Hi, I am having some trouble understanding what I have done here, or if it makes any sense at all.(adsbygoogle = window.adsbygoogle || []).push({});

Recall that

[tex]\frac{d^a}{dx^a}x^k = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}[/tex]

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:

[tex]\frac{d^i}{dx^i} x^k = \frac{\Gamma(k+1)}{\Gamma(k - i + 1)}x^{k-i}[/tex]

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;

[tex]|i!| = \sqrt{\frac{\pi}{\sinh(\pi)}}[/tex]

which is nicer to deal with.

Suppose we wanted to take the ith derivative n times;

[tex]\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)}[/tex]

and letting n = i yields

[tex]\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)}[/tex]

which seems to reduce to

[tex]\frac{d^{-1}}{dx^{-1}}x^k = \frac{x^{k+1}}{k\Gamma(k-i+1)\Gamma(k-2i+1)...}[/tex]

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?

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# Differentiating x^k

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