Half derivative of cos x

[@drpeyam]

Note: There’s a typo at the end; there should be no i in sinh(pi/2 i), it should just be sinh(pi/2)

[@yoylecake313]

it would be sin(pi/2 i)

[@adamcummings20]

Dr Peyam is basically a chalkboard himself now

[@danielmilyutin9914]

I've noticed that, too. That's usual for Doctors and PhD-s. :)

[@michalbotor]

wonder why they might want you to think that. wink, wink, nudge, nudge. ;p

[@edgardojaviercanu4740]

He has cleaned the blackboard with his t-shirt.

[@mathevengers1131]

You know, he is something of a blackboard himself.

[@steppindown6874]

What do you mean by that 🤨

[@blackpenredpen]

@19:36 , that wasn't easy to do...

[@adblockturnedoff4515]

Do a barrel roll!

Why? Just changing theta in the polar world ... :-P

[@alanturingtesla]

So good. I always love it when imaginary numbers are included.

[@blackpenredpen]

So good!

[@Fire_Axus]

your feelings are irrational

[@OonHan]

2=1+1

[@nchlhkhfif3143]

unless we're talking relativity

[@iwacher]

Thats debatable.

[@yoylecake313]

that is trivial

[@savajevtic8040]

As the co-founder and admin of the Facebook group Aesthetic Function Graphposting, I want to thank you for joining the group and I encourage you to share some non-integer derivative visualizations in the group!

[@drpeyam]

OMG, I loooooove that group, thanks so much!!!

[@Manuel48437]

I love your fractional derivative videos, can't wait for the next one :P:P:P

[@Fire_Axus]

your feelings are irrational

[@ribozyme2899]

Note: the power rule for half derivatives gives a completely different result! That's because fractional derivatives are non-local. The power uses the base point x=0 implicitly, and this video uses the base point x=-infinity. For a more detailed explanation see: www.mathpages.com/home/kmath616/kmath616.htm

[@yuvalpaz3752]

oh my cosh!!!

[@rextransformation7418]

😂😂😂

[@SlyEcho]

10:40 Mind blown and it makes so much sense too!

[@andreapaps]

This guy is covered in chalk doing maths with a fat smile on his face... What a champ, really enjoyable to watch.

[@ethancheung1676]

This is interesting and without the need to see all the prove why all things work extending to complex number. Just Do It!

[@JBaker452]

I do love the idea of a half-derivative.

[@Thalesfreitas96]

Great video. On the last derivative, the i-th derivative of cosine, you could have expressed the cosine as u(x,y)+i*v(x,y), u being cos(x)cosh(y) and v being -sin(x)sinh(y).

[@papajack2205]

I love how Dr. Peyam was already covered in chalk dust, just before any writing on the board. It was also a really exciting video - thank you.

[@paulg444]

the guy has a fantastic attitude !.. great energy, great teacher.

[@SchrondingersCat]

This channel is bomb! I used to feel like fractional derivatives were so abstract.

[@Fire_Axus]

your feelings were irrational

[@mst.sharminakter4149]

Sir, your lecture is very interesting and there is no confusion to understand.

[@Sam_on_YouTube]

So the rate at which the x component of the unit circle changes in the direction perpendicular to the plane is related to the value of the x component of the unit hyperbola and the value of the y component of the unit hyperbola rotated into the i direction. Holy conic sections batman, that's awesome!

[@spencertaylor6910]

Yeah!!!! Awesome job, as usual. Are you going to do fractional derivatives of logarithms and such? You should. Also, who is your favorite mathematician of all time? Thanks you're the best!

[@drpeyam]

It’s on my countably infinite to-do list, but the answer is easier than you think! The half derivative of ln is a constant times x^(-1/2). Oh, and for mathematicians, so many to choose from!!! I like Euler a lot, and Laplace :)

[@spencertaylor6910]

Dr. Peyam's Show Sweet! Both are legends! Thanks so much!

[@savajevtic8040]

Extending factorials to non-integers, and even to complex numbers, is truly an amazing thing.

[@ehsankharazmi3148]

Nice and passionate video! well explained. Just note that these fractional derivative formulas are valid in unbounded domain, where x belongs to (-infinity, +infinity). In other words, the fractional derivatives are taken from -infinity to x. If the domain is bounded, then there exist additional extra terms.

