years ago

I agree, there are usefull generalization of the derivative. But an important detail is that the operator that is discussed in this post is not linear, i.e. D(x+y) != D(x) + D(y), so it doesn't have all the expected ("intuitive") properties that the usual derivative has.

I don't know the origin of this notation, but I can make up a possible explanation. If you think that every prime is a function of an abstract variable x, so 2 = two(x) = 2 + x, 3 = three(x) = 3 + x, 5 = five(x) = 5 + x, ... evaluated in x=0.

Then, for example, 60 = sixty(x) = two(x) * two(x) * three(x) * five(x) = two(x)^2 * three(x) * five(x)

A number is a function of the primes of the factorizations. You must change the primes into the functions, but you must leave alone the exponents.

Then D is the standard derivative, plus evaluation.

D(60) = sixty'(x) = 2 * two(x) * three(x) * five(x) + two(x)^2 * five(x) + two(x)^2 * three(x) = 2 * 2 * 3 * 5 + 2^2 * 5 + 2^2 * 3 = 60 + 20 + 12 = 92

(I'm mixing the functions and the values when they are evaluated. It's usually not a good idea. If you do that in a Calculus exam the TA will be rightfully angry. But the notation in only text is horrible, so please forgive the technical details.)

With this idea if you have numbers A, B, C such that A = B * C, then A(x) = B (x) * C(x). But if A = B + C then A(x) != B(x) + C(x). This "explains" why the operator D follows the multiplication rule, but not the sum rule.

For example, ten(x) = two(x) * five(x) and six(x) = two(x) * three (x) then sixty(x) = two(x) * two(x) * three(x) * five(x)

But fifteen(x) = three(x) * five(x) and ten(x) = two(x) * five(x), but twenty-five(x) = five(x)^2 and clearly three(x) * five(x) + two(x) * five(x) != five(x)^2.

Edit: Don't take this "explanation" very literally, for example this idea doesn't extend to D^2 directly

D^2 (p^2) = D(p+p) = D(2p) = D(2)p+2D(p) = p+2

D^2(p^2) = D(p+p) != D(p)+D(p) = 2D(p) = 2

D^2(7^2) != (seven(x)^2)'' = (seven(x)+seven(x))' = seven'(x)+seven'(x) = 2seven'(x) = 2

The implicit transformations of numbers into functions and the evaluations in 0 make cause many problems.

clebio• 12 years ago

What an excellent reply. All spot on, but worth it just for this bit alone:

Apparently, more than no sense at all

tel7• 12 years ago

To tie some intuition back, here's how

      I(region, derivative(quantity)) 
    = I(boundary(region), quantity)

is just the Fundamental Theorem of Calculus.

What you need to do is take "region" to be an interval of the real line like [a, b]. Now, the boundary of "region" is the set of the end points, {a, b}. Thus, we've now transformed this equation to

    I([a,b], d(quantity)) = I({a, b}, quantity)

If we see I as being a sum, it's clear that the right-hand-side must be the same as a regular sum, though we need to account for the idea that we're summing from a and to b. Ultimately, we do this by negating where we're coming from[1]

    I([a,b], d(q)) = -q(a) + q(b)

Then we just have to recognize I([a,b], _) as the definite integral from a to b

    DefiniteIntegral(a, b, d(q)) = q(b) - q(a)

and this is a statement that the definite integral of a function on an interval [a,b] is the difference of the antiderivative of that function evaluated at the endpoints---the standard fundamental theorem of calculus!

But notice that in this transformation we've destroyed some information. No longer is it so apparent that "boundary" and "derivative" have some kind of kinship. We also cannot easily generalize this notion to higher dimensional spaces (unless we already know the Stokes Law trick).

[1] This is a bit arbitrary. If we chose it the other way the equation would still hold, it'd just not be the standard convention and thus less recognizable. The reason we choose this is because calculus, it turns out, depends upon a notion of orientation. We have to know whether we're going with or against "the flow".

kvb• 12 years ago

Great answer, but on a pedantic note, it's not quite true that the derivative of a data type is its zipper. The derivative is the one-hole context, but the zipper is more like a one-substructure context instead. There's a bit more elaboration here: [1].

[1] http://en.wikibooks.org/wiki/Haskell/Zippers

JadeNB2• 12 years ago

Likewise, it's often valuable to define integration as nothing more than the relationship such that

    >  I(region, derivative(quantity)) 
    > = I(boundary(region), quantity)

> which is known as the Generalized Stokes Rule.

