The craziest definition of the derivative you have ever seen!

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[Nikolas_Davis]

This reminds me of Charles Babbage famously explaining the operation of his Difference Engine as
"Δ^7 U_z = 0". His Engine worked by calculating up to the 7th order finite difference (derivative) on an initial "seed" of tabulated values of any given function, then worked backwards to produce more tabulated values at fixed intervals - essentially a finite element mechanical integrator.

[carstenmeyer]

10:49 There is a big assumption hidden -- for that expression to make sense, *"f(x)"* must be represented by a power series around "x = 0" that still converges at "x = 1".

While that is a quite a large function set in practice, it does not include all infinitly smooth functions, like e.g. bump functions.

[the-avid-engineer]

12:18 There’s a mistake.. you start with (-1)^n and switch to (-1)^(n+1) with no explanation

[CraigNull]

Sampling f(x) at only integer differences from x simply cannot work as a universal definition of f '(x) due to examples like f(x) = cos(x\*pi). An analytic function no less! So in this derivation where did assumptions about f(x) slip by unnoticed, and what are those assumptions?

[joshuagrumski]

15:03 Isn’t this is very, very roundabout way of doing this proof? Like you already had to know the ln(x+1) series expansion centered at 0 to do the proof that this works, but that means you already have proven this result. The only way that I can imagine that you would get something new out of this would maybe to prove the one result that he challenged the rest of us do because he couldn’t find a nice form for the proof, but even then, I still don’t know if that proof is proving anything new or is circular

[ianrobinson]

This type of relationship of operators is actually the starting point for the traditional subject of Finite Calculus. Elements of it can be found in its modern and broader subject of Numerical Analysis most commonly forming the basis of computer science methods of approximations. It was well developed by the mid-1850s in texts of that time, eg Boole but all but “forgotten” these days.

[diribigal]

The ln(2) expansion is kind of circular. You use the general ln(1+x) series to set up these operator calculations, but that series includes the x=1 case that would have immediately given the ln(2) expansion.

[slowfreq]

Do you have a video on operators like delta and how you manipulate them? I've never really understood that at all, and manipulation of operators like that has always bewildered me.

[Hhsksjbdl]

Here is a formal proof that ln(1+Δ) is the derivative for polynomials:
Denote for short L = ln(1+Δ), D = d/dx the derivative operator (both are linear).
we’ll use induction on the degree of the polynomial.
for constant polynomials the result is trivial.
Assume it is true for polynomials of degree at most n-1.
note that DΔ = ΔD and so (on the space of polynomials:) DL = LD
the space of polynomials p of degree at most n for
which Dp = Lp is a vector space and we assumed it contains
all polynomials with degree <= n-1. therefore it’s
enough to show it contains a single polynomial of degree n,
and then it must contain all of them by linearity.
now, define polynomials f_0(x) = 1, f_k(x) = x(x-1)…(x-(k-1)).
then f_k(0) = 0 unless k = 0 (f_0(0) = 1) and
Δf_k(x) = k*f_(k-1)(x). hence (Δ^k)f_n(0) = 0 unless k = n
and (Δ^n)f_n(0) = n!. this implies L(f_n)(0) = (-1)^(n+1)*(n-1)! = D(f_n)(0)
In addition, by induction we get (D(f_n) is of degree n-1)
D(L(f_n)) = L(D(f_n)) = D(D(f_n))
This is enough to prove that D(f_n) = L(f_n),
that is f_n is a polynomial of degree n in
the previously defined vector space, as needed.

[dalek]

I'm confused shouldn't the (-1)^(n+k+1)=(-1)^(k+m) and the (x+k)^m=(x+n+1)^m. That's what you get when applying the formula.