[@simplyyummy92]

Awesomeness! I love crazy maths things. Thanks πam!

[@61rmd1]

We should love these stuff, it's a wonderful way to see our life. I guess that it should be valid for partial derivatives too...Thanks for sharing all these beauties. Greetings from Italy

[@DancingRain]

I adore your enthusiasm! So validating :) I'm glad I'm not the only one who gets excited about mathematics.

[@EngMorvan]

The last expression can also be simplified to D^i cos(x) = cosh((π/2)-ix)

[@MF-lg8mt]

This video is interesting. However, there are many operators for fractional derivatives and the semi group property(D^aD^b f=D^{a+b}) work only with under assumptions for the funtions. I work on control theory and fractional differential equations. Best regards from Mexico! 😊 I your follower now.

[@MrRyanroberson1]

Just as we have the factorial expansion, we might need the same type of generalization for the nth derivative specialized for things like y^n for y is f(x), instead of raw x^n

[@Galileo2pi]

Each time I watch this demo I like it more, thanks

[@goofypettiger]

Excellent lecture. Thank you for sharing this.

[@johnny_eth]

Ah at 12:00 very interesting. I paused the video and did the fractional derivate of sine on my head. I got to D^α(sin(x)) = sin(x) * cos(π/2)^α + cos(x) * sin(π/2)^α This seems quite similar to your cosine expansion.

[@g3452sgp]

This is great. I could understand.

[@stevenwilson5556]

The one part missing from this video is what the hell you use fractional and alpha derivatives for. But the calculation part is genius. There's a trig identity about cos(x - pi/2) = sin(x), so some of this video would have been simplified by citing it.

[@aymankarim3206]

Thank you mr Dr Peyam

[@VSP4591]

Dear Dr. Peyam. We all know that the first derivative is the tangent to the function. What would be the geometrical signification of a half derivative of a function??? Thank you.

[@adamur2]

graph of that ( or of absolute value of that) expression would be very interesting, to compare with cos

[@SJJR1]

Excellent explanation, very enthusiastic!!!!!!

[@PedroHenrique-zy3uh]

This is really REALLY AWESOME! Thank u so much for doing this videos

[@twwc960]

In a previous video, you showed how to calculate the half derivatives for powers of x. If you expand exp(x), or sin(x) or cos(x) as power series and apply the half derivative operator to each term, do you get the same results as the ones you define in this video? Since D^(1/2) is linear, it seems that you should, but I'm not sure how to derive that.

[@szturmix]

This guy is brilliant!

[@jarikosonen4079]

This would make sense if taking 1/2 derivative twice gives the derivative. Can you take negative derivate with this then to get integral? D(-1){cos(x)}=sin(x)+C ? Maybe this can be used to prove the D(-1) does not need the 'C'... Also this makes sense as smoothly 'shifting' the curve between the original and the derivative, but does it have any application outside the mathematics or can this be used to prove some other mathematical theories, etc?

[@farstar31]

Pretty good stuff, I like this channel. By the way, it seems you put an "i" in the hyperbolic sine at the end.

[@gnikola2013]

According to the Fundamental Theorem of Calculus, derivation is the inverse operation of integration, and vice versa. However, the concept of the integral was developed at first from a more geometric-like view, calculating the area underneath a curve, thinking about it like the sum of the area of rectangles. Would there be a way for you to develop the concept of a "fractional integral"? I don't think it would have a geometric meaning such as calculating an area, maybe you may use the gamma function like you did with the fractional derivative. After all, you moved away from its geometric point of view, you could do the same with integration.

[@gnikola2013]

I actually hadn't reach to the end of the video lol. It's amazing how you can define the fractional integral just by considering a negative value of alpha. It's really astounding how all these definitions become kinda trival by thinking about the derivative as a linear transformation. I need to learn much more!!

[@scitwi9164]

Neat, eh? Makes you think of all those functions that were hard to differentiate or integrate with whole steps. Now we can think whether some of them could be calculated if the multiple of that step is known ;)

[@ChefSalad]

Actually D(-1/2)(cos(x)) being a half-integral, should have an additional term, I think. Specifically it should be cos(x+π/2*α)+2*√(x/π).

