Recursive Derivatives

[insane_alien]

Okay, so i've manager to lump a mathematical model into the one mega equation but i've come up against a slight problem, getting the derivative(not essential but makes my work a lot easier).

 

see, the first derivative is a function of the second derivative which is a function of the third which is a function of the fourth which is... you get the picture.

 

as i have a program that can go through them automatically, i got the 1000th derivative and it was still a function of the 1001th derivative.

 

now, i did encounter such stuff in my math classes, but only there. i've never had to deal with one in real life never mind this one thats so complex.

 

how is it you deal with these things again? google only seems to return eejits who fail to understand what the second derivative test is.

[timo]

I do not fully understand your question but something akin to this is a "differential equation" where the derivative of a function depends on lower derivatives and the function, e.g. f'(x) = a*f(x). I am not sure to what extent that suits what you described. There is no general way to solve such equations analytically and the solutions -so they exist- are not unique. A special case are linear differential equations which can be solved. For example, the solution to the diff. eq. I took as an example is any function f(x) = C exp(ax) with arbitrary C.

[insane_alien]

right, the differential comes out such that

 

dy/dx = a*d^2y/dx^2

 

where a is the rest of the whole equations

 

its a bit more complicated than that but thats pretty much what mathematica says and its what maxima says too.

[timo]

Well, then with g(x) := dy/dx that is exactly my example and the solution is dy/dx = g(x) = C exp (x/a).

 

EDIT: Comment for the sake of completeness: It is not obvious but can be proven (I just don't know the proof anymore) that these functions (parameterized by the C) are indeed the only solutions. Note that if a depends on x then you are screwed.

[insane_alien]

yes, i remember that but it doesn't really apply to what i'm meaning.

 

i'll try again to get the point across

 

f'(x)=a*f''(x)

f''(x)=b*f'''(x)

f'''(x)=c*f''''(x)

f''''(x)=d*f'''''(x)

f'''''(x)=...

 

and so on ad infinitum

[Mr Skeptic]

Are the a, b, c, d... constants?

[insane_alien]

not quite, i was just using them to represent the rest of the equation.it changes slightly each time you differentiate.

[timo]

He probably was not asking if a=b=c=... but if they are functions of x.

[insane_alien]

oh i know they are functions of x, thats not a problem. its the fact that they are also functions of the the (n+1)th derivative that is causing me issues here.

[Cap'n Refsmmat]

I don't think you get what's being asked. Are the a, b, c, d and so on functions of x, or just constants?

 

i.e. is [math]f^{(n)}(x) = a_n(x) f^{(n+1)}(x)[/math] or just [math]f^{(n)}(x) = a_n f^{(n+1)}(x)[/math]?

 

([math]f^{(n)}[/math] means the nth derivative of f.)

 

Or, simpler still, does a have x in it?

[insane_alien]

an(x)f^(n+1)(x)

 

more or less.

 

it would probably be easier if i was allowed to tell you the function butit falls under the confidentiality agreement.

[Bignose]

These things aren't all that uncommon when doing an expansion via a perturbation method. Basically, at some point you have to truncate or approximate the expansion. I.e. decide that the 6th derivative and further are constants or negligible or some other assumption. Knowing the pathology of the solution can be very useful to know what and when such approximations are needed/appropriate.

 

Nayfeh's Perturbation Methods is a absolute classic on perturbation methods and is the standard by which all other books on the subject are judged. You may want to poke through that book and see if it can help.