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Second Order Derivative Notation (mingled with)

  1. Dec 3, 2014 #1
    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.
     
  2. jcsd
  3. Dec 3, 2014 #2

    Mark44

    Staff: Mentor

    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.
    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.
    They don't. You can combine two operators to get another operator. In both cases, the operators have to be applied to a function.
     
  4. Dec 3, 2014 #3
    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.
     
  5. Dec 3, 2014 #4

    Mark44

    Staff: Mentor

    Your experience wasn't evident to me when you asked whether dx/d had any meaning.
     
  6. Dec 3, 2014 #5
    ??? 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.
     
  7. Dec 3, 2014 #6

    Mark44

    Staff: Mentor

    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.
    Yes, I have heard of fractional derivatives.
     
  8. Dec 3, 2014 #7

    Stephen Tashi

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    Science Advisor

    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?".
     
  9. Dec 3, 2014 #8
    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 similiar to the chain rule for the first derivative, where the notation is used to represent a rather complicated operation as simple algebra.
     
    Last edited: Dec 3, 2014
  10. Dec 3, 2014 #9
    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. Dec 3, 2014 #10
  12. Dec 3, 2014 #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 cant imagine what it could possibly mean

    lol my 2cts literally beat me to it by a second
     
    Last edited: Dec 3, 2014
  13. Dec 3, 2014 #12
    This argument isn't actually about fractional derivatives, that was just an example I gave of a strange notation which nevertheless has a meaning.
     
  14. Dec 3, 2014 #13
    jeez if we can do fractional derivatives whats next?
    da+biy/dxa+bi
    :p
     
    Last edited: Dec 3, 2014
  15. Dec 3, 2014 #14
  16. Dec 3, 2014 #15
  17. Dec 3, 2014 #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.
     
  18. Dec 3, 2014 #17

    Stephen Tashi

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    Science Advisor

    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].
     
  19. Dec 3, 2014 #18
    really i think it would be nice if someone put up a list of "acceptable manipulations of leibniz notation"
     
  20. Dec 3, 2014 #19

    Mark44

    Staff: Mentor

    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.
     
  21. Dec 3, 2014 #20
    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.
     
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