I like your explanation very much, but I think that calling the integral 'the relationship' (emphasis on the definite article) satisfying Stokes is an overstatement. There are many such; for example, I(_, _) = 0.

Chinjut4• 12 years ago

If you like, think of this as first turning a number into the unique monic polynomial with negated prime roots whose value at 0 is that number, then taking the derivative of that polynomial at 0.

...Not that there's any particular reason you should like this.

cperciva• 12 years ago

In a sense this is d/d{primes}. D(8) is how fast 8 changes with respect to 2, while D(60) is how fast 60 changes if 2, 3, and 5 simultaneously increase (at the same rate).

e12e• 12 years ago

So... we can use it to determine the age of the universe, if we know at which point in time 6 * 9 = 42 held, by extrapolating back to at what point 6 * 9 = 0? ;-)

[ed: looks like my multiplication signs got eaten by hn]

joeframbach• 12 years ago

Think of Derivative as a function which inputs a function and outputs a function. When you think "derivative of 5 is 0", you imply "derivative of f(x)=5 is f'(x)=0"

This article seems to define a function called Derivative which inputs a number and outputs a number. In this case, "derivative of 5 is 1" actually translates to "derivative of 5 is 1".

scythe• 12 years ago

Here's a lemma that helps to prove D(n) is well-defined:

LEMMA: D(pq•r) = D(p•qr):

D(pq•r) = pq D(r) + r D(pq)

= pq D(r) + rp D(q) + qr D(p)

= p (q D(r) + r D(q)) + qr D(p)

= p D(qr) + qr D(p)

= D(p•qr)

There are a number of generalizations of the notion of "derivative"; in particular, you can take the derivative of a regular expression:

https://github.com/scythe/dreg/blob/master/re-deriv.pdf

...which I've been slowly turning into a regex matcher.

gergoerdi• 12 years ago

If you're interested, regex matching via derivatives have been done before, see e.g. http://matt.might.net/papers/might2011derivatives.pdf

dead10ck2• 12 years ago

I feel like I'm missing something. Isn't the derivative of a constant just defined to be 0? Why does the definition in the article restrict it to primes?

j2kun• 12 years ago

This article is defining a new notion which is different from the derivative of a function. It's called derivative because of its inspirations and similarities to the classical idea of a derivative. It's interesting because it appears to have close relationships to important open problems in number theory, such as the Goldbach conjecture.

And it's not restricted to primes, it's just defined first for primes and then extended to all integers later.

dead10ck2• 12 years ago

Ahh, I see. That makes more sense.

However, I'm not sure I understand how the definition extends to all numbers. Edit: nvm, I'm dumb.

glandium• 12 years ago

What is confusing is that the article introduces both derivatives (of a function and of a number) as D, when both are definitely not the same thing.

EDIT: that is, the fact that they have the same name and the same symbol is what's confusing.

martincmartin2• 12 years ago

You're right, it's not the normal definition of derivative. He's defining a function on numbers that has some similarities to the derivative (the multiplication rule is the same, and therefore so are some other properties), but isn't the derivative.

It's common in mathematics to take a word that already means something, and re-use it for a different but related concept. For example, in group and field theory, you talk about addition and multiplication, but they're not necessarily the standard definition.

robzyb• 12 years ago

It's common in mathematics to take a word that already means something, and re-use it for a different but related concept. For example, in group and field theory, you talk about addition and multiplication, but they're not necessarily the standard definition.

That's true, but I feel like the article mislead me because it didn't give me a proper and timely explanation of what it means by "derivative".

thomasvarney723• 12 years ago

I find I have the same problem explaining computer technologies to people.

martincmartin2• 12 years ago

Barbeau was a professor of mine back in the late 1980s, great to hear that he's still around. It was a wonderful class, with Ravi Vakil [1] and Nima Arkani-Hamed [2] in it.

[1] http://en.wikipedia.org/wiki/Ravi_Vakil [2] http://en.wikipedia.org/wiki/Nima_Arkani-Hamed

adamgravitis• 12 years ago

He taught our Calculus II course a few years ago, which had rather famous exams.

A couple years later I was poking around the library and found a book he'd just published entitled something like "50 challenging undergraduate mathematics problems", of which I was annoyed to recognize several :-)

tempodox• 12 years ago

Wonderfully strange. I had to implement it as a shell command:

https://github.com/tempodox/NumberDerivative

The Derivative of a Number

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