[@xy9439]

Does this have any practical applications? Still, it would be interesting to elaborate a whole theory around this subject

[@robsternquist4406]

Adàlia Ramon I doubt there is an obvious application but as with allot of math at the time it doesn't have a obvious use but later we do eventually find one. However math need not always have a use sometimes it's beauty is reason enough to me.

[@mht5749]

It's actually used in some parts of engineering

[@postbodzapism]

en.wikipedia.org/wiki/Fractional_calculus#Applications

[@youmah25]

it can be used in control theory "fractionnal pid"

[@MF-lg8mt]

Yes. There are many applications. I work on control theory and fractional differential equations.

[@WerIstWieJesus]

Wonderful: for the fractional derivative of cos and sin you get the same result as I posted yesterday to you. My intuition went the other way around: In the integer derivations of cos I substituted sin as cos(x+pi/2) and got the same regularity I generalized. ... and BTW: With the addition theorems we can now indicate directly any integral or derivation of a Fourier analysis as another Fourier analysis. Is this perhaps a deeper understanding of Integration and Derivation: The rotation of the basis in a Fourier analysis?

[@martind2520]

That is incredible stuff. This is the reason I love maths. Speaking of which, I have a question I set myself that I haven't been able to solve. What is lim(n->infinity)(sum(1/m) - ln(n)) where the sum is from m=1 to m=n? I've googled it and found nothing, the best I've been able to do is prove that it is between 0.5 and 1. Do you have any methods? Thanks.

[@redvel5042]

It is the Euler-Mascheroni Gamma constant.

[@martind2520]

Wow, thank you, awesome.

[@redvel5042]

You're welcome. Gamma is actually quite the fascinating constant. As far as I remember, it's still unknown whether it is irrational or not, but if it is rational, then the numerator would be huge. It also shows up in the Riemman Zeta function, in the approximation for it around s = 1, if I remember correctly. Moreover, it seems to be related to the Gamma function, too. I don't really remember how it is related to it, though. Regardless, it is rather interesting.

[@drpeyam]

Wow, I learned something new today :P

[@mirkorokyta9694]

There really is a connection between gamma (Euler-Mascheroni constant) and Gamma (function). There are several of them, e.g. comparing the derivative of Gamma in whole numbers to gamma, but in your context one of the most interesting would be lim(x->0) (1/x - Gamma(x)) = gamma, showing the behavior of Gamma around zero.

[@metrogman2409]

That was exciting and great the whole way through. Thanks

[@rextransformation7418]

Is there any practical application from all this information?

[@Fire_Axus]

you can also use the taylor expansion

[@GinoGiotto]

Really cool!! I don't understand very well the linearity of fractional derivatives, can you demonstrate that? Which fractional derivatives are linear and which not?

[@drpeyam]

All fractional derivatives are linear, it’s sort of a requirement of fractional derivatives. That is we always have D^a (f + g) = D^a f + D^a g, and D^a cf = c D^a f for any constant c, and this is valid for all values of a

[@JBaker452]

Dr. Peyam is now a new super hero named: Chalk Man! :-) Easily recognized by his back shirt and shorts covered in ever-powerful chalk dust. Maybe he alone may someday solve the deepest problems surrounding the non-polynomials - hmmm...

[@scitwi9164]

Chalk Man. Half-man half-chalk :) (and half-derivative is his superpower)

[@yamsox]

I've been waiting for this my entire life

[@georgehnatiuk5806]

