The craziest definition of the derivative you have ever seen! 

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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.

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.

Finally a good place to Stop

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?

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

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

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

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.

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.

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.

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.

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

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

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

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?

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.

I never formally kearned pretty much everything in this video yet you managed to make me understand it. Thank you.

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

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

But for what norme because for the 1 norme, norme of delta is 1 so ln(1+delta) is it always defined ?

It would be nice knowing how does convergence work with operators, the expansion of the logarithm has a limited radius of convergence

It would seem that applying the delta operator to any general (ln a) a^x will yield a^x(a-1)

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.

Great video! Another cool way to define the derivative is by using the symmetric difference: Δs f(x) = ( f(x+1) - f(x-1) ) / 2 = ( ( exp(d/dx) - exp(- d/dx) ) / 2 ) f(x) = Sinh(d/dx) f(x) d/dx = ArcSinh(Δs) These formal expansions are intuitive and systematic ways of finding finite difference coefficients at k-th order. en.wikipedia.org/wiki/Finite_difference_coefficient

You forgot the most important application, which is for deriving the formulas for numerical calculation of derivative much easier. You can easily derive Newton's forward difference formula from ln(1+delta). For the backward and central difference formulas, you can try to derive one by defining d/dx in terms of either delta f(x) = f(x) - f(x-1), or delta f(x) = (f(x+1)-f(x-1))/2.

Are there any type of functions where it is easier to solve/compute the derivative using ln(1+delta) definition than using the 😂conventional d/dx operator?

I must be missing something very basic about this definition. Here are 2 apparent contradictions, extremy simple ones: 1. Take f(x) = (x-0.5)^2. Then clearly f(1) = f(0) ==> delat(0) = 0 ==> ln(1+delta) = ln 1= 0 ==> f'(0) = 0; but this contradicts the regular calculus result f'(0) = -1. 2. More generally, f'(x) depends according to this definition only on the values f(x+1) and f(x). e.g. f'(0) depends only on f(1) and f(0). But clearly you can define infinite number of fuctions that have the same f(0) and f(1), but different slopes at zero. So what did I miss?

Isn’t that ln2 identity derivation circular since it relies on using the talyor expansion of ln(1+x), from which you could directly write down the ln2 identity

Your editor keeps missing more and more times when you start a new take

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.

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 Δ^(n+1)(x^m)=Δ^n((x+1)^m)-Δ^n(x^m) 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 Δ(x^m)=(x+1)^m-x^m=1+x+...+x^(m-1) is a degree m-1 polynomial, by linearity along with induction, Δ^n(x^m) has degree m-n for nm, Δ^n(x^m)=0.

At 5:30, is there a textbook that explains the delta**2, **3, expansion?

19:15 this should be (k+2) since x is approaching 1

Hi, On shorts (where there is no comment box), when I click on "Playing two gammas", I get "the how and why of set exponents" 😁

9:06 I took you up on it. It isn't terrible, but it isn't easy, and I needed to use induction more than once. To give an idea, first I expressed the nth forward difference on x^(m+1) in terms of the nth forward difference on x^m (which introduced a factor of (x + n) and an additional term), and then I had to show that an alternating sum of powers of forward difference operators all acting on (x + 1)^m is precisely x^m, which took the bulk of the effort. I thought that was interesting, since you've got this relatively simple operator composed of a finite sum of powers of the forward difference, and its operation on a polynomial shifts its argument by -1, so the operator is in fact e^{- d/dx). It turns out that the nth derivative of x^m introduces the coefficient n! times m choose n, whereas the coefficients of the nth power of the forward difference operator applied to x^m are also m choose n.

That's crazy amazing! I didn't think you could compute the derivative at one point by combining values from allover the function domain!

If i understand this correctly, that derivative operator calculates f'(x) by sampling the function f(x) with integer spacing. Surely there are functions where this doesn't work. For example, f(x) = sin(2nπx). f'(x) = 2nπ cos(2nπx) but ln(1+∆) f(x) = 0 since ∆ⁿf(x) is always 0 (this follows from ∆sin(2nπx) = 0).

