Question

Noninteger derivative question.

Answer 1

It is known that for positive integers, this is true. \[\Large n!=\int\limits_0^\infty x^n e^{-x}dx\] And by allowing ourselves to plug in non-integer values, we can extend the factorial function to non-integer values, which we call the gamma function, and get interesting results like \[\Large \frac{1}{2} ! = \frac{\sqrt{\pi}}{2}\]
So now here's where my question begins. If you play with derivatives of polynomials, it is fairly obvious that the nth derivative of x raised to the n power will be: \[\Large \frac{d^n}{dx^n} \left( x^n \right) = n!\]
So is it possible to show that there are non-integer derivatives and if so, is this consistent with the gamma function? For example, \[\Large \frac{d^{1/2}}{dx^{1/2}} \left( x^{1/2} \right) = \frac{1}{2}!\]

Answer 2

theoretically i think it's possible , but in that case what is the definition of the derivatives ?

Answer 3

I guess that's what I'm looking to find out. I've heard that noninteger derivatives have applications in electrochemistry, and I've read some things that I didn't understand about them in the past.

I like to think that x^3 is x multiplied by itself 3 times, so x^(1/2) must be x multiplied by itself a half of a time right? Ok, obviously that's ridiculous but the point is we use square roots even though the exponent is sort of weird.

Answer 4

One problem is that \((1/2)!\) is not defined.

Answer 5

@wio scroll up. http://www.wolframalpha.com/input/?i=%281%2F2%29%21

Answer 6

Yeah, in this case \(1/2\) not a positive integer.

Answer 7

wio its not defined on n! of 1.2.3....(n+1)!
but with gamma function its ok

Answer 8

i mean there might me a space ( wich is not real )
that hold that value

Answer 9

but again the problem is with the derivative definition , which i have no clue what it could be ...

Answer 10

The problem is that the \(n\)th derivative equation provided applies for positive integers, and it's not shown that it applies for the gamma function.

Answer 11

We know \[
\frac{d^{1/2}}{(dx)^{1/2}}\frac{d^{1/2}}{(dx)^{1/2}} x = \frac{d}{dx}x = 1
\]We could say \[
\frac{d^{1/2}}{(dx)^{1/2}}x = f(x), \quad \frac{d^{1/2}}{(dx)^{1/2}}f(x) = 1
\]

Answer 12

Now, if \(f(x)=c\), then that means the half derivative can turn constants into constants...

Answer 13

Yeah, but you have to show that f(x) is a constant.

Answer 14

yes we could start by writing a general k-th derivative for the function x^n
then plug in k = 1/2 and n = 1, for example, to obtain a half-derivative

Answer 15

if we wanted f(x) to be constant, we could use k = 1/2 and n = 0
Since x^0 = 1

Answer 16

Here's a fun fact:
\[\large \frac{d^n}{dx^n}(e^{ax})=a^n e^{ax}\]
replace a with i for added fun.

See what happens if we try to use this for non-integers?

Answer 17

The reason that interests me is because e^x shows up in the gamma function and the power series of e^x has a lot of factorials in it.

Answer 18

Well, you can always verify it.

Answer 19

Suppose \[
\frac{d^{1/2}}{(dx)^{1/2}}e^{ax} = a^{1/2}e^{ax}
\]Then we can say: \[
\frac{d^{1/2}}{(dx)^{1/2}}a^{1/2}e^{ax} = a^{1/2}\frac{d^{1/2}}{(dx)^{1/2}}e^{ax} = a^{1/2}a^{1/2}e^{ax} = ae^{ax}
\]

Answer 20

So \[
\frac{d}{dx}e^{ax} = \frac{d^{1/2}}{(dx)^{1/2}}\frac{d^{1/2}}{(dx)^{1/2}}e^{ax}
\]

Answer 21

This only works assuming that normal derivative rules work for half derivatives.

Answer 22

Fractional calculus interesting ...

