Second Order Derivative Notation (mingled with)

In summary, the conversation discusses various notations and their meanings in relation to derivatives. The speaker brings up an alternative notation for the second derivative, which is not the standard notation. The other speaker explains why this notation is not valid and discusses the importance of indicating the independent variable in notation. The conversation also touches on the idea of fractional calculus and the potential for unusual notations to have defined meanings. However, it is emphasized that notation without a clear definition does not hold any mathematical significance.
  • #1
BHL 20
66
7
I've been thinking about something recently:

The notation d2x/d2y actually represents something as long as x and y are both functions of some third variable, say u. Then you can take the second derivatives of both with respect to u and evaluate d2x/du2 × 1/(d2y/du2).

Now I think it's also reasonable to express d2x/d2y as the product of dx/d and d/dy. Although dx/d is a notation I've never seen before I assume it's an operator. So how can the combination of two operators give an actual function? I think what I'm assuming is wrong so please explain why this does not work. If the notation dx/d isn't defined, why has no one defined it?

If dx/d is not an operator, what is it? Because d/dy is certainly one, I've found it possible to calculate d/dx for various functions x and y by integrating the expression for d2x/d2y. There doesn't seem to be any sort of pattern, but I probably haven't looked hard enough.
 
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  • #2
BHL 20 said:
I've been thinking about something recently:

The notation d2x/d2y actually represents something as long as x and y are both functions of some third variable, say u.
This isn't the standard notation for the second derivative of x with respect to y. That notation looks like this:
$$ \frac{d^2 x}{d y^2}$$
In this case, x is considered to be a (twice-differentiable) function of y. No third variable is required.
BHL 20 said:
Then you can take the second derivatives of both with respect to u and evaluate d2x/du2 × 1/(d2y/du2).

Now I think it's also reasonable to express d2x/d2y as the product of dx/d and d/dy.
I don't. Both of these are meaningless, since they don't convey the information about what variable we're differentiating with respect to. IOW, they don't indicate which is the independent variable. The reasoning behind the notation that I showed above is this:
$$\frac{d}{dy} \frac{dx}{dy} = \frac{d^2 x}{dy^2}$$
It's important to remember that this is a notational device.
BHL 20 said:
Although dx/d is a notation I've never seen before I assume it's an operator. So how can the combination of two operators give an actual function?
They don't. You can combine two operators to get another operator. In both cases, the operators have to be applied to a function.
BHL 20 said:
I think what I'm assuming is wrong so please explain why this does not work. If the notation dx/d isn't defined, why has no one defined it?

If dx/d is not an operator, what is it? Because d/dy is certainly one, I've found it possible to calculate d/dx for various functions x and y by integrating the expression for d2x/d2y. There doesn't seem to be any sort of pattern, but I probably haven't looked hard enough.
 
  • #3
Mark44 said:
This isn't the standard notation for the second derivative of x with respect to y. That notation looks like this:
$$ \frac{d^2 x}{d y^2}$$
In this case, x is considered to be a (twice-differentiable) function of y. No third variable is required.

Oh please, I've studied calculus in both high school and college, and I've been applying it to problems in physics for the past 3 years. I think I know that much.

I used a different notation on purpose because I was wondering if it had any meaning.
 
  • #4
BHL 20 said:
Oh please, I've studied calculus in both high school and college, and I've been applying it to problems in physics for the past 3 years. I think I know that much.
Your experience wasn't evident to me when you asked whether dx/d had any meaning.
BHL 20 said:
I used a different notation on purpose because I was wondering if it had any meaning.
 
  • #5
Mark44 said:
Your experience wasn't evident to me when you asked whether dx/d had any meaning.

? Experience doesn't rule out the possibility of an unusual notation being defined. I don't know if you have heard of fractional calculus but the notation d1/2x/dt1/2 has actually been defined, and it has some esoteric applications in science, despite looking absurd to most maths students.
 
