Looking Again At Fractional Calculus

 Recall that in ordinary, single variable differential calculus one is used to seeing the symbolism:

d ⁿ f/ dx ⁿ  

Or the differential operator equivalent:  (D _x y) ⁿ

For the nth derivative of a function f with respect to x when n is a non-negative integer. We also know integration and differentiation are inverse processes so it is natural to associate the particular symbolism

d  ^-1 f/ [dx]^-1    

with the indefinite integration of f with respect to x. However. one must stipulate a lower limit of integration in order that the indefinite integral be completely specified. For the purpose of this blog post on fractional calculus I will associate the following:   

d  ^-1 f/ [dx]^-1       =     ò ^x _o   f(y) dy

The preceding can be generalized to the case of multiple integration, whereby, for example:

d  ^-2 f/ [dx]^-2       =  ò ^x _o    dx ₁  ò ^x1 _o   f(x _o)   dx _o

d  ^-n f/ [dx]^-n    =  ò ^x _o    dx _n-1  ò ^xn-1 _o  dx _n-2  .....ò ^x2 _o    dx ₁ ò ^x1 _o   f(x _o)   dx _o

And use is made of the identity:

ò ^x _a   f(y)   dy   =   ò ^x-a ₀   f(y + a)   dy

To extend the formalism to lower limits, i.e. other than zero. Thus we may define:

d  ^-1 f/ [d(x - a) ]^-1     =   ò ^x _a   f(y)   dy

d  ^-n f/ [d(x - a) ]^-n     =  

 ò ^x _a    dx _n-1  ò ^xn-1 _o  dx _n-2  ò ^x2 _o    dx ₁   ò ^x1 _o   f(x _o)   dx _o

Caution obviously needs to be exercised when applying the contracted equivalent form, e.g.

d ⁿ f/ [d(x - a) ]ⁿ     =  d ⁿ/ dxⁿ 

characteristic of a local operator, say to negative orders, or to fractional orders of either sign, given:

d ^-n / [d(x - a) ]^{- n}   ≠  d ^-n/ [dx] ^-n 

The key point to bear in mind here is that the appearance of fractional orders is what defines fractional calculus. This use also marks the emergence of what we call differintegrals. From the simplistic basis provided here it is then possible to venture into the realm of differintegral operators and their application. 

We now look at distinct examples of fractional calculus applications:

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 reckon in the Gamma function:

G (x) = (x - 1)!

 along with 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 Calculus 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.}