Fractional Calculus Continued: Some Basic Operations, Examples

 When last we looked at fractional calculus, e.g.

• An Introductory Look At Fractional Calculus

It was in terms of the most elementary basics, and definitions.  The nature of the fractional derivative - from Part 1 - was that in this special calculus one had to think 'outside the box' i.e. as opposed to simply differentiating a function successive times and getting new, simpler functions, i.e.  d(3x ²)/ dx =  6x,  d/dx (6x) = 6.  In other words  d ⁿ/ dxⁿ   - taking the nth derivative of a function - only makes sense if dealing with positive integers n.  But to extend this to other n values, say n= 1/2, n = 3/2 etc. we need to think of it  instead as dealing with an operator, which effects a transformation-  i.e. something that takes in a function f as an input and gives a function f' as an output.  As opposed to repeated differentiation yielding simpler forms of the same initial function.  

 With that in mind we look in this post at more elaborate operations which kind of set the stage for doing actual fractional calculus as well as seeing applications later. So we now look at  d ⁿ/ dxⁿ  etc. as operators which transform a function to a new function.   We will see that in expanding our scope to do fractional derivatives (and integrals) we set the stage for later applications that are bey9ond the reach of classical calculus.  

1.  Composition Rule for Mixed Integer Orders:

The identities:

d ⁿ/ dxⁿ   [d ^N f/ [dx] ^N] =   d ^n+N f/ [dx] ^N+n]  

= d ^N/ dx^N   [d ^N f/ [dx] ^N]

And:

d  ^-n f/ [d(x - a) ]^-n   [d  ^-N f/ [d(x - a) ]^-N   = d ^-n-N f/ [d(x - a) ]^-n-N

=  d  ^-N f/ [d(x - a) ]^-N  [d  ^-n f/ [d(x - a) ]^-n

Are obeyed when n and N are non-negative integers.  Indeed, these identities are basic to the concept of multiple differentiation and integration.

2. The Chain Rule for Differentiation:

The chain rule:

d g(f(x)  / dx = d(g(u)/ du  [ d/dx (f(x) ] = g  ⁽¹⁾  f  ⁽¹⁾

Is that which enables g(u) to be differentiated with respect to x if the derivation of g(u) with respect to u and of u with respect to x are known, is one of the most useful in differential calculus.   The rule can be extended to higher orders of differentiation as well as fractional orders.  For economy we can use:   

 g ⁽ⁿ⁾   and     f  ⁽ⁿ⁾

To denote:  d ⁿ/ duⁿ   g(u)   and:   d ⁿ/ dxⁿ  f(x)  respectively.  

 By approximation of the basic chain rule and Leibniz' rule to differentiate a product we have:

d ²/ dx²  g (f(x)) = d/dx [d/ dx g (f(x)) ]  =  

d/dx [d/ du g(u)] [d/dx f(x)]

= [ d/dx d/du g(u)] [d/dx f(x)] +  [d/du g(u)] [d ²/ dx² f(x)]

= [du/dx  d ²/ du ²  g (u)  ]  [d/ dx f(x)] + [d/du g(u)][d ²/ dx² f(x)]

=  g  ⁽¹⁾  f  ⁽¹⁾    +    g  ⁽¹⁾  [ f  ⁽¹⁾]²

The expression of a similar procedure yields successively:

d ³/ dx³  g (f(x)) = g  ⁽¹⁾  f  ⁽³⁾    +   3 g  ⁽²⁾  f  ⁽¹⁾  f  ⁽²⁾   +    g  ⁽¹⁾  [ f  ⁽¹⁾]²

d⁴/ dx⁴  g (f(x)) = g  ⁽¹⁾  f  ⁽⁴⁾    +   4 g  ⁽²⁾  f  ⁽¹⁾  f  ⁽³⁾   +    6g  ⁽²⁾  [ f  ⁽²⁾]²   +    

