Half Derivative as a limit
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- Опубликовано: 28 окт 2024
- In this video, I present the Grunwald-Letnikov formula, which expresses the half derivative in terms of a limit. It is very reminiscent of (and in fact motivated by) the Leibniz formula for derivatives, and gives a geometric interpretation of half derivatives. Enjoy!
Check out my fractional derivatives playlist: • Fractional Derivatives
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So you also have a phd in derivatives. I see.
Hahaha, a Professor of the Graduate School in Derivatives 😉
@@drpeyam
I knew it!
captain america!
captain marvel!
doctor strange!
doctor derivative!
Quantum mechanics' students do not approve this video...
h bar is 1,05457*10^(-34) J*s approx
Dr Peyam, the most awesome thing is that in Russian "leto" means summer. So Letnikov surname has the same root as summer.
Whoaaaaa!!!!
Next, recall that we define the regular integral be the limit of Riemann sum. Can you express fractional integral as a limit of some kind of sum? If so, prove that the half integral of the half integral is the regular integral using this definition
That would be cool!!!
15:02 l from k to infinity. 17:24 k from 0 to l.
Wow, last video you took was for this! How brilliant appoach to 'real' derivative! I really thank you to show us these kind of interesting topics. And I really like your way to give lectures. I think there must be more abstract secrete in these kind of topics. The formula you derived was only by observing and appreciating patterns. As abstract algebra was started by understanding a bunch of permutations, there must be something which I might overlooked in binomial expansion. Who made this beautiful distraction!
What about a video on the square root of the Laplacian? The way Dirac did it is amazing
Great idea
If there are fractional derivatives, is there such a thing as a fractional integral?
just draw a fraction of the integral tail and thats all
i believe that for a noninteger α the two are intertwined.
The -1st derivative is the integral. Negative fractions represent fractional integrals. en.wikipedia.org/wiki/Differintegral
There are some problems on fractional integrals in Folland's textbook on Real Analysis.
Hey in the k-l double summation, shouldn't k go from 0 to l instead of 0 to infinity? Because m never goes below 0 so k will never go above l.
Thank you Dr Peyam, these topics are so fascinating...but when i was youg I studied functional analysis, generalised functions, complex analysis on different books (i.e. Rudin, Zuily, Hormander...), but I've never found a single chapter of them dealing with non-integer derivative and similar. Could you give me some reference? many thanks again...
Great video! One question. Why alfa choose m does not involve the gamma function?, because according to the regular definition of the combinatory number, the numerator would be alfa! And since alfa is rational, the gamma function would be involved.
Your way of writing "f" always fascinates me.
Another great video. Would you have any comment on how to apply this to modeling real world processes? What I mean is that an integral can be modeled as:
y(k+1)=y(k)+u(k)*dt : where y is the integral output and u is the integral input.
Can we make a similar formula for the half integral or half derivative? If the answer is an infinite series that makes it much harder to model than the above formula which is a very compact formula.
If you're curious, the half derivative comes into play when modeling the photon counter used on very sensitive optical devices such as those on the Hubble space telescope or Gemini ground-based telescope.
Thank you Dr Peyam i love this channel
An obviously drunk H walks into a bar. The bartender asks if H hasn't had too many, so H replies, I haven't reached my limit!
and the bar was named after him as well… H bar
Nice.
Is there a way to generalize derivation to any fractional derivative (or indeed to complex derivative) by using fourier series or fourier transform ?
After all, exp' (ikt)=k*exp (i(kt+pi/2))
So, maybe :
Dalpha exp(ikt) = k^alpha*exp(i(kt+pi*alpha/2))
Half derivative and Fourier transform ruclips.net/video/jifnh7XNtfo/видео.html
Very clear steps, thanks
I kinda giggled at that "bar-bar...this is barbaric", because that's exactly how the Greeks coined the term "barbaros"(=barbarian)
Does this limit even converge for functions like f(x)=x?
I think I got that this limit doesn't exist when I tried this definition.
I saved your fractional derivatives playlist :-)
Thank you!!! 😄
I was wondering if someone could explain more in detail how we moved from having h and hbar in the two separate limits to just h in the next step, expressed as one limit. The time stamp is 02:17 of the video. Thank you so much it would really help with my project
In general if a limit in (x,y) exists, then it exists along every direction, in particular the direction y = x, which becomes the limit in (x,x)
Here it’s the same but with h instead of x and hbar instead of y
Don't you have to use the Cauchy Product at 22:00?
I am very curious what the potential applications of this are.
19:00 looks a bit like the Vandermonde Identity
Grunwald is also a name of a certain village in Poland. Name of the battlefield where we, Poles, kicked those German asses in 1410 :P
Gosh, I love your videos!
What is the result of the sum:
(D^1/2 + D^1/4 + D^1/8 + ...) 1/x = ???
exp(i*Pi) = -1 so ----> D^exp(i*Pi) f(x)= D^-1 f(x) = Integral f(x) ????
What if I'm a physicist and I have hbar going to 1 instead of 0?
Plzz can you integrate log(sin x) from 0 to pie
Already done ✅
Peyams video about it: ruclips.net/video/iNaiq_IETEs/видео.html
Started to lose me at 2:10 … what's this ”diagonal argument”?
i've calculated 0 < α < 1 derivative of the constant function and i've got oo. is that really correct? if so: brr!! does that mean that D^ α ∘ D^(1-α) != D?
No the you are doing it wrong.the def of D^a is supposed to be consistent with integer derivatives
@@avdrago7170 let 0 < α < 1 and f = c, then
D^α(c) := lim as h -> 0 of 1/h^α sum(n=0, oo) (-1)^n (α choose n) c.
notice that by the definition of the falling factorial (α choose 0) := (α)_0/0! = 1/1 = 1, and by the definition of the binomial coefficient (α choose n) = 0 for n ≥ 1. therefore the above series reduces to just one term, the zeroth one
D^α(c) = lim as h -> 0 of 1/h^α (-1)^0 (α choose 0) c = lim as h -> 0 of 1/h^α 1 1 c = lim as h -> 0 of c/h^α = ±oo.
I've got 0 , maybe we should find a middle ground or something?
@@michalbotor @michał botor For 0 < α < 1 (α choose n) is not equal to 0 if n ≥ 1. So the the sum is not a finite sum.
D^α(c) = lim as h -> 0 of 1/h^α ∑ₙ₌₀ᵒᵒ (α choose n) (-1)ᵏ c
Note that by taylors theorem we have:
(1+x)^α = ∑ₙ₌₀ᵒᵒ (α choose n) xⁿ
So we get:
D^α(c) = lim as h -> 0 of 1/h^α * c (1+(-1))^α = 0
as expected