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Because Zero can be expressed as $\frac {x^{-n}}{\Gamma (-n+1)} , x \neq 0$ where $n$ can be any natural number.

In the same manner that we represent any constant as $C \frac {x^0}{0!}, x \neq 0$ when dealing constants with fractional calculus.

And from the generalized equation $D^{1/2} x^{m} = \frac{\Gamma\!\left(m+1\right) \, x^{m-1/2}}{\Gamma\!\left(m+\frac{1}{2}\right)} , m \in \Bbb R$

I can say $D^{1/2}(0)=\frac {d^{\frac 1 2}}{dx^{\frac 1 2}}(\frac {x^{-n}}{\Gamma (-n+1)} )=\frac {x^{-n-\frac 1 2}}{\Gamma (-n+\frac 1 2 )}$ , where $n \in \Bbb N$

So the result is "unlimited values for Zero".

Now there is a problem that any function $f(x)$ can be also expressed as $(f(x)+0)$ or as $(f(x) +0+0+0+\cdots)$ etc...

Therefore $\frac {d^{\frac 1 2}}{dx^{\frac 1 2}}(f(x))$ looks undifined

which seems to me that the half order diffrential is completely undifined.

I am new to this field and the picture is not so clear to me.

So is this problem exist? and why the generalized equation give these so many values to Mr. Zero?

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  • $\begingroup$ the fractional derivative of $x^{-n}, n \in \mathbb{N}^*$ isn't well-defined $\endgroup$ – reuns Aug 31 '16 at 1:37
  • $\begingroup$ @user1952009 So what the fractional derivative of Zero? $\endgroup$ – Pentapolis Aug 31 '16 at 8:36
  • $\begingroup$ And why $D^{1/2} x^{m} = \frac{\Gamma\!\left(m+1\right) \, x^{m-1/2}}{\Gamma\!\left(m+\frac{1}{2}\right)}$ is generalized for $m \in \Bbb R$ if $\Bbb Z^-$ in not included. $\endgroup$ – Pentapolis Aug 31 '16 at 8:42
  • $\begingroup$ @Pentapolis Same general idea: math.stackexchange.com/questions/1811780/… $\endgroup$ – Simply Beautiful Art Sep 3 '16 at 1:13
  • $\begingroup$ May be of interest: xuru.org/fc/Intro.asp $\endgroup$ – Simply Beautiful Art Sep 3 '16 at 1:14
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Because Zero can be expressed as $x^{−n}\over\Gamma(−n+1)$,$x≠0$ where $n$ can be any natural number.

Well, no, that's undefined. You probably meant the limit as $n$ approached any natural number, which is quite a hinderance.

In the same manner that we represent any constant as $C\frac{x^0}{0!}$, $x\ne0$ when dealing constants with fractional calculus.

Well, consider the derivative to be a linear operator since we have

$$\frac{d}{dx}Cf(x)=C\frac{d}{dx}f(x)$$

$$\frac{d}{dx}f(x)+g(x)=\left(\frac{d}{dx}f(x)\right)+\left(\frac{d}{dx}g(x)\right)$$

Similarly, the antiderivatives follow the same rules above, so let us assume that it holds for fractional derivatives (or else things get messy and stop making sense)

$$D^a_xCf(x)=CD^a_xf(x)\tag1$$

$$D^a_xf(x)+g(x)=\left(D^a_xf(x)\right)+\left(D^a_xg(x)\right)\tag2$$

where $D^a_x$ is the $a$th derivative with respect to $x$.