Hello, Generalizing the operation of differentiation and integration to non-integer orders can be performed in various ways. The exponential approach seems to give a very satisfactory way of defining fractional derivatives but it is very limited and not very useful. In fact, it does not lead to a consistent mathematics outside of the realm of exponentials where the rth derivative of e^ax is simply defined as (ar)e^ax. Such a definition is consistent with the purely exponential approach and you can combine complex exponentials to form periodic sinusoids and represent periodic functions. However, the method fails when considering power functions. There is no Fourier representation of open-ended functions such as polynomials, so they have no well-defined spectral decomposition. Of course, we can find the Fourier representation of x over some finite interval, but what interval should we choose? This method you outlined does not work with transform theory such as the Laplace transform. The exponential method fails and leads nowhere. Defining fractional calculus operations can be accomplished in a number of ways for such is not unique. A very useful method in defining fractional derivatives and integrals is through the Laplace Transform whereby you perform the operations in the complex frequency "s" domain by multiplying or dividing the transformed function f(x) by s to a fractional power , r, that represents the fractional degree of integration or differentiation. For the (r-th derivative) or (r-th integral): differentiation: s^r * F(s) integration: F(s) / s^r Using this method we obtain the following: 1/4 derivative of "x" is g1(x) = [ x^3/4 ] / Γ(7/4) 3/4 derivative of "x" is g2(x) = [ x^1/4 ] / Γ(5/4) 1/4 integral of "x" is g3(x) = [ x^5/4 ] / Γ(9/4) 1/2 integral of "x" is g4(x) = [ x^3/2 ] / Γ(5/2) NOTE: Γ(5/2) = [ (3•√π) / 4 ] Remember, Γ(p + 1) = p Γ(p) The use of this relationship will result in many cancellation of factors with a much simplified expression. ======================================================================== This approach leads to consistent results when performing a sequence of operations: For Example: derivative of 1/4 integral of "x" is the 3/4 derivative of "x" d [g3(x)] /dx = d [ ( x^5/4 ) / Γ(9/4) ] /dx = 5/4 [ x^1/4 ] /Γ(9/4) = [ x^1/4 ] / Γ(5/4) = g2(x) ======================================================================== Where p is any REAL number such that p > - 1/2 , the general 1/2 derivative for x^p is: h(x) = [ ​1⁄2 [d x^p ] / dx​1⁄2] = [ Γ(p + 1) / Γ(p + 1/2) ] x^(p - 1/2) NOTE: when p = - 1/2, h(x) = [ ​1⁄2 [ d (1/√x ) ] / dx​1⁄2] = √π 𝛅(x) The 1/2 derivative of a constant, A, is not zero: letting p = o [ ​1⁄2 [ dA ] / dx1⁄2 ] = A / √(π•x) ################################################################ The general 1/2 integral for x^p where p > - 3/2 is any REAL number: h(x) = [ ​1⁄2 ∫​ ] (x^p) dx = [ Γ(p + 1)/Γ(p + 3/2) ] x^(p + 1/2) + A/√x where "A" is an arbitrary constant NOTE: p > - 3/2 If you 1/2-integrate a second time, you will find that the double 1/2-integral on x^p is the full integral as should be the case. g(x) = [ ​1⁄2 ∫​ ] h(x) dx = [ ​1⁄2 ∫​ ] { [ Γ(p + 1)/Γ(p + 3/2) ] x^(p + 1/2) + A/√x } dx g(x) = x^(p+1)/(p + 1) + A√π = x^(p+1)/(p + 1) + C The arbitrary constant of integration "C" is A√π from the first 1/2-integration, which is just another way of writing the arbitrary constant. Since "A" is arbitrary, C is also arbitrary: C = A√π is arbitrary since "A" is arbitrary If you take the 1/2 integral of "x" wrt "x" n times, the result is: [ x^(1 + n/2)] / Γ(2 + n/2 ) = [(√x)^(2 + n)] / [ (1 + n/2)(n/2)Γ(n/2) ] ------------------------------------------------------------------------------------------------------------------------------------------------ When you indefinitely integrate a function, f(x), wrt "x" you get the function, F(x) + constant, which is the (antiderivative). The constant is added because you get the SAME original function, f(x), you started with when you differentiate the antiderivative since the derivative of a constant is ZERO. So the antiderivative is determined to within a CONSTANT when integrating indefinitely. Can you add a constant to the indefinite fractional integration? Well, it depends on what the fractional derivative of a constant is. If the 1/2 derivative of a constant is ZERO then an arbitrary constant can be attached to the half-integral (anti-half derivative) without affecting the half-derivative of the anti-half derivative. However, that is not the case. The 1/2 derivative of a constant, A, is not zero as given above: [ ​1⁄2 [ dA ] / dx1⁄2] = A/√(π•x) That being the case, one has to ask, is there a function whose 1/2 derivative is ZERO or actually an impulse function that is zero everywhere except between x = 0- and x = 0+ that can be added to the 1/2 integral so that it will not affect the result when the half-derivative of the anti-half derivative is taken? The answer is YES there is a function and that function is: g(t) = 1/√x = x^(-1⁄2) where [ ​1⁄2 [ dg(x) ] / dx​1⁄2] = √π • δ(x) δ(x) is the impulse function which is ZERO everywhere except x = 0 so we can write: 1/2 integral of "x" is: [ ​1⁄2 ∫​ ] x dx = [ 4/(3•√π) ] • [ x^ 3⁄2 ] + A/√x A = arbitrary constant ------------------------------------------------------------------------------------------------------------------------------------------------ Check results: [ ​1⁄2 ∫​ ] x dx = [4/(3•√π)] • [ x^ 3⁄2 ] + A/√x [ ​1⁄2 [d(x^ 3⁄2 )] / dx​1⁄2 ] = [ (3•√π) / 4 ] • x [ ​1⁄2 [ d(A/√x ) ] / dx​1⁄2 ] = A √π • δ(x) Take 1/2 derivative of the 1/2 integral of "x" which should return "x" back: [ ​1⁄2 [ [ ​1⁄2 ∫​ ] x dx ] / dx​1⁄2 ] = [4/(3•√π)] • [ ​1⁄2 [ d (x^ 3⁄2 ) ] / dx​1⁄2 ] + [ ​1⁄2 [ d( A/√x ) ] / dx​1⁄2 ] = [4/(3•√π)] • [ (3•√π) / 4 ] • x + A • √π • δ(x) = x + c • δ(x) where c is an arbitrary constant c = A√π δ(x) = delta function which is ZERO everywhere x≠0 It checks, the function "x" is returned back after performing two fractional operations of integration and differentiation. When other than simple power functions are integrated or differentiated fractionally, the results can be rather messy. Consider the 1/2-integral of Sin(x): [ ​1⁄2 ∫​ ] Sin(x) dx = √2 [ C(√(2x/π) Sin(x) - S(√(2x/π) Cos(x) ] where C(u) is the Fresnel "C" integral S(u) is the Fresnel "S" integral The 1/2-derivative of the exponential function, e^(kx): [ ​1⁄2 [ d (e^(kx) ] / dx​1⁄2 ] = √k * e^(kx) * Erf(√(kx) + 1 / √(πx) GH I just uploaded a video that discusses this: kzbin.info/www/bejne/aHe9na1qYtWZY68 GH