[atreidesson]

"If you go high, you always get zero" © Michael Penn

[dyld]

Here's the outline of my proof for the challenge:

1. Show by induction that
D^n(x f(x)) = x D^n f(x) + n (D^n f(x) - D^(n-1) f(x))

2. Using the above result, we can show that

ln(1 + D) (x f(x)) = x ln(1+D) f(x) + f(x) - (-1)^m D^m f(x)

3. Let f(x) = x^(m-1). Then D^m x^(m-1) = 0. Finish by induction over m.

[carloseliasmartinez]

I love Michael's videos. I use to watch them with a notebook and pen in hand since they always inspire me to explore what the concepts he's presenting. I've applied this definition of derivative to the exponential function and found a quite... "weird" result: a series that should diverge gives a finite result. I've analysed the same series by a couple of different methods and got the same result, but by employing "forbidden" methods (like applying the equation of the geometric series to a reason that's larger than one in absolute value) and found the definition of polylogarithmic functions. I still cannot fully believe Li1(1-e) = -1, but I guess it's some sort of analytic continuation.

Is there any way I could contact Michael about this result in case he might consider interesting to make a video on this topic?

[yoav]

In the exercize it should be (-1)^(k+r) not(k+m), and also it should be (x+k)^m not (x+r)^m.

[clearnightsky]

It's a dream! A neet way to compute the derivative of functions that are not continuos but for which you know the values at an infinite no. of points. :D

[il_caos_deterministico]

It can be shown that
Δ^n f(x) = sum{k=0}^n (-1)^k (n over k) f(x+n-k)
So if r=n-k then it is equal to

sum{r=0}^n (-1)^(n+r) (n over r) f(x+r)
(we used (n over k)=(n over n-k) and (-1)^(n-r)=(-1)^(n+r) )

Reindexing this is

sum{k=0}^r (-1)^(k+r) (r over k) f(x+k)
The last formula of the exercise has probably a minimal error in indexes.

[Calcprof]

I like this better as e^D = 1 + Δ . Which gives translation in terms of derivative, which is how you prove this anyway.

[divide]

15:44 exercise solution attempt- I was actually trying to do the challenge but ended up accidentally solving this one instead.

First, we claim that Δ^n(x^m)=Σ(-1)^k*C(n, k)*(x+n-k)^m, the sum being taken over k=0, ..., n, as long as n is a positive integer not exceeding m.

We may prove this by induction. The base case n=1 is trivial. Expanding via the above equation gives

(x+1+n)^m+Σ[(-1)^(k+1)*C(n+1, k+1)(x+n-k)^m]-(-1)^n*x^m, where the middle sum is taken over k=0, ..., n-1 and has been simplified via Pascal's identity and re-grouping. We may notice that (-1)^0*C(n+1, 0)*(x+n+1-0)^m=(x+1+n)^m, and (-1)^(n+1)*C(n+1, n+1)*(x+n+1-(n+1))^m=-(-1)^n*x^m, so reindexing the above sum gives us the desired result. By induction, the claim is proven. 

Reindexing n-k to k, Δ^n(x^m)=Σ(-1)^(n-k)*C(n, k)*(x+k)^m=Σ(-1)^(n+k)*C(n, k)*(x+k)^m, the sum being taken over k=0, ..., n.

Since is a degree m-1 polynomial, by linearity along with induction, Δ^n(x^m) has degree m-n for n<=m, and for all n>m, Δ^n(x^m)=0.

[polyhistorphilomath]

Here's a direct approach to the problem posed at 9:24 .
Δ (f)(x) = f(x+1) - f(x) by definition.
By definition the shift operator is e^tD (f)(x) = f(x+t).
Substitute a shift by t=1 into the definition of the forward difference.
e^D-1= Δ. (add 1 to both sides)
e^D = Δ+1. (take logarithms)
ln(e^D) = ln(Δ+1). Now we have the LHS of the equation provided (on the RHS here).
Take down the power on the left, then remove the logarithm of the natural number (=1).
D ln(e) = RHS. D = RHS.
The polynomial rule is Dx^m = mx^(m-1).

ln(Δ+1)x^m = mx^(m-1).

[Edit (added f where applying the t-shift)]
I just watched the rest of the video. Looks like I wasn't the only one with this idea.

[agytjax]

At 1:54, how did you transform to geometric series sum ? The sum 1/(1+t) is applicable only when |t| < 1. There is no assumption that |t| < 1