Imaginary/Complex and Negative Derivatives?

[FeDeX_LaTeX]

Hello;

You can have positive integer derivatives, such as this:

\frac{d^{2}}{dx^{2}}(x^{2}) = 2

You can have fractional derivatives too;

\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}(x) = \frac{2\sqrt{x}}{\sqrt{\pi}}

But what about negative derivatives?

\frac{d^{-2}}{dx^{-2}}(x^{2})

Or even imaginary or complex derivatives?

\frac{d^{i}}{dx^{i}}(x^{3})

\frac{d^{3 + 2i}}{dx^{3 + 2i}}(3x^{2})

Are these defined in any way?

Thanks.

EDIT: Just had a re-think, aren't negative derivatives just integrals? Or are they something else?

 

[mathman]

Where did you see these ideas (integer derivative, fraction derivative, etc.)?

 

[FeDeX_LaTeX]

Hello;

I can't remember where I read them, it was probably a wiki article. I recall that a fractional derivative can be found if you use;

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

And by integer derivative I just mean an ordinary derivative, i.e. the 2nd derivative of the function f(x) = x would be an integer derivative.

EDIT: I found an article; http://en.wikipedia.org/wiki/Fractional_calculus

 

[AlephZero]

EDIT: Just had a re-think, aren't negative derivatives just integrals? Or are they something else?

For some calculus operations, it is useful to write D for the differential operator d/dx, and then treat "D" as if it was a "number".

When doing this, positive and negative integer powers of D make sense, and negative powers correspond to integration. Fractional or complex powers don't have any meaning.

As well as negative powers like 1/D or 1/D², expressions like 1/(D+a) and 1/(D²+a²) also have useful interpretations in terms of integration.

WARNING: using this D notation doesn't introduce any new math concepts, it is only a shorthand way of writing things down. If you misuse it, the final result will be wrong!

 

[FeDeX_LaTeX]

Hello;

Thanks for the replies. I have seen the use of D as the differential operator.

So, have you heard of any ways to find imaginary or complex derivatives of simple functions?

Also, you said that one can have expressions like \frac{1}{D + a} and \frac{1}{D^{2} + a^{2}}. What does the 'a' represent?

Thanks!

 

[chiro]

Hello;

Thanks for the replies. I have seen the use of D as the differential operator.

So, have you heard of any ways to find imaginary or complex derivatives of simple functions?

Also, you said that one can have expressions like \frac{1}{D + a} and \frac{1}{D^{2} + a^{2}}. What does the 'a' represent?

Thanks!

Hey there.

When we refer to say D^-1 or 1/D we are actually referring to the anti-derivative. When we refer to D^-2 or 1/D^2 we are referring to the anti-derivative of the anti-derivative of the function.

In general however, if we get something like say 1/(D+1) we have to use some tools to get the right expansion.

If our expression is a polynomial, then a binomial expansion, or a long division will suffice where we only take the appropriate terms (as a polynomial has a finite number of terms and powers).

You will come across problems like these and techniques to solve them in an intro calc class and in a differential equations class.

With regards to what 'a' means, its probably for most cases going to represent a numeric value and not a function, independent variable, or operator.

From what I recall, I think it was Heaviside that introduced this kind method to solve differential equations by using methods like binomial and long division by treating D as a linear operator.

 

[chiro]

For some calculus operations, it is useful to write D for the differential operator d/dx, and then treat "D" as if it was a "number".

When doing this, positive and negative integer powers of D make sense, and negative powers correspond to integration. Fractional or complex powers don't have any meaning.

As well as negative powers like 1/D or 1/D², expressions like 1/(D+a) and 1/(D²+a²) also have useful interpretations in terms of integration.

WARNING: using this D notation doesn't introduce any new math concepts, it is only a shorthand way of writing things down. If you misuse it, the final result will be wrong!

That is not true that fractional powers have no meaning. One application is an "interpolation like" perspective of basis functions in Fourier transforms.

 

[FeDeX_LaTeX]

Hello;

Thanks, that makes sense.

I was just experimenting with complex number derivatives;

We know that:

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

We can observe through calculation that:

\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}x = \frac{d}{dx}x = 1

Can we extend this to complex numbers? Let's try:

\frac{d^{i}}{dx^{i}}\frac{d^{1-i}}{dx^{1-i}}x = \frac{d}{dx}x = 1

We haven't done any calculations for this one so we'll do this below:

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

Simplifying:

= \frac{x}{\Gamma(2-i)\Gamma(1+i)}

But we know that the FIRST derivative of x is just 1, so:

\frac{x}{\Gamma(2-i)\Gamma(1+i)} = 1

But this implies that:

x = \Gamma(2-i)\Gamma(1+i)

So where to go from here? How do I show that:

\frac{x}{\Gamma(2-i)\Gamma(1+i)} = 1

?

The denominator must be equal to x, but x is a variable rather than a fixed value.