[@danieldelacruz7038]

cool

[@ekadria-bo4962]

how about chen lu for fractional derivative? and how about for tan x ?..

[@drpeyam]

No Chen Lou, at least as far as I know! Hahaha, and tan is a completely different problem since the quotient rule doesn’t necessarily hold! Oh, and it’s not tan(x + pi/2 alpha) :P

[@ekadria-bo4962]

Its very sad news..

[@Lavastine]

In this, you created a function that allows you to calculate the alpha-th derivative of e^kx, including negative numbers giving integrals (and fractional integrals lol). Do you think it would be possible to do the same to functions that are differentiable, but not integrable by standard means, to create a method of integration for them?

[@drpeyam]

I’m not really sure, but that’s a great idea! Like instead of integrating a function you fractionally half-integrate them twice!

[@Lavastine]

Other ideas for further exploration: What do the critical points of a fractional derivative tell us about the original function? Fractional gradients to weight importance of certain variables to a functions results (like a weighted average maybe, or perhaps like the sides of a right triangle)? Fractional optimization with lagrange multipliers?

[@paulomartins5131]

is this definition equivalent to the fractional Riemann-Liouville derivative or Caputo derivative?

[@OnTheThirdDay]

6:23 is great.

[@The1RandomFool]

I like the sound of this chalkboard.

[@Kapomafioso]

Is there something like (1/2)-th order differential equation? Like, (d1/2/dx1/2) f(x) = f (x). If yes...how many initial conditions do you specify? For the first-order ODE, you specify one IC. For second-order ODE you have to specify two IC's. (1/2)-the order ODE...? :D

[@drpeyam]

Absolutely! You might want to check out Fractal Derivative kzbin.info/www/bejne/javblWSLed96iqM

[@wahyuhidayat7042]

I never thought before that it will be so cool.

[@rodge4411]

if you taylor expand cos and take the fractional derivative (the polynomial way) term by term, will it converge to your exponential definition?

