we use these in physics in a lot of stuff, the typical example is the operator momentum p = (-ih/2π) d/dx in quantum mechanics. The translation operator we define roughly T(a)= exp(iap)
@@knivesoutcatchdamouse2137 I use "a" to describe the distance of translation, e.g. if ψ(x) is the wave function (usually L² integrable) the translation is T(a)ψ(x) = ψ(x+a). I think.
Hi, The first time I saw that was in automatic. We define the Heaviside symbol s to be equal to d / dt . And the derivative of a function f(t) is given by s multiplied by f(t). And we derived easily that e^s times f(t) is equal to f(t+1), more generally e^(s T) f(t) = f(t+T) This is used in automatic to calculate the transfert function of a "pure delay".
Dude at this point I'm not even surprised that Peyam left the rest of math TH-cam in the dust half a decade ago...I remember doing a video on f'=f^(-1) and someone in the comments pointed out that Michael Penn did a video on that a year or 2 ago ....after one TH-cam search I found out that Dr. Peyam did that 5 years before even Penn!! That guy's criminally underrated and definitely one of the best channels on TH-cam.
Majik. I like that little illustrative chart showing the relationship between N and k and how their indices are related in the last example.. Nice visualization tool. I'm getting voice.
At 8:28 , it should be k (k - 1) ... (k - (N - 1)) in the numerator, wheras you have k (k - 1) ... (k - (N + 1)). Not a big deal in the end, and I like the video!
Since the McLaurin expansion is the so-called exponential generating function (egf) of f, your work shows the (well known) fact that multiplying the egf with e^x shifts the expansion to the left. (I think)
You have the right idea, but a generating function (exponential, ordinary, etc.) is a power series generated from a *sequence*. In the case of the Maclaurin series, it is the EGF that comes from the sequence of derivatives of f.
Seems like in general, e^D(f(x)) = T_x(x+1) where T_x is the Taylor series approximation of f at x. Because T_x(t) = f(x) + (t-x)^1/1! f'(x) + (t-x)^2/2! f''(x) + ... And for t = x+1, that's just f(x)/0! + f'(x)/1! + f''(x)/2! + ... = e^D(f(x))
so does that mean that, like limit a->0 ((e^(a d/dx) - e^(-a d/dx))/(2 a) f(x)) = limit a-> 0 ((sinh(a d/dx) / a) f(x) ) = f'(x)? (because e^(a d/dx) f(x) = f(x+a) and so on)
This is probably obvious but e^(-D)(f) = f(x-1) right? Interesting that the product rule here is what you might call the “freshman” product rule (apply the operator on both functions) and then multiply them. I guess since all you are doing is a horizontal translation to the function, that should make sense?
really nice video! but i want to ask, why does D^n equals the nth order derivative? isn’t ^n just a notation, which doesn’t mean exponentiation? (sorry for bad english)
we use these in physics in a lot of stuff, the typical example is the operator momentum p = (-ih/2π) d/dx in quantum mechanics. The translation operator we define roughly T(a)= exp(iap)
th-cam.com/video/LlSHsqJHD0o/w-d-xo.htmlsi=jat3Ag06jtAFRCdj
What is the 'a' in this context?
@@knivesoutcatchdamouse2137 I use "a" to describe the distance of translation, e.g. if ψ(x) is the wave function (usually L² integrable) the translation is T(a)ψ(x) = ψ(x+a). I think.
Okay, thanks. I assumed it was translation but wasn't sure! I really wish I knew more about the physics of quantum mechanics.
@@knivesoutcatchdamouse2137 I started studying Quantum Mechanics from online courses and Zettilli's book. Maybe you can start there as well :)
Hi,
The first time I saw that was in automatic. We define the Heaviside symbol s to be equal to d / dt . And the derivative of a function f(t) is given by s multiplied by f(t).
And we derived easily that e^s times f(t) is equal to f(t+1), more generally e^(s T) f(t) = f(t+T)
This is used in automatic to calculate the transfert function of a "pure delay".
I love this kind of operational calculus. Anyone studying Lie groups knows about this result.
In the last step,if we assume h is holomorphic with h=fg then e^D (h)=h(x+1)=fg(x+1)=f(x+1)*g(x+1)
I just so happen to have watched Dr. peyam’s half decade old video on this exact thing earlier today, then you drop this.
Dude at this point I'm not even surprised that Peyam left the rest of math TH-cam in the dust half a decade ago...I remember doing a video on f'=f^(-1) and someone in the comments pointed out that Michael Penn did a video on that a year or 2 ago ....after one TH-cam search I found out that Dr. Peyam did that 5 years before even Penn!! That guy's criminally underrated and definitely one of the best channels on TH-cam.
Excellent job! I have seen this before, but not the last part about the product rule! Never knew it has such a "nice" Product rule!!!
They actually taught me this derivative.
Very useful analysis. Thank you
Majik. I like that little illustrative chart showing the relationship between N and k and how their indices are related in the last example.. Nice visualization tool. I'm getting voice.
At 8:28 , it should be
k (k - 1) ... (k - (N - 1))
in the numerator, wheras you have
k (k - 1) ... (k - (N + 1)).
Not a big deal in the end, and I like the video!
Very nice! And at the end abonus, the product rule that we've always dream'ed about😊
Since the McLaurin expansion is the so-called exponential generating function (egf) of f, your work shows the (well known) fact that multiplying the egf with e^x shifts the expansion to the left. (I think)
You have the right idea, but a generating function (exponential, ordinary, etc.) is a power series generated from a *sequence*. In the case of the Maclaurin series, it is the EGF that comes from the sequence of derivatives of f.
Recently watched Supwares video on the topic. Great minds truly think alike!
thank you very much, i wanted to understand this since i First saw It
Seems like in general, e^D(f(x)) = T_x(x+1) where T_x is the Taylor series approximation of f at x.
Because T_x(t) = f(x) + (t-x)^1/1! f'(x) + (t-x)^2/2! f''(x) + ...
And for t = x+1, that's just f(x)/0! + f'(x)/1! + f''(x)/2! + ... = e^D(f(x))
so does that mean that, like
limit a->0 ((e^(a d/dx) - e^(-a d/dx))/(2 a) f(x)) = limit a-> 0 ((sinh(a d/dx) / a) f(x) ) = f'(x)?
(because e^(a d/dx) f(x) = f(x+a) and so on)
Q: if we have f(x+1)=exp(f(x)) can we write f(x)^n= n-th derivative with f(x) for n to infinity?
This is probably obvious but e^(-D)(f) = f(x-1) right?
Interesting that the product rule here is what you might call the “freshman” product rule (apply the operator on both functions) and then multiply them. I guess since all you are doing is a horizontal translation to the function, that should make sense?
very nice thank you 😊
really nice video! but i want to ask, why does D^n equals the nth order derivative? isn’t ^n just a notation, which doesn’t mean exponentiation? (sorry for bad english)
We're defining it here as the order of the derivative
Hey man! I admire all your work! ¿What resources you use to learn about this topics?
Pls take this integral
Int( sin ( (cotx)^2) * (( secx)^2) ) dx
Limits 0 to pi/2
There seems to be some error in the vid
???
it took infinite derivatives before e^x was finally changed...
Since you don't have e^D(fg) = e^D(f)g+fe^D(g), I don't think e^D can be called a derivative, no?
My favourite way of switching the order is the following
We have 0 ≤ k ≤ n ≤ oo
Then just change the order in which you construct this chain