So, assuming that I just had

\frac{x}{\Gamma(2-i)\Gamma(1+i)}

as my first derivative, is there any way to simplify this further?

I mean, I'm just trying to see how you can arrive at the first derivative of x by using;

\frac{d^{i}}{dx^{i}}\frac{d^{1-i}}{dx^{1-i}}x

You should get 1, but I'm not sure how you can arrive at that from here?

 

[Char. Limit]

I calculated each derivative separately, using your formula. For the first one, I got...

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

Then...

\frac{d^i}{d x^i} \frac{x^i}{\Gamma(1+i)} = 1

So I don't see the problem.

 

[Char. Limit]

Wait... sorry about the double post, but does this mean that complex derivatives exist AND ARE COMPUTABLE?

 

[oddcitations]

Wait... sorry about the double post, but does this mean that complex derivatives exist AND ARE COMPUTABLE?

Sorry about posting, I have no idea what derivatives are and can't help in any way, shape, or form. BUT.
If you are going by that logic, the concept of math doesn't really exsist. But, in hindsight, you can make a 3D representation of the complex plane from a printer that prints in 3D, using plastic instead of paper. It's almost as cool as having a laser in your bathroom.

But if you are going by another logic, they exsist because they are computable, and can be proven. Hm. Deja vu.

 

[Phrak]

Complex derivatives are an interesting idea. Glad you brought it up. Come to think of it, taking the exterior derivative of a complex tensor on a complex manifold nearly demands that the derivative operator should also be complex. The reasoning is that the exterior derivative acting on a tensor looks just like a dual vector wedged with the tensor, which is just multiplication in funny ways.

I should have to see if this notion has any relationship to your scalar complex derivative--if it can be made to work at all...

 

[Char. Limit]

Are there any proofs we could make that generalize the fractional derivative to complex numbers?

 

[chiro]

Are there any proofs we could make that generalize the fractional derivative to complex numbers?

Given a standard function from C to C (or R^2 -> R^2 with the i^2 = -1) you basically use the standard multivariable definitions and use the fractional definition of the derivative for single variables where appropriate. So all the partials basically use the generalized notion of the derivative instead of the standardized integral definition of the derivative.

 

[FeDeX_LaTeX]

Hello;

Thank you for the replies.

\frac{d^i}{d x^i} \frac{x^i}{\Gamma(1+i)} = 1

How did you compute this?

 

[Char. Limit]

Hello;

Thank you for the replies.



How did you compute this?

With your equation involving the gamma function. First note that...

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

Thus, by then dividing by the constant \Gamma(1+i) that I left out (it's a constant, so I can do that), we get 1.

 

[JJacquelin]

Thanks to the Riemann Liouville transform :

 

[Char. Limit]

Hmm... I was trying, on a whim, to calculate the half-derivative of sin(x). However, I'm stuck because I don't know the derivative of this:

_1F_2(1;\frac{5}{4}, \frac{7}{4};-\frac{x^2}{4})

Can anyone familiar with hypergeometric series help me with this derivative? I can't even find it documented.

 

[JJacquelin]

Fractional derivatives of sine and cosine functions involves the generalized Fresnel integrals. (possible reduction in some particular cases)

 

[JJacquelin]

From : J.Spanier, K.B.Oldham, "An Atlas of Functions", Hemisphere Pubishing Corporation, Springer-Verlag, 1987, p.160 and p.382

 

[JJacquelin]

Half derivative of sin(x), from :
J.Spanier, K.B.Oldham, "An Atlas of Functions", Hemisphere Pubishing Corporation, Springer-Verlag, 1987, p.304

 

[FeDeX_LaTeX]

Hello;

Thanks for the replies.

I have been doing some more thinking, and I was wondering whether or not this is true;

If the series \sum_{n=1}^{\infty}\frac{1}{2^n} = 1,

then for any function f(x),

will taking the 1/2 derivative, then the 1/4 derivative, then the 1/8 derivative, then the 1/16 derivative (and so on) yield the 1st derivative, since the series converges? Just a thought.

EDIT: I tried writing this out as an infinite series; you end up with the terms x^(1/2), x^(3/4), x^(7/8) ... x^{\frac{2^{n-1}}{2^n}}. Since you are multiplying these terms, you end up with x^{\infty} since the series diverges... so does that mean that taking the (1/2^n)th derivative repeatedly (infinite times) yields infinity? Strange... I would have thought it to be 1, because of the convergence of the summation of 1/2^n from n = 1 to infinity.

 

[JJacquelin]

I tried writing this out as an infinite series; you end up with the terms x^(1/2), x^(3/4), x^(7/8) ... . Since you are multiplying these terms, you end up with x^infinity since the series diverges...

Why multiplying these terms ? I cannot understand what you mean.
When we proceed successive derivatives, we don't multiply them. For example, if we derivate two times a function in odrer to obtain the second derivative, we don't multiply the first derivative.