Oliver Heaviside Used the 1/2 derivative to sole some partial differential equations problems related to what we now call the heat equation. In particular the problem of a 1/2 plane, say y
That would make sense, since if the factorial of a negative integer was defined, taking an integral of a function f(x) negative one times would give you f’(x).
In fact that's where that formula for fractional derivative comes from. Using Cauchy's formula for repeated integration and extending it to fractional values
For completeness of testing, it would have been good to show that D^-1/2 D^1/2 f(x) = f(x) + C It would also be good to show whether associativity and commutativity hold.
Incredibly interesting, I'd love to know who originally worked on this? I enjoy learning a little of the history behind these things so I know they didn't just pop out of the sky.
I tried the half derivative of the function f(x)=x^n and show that D^(1/2)(D^(1/2)f(x))=Df(x)=nx^(n-1) for every positif integer n, and it turned out pretty fun. I must say, I used a result for evaluating a even power of a sine function but it was nice!
From what I understand, there isn't just one definition for fractional derivatives/integrals. I believe several people have done them, one of them being Cauchy
Could you use this to determine the integrals of functions with no elementary solution or even the integrals of functions that have a strange but difficult to calculate solution (e.g. integral of sqrt((sin x)^2)dx)?
To the contrary, it can be fairly easily demonstrated that this does not hold in general. For example, let α = 1 and β = 1/2, and let f(x) = 1 everywhere. Then, it follows that (D^β)(D^α)(f) is equal to the zero function, but (D^α)(D^β)(f) is not equal to the zero function, but is instead a constant multiple of g, where g(x) = 1/sqrt(x). Actually, there are far bigger problems with the definition used in the video, which I will elaborate in my comment to the video.
It really depends on which operator you are using, since there are many different operators in fractional calculus. The one in this video is called the Riemann-Liouville differintegral operator, which is obtained by using repeated derivatives of the repeated fractional integrals, and the fractional integrals are obtained from Cauchy's repeated integration formula.
With all due respect, the way these operators are introduced and presented in this video is misleading. The problem here is that the video implies that, with the way the operators D^α are defined for real α, it would seem to be true that (D^α)(D^β) = D^(α + β) for all real α, β, and yet, this is false. As a counterexample, let α = 1/2, β = 1, and let f = 1. In fact, there is a bigger fundamental issue here: there cannot exist a linear operator H such that H^2 = D. Therefore, there is no such a thing as a half-derivative, strictly speaking. Here is a proof: we know that D(1) = 0. Hence H(D(1)) = H(0) = 0. H(D(1)) = D(H(1)), so H(1) = c, for some constant function c. Now, H(H(1)) = H(c) = cH(1), but H(H(1)) = D(1) = 0, so cH(1) = 0, and so, c = 0 or H(1) = 0, and if c = 0, then H(1) = 0, thus H(1) = 0 in all cases. Furthermore, remember that D(x) = 1, so H(D(x)) = H(1) = 0. However, H(D(x)) = D(H(x)), so D(H(x)) = 0, hence H(x) = k, for some constant k. Thus, H(H(x)) = H(k) = kH(1) = 0, but H(H(x)) = D(x), so D(x) = 0. Finally, since D(x) = 1, it follows that 0 = 1. This is false, and so, we have a contradiction. Since every algebraic step is correct, then the assumption that there exists linear H such that H^2 = D must be false. Q. E. D. I understand that the operator you presented in the video is a legitimate mathematical object: it is called the Riemann-Liouville differintegral operator, and it is a generalization of Cauchy's repeated integral operator, and it does have many applications in physics. I have no problem with you introducing the operator, discussing its properties, and teaching its applications. Nonetheless, it is inaccurate to say that it is a fractional derivative. I feel justified in pointing this out, because this is not some minor technicality, and unfortunately, too many other content creators have already spread this same misconception. Here is what I will say, though: there is such a thing as a fractional integral. The integral operator does not have the problem that the derivative operator has when attempting to fractionalize it. The source of the problem is that, contrary to popular misunderstanding, the derivative and integral operators are not mutually inverses of each other, and so, they do not commute. Yes, the derivative operator applied to the integral operator is equal to the identity operator, but it does not work the other way around. Also, it is possible to restrict the domain of the derivative operator D (i.e., restrict the space of functions you are allowed to use) such that, in this restricted space, fractional derivatives can be defined. However, there is a huge drawback here: you have to get rid of all functions of the form x^α, including nonzero constant functions (as well as linear combinations thereof), and of logarithms, products thereof, and all functions resulting from their integration and arithmetic operations between them. You would be getting rid of most useful functions you would want to work with. So, even if this approach were available, in practical applications, you would never use it anyway.
