Are derivatives linearly independent ?

In this video, I show that the differentiation transformations D, D^2, and D^3 are linearly independent (where Df = f' etc.) This result can be easily generalized to n transformations. It's a great exercise in showing that abstract linear transformations are linearly independent, enjoy!
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[@Adraria8]

Thinking of points as vectors is one thing, thinking of functions as vectors is another, but thinking of functions *of* functions as vectors is too far!

[@zac4550]

Is this the only way to show this? Would it be possible to show a=b=c=0 is true for a general smooth function f(x) instead of having to pick specific functions x, x ^2, x^3 like you did in the video?

[@drpeyam]

It’s hard to do in general, I think

[@ta1sty695]

It is impossible because f(x) = e^x and derivatives of f are all lineary dependant But you can proof it for all polynoms with the highest power being bigger than 1

[@cuentafake140]

If you take in consideration that any function can be represented using taylor series (polynomials) then you just have to prove it using polynomials (which Peyam just did). Or maybe fourier series I don't know.

[@jonasdaverio9369]

Well, I will still disagree with you, hahaha! You can totally define a vector space of differential operators. By that way, it totally makes sense to say two operators are LI. But that's not really disagreeing, since that's precisely what you assume in video, however only implicitly.

[@typo691]

Almost thought this was a 3b1b video from the thumbnail lol

[@drpeyam]

Haha, wow, didn’t even realize that!

[@benjaminbrady2385]

Same, that classic 3b1b colour palette and polish!

[@roygalaasen]

Yeah, me too. And it is not the first time lol

[@neilgerace355]

5:29 Invoke the Teletubbies in your next paper

[@jonasdaverio9369]

You probably know me for always disagreeing with you, but this time, I won't! Thank you for all your work by the way.

[@drpeyam]

Thank you!!! 🙂

[@RanEncounter]

But you have another comment that shows a contradiction.

[@jonasdaverio9369]

@@RanEncounter Yes, indeed, but I think I've warned the reader 😂

[@RanEncounter]

@@jonasdaverio9369 yeah 😂

[@lisalisa9706]

I don't get why the proof works. It feels like choosing suiting vectors to make a,b,c equal 0 in R^3

[@sylvesterli1166]

Proof of linear independence should be proof the statement that for all a b and c they only equal to T0 then a=b=c=0. It is for all a and b and c, not there exists a b and c. So confused...

[@Pradowpradow]

Hi, what does it mean that there exists a set of functions such that those derivatives are linearly dependant? (just solve an ODE)

[@giuliovittoriocarassai7276]

But that works only for polynomials because 1, x, x^2,... are the basis.

[@alexmangiapane2258]

Giulio Vittorio Carassai exactly! If you pick e^x then a = 1, b =-1 and they are linearly dependent

[@rkanexwasaby600]

@@alexmangiapane2258 you are smart boi

[@drpeyam]

They are still linearly independent as operators....

[@juandeluna2652]

I learn better with you... greetings dear Dr Peyam

[@ohanneskamerkoseyan3157]

I arrived to the same conclusion with a different approach. Fourier transform of (d/dx)^n is (i*k)^n. Therefore, the FT of a linear combination of derivatives would be a polynomial of i*k. Since polynomials are unique, then derivatives are linearly independent. But this approach may be limited to some normalizable space, since you can take the FT of certain functions. I'm not sure, though.

[@arthurlbn]

This operators are LI for a function f(x)=exp(x)? How is it possible? D(f)=D2(f)=…

[@cuentafake140]

The function: sin(e^(cosx))^In(tanx^2x^x)^arctac(x^x^x^x) is not linearly independant when you differentiate it. smh dislike

[@drpeyam]

We’re not talking about functions being linearly independent, but operators. Now take that dislike back :P

[@cuentafake140]

@@drpeyam That makes more sense. Take like instead

[@gastonsolaril.237]

Peyam! Nice to see you again, bro. I'm a loyal follower of ya, since a few months ago. Well, the procedure is nice. But I think (it's my opinion, maybe I'm wrong) that is kinda incomplete because you've just tried the diff op for three functions and there should be a proof that includes every function in the infinite-dimension function space. I think that you should have perhaps concluded (after "x^3") with a taylor series, since every differentiable function inside a radius of convergence can be expressed as such, and it's kind of the continuation of the thing of trying the linear combination equation with different powers of "x". What is your opinion about it? Anyway, excellent video and excellent teaching skills, as always. Have a great weekend!