[@scitwi9164]

1. Can yoy tell me the sources of your knowledge? You seem to know a lot of things that don't seem to be usually taught in math courses. Some good books or websites? 2. You should now do some videos on tensors. They are usually explained very terribly, with a lot of multi-variable calculus and obfuscated with unnecessary references to physics, mostly Relativity (I guess it's because of its origins: Levi-Civita and Einstein developed it mostly for use in Relativity). 3. Do you know some good resources on calculating the algebraic value of the Gamma function for different fancy arguments? (including the negative or imaginary ones, which I guess is not quite possible if the Gamma function is defined in terms of that funky integral).

[@drpeyam]

I use a variety of sources, some from my knowledge, but mostly from Wikipedia or stackexchange :P Hahaha, I wish I understood tensors, but I’ve been taught the same sloppy definitions as you have! I’ll look into it :) And usually it’s nearly impossible to calculate explicit values of the Gamma function, except for integers and half integers. Wikipedia has all the known values and I’d be happy to make more videos about them!

[@koenth2359]

You know what I find the best of it all? That none of the comments was like: 'what's the use if this?' !

[@drpeyam]

Haha, agreed :)

[@jh8801]

I have a question: Do the usual rules for differentiating (like d/dx(f(x)*g(x)) = ... and d/dx(f(g(x))) = ... and so on) work for fractional or even imaginary derivatives too?

[@drpeyam]

Apparently yes! I was very surprised by that too

[@markopolo6104]

You showed how it works for expotential an triginimetric functions. But how dose it work witk arbitrary funtions?

[@MrAssassins117]

Dr. Peyam can you show an application of Fractional derivatives, i have a teacher that in his papers talks about these derivatives for solving RC Circuits, I'm an Electric Engineer student and i'm very interesting about this subject in particular, so, if you can i'll be very graceful

[@drpeyam]

Please see the pinned comment on my previous fractional derivative video!

[@MrAssassins117]

Dr. Peyam's Show Ok, I'll check it, thanks

[@erfanmohagheghian707]

The half derivative of e^(kx) from Cauchy integral formula is {sqrt(k)*exp(k*x)*erf(sqrt(k*x)) + 1/sqrt(pi*x)}. Why is it so different ?

[@khaledelnour5526]

thank you very much , we are waiting for the new videos.

[@jafetriosduran]

Dr. Peyam, I've been watching your series of videos and would like to see and application of this calculus for an PID controller, it'd be great

[@justcarcrazy]

Are fractional integrals possible as well? If so, what would be the fractional antiderivative of 1/x? Would that be the same as the fractional derivative of ln(x)?

[@drpeyam]

Yes, absolutely! I’m guessing that the half integral of 1/x is C/sqrt(x) for some constant, and yes it’s the same as the half derivative of ln(x)

[@samirgeiger1042]

Such a great video !!!

[@LakshmiRupaMPHD]

sir can you suggest any good book to learn fractional calculus

[@orenzeshani]

Aren't you suppose to also multiply by the fractional derivative of x which is x to the power of alpha divided by gamma of 2 minus alpha?

[@SultanLaxeby]

Is i^i really well-defined? Because for example i=e^(5*i*pi/2), thus i^i=e^(-5*pi/2) as well

[@drpeyam]

Strictly speaking when you compute i^i you calculate e^(i ln(i)) and ln(i) has many values BUT it’s unique if you define it to be the smallest positive number z such that e^(i z) = i. This is what’s called the principal logarithm!

[@pablojulianjimenezcano4362]

This is incredibly beatifull!!!

[@deeptochatterjee532]

If you applied the i'th derivative -i times would you get the first derivative?

[@drpeyam]

I guess you would, except I’m not sure how to apply a fractional derivative -i times :P

[@timh.6872]

If you take the -i th derivative of the ith derivative of f, you should just get f (good sanity check there!)

[@OhannesKamerKoseyan]

This. Is. AWESOME!! ❤️❤️

[@michalbotor]

the form of D^a cos(x) = cos(x + pi/2*a) and D^a sin(x) = sin(x + pi/2*a), where -oo < a < oo is simply beautiful and soo insightful. it always bugged me that sine and cosine seemed clearly intertwined with respect to derivative operation and that for both of them separately the sequences of their consequtive derivatives exhibited a cycle of 4, but at the same time there were this two seemingly symmetry breaking, pesky patterns of 1s and -1s... but now, thanks to you i've finally seen the well hidden magnificent gem behind it all. and it's an incredibly joyful and satisfactory moment that i'm experiencing right now. thank you dr peyam! ;)

[@fstasel]

Dr. Peyam, is there an approximation to calculate half (or fractional) derivative of a function? (something like f'(x) ~= f(x) - f(x+1))

[@drpeyam]

Good question! I think something reasonable would be f(x+h)-f(x) divided by sqrt(h)

[@fstasel]

@@drpeyam Hmm, in this case, if i substitute this formula for f(x) in the same formula, i'll have the approximation for 1st order derivative?