@@abhinavanand9032 H is, by hypothesis, a linear operator, and since it is linear, H(0) = H(0 + 0) = H(0) + H(0). H(0) = H(0) + H(0) implies H(0) = 0, because you can subtract H(0) from both sides of the equation.
@@angelmendez-rivera351 Can you always interchange linear operators. Take the integral operator from 0 to x and the derivative operator if you apply derivative first then integral you get f(x)-f(0) if you do the reverse you get f(x)
From 15:10 to 15:30 can you please explain how you got derivative 0. Since after integration we get 1/√xArcsin(u/√x) after putting u by √x we get π/2√x . It's function of x or is there a mistake in my calculations
no offence isn't it high school level math I mean u r asking for dy/dx , where y=f(t) x=g(t) or something like that 😊 if not please elaborate on ur question 😅
@@aravindakannank.s. Yes, you have the right idea. Strictly speaking, what you get is f'/g', where f' and g' are the derivatives of f and g, respectively, so this is not too interesting.
Oliver Heaviside Used the 1/2 derivative to sole some partial differential equations problems related to what we now call the heat equation. In particular the problem of a 1/2 plane, say y
Theory before watching the video: the negative half derivative somehow turns out to be a half integral
That would make sense, since if the factorial of a negative integer was defined, taking an integral of a function f(x) negative one times would give you f’(x).
In fact that's where that formula for fractional derivative comes from. Using Cauchy's formula for repeated integration and extending it to fractional values
For completeness of testing, it would have been good to show that D^-1/2 D^1/2 f(x) = f(x) + C
It would also be good to show whether associativity and commutativity hold.
Incredibly interesting, I'd love to know who originally worked on this? I enjoy learning a little of the history behind these things so I know they didn't just pop out of the sky.
Check out the Historical Notes section of the Wikipedia entry for "Fractional Calculus".
Oh yes,please more cool mini lectures like this!💯💯
I knew about fractional derivatives but didn't knew the equation is so cool,ouh ma gwad.
I tried the half derivative of the function f(x)=x^n and show that D^(1/2)(D^(1/2)f(x))=Df(x)=nx^(n-1) for every positif integer n, and it turned out pretty fun.
I must say, I used a result for evaluating a even power of a sine function but it was nice!
From what I understand, there isn't just one definition for fractional derivatives/integrals.
I believe several people have done them, one of them being Cauchy
The definition used in the video is indeed suspiciously similar to Cauchy's integral formula.
seems like the formula is taking the (1-nth) integral and differentiating once ?
Could you use this to determine the integrals of functions with no elementary solution or even the integrals of functions that have a strange but difficult to calculate solution (e.g. integral of sqrt((sin x)^2)dx)?
Can you proof generally that the applying the alfath drivative then the betath drivative is the same as doing (alfa+beta)th drivative
For the purpose of proofs....I am a physicist 😂😂😂
To the contrary, it can be fairly easily demonstrated that this does not hold in general. For example, let α = 1 and β = 1/2, and let f(x) = 1 everywhere. Then, it follows that (D^β)(D^α)(f) is equal to the zero function, but (D^α)(D^β)(f) is not equal to the zero function, but is instead a constant multiple of g, where g(x) = 1/sqrt(x). Actually, there are far bigger problems with the definition used in the video, which I will elaborate in my comment to the video.
How does one come up with the definition to the fractional derivative operator? What is the logic behind it?
Check out the book linked in the comments section
@@maths_505 Thanks!
It really depends on which operator you are using, since there are many different operators in fractional calculus. The one in this video is called the Riemann-Liouville differintegral operator, which is obtained by using repeated derivatives of the repeated fractional integrals, and the fractional integrals are obtained from Cauchy's repeated integration formula.
Master, could you please make video on solving auxiliary differential equation please 🙏🙏🙏🙏🙏🙏🙏🙏
What is the application of such exotics?
With all due respect, the way these operators are introduced and presented in this video is misleading. The problem here is that the video implies that, with the way the operators D^α are defined for real α, it would seem to be true that (D^α)(D^β) = D^(α + β) for all real α, β, and yet, this is false. As a counterexample, let α = 1/2, β = 1, and let f = 1.