[@gastonsolaril.237]

PS: Let me add a little debatable fact. If the function space is infinite-dimensional, it should mean that every function is linearly independent from each other. Then, it makes "almost" complete sense that two different differential operators are linearly independent because the result of applying them to one function is obtaining two different LI functions, and therefore the operators should be LI too. The loophole in this (and that's why I said "almost") is precisely the exponential function "e^x" since all derivatives are equal. Noticed it?

[@arthurreitz9540]

We can use f(x)=exp(1/2 x^2) (Because f'(x)= x × f(x) ) If we take a linear combination, we have P(x)×exp(1/2 x^2) = 0 So P=0 The coefficient are mixed up but I think it gives us a triangular system. So the coeff are all zeroes ->Result

[@drpeyam]

Niiiiice!!!!

[@benjaminbrady2385]

Haven't seen the video yet, but can't you treat derivitaves linearly when using the Laplace transform?

[@drpeyam]

Watch the video :)

[@zaki7705]

I'm not a mathematician to argue with such a professor , but ... Am i the only one that think that this doesn't proove that Diffrentiation is LI for all smooth functions !? If we try e^x we can find a non-zero constants (a,b,c) that make the quality true ... I know i missed something just wanted to know what is it .. Thank you sir

[@drpeyam]

It’s linearly independent as an operator

[@Eva-ez1ks]

[aD+bD^2+cD^3](e^x)=T_0(e^x)=0 ae^x+be^x+ce^x=0 a=b=c=0 would work. So would a=b=1, c=-2 So would any solution of a+b+c=0 It is not the case that for all functions f [aD+bD^2+cD^3](f)=T_0(f) implies a=b=c=0

[@Eva-ez1ks]

I think I may have misinterpreted what you said in the video. Were you saying, forallf([aD+bD^2+cD^3](f)=0) --> a=b=c=0 or forallf([aD+bD^2+cD^3](f)=0 --> a=b=c=0)?

[@drpeyam]

The first one!

[@Eva-ez1ks]

@@drpeyam neat!

[@alwysrite]

was watching some of Gilbert Strang MIT lectures on Linear Algebra - very helpful.

[@michaelempeigne3519]

But is it not supposed to be that D is to be applied to a single function for all three derivatives, therefore, I do not agree with your video work?

[@WaveSound123]

Wait... why can you do that? I tought that you have to pick an f for which to show that a=b=c=0. Why does your approach work?

[@ethancheung1676]

Liviu Stoica the statement is “if LHS is true for all smooth function f, then show RHS”. So he picked a few f to show RHS

[@weinihao3632]

a,b and c are constants, independent of the choice of f

[@M.J.D.R]

So we literally mixed linear algebra with calculus ?

[@pierreabbat6157]

You could try f(x)=exp(x)+sin(x).

[@chibigato3x311]

La derivada de orden k,l,m, con k , l, m en algún campo, son linealmente independientes? 🐶

[@dhunt6618]

When you started off, I was wondering about the low power functions (1, x, x^2, etc.) but you immediately made it clear using your previous videos. A sign of an excellent teacher! Thank you!

[@drpeyam]

Thank you! :)

[@Bombelus]

Suppose these operators weren't LI, would take break differential equations in some way? Make them easy maybe?

[@drpeyam]

It would make it almost trivial! Suppose say D3 is a linear combo of D and D2, then any 3rd order ODE is just a second order ODE, and in fact any higher order ODE would just be a 2nd order ODE

[@alanhiguera3484]

what about e^-x? wouldnt D+D^2 applied to e^-x give 0 making it not linearly independent?

[@xy9439]

The point is, the equality aD+bD^2+ ⋯= 0 has to hold for ALL functions in C^∞(ℝ)

[@alanhiguera3484]

Adàlia Ramon but in order for the derivatives to be linearly independent such an equation must imply all of the coefficients must be zero. my example has two nonzero coefficients that satisfy the equation, making it not linearly independent.