6g  ⁽³⁾  [ f  ⁽¹⁾]² [ f  ⁽²⁾]  +  g  ⁽⁴⁾  [ f  ⁽¹⁾]⁴

d⁵/ dx⁵  g (f(x)) = g  ⁽¹⁾  f  ⁽⁵⁾    +   5 g  ⁽²⁾  f  ⁽¹⁾  f  ⁽⁴⁾   +    

10 g  ⁽³⁾  [ f  ⁽¹⁾]²   f  ⁽³⁾   +    

30 g  ⁽³⁾ f  ⁽¹⁾ [ f  ⁽²⁾]²   + 10  g  ⁽⁴⁾  [ f  ⁽¹⁾]³  f  ⁽²⁾  +  g  ⁽³⁾  [ f  ⁽¹⁾]⁵

^{The generalization to large n is accomplished via Faa di Bruno's formula, i.e.}

Faà di Bruno's formula - Wikipedia

^{But that is weighted by complexity and hence of little general utility in its applications.}    

        ³. Doing Fractional Derivatives:         

The key thing here is to grasp we will need the Gamma function (G(x) ) and the Riemann-Liouville fractional integral .(But we cannot use n < 0 because the Gamma function which we will need, is not defined for numbers less than zero).  This means in taking any fractional derivative we will be trying to preserve:

d ^N f/ dx ^N =  I^N f(t)  =   f(t)

Where I denotes an integral containing the Gamma function, i.e. the Riemann-Liouville fractional integral where we can spot the Gamma function present in the denominator preceding the integral from a to x.  Note the condition Re(u) > 0.

The fractional derivative then will have the form:

Which we can use to do simple-   not too complex! - fractional derivatives.  Then the full form for a fractional differentiation can now be expressed.

Where we have used the Cauchy form for repeated integration and regular positive integer differentiation.  

 We now consider discrete examples of how to take fractional derivatives.  For the unit function (f= 1) we have:

d^q (1)/ [d(x - a) ]^{q    =}  lim  _N ® oo  {N/ (x – a)} ^q   G(N - q)/ G(1 - q) G(N) 

 Where the symbol  denotes the Gamma function, see e.g.

Looking Again At The Beta - And Gamma - Functions (brane-space.blogspot.com)

Meanwhile, the fractional derivative of a series is:

d  ^-1 f/ [d(x - a) ]^-1   (å ^¥ _j=0   f _j)  =  å ^¥ _j=0  d  ^-1 f _j / [d(x - a)]^-1   

                  a <   x   <  b                         

provided the series converges uniformly in the interval a <  x  <  b        

Much easier examples include:

The zero function (or any constant) for which:

       d^q (C)/ [d(x - a) ]^q   =   C  d^q (1)/ [d(x - a) ]^q    

= C [x - a) ]^q  / G(1 - q)         

Specifically:  If q = 1/2,    G(1 - q)   =  G(1 - ½) =  1.772

d^q (C)/ [d(x - a) ]^q   =  d^q (0)/1.772   =   0

Other simple examples:   

a)     D ^a  (x ⁿ) = G(n + 1)/ G(n + 1 -  a)  

  b)  D ^1/2 (1)  =  1/ Ö p x                                  

c)  D ^a  (sin (x)) =   sin  (x +   a p / 2)

See Also:

• Fractional Gamma Functions
• 
  

Suggested Problems for the Math Whiz:

1) Let f(x) =  y =   x ³   - 3 x ²  +   5x  - 4

And: g(u) =  x =   u ²  +  u

Use the chain rule to show:  d y/ du =  dy/dx  (dx/du)

2) Find the fractional derivatives: D ^1/2  (x ⁰)  and D ^1/2  (x ²)

3)  Find the fractional derivative: D ^1/2  (sin (x))

4) Is D ^1/2 (1)   the same as: d ^q (1)/ [d(x - a) ]^q 

^{For q = 1 in the 2nd?  Explain.}