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Fractional derivative for polynomial

In most cases, when we are calculating different orders of derivatives of a function, the orders of the derivatives we are facing are non negative integer values. For example, for function , we have:


From formulas above, we can easily acquire the general formula for the -th derivative of .

Recall that where . Therefore, we can use Gamma function to generalize the domain of from to .
Then,

Ex1:

Ex2:

Ex3:

{% note warning, In the result of _Ex3_, we can see that the result function is not continuous over point , while the 0th and first order derivative of the original function is actually continuous over the point. This should due to a definition problem (not sure about whether this problem can be fixable in the further definition of fractional calculus). %}

Deriving general formula for fractional derivative

The Grünwald-Letnikov Fractional Derivative

Recall the very first definition of “derivative”:

Note: Using backward difference here for simplicity in the further steps.
Then,

Goal here is to make . Therefore,

Define for some constant , as , if .
Then,

At this stage, we have defined the formula for the GL derivative:

The Riemann-Liouville Fractional Integral

In order to get even more compact formula, we want to define a new concept called negative binomial coefficients.

Recall the formula for the normal negative binomial coefficients:

Numerator in the formula has terms in total. Therefore, by replacing with in the formula, we can have:

Therefore,

Consider :

Consider :

Similarly:

As integral can be considered as “inversed derivative”, here we can define a new notation called Riemann-Liouville Fractional Integral:

Ex: for

Properties of The Riemann-Liouville Fractional Integral

Proof for the third property:

Stickers by the end of this note

  1. The conversion from GL fractional derivative to RL fraction integral still seems to be unclear to me, need further investigation about the topic.
  2. The formula for RL fraction integral seems to be too good to be true, as it can compute high order integration with only one integration. And the term inside the formula seems to have some underlying connection to taylor expansion.
  3. Work out the proof for the missing step in the course.

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