[@Royvan7]

you've proven that a particular set of derivatives aren't LI. the video was showing that the derivative *operators* are LI.

[@gobberman09]

*equis cuadrado* I loved that!

[@ta1sty695]

Unless i understood something wrong Your proof is false: M = {D, D^2 , D^3} Since you didnt eclide the zero function, set f=0, the zero is always linear dependant Otherwise you said you can do it with n derivatives, im not familiar with english math but what i understand from that is that |M| = n M = { D, D^2 , ... , D^n} which would be false because: Take n = 4, f(x) = sin(x) => asin(x) + bcos(x) - csin(x) - dcos(x) = 0 a=b=c=d for every a in R So you specifically have to exclude the 0 function and make sure |M| = 3

[@ta1sty695]

Or take f(x) = e^x Your proof is false in the general case Cause ae^x + be^x + ce^x = 0 for c = 0 and a= -b But you can say that derivatives of polynoms are linearly independent, wich is true for every polynom with deg(p)>1

[@drpeyam]

The proof is correct. We need to show M is linearly independent, not functions

[@Handelsbilanzdefizit]

Perhaps someone can help me. I'm looking for a matrix A(x,y,z)•∇=R(x,y,z), where length of the vector 'R' is: |R(x,y,z)|=detA(x,y,z) Are there solutions for this?

[@peterjohannsen2183]

try the zero matrix

[@Handelsbilanzdefizit]

@@peterjohannsen2183 Sorry, the second line in my comment was wrong. Instead of: I'm looking for 'a' matrix I meant: I'm looking for matrices. My question "Are there solutions for this?" asks for more than one solution.

[@chrislarson5097]

You've shown that differentiation in a finite polynomial space is always linearly independent, but that doesn't make it so in the space of all smooth curves. The obvious counterexample is the exponential function, or really ANY exponential function e^kx. Since differentiation of this function in the smooth function-space is not linearly independent, you can say for certain that differentiation is not always linearly independent. Since a finite polynomial space is still contained within this smooth function-space, there are also obviously some examples where it is linearly independent, but not always.

[@drpeyam]

They are still linearly independent as operators...

[@TIMS3O]

The proof is correct. Let V be the space of all infinetly differentiable functions on R and let W be the space of all linear functions L:V ->V. Also let D^n denote the n:th derivative operator then the set {D^0,D^1,...,D^n} is an independent set of vectors in W. This amounts to showing that if x_0D^0+...+x_nD^n=0 (the zero vector) for some scalars x_i then x_i=0 for all i. If we call the left hand side T the assumption is that T=0, applying this to 1 it follows that T(1)=0 hence x_0=0. Doing the same with x it follows that x_1=0 and if we continue in this manner it follows that x_i=0 for all i hence the vectors are linearly independent.

[@andres.robles6]

Cuando pondrás subtítulos en español? :c

[@drpeyam]

Puedes añadir subtítulos si quieres

[@andres.robles6]

@@drpeyam cómo así?

[@drpeyam]

No sé cómo, quizás con “Add subtitles/CC?”

[@drpeyam]

support.google.com/youtube/answer/6054623?hl=en

[@shayanmoosavi9139]

Nice. Can you do more videos on linear independence?

[@drpeyam]

There are bunch of them on my linear algebra playlists

[@lathavedachalam89]

Superb sir

[@tgeofrey]

Thank you

[@alexmill8446]

Brother Tumaini? Are you also from Uganda? I am searching for my brother, name Tumani that dissapeared when me forget about him when I was bargening in the market. Was your father a shoe master?

[@JuanLopez-rl7ry]

I hope you made a video on "dx" as how something that infinitesimally small produce multiplied by height give an area?

[@goatmatata2798]

Hi, nice video, would you please create a video on the proof of gauss's multiplication theorem for the gamma function, thanx

[@drpeyam]

I don’t know what that is

[@goatmatata2798]

@@drpeyam the general case for the duplication formulae

[@drpeyam]

I don’t know what the duplication formulae are

[@goatmatata2798]

@@drpeyam you seriously don't know legendres duplication formulae,,, thanx anyway