Question on Integrals

[rabbit]

Also, note, contrary to Rabbit's post, [tex]f(x)\frac{d^2}{dx^2} \ne \frac{d^2f(x)}{dx^2}[/tex]

Also, note, I didn't say that.

[Rule]

[tex]f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex]

Where I assume, f''(x) means the second derivative of f with respect to x?
f''(x)=[tex]\frac{d^2f(x)}{dx^2}[/tex]

[Camel]

a) The dot goes above the function's name, not next to it.
b) I've never heard of \dot notation used for anything other than integration, or for letters of a couple of foreign languages.

[tex]f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex]

Where I assume, f''(x) means the second derivative of f with respect to x?
f''(x)=[tex]\frac{d^2f(x)}{dx^2}[/tex]

It's easier to understand why this is true, but why rabbit's statement is not, if you think of [tex]\displaystyle\frac{d^2}{dx^2}[/tex] as a function of its own, as in:

[tex]\displaystyle f^{\prime\prime}(x) = \frac{d^2}{dx^2}[f(x)][/tex]

However, since the statement [tex]\displaystyle\frac{d^2}{dx^2}[/tex] is indeed a quotient - be it shorthand for a more complex division than it appears to be - it's legal to put the target function on the numerator.

[Rule]

a) The dot goes above the function's name, not next to it.
b) I've never heard of \dot notation used for anything other than integration, or for letters of a couple of foreign languages.

I've never seen it used for integration, but it is almost always used, especially in physics and in differential equations, to mean a derivative -- usually the time derivative.  It's not like you did anything wrong, you just deviated from well-established conventions; if you use the dot notation to mean integration, there will likely be some confusion.  See here.

I was being lazy in my notation, but I was referring to the way you used dot, e.g. above the function.

I don't think of d/dx (by itself) so much as quotient as it is a linear operator.
e.g.
[g(x) * d/dx ]*f(x) = g(x)*df/dx != [d/dx*g(x)]*f(x) = dg/dx*f(x)

[Camel]

The [tex]\frac{d}{dx}[/tex] notation doesn't really fit snugly in to any mathematical concept established before calculus, but if you look at the definition of it:

[tex]\displaystyle \frac{d}{dx}[f(x)] := \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}[/tex] you can plainly see that is is a quotient. Example:

[tex]\displaystyle f^\prime(x) = \frac{df}{dx}[/tex] where [tex]df[/tex] is the "rate of change in f" at value x, and [tex]dx[/tex] is the infinitesimal rate of change in x.

[Rule]

[tex]\frac{d}{dx}[/tex] is just an operator in the same way a matrix is an operator.

It is a non-commutative linear operator:
d/dx (af + bg) = a*df/dx + b*dg/dx
d/dx(f) != f*(d/dx),
where a and b are constants, f and g are functions of x.

It operates on a function in such a way that its output is the limit of a quotient.

[Ender]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

[Camel]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

I never said [tex]\displaystyle\frac{d}{dx} = \frac{df}{dx}[/tex], I said [tex]\displaystyle\frac{df}{dx} = \frac{d}{dx}[f(x)][/tex]

"[tex]\display\frac{d}{dx}[/tex]" is not an operator per se; it is incomplete, but is still a quotient. Just because it operates on a function does not make it a function its self, although it's reasonable to think of it as one - I defined it above the way one would define a function. Saying "[tex]\display\frac{d}{dx}[/tex]" is the same as saying "the change in [what is to follow] as x changes."


[edit] tex

[nslay]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

I never said [tex]\displaystyle\frac{d}{dx} = \frac{df}{dx}[/tex], I said [tex]\displaystyle\frac{df}{dx} = \frac{d}{dx}[f(x)][/tex]

"[tex]\display\frac{d}{dx}[/tex]" is not an operator per se; it is incomplete, but is still a quotient. Just because it operates on a function does not make it a function its self, although it's reasonable to think of it as one - I defined it above the way one would define a function. Saying "[tex]\display\frac{d}{dx}[/tex]" is the same as saying "the change in [what is to follow] as x changes."


[edit] tex

An operator, in functional analysis, loosely speaking, is a function that maps functions to functions (a function that accepts a function, and outputs a function).  The derivative is an operator since it maps functions to functions.  The Calculus 1 definition of the derivative is a beginner's view of the derivative.  When you take a course like numerical PDE, you will see that a derivative can be expressed as a matrix which is pretty damn cool (a matrix can be an operator!).  In math, there are often many interpretations of a single problem...who would think a derivative could be thought of as a matrix?  In my experience, math is learned in coats of paint...Calculus is one of the very first coats.  As you get higher in math, you realize how little you know...and actually, you learn how little is known about math in general.  Just as an example, computational fluid dynamics uses linear approximations to solve the Navier-Stokes equation, which is non-linear...we still work with the linear in this day and age!

You should try to listen to Rule, he's one hell of a mathematician (and yes, he's a REAL mathematician)...he's much better than I am anyways.

[Camel]

My mistake: I was thinking in terms of procedural programming, where you must resolve the state of a function before it can be passed to another function, and an "operator" is a special sub-set of a function.

Also, with a deep understanding of matrices, the analogy isn't exactly a big leap of faith. I recall one particular problem in my AP Calculus class in highschool that I solved using matrices. I didn't get credit for the problem because my teacher didn't learn how to do that until after he got his masters degree -- at least, that's what he said when I asked him why I didn't get credit for a correct answer.