Conference rusure::math

    Easy using Fourier Transforms.  Wish I remembered exactly how, though.
    

    Along the same lines I wonder about a sort of meta-derivative,
    though somebody else must have done this centuries ago.
    One standard definition of the derivative is:
    
    		 lim	f(x+dx) - f(x)
    	f'(x) = 	--------------
    		dx->0	      dx
    
    Suppose we take the same definition and replace x with f(x+dx).
    Then the equation becomes:
    
    		 lim	f( f(x+dx) + dx) - f(x+dx)
    	'f(x) =		--------------------------
    		dx->0		      dx

    It is easy to verify that if f(x) = ax+b then 'f(x) = f'(x) = a.
    But if f(x) = x^2 then 'f(x) = 2x^2, decidedly not the same as f'(x).
    Can anybody give a simple formula for computing 'f(x) for general
    polynomials?  What about 'sin(x)?
    
      John

>    		 lim	f(x+dx) - f(x)
>    	f'(x) = 	--------------
>    		dx->0	      dx
>
>   Suppose we take the same definition and replace x with f(x+dx).
>   Then the equation becomes:
>    
>    		 lim	f( f(x+dx) + dx) - f(x+dx)
>    	'f(x) =		--------------------------
>    		dx->0		      dx

    Don't you mean either:

    		 lim	f( x+dx + dx) - f(x+dx)
    	'f(x) =		--------------------------
    		dx->0		      dx
    Or:
    		 lim	f( f(x+dx) +dx) - f(f(x+dx))
    	f'(x) = 	----------------------------
    		dx->0	             dx

    Or something else, perhaps changing dx as well?

    Re .3:
    
    Judging by the example, 'x^2 = 2x^2, I think he meant:
    
        		 lim	f(f(x+dx) + dx) - f(f(x+dx))
        	'f(x) =		----------------------------
        		dx->0		      dx
    
    If we let y = f(x+dx), this is just the limit of [f(y+dx)-f(y)]/dx as
    dx goes to zero.  This is f'(y), the derivative of f at y.  Having the
    dx in y = f(x+dx) can be ignored; it just means the limit approaches
    f'(y) a bit differently.
    
    In general, 'f(x) = f'(f(x)).  Several examples are:
    
    	f(x) = ax+b, f'(x) = a, f'(f(x)) = a.
    
    	f(x) = x^2, f'(x) = 2x, f'(f(x)) = 2f(x) = 2x^2.
    
    	f(x) = sin(x), f'(x) = cos(x), f'(f(x)) = cos(f(x)) = cos(sin(x)).
    
                
    				-- edp

Not very interesting after all.  Maybe it makes a nice problem for a
calculus course.

< Note 677.0 by ENGINE::ROTH >
                  -< Interpolation between derivative orders >-

    Here's a question related to the one just posed previously.

    You know how to take the n'th derivative of a function for
    various integers n.

    Can you well-define the idea of a fractional derivative that
    interpolates between two orders of the conventional derivative?

    - Jim

>	If we consider representations of time functions in the frequency
>	domain,	differentiation corresponds to multiplication by the
>	transform variable s.  Therefore, multiplication by s raised
>	to an arbitrary fractional number might correspond to
>	'fractional derivatives'.

>	For example, the 0.5'th derivative of f(t) might be:
>	inverse_Laplace( [(s^0.5) * Laplace( f(t), s )], t )
>	where
>	Laplace( f(u), s ) corresponds to the Laplace transform of
>	f(u) with s as the transform variable,
>	and
>	an analogous definition for inverse_Laplace( G(s), u ).

>	Is this what you had in mind ?  Seems like the transforms
>	generated in this manner might be really hard to invert
>	(i.e. analytically).

>			/R Best

    Yes, the replys in .1, .6 show the trick.  In fact, when you transform
    an n'th derivative, you have an n-factorial in the expression, and
    this becomes a general gamma function for fractional derivatives; the
    inverse transform is tabulated.  Fourier transforms may work too.

    Fractional derivatives do have applications, for example they arise
    in the solution of partial differential equations by transform
    techniques but my memory is not clear on how that fits together.

    I haven't seen any 'exotic' types of derivatives that weren't
    just compositions of the 'simple' derivative, but there are a few
    useful examples:

	Logarithmic derivative	f'(z)/f(z)

	Schwarzian derivative	(2*f'''(z)*f'(z)-3*f''(z)^2) / (f'(z)^2)

    - Jim

Newsgroups: sci.math
Path: decwrl!pyramid!batcomputer!cpf
Subject: Area and Volume of z-Sphere
Posted: 15 Mar 87 04:11:42 GMT
Organization: LNS, Cornell University, Ithaca NY
 
Someone asked about the volume and area of a sphere in an arbitrary
(integral) number of dimensions.  Here is the answer for the general
case.
 
 
First, the answers :
 
Area = 2 pi^(z/2)/Gamma(z/2) r^(z-1)
 
Volume =  pi^(z/2)/Gamma(z/2+1) r^z 
 
r is the radius of the sphere, z is the number of dimensions,
pi is pi, and Gamma is the usual Gamma function. 
For integral n, Gamma(n) = n-1!, Gamma (n+1/2) = ((2n-1)!!/(2^n))pi^(1/2).
((2n-1)!! = 1*3*... *2n-1.)
Note that I have used z deliberately -- z may be non-integral, and
even complex (just not zero or a negative integer).  I apologize for 
my tardiness, but I expected one of the mathematicians to answer.  
I guess that only physicists are interested in such practical things 
as non-integral dimensions! (:-))
 
Proof:  First, start out with the only integral a high energy physicist
is able to do:
 
	Int (d^d)x exp (-x^2) = pi^(d/2)	( d is now number of dimensions)
 
		= A Int dx x^(d-1) exp(-x^2)   ( A is area of r=1 d-sphere)
		By spherical symmetry
 
		= A Gamma(d/2)/2  
		Change variables x^2 = y, and definition of gamma function
	A = 2 pi^(z/2), Area = A r^(d-1)
To get V, just integrate A Int dx x^(d-1) between zero and r.
Now analytically continue to z dimesions.  Anyone who mentions that
this is arbritrary for non-integral dimesions (one can add any function
with zeros at all the integers) will be totally ignored.  In practice
one is interested only in behavior near integers.
 
Both high energy and solid state theorists use non-integral numbers 
of dimensions.  High energy physicists use them to deal with various 
divergent integrals (the incantation is `dimensional regularization'), 
while solid state physicists  calculate exact answers to certain problems 
in four dimensions, and the first few terms of a Taylor series in the 
number of dimensions, and try to extrapolate the answers down to 3 
dimensions (`the epsilon expansion').
-- 
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