I don't think that "Formal Taylor Theorem" link appeared on the end of the video as was intended.

You gotta love any connection between discreteness and continuity - particularly in view of modern science which is begging the scale.

What area of Mathematics does this have to do with? I'm a 2nd year bachelor student and I have never seen this come up, I'm wondering if I'll get to see more of this type of thing!

The application of this result to calculate the series representation of ln(2) uses circular logic since we already used the Taylor expansion of ln(1+x) as a given.We could just plug x=1 from the beginning.Other than that,great insight on a topic I've never heard of before!

The double sum should have (n+1)^m not (k+1)^m.

Does this have to do with e^d/dx(f(x))=f(x+1)?

It's a definition without limits.. incredible.. the derivative becomes a pure algebric operator

Is there an area of operator equations where instead of solving for a function like in differential or functional equation you solve for an operator like the derivative operator?

That was fresh.

This is fantastic

How do you take ln of e^(d/dx)? I am rusty on my operator theory, but ln does not have a valid expansion around x=0.

For the "direct" way, why not use induction? You already proved it for the x^3 case, so it shouldn't be too tricky to show it works for higher powers. Especially because there's a Pascal relation going on. I'll try it.

One needs to credit Leonard Euler for contributing to developing this formula.

Q&A @Michael: Is this limited to whole integers, only? If not, then why is delta^m(x^3)=0, m>/=4 rather than >3 (e.g.: any rational or irrational number greater than, but not equal to, 3)

Greetings, Michael! Did you think about finding inverse of it? Look, if ln(1+delta) defines derivative, then (ln(1+delta))^(-1) defines anti-derivative, right? If the inverse operator is found, will integration be easier? How many new integrals can be found in closed form? I think, it's a cool concept, it can simplify computations much, what do you think?

Wow , i dont think anyone dealing with math gets bored

Now consider f(x)=\begin{cases} exp(-1/x^2) if x>0 \\ 0 if x

thats cool

Hi! could you recommend me literature about this topic of finite calculus? Im really interested. In a previous video you mentioned there is a stoke’s theorem for finite calculus. I’m struggling to find that online. Also, if you know about some book that covers the relation between finite and infinitesimal calculus, as this video is about in some extent, i would really appreciate it. Greetings from Argentina!

Can delta be used to define a fractional derivative? Will it give different results from previous video technique with the Laplace transform (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-LfcVxbJzpRw.html)?

Just wondering. This is only true if f(x) is infinitely differentiable, or?

How does the spectral theory of functions account for power series with finite radii of convergence?

Delta = exp(D)-1; ln(1+Delta)=ln(exp(D)) =D

10:00 cursed notation warning haha

what if f(x)=f(x+1) for some x?

I want to know the integral of floor x.

Surprising! could it be in anyway usefull to compute the derivative of discrete data ? If so would it be relevent to use it instead of existing discrete derivative algorithms ?

The power of -1 should be n instead of n+1 in the first application I think

Whoa! BUT if that is the derivative, what is the anti-derivative?

log(1+Δ) exp( ik x ) = ik exp( ik x ).

Yeah I made very beautiful formulas, still I don't published, already pass 23 years.

Your first application is not interesting, because the expansion of ln(2) as a rational series follows directly from the same expansion necessary to calculate ln(1+Δ) to begin with.

what the heck is that

This settles it then. A particle's momentum really depends on its future behavior! Therefore everything in the universe is predetermined. Unless maybe uncertanty principle?

I think yu don't have to put one but thé identité of derivative

If for some reason we dont know the value of m but do know the value of m-1, this formula will not be helpful to find it because it requires the binomial expansion of (k+1)^m so the definition is circular and non constructive

13:18 and from inspection, 2 = 1 + Δ application concluded

Then derivative becomes not local character of function. And this is problem for many cases. It uses nonlocal operator Δ.

Crazy!!!!!!!!

Nice identity!

Never been this early