Answer 23

Fractional differential equations! =P

Answer 24

yeah lol i never touched that subject =)

Answer 25

Let \(a=0\), this implies that \[
\frac{d^{1/2}}{(dx)^{1/2}}1 = 0
\]Likewise \[
\frac{d^{1/2}}{(dx)^{1/2}}c=c\frac{d^{1/2}}{(dx)^{1/2}}1 =c( 0)=0

\]

Answer 26

I like that @wio , very interesting. Something to note, it seems that this is a case where 0^0=1\[\large \frac{d^0}{dx^0}(e^{ax})=0^0e^{ax}\]

Here is another thing I'm thinking about.

\[\large \frac{d}{dx}(x^n)=nx^{n-1}\]\[\large \frac{d^2}{dx^2}(x^n)=n(n-1)x^{n-2}\]
It seems like there should be remain some sort of constant "distance" between derivatives.
\[\large \frac{d^{1/2}}{dx^{1/2}}(x^n)=\alpha x^{\beta}\]\[\large \frac{d}{dx}(x^n)=nx^{n-1}\]

So maybe we can kind of equate the first parts and second parts some how. Maybe replace the second one's n's with m's and work something out.

Answer 27

http://en.wikipedia.org/wiki/Fractional_calculus

Answer 28

Some of that stuff looks familiar. But I kinda want to figure this out without any help, it's more fun this way.

Answer 29

Maybe using product rule, and differentiating \(xe^{ax}\) would give us info about the half derivative of \(x\).

Answer 30

\[
\frac{d^{1/2}}{(dx)^{1/2}}xe^{ax} = e^{ax}\frac{d^{1/2}}{(dx)^{1/2}}x+\frac{d^{1/2}}x{(dx)^{1/2}}e^{ax} = e^{ax}\frac{d^{1/2}}{(dx)^{1/2}}x+xa^{1/2}e^{ax}
\]

Answer 31

Half derivative on left part gives \[

\frac{d^{1/2}}{(dx)^{1/2}} e^{ax}\frac{d^{1/2}}{(dx)^{1/2}} = a^{1/2}e^{ax}\frac{d^{1/2}}{(dx)^{1/2}} x + e^{ax}

\]

Answer 32

Notice the right part is the left part with a constant... \[
y = a^{1/2}e^{ax}\frac{d^{1/2}}{(dx)^{1/2}} x + e^{ax} + a^{1/2}y


\]

Answer 33

I think this means at most that the product rule doesn't work for non-integer derivatives. Now that I think about it, the product rule is specifically for first derivatives, not half or second derivatives if that makes sense.

Answer 34

Hmm, that is true.

Answer 35

Generalized product rule would be \[
\frac{d^n}{(dx)^n}f(x)g(x) = \sum_{k=0}^n\binom nk f^{(k)}(x)g^{(n-k)}(x)
\]However, this uses binomial theorem.

Answer 36

Possibly useful, I noticed the other day that: \[(f+g)^n\] has the same form as \[(fg)^{(n)}\] Or to sort of rephrase it as the binomial theorem: \[\frac{d^n}{dx^n}[f(x)*g(x)]=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right) \frac{d^{n-k}f(x)}{dx^{n-k}} \frac{d^{k}g(x)}{dx^{k}}\] I think the binomial coefficients hide the most interesting part, the factorials.

It's starting to get clunky, but this is sort of how you're thinking @wio with using the product rule, this is sort of a generalized product rule and it's based on factorials. So it sort of is a problem that won't go away for us unfortunately.

Answer 37

lol well you got it.

Answer 38

Okay, so we can let \(f(x) = 1\) and \(g(x)=x\).

Answer 39

The summation breaks for n=1/2. How do we add up all the terms from k=0 to k=1/2?

Answer 40

Well there is this: http://en.wikipedia.org/wiki/Binomial_series

Answer 41

Hmm, I'm stuck. I think I need to sleep on it.

Answer 42

Oh hey, I just got 99 lol.

