Differentiating x^k: Help Understanding Complex Derivatives

[FeDeX_LaTeX]

Hi, I am having some trouble understanding what I have done here, or if it makes any sense at all.

Recall that

$$\frac{d^a}{dx^a}x^k = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}$$

We can extend this to complex numbers if we let a be of the form p + qi, where p and q are real, and i is the imaginary unit. And it follows that when one finds the ith derivative of a function, and then finds the (1-i)th derivative of THAT, then you end up with the first derivative of the function, since i + (1-i) = 1.

Taking the ith derivative is itself doable:

$$\frac{d^i}{dx^i} x^k = \frac{\Gamma(k+1)}{\Gamma(k - i + 1)}x^{k-i}$$

But supposing we wanted to do that i times. Since i^2 = -1, finding the ith derivative i times *should* be the same as performing one integration.

Unfortunately finding ith derivatives often requires the use of i! (i factorial), which gives an answer of the form a + bi where a and b are irrational and only expressible through the use of infinite series. But it just so happens that evaluating the modulus of i factorial gives;

$$|i!| = \sqrt{\frac{\pi}{\sinh(\pi)}}$$

which is nicer to deal with.

Suppose we wanted to take the ith derivative n times;

$$\frac{d^{ni}}{dx^{ni}}x^k = \frac{\Gamma(k+1)x^{k-ni}}{\Gamma(k-i+1)\Gamma(k-2i+1)...\Gamma(k-ni+1)}$$

and letting n = i yields

$$\frac{d^{-1}}{dx^{-1}}x^k = \frac{\Gamma(k+1)x^{k+1}}{\Gamma(k-i+1)\Gamma(k-2i+1)...\Gamma(k+2)}$$

which seems to reduce to

$$\frac{d^{-1}}{dx^{-1}}x^k = \frac{x^{k+1}}{k\Gamma(k-i+1)\Gamma(k-2i+1)...}$$

since gamma(k+1)/gamma(k+2) = k!/(k+1)! = 1/k.

But this then seems quite strange, because the denominator should equate to k+1 if this is true (we are integrating x^k here using differentiation). But I do not see how that infinite product will approach that.

Can anyone help me out here?

 

[pwsnafu]

That's not how the i-th order derivative is usually defined. We start with
$I^\alpha f(x) = \frac{1}{\Gamma(\alpha)}\int_0^x (x-t)^{\alpha-1} f(t) \, dt$
for $Re \alpha > 0$.
To get the derivative of order i, you calculate the integral of order 1-i, and then differentiate, so we obtain
$D^i f(x) = \frac{d}{dx} \frac{1}{\Gamma(1-i)} \int_0^x (x-t)^{-i} f(t) \, dt$
this integral requires a branch cut, and the default choice is to place it so it goes through the origin.

Also I want to point out

And it follows that when one finds the ith derivative of a function, and then finds the (1-i)th derivative of THAT, then you end up with the first derivative of the function, since i + (1-i) = 1.

is false for fractional derivatives. That means $D^\alpha D^\beta \neq D^{\alpha+\beta}$ for general α and β.

But supposing we wanted to do that i times. Since i^2 = -1, finding the ith derivative i times *should* be the same as performing one integration.

This is a different concept. You are taking a fractional derivative and fractionalising it, so you need far more sophisticated tools. I'd say you won't get simple integration back.

 

[Mute]

Hi, I am having some trouble understanding what I have done here, or if it makes any sense at all.

Recall that

$$\frac{d^a}{dx^a}x^k = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}$$

If you take this form and set $a = i^2 = -1$ you get $x^{k+1}/(k+1)$, as desired.

We can extend this to complex numbers if we let a be of the form p + qi, where p and q are real, and i is the imaginary unit. And it follows that when one finds the ith derivative of a function, and then finds the (1-i)th derivative of THAT, then you end up with the first derivative of the function, since i + (1-i) = 1.

Taking the ith derivative is itself doable:

$$\frac{d^i}{dx^i} x^k = \frac{\Gamma(k+1)}{\Gamma(k - i + 1)}x^{k-i}$$

But supposing we wanted to do that i times. Since i^2 = -1, finding the ith derivative i times *should* be the same as performing one integration.

Unfortunately finding ith derivatives often requires the use of i! (i factorial), which gives an answer of the form a + bi where a and b are irrational and only expressible through the use of infinite series. But it just so happens that evaluating the modulus of i factorial gives;

$$|i!| = \sqrt{\frac{\pi}{\sinh(\pi)}}$$

which is nicer to deal with.

Suppose we wanted to take the ith derivative n times;

$$\frac{d^{ni}}{dx^{ni}}x^k = \frac{\Gamma(k+1)x^{k-ni}}{\Gamma(k-i+1)\Gamma(k-2i+1)...\Gamma(k-ni+1)}$$

and letting n = i yields

$$\frac{d^{-1}}{dx^{-1}}x^k = \frac{\Gamma(k+1)x^{k+1}}{\Gamma(k-i+1)\Gamma(k-2i+1)...\Gamma(k+2)}$$

You can't do that. The way you have written things, n is explicitly an integer. You can't just set n = i. You would need to write your expression in a form that can be legitimately generalized to non-integer n. The reason the formula for a fractional derivative of $x^k$ works is that it reduces to the usual result when $a$ is an integer, but also the Gamma function form of writing the result easily generalizes to non-integer derivative orders because the Gamma function has an analytic continuation into the complex plane.