Discrete Integral?

Inspired by this quiz about discrete derivatives.

When I found this quiz, I remembered a thought I had had when I had heard about the formula \( \sum_{k=1}^n k = \frac{n(n+1)}2 \).

If you expand the fraction, you get \( \frac 1 2 n^2 + \frac 1 2 n \), which almost looks like the antiderivative of \( k \) except for the \( \frac 1 2 n \) term. When I did this for other sums, I got the same result:

\( \sum_{k=1}^n {\color{red} k^2} = {\color{red} \frac 1 3 n^3} + \frac 1 2 n^2 + \frac 1 6 n \)

\( \sum_{k=1}^n {\color{red} k^3} = {\color{red} \frac 1 4 n^4} + \frac 1 2 n^3 + \frac 1 4 n^2 \)

\( \sum_{k=1}^n {\color{red} k^4} = {\color{red} \frac 1 5 n^5} + \frac 1 2 n^4 + \frac 1 3 n^3 - \frac 1 {30} n \)


Because of the definition of these sums, their discrete derivatives are always \( n^a \) (a is some natural number), but even for discrete derivatives of simple polynomials the fundametal theorem of calculus still almost holds; at least for the highest order term.

So now I'm wondering if

  1. this also holds for functions that aren't polynomials, for example exponential functions or trigonometric functions
  2. the discrete antiderivative of the discrete antiderivative is also similar to the normal second antiderivative
  3. there are some polynomials for which antiderivative and discrete antiderivative are exactly the same

EDIT: I've found the thing I'm calling "discrete antiderivative" as finite difference on Wikipedia and the article points out Faulhaber's formulas and the difference between odd and even degree polynomials. However, I haven't found what I'm searching for – a comparison/link to standard calculus integrals.

Note by Henry U
4 months, 2 weeks ago

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The formulas you have provided are called Faulhaber's formulas and are linked to Bernoulli numbers. There is a result by John Conway recently (2016, I think), where he is able to compute "derivatives" and "integrals" of these Faulhaber formulas, which distinguish between odd degrees and even degrees of the polynomials. Note that here, the notions of "derivative" and "integral" are merely just linear functions on polynomials in \(n\), which sends one polynomial to another with the linearity properties satisfied; no notion of calculus is required here. Because of this, we need to be precise as to what a "derivative" and an "integral" is here, as well as what a "discrete derivative" and a "discrete antiderivative" here. None of your questions can be answered without first establishing this point; you cannot simply input the framework of calculus into something that is only applied in a discrete way.

P.S. As for "exponential" functions and "trigonometric" functions, I don't see the relevance of Faulhaber's results to them as they are not properly defined currently.

A Brilliant Member - 4 months, 2 weeks ago

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I would say,

  1. Derivative and Antiderivative are defined by the power rule from calculus, so nothing discrete here
  2. The discrete derivative is defined as \( f(x+1)-f(x) \) – just like in the quiz.
  3. The discrete integral is defined by Faulhaber's formula – thank you for pointing that out – as \( \sum_{k=1}^n k^a \)

Then all those four functions applied to polynomials are also polynomials and satisfy linearity (because of the commutative property for 2. and the linearity of sums for 3.)

So with these definitions, I hope I have stated my questions unambiguously, but I also think that they don't really make sense for other functions like exponential or trigonometric functions. But Conway's result sounds interesting, I think I'll take a look at it.

Henry U - 4 months, 2 weeks ago

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What power rule from calculus? It's simplest to just define the Derivative as a linear map, where for integers \( n \geq 1 \),

\( D(x^n) \equiv nx^{n-1} \)

and the Integral as the inverse mapping of the Derivative, i.e. for integers \( n \geq 0 \),

\( I(x^n) \equiv D^{-1}(x^n) = \frac{x^{n+1}}{n+1}. \)

This was how Conway presented his definitions; with this, you don't need to be sloppy with them and avoids having to define a "discrete" version of them. As I said, no notion of calculus is ever involved; we are merely defining the Derivative and Integral as linear maps/functions which map one polynomial to another.

My main point was to ask you why you care to link these to the "standard" framework of calculus, when we have a nice stand-alone framework to study Faulhaber polynomials which only involve algebra & combinatorics.

P.S. I urge you again to look up Bernoulli numbers; I think this should answer your question regarding the coefficient of the largest-degree term.

A Brilliant Member - 4 months, 1 week ago

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@A Brilliant Member My main reason to link these to calculus was because I am more familiar with (anti-)derivatives and I thought they don't look as complicated (The antiderivative of some polynomial involves only one term for each term of the polynomial whereas Faulhaber's formula has many more terms of lower degree).