  • #6
BHL 20 said:
? Experience doesn't rule out the possibility of an unusual notation being defined.
That's true, but we get a lot of posts here from people who have little or no experience with calculus. I had no way of knowing what your background was, and what you wrote (for example, about the need for both x and y to be functions of a third variable, as well as your dx/d and dy/d operators) suggested to me that you didn't have much experience.
BHL 20 said:
I don't know if you have heard of fractional calculus but the notation d1/2x/dt1/2 has actually been defined, and it has some esoteric applications in science, despite looking absurd to most maths students.
Yes, I have heard of fractional derivatives.
 
  • #7
BHL 20 said:
Oh please, I've studied calculus in both high school and college, and I've been applying it to problems in physics for the past 3 years. I think I know that much.

The way you phrased your question gives the impression that you think notation has hidden properties and that certain strings of symbols have specific meanings even though they have not been given definitions. This is an imaginative way to think and occasionally it may lead to inventing useful definitions, but it is only a way of daydreaming about things. Thinking that strings of symbols must have meaning is the typical thought pattern for people who have no understanding of the logical structure that mathematics requires - for example, people who start threads with questions like "Is 0/0 equal to 1" or "0.9999... is not equal to 1" Such questions begin with the fallacy that strings of symbols have a specific meaning even when no definition for them has been stated. If you want to be recognized as a sophisticated thinker, don't say things like "Although dx/d is a notation I've never seen before I assume it's an operator." What you should ask is "Is there a useful mathematical definition for the notation dx/d?".
 
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  • #8
Stephen Tashi said:
The way you phrased your question gives the impression that you think notation has hidden properties and that certain strings of symbols have specific meanings even though they have not been given definitions. This is an imaginative way to think and occasionally it may lead to inventing useful definitions, but it is only a way of daydreaming about things. Thinking that strings of symbols must have meaning is the typical thought pattern for people who have no understanding of the logical structure that mathematics requires - for example, people who start threads with questions like "Is 0/0 equal to 1" or "0.9999... is not equal to 1" Such questions begin with the fallacy that strings of symbols have a specific meaning even when no definition for them has been stated. If you want to be recognized as a sophisticated thinker, don't say things like "Although dx/d is a notation I've never seen before I assume it's an operator." What you should ask is "Is there a useful mathematical definition for the notation dx/d?".

Sorry for asking it in this way then, but I did say that what I think what I'm assuming is wrong.

I admit you're right, I like to think that there's more to mathematical notation than meets the eye. I expect whoever came up with the notation to have thought it out, so that it makes sense. The notation for the first derivative is quite straightforward, dy/dx can be understood simply as the infinitesimal change in y divided by the infinitesimal change in x, at the point at which it's evaluated. The logic behind the notation for the second derivative is not as apparent. Of course, it's just the derivative of the first derivative, but why did someone decide to write it in that way?

The first derivative notation is very handy with the way it can be inverted to get the derivative of the bottom variable with respect to the top one. The reciprocal of d2y/dx2 is not equal to d2x/dy2 and the notation certainly implies that. The purpose of this exercise was to see if the "meaning" ends there, i.e. if the notation does anything more than tell us the derivative can't be inverted in this way. For example, does the second derivative have anything similar to the chain rule for the first derivative, where the notation is used to represent a rather complicated operation as simple algebra.
 
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  • #9
BHL 20 said:
Oh please, I've studied calculus in both high school and college, and I've been applying it to problems in physics for the past 3 years. I think I know that much.

I used a different notation on purpose because I was wondering if it had any meaning.

This was not clear at the onset.
The only meaning of dy/d that I can come up with is (d/dy)^-1 in other words, integration.
 
  • #11
at first i thought this was an argument over semantics/notation, but i looked up
http://en.wikipedia.org/wiki/Fractional_calculus
very interesting.. seems like a sound idea but i can't imagine what it could possibly mean

lol my 2cts literally beat me to it by a second
 
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  • #12
bluntwcrackrap said:
at first i thought this was an argument over semantics/notation, but i looked up
http://en.wikipedia.org/wiki/Fractional_calculus
very interesting.. seems like a sound idea but i can't imagine what it could possibly mean

lol my 2cts literally beat me to it by a second

This argument isn't actually about fractional derivatives, that was just an example I gave of a strange notation which nevertheless has a meaning.
 