[@ashes2ashes3333]

Sorry I’m not quite convinced that this is well defined for more difficult functions. It seems like this fractional derivative only works for particularly nice functions whose Taylor series have an extremely regular form... What about say the half derivative of erf(x), or say erf(Li(x)) or cosh(arctan(x))? These have well defined derivatives (on their natural domains) but how would you get a half derivative?

[@ashes2ashes3333]

Actually no perhaps I’m on board... for functions nice enough to admit Taylor series, a half derivative (assuming its linear) could conceivably act on each term in the Taylor expansion term by term. This seems like it could work

[@pco246]

Since you uploaded the first fractional derivative video I've been wondering, could you take a similar approach to raise matrices to non-natural powers?

[@drpeyam]

It’s actually more straightforward than you think :) If A is diagonalizable, then A = PDP^(-1) and so A^n = PD^n P^(-1) and this works for ANY n, even real or complex numbers

[@nickstenerson6310]

More please. More... More fractional Calculus

[@georgehnatiuk5806]

For more fractional calculus you might look at: kzbin.info/www/bejne/aHe9na1qYtWZY68 GH

[@Sam_on_YouTube]

Can you take the ith derivative twice to get an integral? Let's see: Di(2^i*e^2x)=2^i*Di(e^2x)=2^i*2^i*e^2x=(e^2x)/2 D((e^2x)/2)=2(e^2x)/2=e^2x Checks out for that example. Does anyone know if this works generally? Is this another way to do an integral? I've never seen anything quite like this in the 3 1/2 years of calculus I took between high school and college. I wonder why, it isn't that complicated and it is pretty interesting.

[@drpeyam]

Unfortunately 2^i times 2^i is 2^(2i), so taking the imaginary derivative twice doesn’t give the integral! It would be cool to have a rule * whereby Da * Db = Dab, because then we’d have Di * Di = D-1, which would be the integral!

[@tonykarp5981]

Are there any integration techniques using i’th derivatives or something like that?

[@drpeyam]

Not that I know of :) I’m not even sure where imaginary derivatives appear in math

[@cubicardi8011]

9:20 Let's do the same Spiel again 😂

[@陳柚子-h8n]

what is half derivative of tanx Is chain rule still hold?

[@ronraisch510]

for rational alpha i get how you define D^alpha but what is the definition for irrational(complex is kind of obvious from all real)

[@Risu0chan]

Is this expression of the fractional derivative of exp(kx), when written as an infinite series Sum (kx)^j/j! , consistent with the sum of the fractional derivative of monomes (kx)^j (featuring the Gamma function)? I can't find a proof of it.

[@drpeyam]

You can try it out using the formula I gave in my previous fractional derivative video :)

[@Kavukamari]

my guess is that the trig functions will drift leftwards linearly between sin(x), cos(x), -sin(x), and -cos(x)

[@Kavukamari]

:)

[@DPets]

Amazing math!!!

[@fonzi102]

Is it linear?

[@josir1994]

is this consistent with the half derivative of x^n?

[@moorthysathishkumar1529]

How to calculate fractional derivatives for cos(θt) and sin(θt)?

[@chaoticoli09]

Sorry if this is already answered in the video somewhere, but could alpha be irrational?

[@drpeyam]

Yep

[@Olydis]

Check out medium.com/@olydis/fractional-derivative-playground-74e61c28721f if you want to play with fractional derivatives interactively :)

[@mathphys685]

OMG THIS IS BEAUTIFUL

[@milindranjan2253]

"Your whole life you've been lied in Calculus" if thats not sad then what is!!

[@manuelgnucci7760]

1:53 sqrt(2) is actually irrational.

[@daikithereal]

Hellooo!! Can alpha be a matrix?????

[@bernardz2002]

What about some other valued derivatives other than fractions like e-th, pi-th derivatives?

[@drpeyam]

The exact same process works for alpha = e, pi, whatever you want