No, according to $(1)$, if we have $C=0$, we will always get

$$D^a_x0=0$$

Now, from the above link, the fractional constant of differintegration can be considered, but it is a mighty difficult and confusing task, which is why linear operators cannot agree with the communicative nature that fractional derivatives should have. In other words,

$$D^aD^bf(x)\ne D^{a+b}f(x)$$

must not hold if we consider both the fractional constant of integration and the derivative to be a linear operator $(1,2)$. An example of why is presented in the body of your paragraph:

$$D^{1/2}_xf(x)+0+0+0+\dots$$

And from the generalized equation $D^{1/2}_xx^m=\frac{\Gamma(m+1)x^{m-1/2}}{\Gamma(m+1/2)}$, $m\in\mathbb R$

To show you how convoluted fractional calculus is, see that the above formula is not entirely true, for example

$$D^{1/2}_xx^{-1}=\frac{x^{-3/2}\left(\ln(4x)-2\right)}{-2\sqrt\pi}$$

and certainly $-1\in\mathbb R$. (I would try to compare this to what would happen if you used your formula)

So the result is "unlimited values for Zero".

Well, by now you probably noticed that fractional calculus isn't all that set in stone and that your reasoning was probably wrong (but intuitive). One could simply point out that you can't differentiate a function and get more than one answer. If you did, you probably didn't split the problem into its respective branches, and thus ended up solving for the general case of a problem you probably didn't mean to do in the first place. (for example, solving for $y'$ in $x^2+y^2=r^2$ gives two different answers, which can be seen by a graph).

Another counter-argument relates to the top of this question, stating that $0$ is not representable in the form you had, but rather is the limit of the expression as $n$ approaches any natural number. So what you used is something far more complicated than any linear operator;

$$D^a_x\lim_{n\to u}f(x,n)=\lim_{n\to u}D^a_xf(x,n)\tag3$$

which is the assumption that limits and fractional derivatives are communicative.

I wish it could be as simple as $(3)$, but as you continue on with that type of thinking, you run into paradoxical answers that simply shouldn't be.

The second link I gave you has knowledgeable content on fractional calculus in general, and you may learn a thing or two from it. It is designed somewhat like course notes, and you may find it interesting.

It also presents problems (similar to yours) that fractional calculus faces and if you read far enough into it, you might find answers to questions you have and questions you didn't even think about. (fair warning: you may get disappointed that fractional calculus isn't really 'complete' and that many problems like yours are usually ignored when attempting problems)

EDIT:

If you consider the Grünwald–Letnikov derivative, you would get

$$D^qf(x)=\lim_{h\to0}\frac1{h^q}\sum_{0\le m\le\infty}(-1)^m\binom qmf(x+(q-m)h)$$

For $f(x)=0$, this is equivalent to $0$ for all $q$.

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  • $\begingroup$ Thank you Simple Art for your feedback I will see your links. $\endgroup$ – Pentapolis Sep 3 '16 at 22:59
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    $\begingroup$ @Pentapolis Well, I'm sorry if I couldn't be of as much assistance as you might have hoped. This might help?: xuru.org/Contact.asp $\endgroup$ – Simply Beautiful Art Sep 3 '16 at 23:21
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    $\begingroup$ also note that $D^{-n} (0) =\sum_{k=0}^{n-1} c_kx^k, n \in \Bbb N$ $\endgroup$ – Pentapolis Sep 7 '16 at 15:19
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    $\begingroup$ @Pentapolis Yes, I understand, but one could just as easily use $0=\lim_{n\to\infty}\frac xn$ or any number of things. So limits should not be used as the basis of what something is. Limits are close. $\endgroup$ – Simply Beautiful Art Sep 8 '16 at 21:10
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    $\begingroup$ @Pentapolis And yes, it is agree-able that the Grünwald–Letnikov derivative is like your $\sin(n\pi)$ example, but isn't everything you did above like the $\sin(n\pi)$ example? Almost every definition of a fractional derivative comes like this. Its like the factorial$\to\Gamma$. There are different definitions of the fractional factorial that all agree with positive integers, but given different properties desired, the result may be different. Some comments and answers here explain better: math.stackexchange.com/questions/1847678/… $\endgroup$ – Simply Beautiful Art Sep 8 '16 at 21:14

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