And then I also thought, if we have one mapping \( I(x^n) = D^{-1}(x^n) = \frac{x^{n+1}}{n+1} \) and then another mapping \( F(x^n) = (Faulhaber's formula) \), of course it will be interesting if there exist polynomials for which these two mappings give the same result.

Henry U - 4 months, 1 week ago

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@Henry U I understand, but you must let go of the standard calculus framework when you venture into the world of discrete mathematics, as nothing from there will work out well. One clarification: Faulhaber's formula is a polynomial, and hence cannot be considered a mapping.

As to your final sentence, I vehemently disagree; there is no polynomial that is invariant under the Derivative or Integral linear map (it is easy to show this). You will have to venture into the realm of fiction to find such a case, by relying on "infinite processes" which would go against everything in discrete mathematics.

EDIT: To save you the trouble, note that what you would call "exponential" or "trigonometric" functions are really just "infinite sums".

A Brilliant Member - 4 months, 1 week ago

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@A Brilliant Member Could you maybe describe a little proof why there exists no polynomial \( p(x) \) for which \( I(p(x)) = F(p(x)) \) is true?

I don't directly see an intuitive reason why this is true, but I also haven't found an example.

Henry U - 4 months, 1 week ago

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@Henry U (I thought I replied to this already... hmm.)

I think you are asking the wrong questions. Did you read up on Faulhaber polynomials? They simply give the sum of the \(k^{\text{th}}\) powers of the first \(n\) numbers... so it does not make sense to talk about the Faulhaber polynomial of a polynomial.

I firmly believe you are trying to muddy the waters between the standard framework of calculus and what one would call "discrete calculus", which relies more on difference equations than limiting processes. The two areas, while not mutually exclusive, are highly divergent from one another and thus it's important to separate both frameworks.

A Brilliant Member - 4 months ago

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@A Brilliant Member By the faulhaber polynomial of a polynomial, I meant the sum of the faulhaber polynomials of the terms of the polynomial (too complicated). For exapmle,

\( p(n)=n^2-2n \)

Of course

\( F(n^2) = \frac 1 3 n^3 + \frac 1 2 n^2 + \frac 1 6 n \)


\( F(n) = \frac 1 2 n^2 + \frac 1 2 n \)

Now I thougut, every polynomial (like \( p(n) \)) is a linear combination of all powers of \( n \). So the Faulhaber polynomial could be the same linear combination of the individual Faulhaber polynomials.

\( F(n^2-2n) = F(n^2) - 2F(n) \)

This makes sense because \( F(n^a) \) is just a partial sum of \( n^a \) and, since sums are linear, we can split one sum of a polynomial into many sums, one for each term of the polynomial.

In my example, I would argue

\( F(n^2-2n) = F(n^2) - 2F(n) = (\frac 1 3 n^3 + \frac 1 2 n^2 + \frac 1 6 n) - 2(\frac 1 2 n^2 + \frac 1 2 n) = \frac 1 3 n^3 - \frac 1 2 n^2 - \frac 5 6 n \)

Henry U - 4 months ago

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@Henry U But it’s not a linear mapping...

A Brilliant Member - 4 months ago

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@A Brilliant Member I think it is. If you double the left side, you have \( 2F(n^2-2n) \) and I think this is equal to t he right side with doubled coefficients. Or isn't that the definition of linear?

Henry U - 4 months ago

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@Henry U I think you need to revisit the definition of a linear mapping... in any case, you are overcomplicating your problem & question. Go five steps back and take a deep breath.

A Brilliant Member - 4 months ago

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@A Brilliant Member I'm haven't any experience with vector spaces, but I think I understand what a linear mapping is, at least well enough to say why faulhaber's formula is a linear mapping. Here's what I think:

Wikipeda says

[…] A linear map […] is a mapping \( V \to W \) between two modules (including vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication.

Let \( V \) and \( W \) be vector spaces over the same field \( \mathbf{K} \). A function \( f : V \to W \) is said to be a linear map if for any two vectors \( \mathbf {u} ,\mathbf{v} \in V \) and any scalar \( c \in \mathbf{K} \) the following two conditions are satisfied:
\( f(\mathbf{u}+\mathbf{v}) = f(\mathbf{u})+f(\mathbf{v}) \)
\( f(c\mathbf{u}) = c f(\mathbf{u} \)

  • Applying Faulhaber's formula to a polynomial is a mapping between two vector spaces since the set of polynomials is a (infinitely-dimensional) vector space
  • Faulhaber's formula gives thhe result of a summation, which is linear, so the formula is also linear

Henry U - 4 months ago

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@A Brilliant Member But of course, Discrete Mathematics and calculus don't really fit together, so maybe this won't lead to anything

Henry U - 4 months ago

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