  • #13
jeez if we can do fractional derivatives what's next?
da+biy/dxa+bi
:p
 
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  • #16
Setting the exponent to -1 is the answer to your original question .
I do not like the notation dy/dx, I prefer d_xy.
Thus you can write d_x^p, the partial derivative wrt x to the power p.
 
  • #17
Returning to the original post, if we seek to interpret [itex] \frac {d^2 y}{d^2 x} [/itex] as the application of an operator, there are some difficulties. It wouldn't be an operator that operated on a single function. It requires a pair of functions. if we denote the operator with the single symbol [itex] W [/itex] then [itex] W [/itex] is a mapping from a a pair of functions [itex] (f,g) [/itex] to the single function [itex] \frac{f'}{g'} [/itex]. Trying to define what the operator [itex] W [/itex] does to a single function written as a fraction would required overcoming the fact that same function can be denoted by different fractions. For example, [itex] f(u) = \frac{u^2}{u} = \frac{u^4}{u^3} [/itex].
 
  • #18
really i think it would be nice if someone put up a list of "acceptable manipulations of leibniz notation"
 
  • #19
bluntwcrackrap said:
really i think it would be nice if someone put up a list of "acceptable manipulations of leibniz notation"
There's really not much to say, at least if we limit the discussion to positive integer indexes. The motivation for second- and higher-order derivatives in Liebniz notation is this, I believe:
$$ \frac{d}{dx} \frac{dy}{dx} = \frac{d^2y}{dx^2}$$
We're applying the ##\frac{d}{dx}## operator to the derivative ##\frac{dy}{dx}##. As a notational convenience to be able to write the derivative in a more compact form, someone thought that d2y should be what appears in the "numerator". The notation dx2 that appears in the "denominator" is suggestive of dx * dx, or shorthand for (dx)2.

In any case, it's just notation.
 
  • #20
Mark44 said:
There's really not much to say, at least if we limit the discussion to positive integer indexes. The motivation for second- and higher-order derivatives in Liebniz notation is this, I believe:
$$ \frac{d}{dx} \frac{dy}{dx} = \frac{d^2y}{dx^2}$$
We're applying the ##\frac{d}{dx}## operator to the derivative ##\frac{dy}{dx}##. As a notational convenience to be able to write the derivative in a more compact form, someone thought that d2y should be what appears in the "numerator". The notation dx2 that appears in the "denominator" is suggestive of dx * dx, or shorthand for (dx)2.

In any case, it's just notation.

At this stage, I agree with you. I suppose it can't get any more profound than this because people prefer to take derivatives and integrals one at a time. Maybe if we had a set of rules for taking second derivatives, derived from:
$$lim_{h→0}\ \dfrac{f(x+2h)-2f(x+h)+f(x)}{h^{2}}$$
it would be possible to invent a notation that incorporates at least one of those rules in a clever way. But yes, this approach would be impractical so the official second derivative notation is nothing more than a notation.
 

1. What is a second order derivative?

A second order derivative is the rate of change of a rate of change. It measures how fast the rate of change is changing, or the curvature of a function at a specific point.

2. How is a second order derivative notated?

A second order derivative is typically notated using either d²y/dx² or f''(x). The notation d²y/dx² is read as "d squared y by d x squared" and indicates the second derivative of y with respect to x. The notation f''(x) is read as "f double prime of x" and is another common way to represent the second derivative of a function f.

3. What does the notation "mingled with" mean in reference to second order derivatives?

The phrase "mingled with" is often used to describe a second order derivative with respect to multiple variables. For example, if a function has two independent variables x and y, the second order derivative would be notated as d²z/dx² mingled with d²z/dy². This means that the second order derivative is taken with respect to both x and y simultaneously.

4. How is the second order derivative calculated?

The second order derivative can be calculated by taking the derivative of the first order derivative. This process is known as taking the derivative twice. For example, if a function is given as y = x², the first order derivative would be y' = 2x, and the second order derivative would be y'' = 2. In other words, the second order derivative is the derivative of the slope of a function.

5. What is the significance of the second order derivative in calculus?

The second order derivative is significant in calculus because it provides information about the curvature of a function. It can help determine the concavity and inflection points of a curve, as well as the rate at which the slope is changing. Additionally, the second order derivative is used in optimization problems and is essential in understanding the behavior of complex functions.

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