Answer 43

nice topic , and it make me think nw since there is fractional derivatives
is there complex derivatives mmmm lolz or irrational derivatives

Answer 44

Another repetitive derivative is: \[
\frac{d^{2n}}{(dx)^{2n}}\sin(ax) = -1^{n}a^{n}\sin(ax)
\]

Answer 45

We could consider \(\lim_{n\to 1/4}\)

Answer 46

Which gives you \[
\frac{d^{1/2}}{(dx)^{1/2}}\sin(ax)=- a^{1/4}\sin(ax)
\]

Answer 47

But I'm not sure if that checks out...

Answer 48

It doesn't really work even when \(n\to 1/2\).

Answer 49

I just had a thought as I shut my eyes to sleep! Also, I have an answer for you wio. \[\Large \frac{d^n}{dx^n}(\sin(ax))=a^n \sin(ax+n \frac{\pi}{2})\]

Now I haven't checked this next thing: \[\Large \frac{d^n}{dx^n} ( x^p)= \frac{p!}{(p-n)!}x^{p-n}\]

That should work though. I'll check it in the morning.

Answer 50

ill try to work on it too , i might start from integral inverse , i saw something like this one before mmm

Answer 51

let \(p=1\) and \(n=1/2\).

Answer 52

then you get \[
\frac{x^{1/2}}{(1/2)!}
\]

Answer 53

Ok I lied I couldn't wait. It works. =)

Answer 54

The half derivative of that will then be (1/2)! divided by (1/2)! which is 1, which is also the answer of 1! like we expect.

Answer 55

\[\Large \frac{d^{a}}{dx^{a}}(\frac{d^{b}}{dx^{b}}x^p)=\frac{d^{a}}{dx^{a}}(\frac{p!}{(p-b)!}x^{p-b}) =\\ \Large \frac{p!}{(p-b)!}\frac{d^{a}}{dx^{a}}(x^{p-b})=\frac{p!}{(p-b)!}\frac{(p-b)!}{(p-b-a)!}x^{p-b-a}= \\ \Large \frac{p!}{(p-b-a)!}x^{p-b-a}=\frac{d^{a+b}}{dx^{a+b}}(x^p)\]

That's pretty general.

Answer 56

Thanks @wio and @ikram002p you helped more than you know! Thank you thank you! I'm going to go to sleep; or at least try to. If you find something wrong with my reasoning tell me so I can figure it out haha. Thanks again, and good night! =D

Answer 57

good night !

Answer 58

The reason why \[
\frac{d^a}{(dx)^a}\frac{d^b}{(dx)^b} f(x)= \frac{d^{a+b}}{(dx)^{a+b}} f(x)
\]is because that is the rule for derivative of natural numbers, and so when extending to rational number, such derivatives must be defined in a way that obeys the rule.

Answer 59

Now I think there are some interesting difficulties. \[\Large \frac{d^{1/2}}{dx^{1/2}}(e^x)=\frac{d^{1/2}}{dx^{1/2}}(\sum_{n=0}^{\infty}\frac{x^n}{n!})\] \[\Large e^x = \sum_{n=0}^{\infty}\frac{x^{n-1/2}}{(n-1/2)!}\] Now there are two options here.

We can either say that the term that was n=0 was a constant and so we raise the index by 1 since we had determined that the half derivative of a constant is 0. (In fact it seems obvious that any order derivative of a constant should be zero).

So now if we plug in x=0 we should get 1. However there are only essentially square roots here, so plugging it in should give us 0 which is wrong. So perhaps e^x is only its own integer derivative, not necessarily fractional derivative. Alternatively maybe the limit of this infinite sum ends up approaching 1 at x=0.

Maybe it's not safe to assume that a half derivative of a constant is 0 Maybe there is some room in there to wiggle around, I don't know. Another thing to keep in mind is that we sort of got the half derivative of a constant equal to 0 based on our assumption of the derivative of e^ax which might have been wrong to begin with.

Anyways, something to think about I guess I just wanted to write it down while it was still in mind.