In fact, there is a bigger fundamental issue here: there cannot exist a linear operator H such that H^2 = D. Therefore, there is no such a thing as a half-derivative, strictly speaking. Here is a proof: we know that D(1) = 0. Hence H(D(1)) = H(0) = 0. H(D(1)) = D(H(1)), so H(1) = c, for some constant function c. Now, H(H(1)) = H(c) = cH(1), but H(H(1)) = D(1) = 0, so cH(1) = 0, and so, c = 0 or H(1) = 0, and if c = 0, then H(1) = 0, thus H(1) = 0 in all cases. Furthermore, remember that D(x) = 1, so H(D(x)) = H(1) = 0. However, H(D(x)) = D(H(x)), so D(H(x)) = 0, hence H(x) = k, for some constant k. Thus, H(H(x)) = H(k) = kH(1) = 0, but H(H(x)) = D(x), so D(x) = 0. Finally, since D(x) = 1, it follows that 0 = 1. This is false, and so, we have a contradiction. Since every algebraic step is correct, then the assumption that there exists linear H such that H^2 = D must be false. Q. E. D.
I understand that the operator you presented in the video is a legitimate mathematical object: it is called the Riemann-Liouville differintegral operator, and it is a generalization of Cauchy's repeated integral operator, and it does have many applications in physics. I have no problem with you introducing the operator, discussing its properties, and teaching its applications. Nonetheless, it is inaccurate to say that it is a fractional derivative. I feel justified in pointing this out, because this is not some minor technicality, and unfortunately, too many other content creators have already spread this same misconception.
Here is what I will say, though: there is such a thing as a fractional integral. The integral operator does not have the problem that the derivative operator has when attempting to fractionalize it. The source of the problem is that, contrary to popular misunderstanding, the derivative and integral operators are not mutually inverses of each other, and so, they do not commute. Yes, the derivative operator applied to the integral operator is equal to the identity operator, but it does not work the other way around.
Also, it is possible to restrict the domain of the derivative operator D (i.e., restrict the space of functions you are allowed to use) such that, in this restricted space, fractional derivatives can be defined. However, there is a huge drawback here: you have to get rid of all functions of the form x^α, including nonzero constant functions (as well as linear combinations thereof), and of logarithms, products thereof, and all functions resulting from their integration and arithmetic operations between them. You would be getting rid of most useful functions you would want to work with. So, even if this approach were available, in practical applications, you would never use it anyway.
Why H(0)=0
My jaw has been dropped at this simple but intuitive proof. Thank you for presenting and explaining so extensively!
thanks bro
@@abhinavanand9032 H is, by hypothesis, a linear operator, and since it is linear, H(0) = H(0 + 0) = H(0) + H(0). H(0) = H(0) + H(0) implies H(0) = 0, because you can subtract H(0) from both sides of the equation.
@@angelmendez-rivera351 Can you always interchange linear operators. Take the integral operator from 0 to x and the derivative operator if you apply derivative first then integral you get f(x)-f(0) if you do the reverse you get f(x)
What is the half derivative of the exponential function? Is it the same as itself?
e^x*erf(√x)+1/√(pix)
It depends on how you choose the lower bound of the integral.
Is derivative inverse of the Integral?
Yup
how to prove that formula at the beggining
It can't be proven. One just accepts it as a definition.
Bro, why aren't you uploading shorts😢?
From 15:10 to 15:30 can you please explain how you got derivative 0. Since after integration we get 1/√xArcsin(u/√x) after putting u by √x we get π/2√x . It's function of x or is there a mistake in my calculations
We don't get a 1/sqrt(x) term there.
@@maths_505 but sir our integrand is 1/√(x-u^2 and not merely 1/√(1-u^2 so after integration we get 1/√x Arcsin(u/√x) then apply limits isn't it?
@@vcvartak7111 nah mate we don't have the factor for arcsine integral
@@maths_505You are right, it was mistake from my part
is it just me or the thumbnail changed
yup it's changed
May I ask why you believe in a god? Not hating just wondering about your view.
B😴😴😴😴😴RING ........Try taking derivative w.r.t to a FUNCTION 😈😈😈😈
no offence isn't it high school level math
I mean u r asking for
dy/dx , where
y=f(t)
x=g(t)
or something like that 😊
if not please elaborate on ur question 😅
@@aravindakannank.s. Yes, you have the right idea. Strictly speaking, what you get is f'/g', where f' and g' are the derivatives of f and g, respectively, so this is not too interesting.
@@angelmendez-rivera351 exactly thanks for clarifying me 😊