nth derivative is itself (1 Viewer)

[Sy123]

Are there any elementary functions with its 3rd, 4th etc derivative is itself? (i.e. f''''=f)

If so how would we find these functions?

(except for f(x) = 0)

 

[seanieg89]

So are you just looking for a function with f^{(n)}=f for some fixed n > 2? Or for a function with f^{(n)}=f for all n > 2?

 

[Sy123]

Some fixed n > 2

(except for trivial cases like f(x) = 0 )

EDIT: I think you only need functions like f^(n) = f for prime n since if you have a function g with g^(3) = g and f^(2) = f, then h = f + g would be such that h^(3*2) = h

 

[seanieg89]

Some fixed n > 2

(except for trivial cases like f(x) = 0 )

Then sure, all solutions will in fact be linear combinations of:



for k=0,1,...,n-1.

(Think about what you are multiplying your function by with every differentiation, and the involvement of the roots of unity becomes clear. This works in greater generality to reduce solving linear ODE with constant coefficients to solving polynomial equations.)

 

[Sy123]

Then sure, all solutions will in fact be linear combinations of:



for k=0,1,...,n-1.

Oh ok cool

Thanks!

 

[seanieg89]

Oh ok cool

Thanks!

No worries. You might find it a fun/interesting linear algebra exercise to think about why the kernel of such a linear operator (an n-th order linear constant coefficient ode defined on the vector space of C^\infty functions) should have dimension n.

Otherwise there's no reason why there cannot be other smooth solutions, let alone non-smooth ones.

 

[Carrotsticks]

Sy123, you learn this in 2nd year L.A @ USYD + bit of knowledge of 2nd sem 1st year, incase you were wondering when you do this at uni.

 

[pHyRe]

why didn't you say just say e^x instead of some complicated crap like that haha or is what you said some kind of general form?

 

[seanieg89]

why didn't you say just say e^x instead of some complicated crap like that haha or is what you said some kind of general form?

1. Because we explicitly know ALL functions whose n-th derivatives are equal to themselves, and the expression isn't much more complicated. Why specify one object that satisfies a property when it is just as easy to specify all of them? Also, just stating e^x gives us no understanding of how this problem depends on n.

2. For pretty much all physical applications, we will need to use this basis of solutions to construct the unique one that matches initial conditions. One will not generally be enough.

3. Because the idea of looking at linear combinations of functions of the form e^{rx} kills ALL constant coefficient ODE, a much more general problem.

 

[anomalousdecay]

I did something mid-last year which I found to be an awesome little idea:

The result was:



Or for a more general case: