Explicit definition for antiderivative?

[Jhenrique]

Accidentally I wrote in the wolfram f(x) = f(1/x) the the wolfram give me the solution for this equation (f(x) = Abs(log(x))). Hummmm, nice! Thus I thought: given the definition of derivative, $f'(x) = \frac{f(x+dx)-f(x)}{dx}$, is possible to isolate f(x) in this equation? If yes, how?

I realized that exist a pattern for the expression of derivative: $$\\ f^{(1)}(x) = \frac{f(x+dx)-f(x)}{dx} \\ \\ f^{(2)}(x) = \frac{f(x+2dx) - 2 f(x+dx) + f(x)}{dx^2} \\ \\ f^{(3)}(x) = \frac{f(x+3dx) - 3 f(x+2dx) + 3 f(x+dx) - f(x)}{dx^3} \\ \\ f^{(4)}(x) = \frac{f(x+4dx) - 4 f(x+3dx) + 6 f(x+2dx) - 4 f(x+dx) + f(x)}{dx^4}$$ So maybe could exist a pattern too for f^(-1), f^(-2), f^(-3)... But I haven't sure if is possible to develop this pattern retroactively, I think that would be necessary to define binomial coefficients for negative argument... In the end, what you think about?

 

[Simon Bridge]

I don't think so. The definition of the derivative I know is:
$$f^\prime(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ ... therefore:
$$f^{(2)}(x) = \lim_{h\rightarrow 0} \frac{f(x+2h)-2f(x+h)+f(x)}{h^2}= \lim_{h\rightarrow 0} \frac{f(x-h)-2f(x)+f(x+h)}{h^2}$$... you can work out an expression for the nth derivative by following the pattern... lots of people kinda discover this sort of thing when the get used to the limits: i.e. http://math.stackexchange.com/questions/702185/definition-of-the-nth-derivative-first-post
... but I don't think that the pattern goes backwards.
When you take an antiderivative, you are starting with a function f^(n>0) already.

re: http://www.proofwiki.org/wiki/Definition:Derivative

What I think you are looking for is more like:
http://www.proofwiki.org/wiki/Definition:Primitive_(Calculus )

 

[HallsofIvy]

Accidentally I wrote in the wolfram f(x) = f(1/x) the the wolfram give me the solution for this equation (f(x) = Abs(log(x))). Hummmm, nice! Thus I thought: given the definition of derivative, $f'(x) = \frac{f(x+dx)-f(x)}{dx}$, is possible to isolate f(x) in this equation? If yes, how?

That is NOT the "definition of derivative" unless you are taking dx to be an "infinitesimal". And if you re doing that you would have to be using some deep logical concepts to define "infinitesmals" to begin with! Are you working with "non-standard Analysis"?

I realized that exist a pattern for the expression of derivative: $$\\ f^{(1)}(x) = \frac{f(x+dx)-f(x)}{dx} \\ \\ f^{(2)}(x) = \frac{f(x+2dx) - 2 f(x+dx) + f(x)}{dx^2} \\ \\ f^{(3)}(x) = \frac{f(x+3dx) - 3 f(x+2dx) + 3 f(x+dx) - f(x)}{dx^3} \\ \\ f^{(4)}(x) = \frac{f(x+4dx) - 4 f(x+3dx) + 6 f(x+2dx) - 4 f(x+dx) + f(x)}{dx^4}$$ So maybe could exist a pattern too for f^(-1), f^(-2), f^(-3)... But I haven't sure if is possible to develop this pattern retroactively, I think that would be necessary to define binomial coefficients for negative argument... In the end, what you think about?

 

[DrewD]

Accidentally I wrote in the wolfram f(x) = f(1/x) the the wolfram give me the solution for this equation (f(x) = Abs(log(x))). Hummmm, nice! Thus I thought: given the definition of derivative, $f'(x) = \frac{f(x+dx)-f(x)}{dx}$, is possible to isolate f(x) in this equation? If yes, how?

I realized that exist a pattern for the expression of derivative: $$\\ f^{(1)}(x) = \frac{f(x+dx)-f(x)}{dx} \\ \\ f^{(2)}(x) = \frac{f(x+2dx) - 2 f(x+dx) + f(x)}{dx^2} \\ \\ f^{(3)}(x) = \frac{f(x+3dx) - 3 f(x+2dx) + 3 f(x+dx) - f(x)}{dx^3} \\ \\ f^{(4)}(x) = \frac{f(x+4dx) - 4 f(x+3dx) + 6 f(x+2dx) - 4 f(x+dx) + f(x)}{dx^4}$$ So maybe could exist a pattern too for f^(-1), f^(-2), f^(-3)... But I haven't sure if is possible to develop this pattern retroactively, I think that would be necessary to define binomial coefficients for negative argument... In the end, what you think about?

Your formulae are wrong, but the idea of extending the $n^{th}$ derivative is an idea that apparently has some use. It has been done even for non integer values, but it is not clean. This blog is where I first learned about fractional derivatives. Derivatives stop having a nice clear interpretation though.

But more to the specific question about "negative" derivatives. You can think about the operation of taking the $n^{th}$ derivative of a function of real variables as a function itself $D^n:C^{\infty}\rightarrow C^{\infty}$. But if you try to extend this to negative values of $n$ using the FTC, you have to fix the lower limit of the integral or else you do not have a function (the antiderivative is a set of functions). There is another issue. If we say that $D^{-1}=\int_0^xf(t)\,dt$, it is not true in general that $D^a\left(D^bf(x)\right)=D^{a+b}f(x)$. In particular, $D^{-1}$ is not the inverse of $D^1$. Take $f(x)=2x+2$. Then you have $D^1(f(x))=2$ but $D^{-1}2=2x$. So the idea of thinking about "negative" exponents seems (to me) to have limited use.

 

[Jhenrique]

That is NOT the "definition of derivative" unless you are taking dx to be an "infinitesimal". And if you re doing that you would have to be using some deep logical concepts to define "infinitesmals" to begin with! Are you working with "non-standard Analysis"?

This "definition" of derivative that I use is for my own use and consumption...

 

[Mark44]

This "definition" of derivative that I use is for my own use and consumption...

For your own use is one thing, but when you post something like this--$f'(x) = \frac{f(x+dx)-f(x)}{dx}$
instead of this--$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, you run the risk of not being understood.

 

[Simon Bridge]

I was going to say:

This "definition" of derivative that I use is for my own use and consumption...

You seem, therefore, to be using the word "derivative" to mean something different from what the rest of the mathematical World does. That makes it difficult to help you.

... If what you have to say is important and/or difficult to follow, use the simplest language possible. If the reader doesn't get it then, let it not be your fault.
-- Larry Niven.​

 

[Jhenrique]

For your own use is one thing, but when you post something like this--$f'(x) = \frac{f(x+dx)-f(x)}{dx}$
instead of this--$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, you run the risk of not being understood.

Yeah, sorry!

I was going to say:
You seem, therefore, to be using the word "derivative" to mean something different from what the rest of the mathematical World does. That makes it difficult to help you.

... If what you have to say is important and/or difficult to follow, use the simplest language possible. If the reader doesn't get it then, let it not be your fault.
-- Larry Niven.​

Another problem is that english isn't my natural idiom, so, I have few problems with semantic.

Thinking better, my definiton for derivative isn't incorrect:

$$f'(x) = \frac{df(x)}{dx} = \frac{f(x_1)-f(x)}{dx} = \frac{f(x+dx)-f(x)}{dx}$$

 

[Stephen Tashi]

I think jhenrique's personal definitions of derivatives are related to the world's definitions for the "n point approximations" for derivatives that are used in numerical analysis. ( For example, eq 4.9 of the PDF http://www.google.com/url?sa=t&rct=...o5acY0QrfDXXvTQ&bvm=bv.64764171,d.aWw&cad=rja) So perhaps he can find the symbolic expression of what he wants by looking at the world's work on numerical integration.

 

[Mark44]

Thinking better, my definiton for derivative isn't incorrect:

$$f'(x) = \frac{df(x)}{dx} = \frac{f(x_1)-f(x)}{dx} = \frac{f(x+dx)-f(x)}{dx}$$

For any nonzero value of dx above, the three expressions to the right give only approximate values for the derivative, so the = signs are not applicable here.

I think jhenrique's personal definitions of derivatives are related to the world's definitions for the "n point approximations" for derivatives that are used in numerical analysis. ( For example, eq 4.9 of the PDF http://www.google.com/url?sa=t&rct=...o5acY0QrfDXXvTQ&bvm=bv.64764171,d.aWw&cad=rja) So perhaps he can find the symbolic expression of what he wants by looking at the world's work on numerical integration.

Equation 4.9 of the book you linked to adds a correction term of (h²/12)f⁽⁴⁾($\xi$), which makes the approximation exactly equal to f''(x₀). Jhenrique's formula doesn't do that, nor does he say what dx is.

The business of the correction term might be something that Jhenrique is unaware of.

 

[pwsnafu]

Jhenrique, do you know about the Cauchy formula for repeated integration?

If $f$ is continuous then
$\int_a^{x} \int_{a}^{t_{1}} \int_{a}^{t_{2}} \ldots \int_{a}^{t_{n-1}} f(t_n) dt_{n} \dots dt_3 dt_2 dt_1 = \frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t)\, dt$

 

[Simon Bridge]

Another problem is that english isn't my natural idiom, so, I have few problems with semantic.

...your English is pretty good actually, which means you may get misunderstood when someone thinks you are being subtle.

On PF, we are used to non-native speakers and usually do a fair job of compensating.
You'll know the process from when you have to talk to people who are not native to your language, but must speak it: you use your fluency to compensate for their inexperience. Just let us know if something reads strangely to you.

One of the advantages in mathematics is that we have a common language that transcends the written word - so long as we stick to standard definitions, we should be able to communicate.

If you use the "limit" form of the definition of the derivative instead of the approximate form you have been using, you should be able to make more headway with your inquiry. However, I think the main issue will be working out what would be meant by a negative (or fractional) index for differentiation. You've got some resources for that now. There is also a link for the proper use of the approximate form of the definition you have already been using.

Once you've addressed these issues your mathematical expressions will better match the ideas you are trying to communicate.

Good luck :)