Wikipedia:Reference desk/Archives/Mathematics/2012 April 6

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April 6

Notation question

I need to find a compact way of signifying that a variable is constructed by one or more derivatives with respect to some other variable. To qualify with an example, I would like to distinguish

$z(x)={\frac {d}{dx}}y(x)$

from

$z(x)={\frac {d}{du}}y(x){\text{ , }}x=x(u)$.

I have come up with various conventions, only to discover that they were already used for other things. Is there a convention for this? Thanks!--Leon (talk) 12:40, 6 April 2012 (UTC)

In the second example, z depends not just on x but on the derivative of x with respect to u. Is that the issue here? Sławomir Biały (talk) 12:43, 6 April 2012 (UTC)

Well, in part. Perhaps I should provide more examples.

$z(x)={\frac {d}{dx}}y(x)$

$z(x)={\frac {d}{du}}y(x){\text{ , }}x=f(u)$

$z(v)={\frac {d}{dx}}y(x){\text{ , }}x=g(v)$

$z(v)={\frac {d}{du}}y(x){\text{ , }}x=f(u)=g(v)$.

I want to notationally distinguish between all of these in a concise manner. Is there an accepted way of doing this?--Leon (talk) 12:54, 6 April 2012 (UTC)

Take a look at the Total derivative article. What you're actually doing is composing functions and then applying the chain rule. You can see derivatives, and indeed differential expressions, as functions defined on a jet space. You're just composing a function from the jet space with another function. Why not just make these domains and these compositions explicit? — Fly by Night (talk) 12:59, 6 April 2012 (UTC)

In the first case you'd write $z=y_{x}(x)$, in the second $(y\circ f)_{u}(u)$, in the third $(y\circ g)_{v}(v)$. I'm not sure what the fourth one means. — Fly by Night (talk) 13:09, 6 April 2012 (UTC)

Yes, I am composing functions and applying the chain rule. But the reason I don't want to write this out explicitly each time is that the functions are not always as simple as these. I chose these examples for the sake of clear illustration alone. Here are some different examples, using a function of two variables $Q$, which I need not define.

$z(x)=Q(y(x),{\frac {d}{dx}}y(x))$

$z(x)=Q(y(x),{\frac {d}{du}}y(x)){\text{ , }}x=f(u)$

$a(v)=z(g(v))=Q(y(x),{\frac {d}{dx}}y(x)){\text{ , }}x=g(v)$

$a(v)=z(g(v))=Q(y(x),{\frac {d}{du}}y(x)){\text{ , }}x=f(u)=g(v)$.

In response to Fly by Night's comment, I have amended my notation. Each $a(v)$ is simply a reparametrization of the corresponding $z(x)$.

I would like to distinguish these concisely. Any ideas?--Leon (talk) 13:28, 6 April 2012 (UTC)

I feel like my original question has not been adequately addressed. In each of these expressions, on the left hand side you have the x variable, and on the right you have derivatives of x. Do you intend that derivatives of x can be expressed as functions of x? More examples will not help. You need to say what you mean by these equations. Sławomir Biały (talk) 13:40, 6 April 2012 (UTC)

Yes, I do intend that the derivatives of $y$ with respect to either $x$ or $u$ can be expressed as functions of either $x$ or $v$. Thus the first two expressions are functions of $x$ and the latter two functions of $v$. $f(u)$ and $g(v)$ are smooth functions, and their inverses always exist and are smooth also.--Leon (talk) 13:54, 6 April 2012 (UTC)

This will only make sense if you have a particular function x in mind. Otherwise, as variables the x on the left and x on the right signify different things. On way to resolve such difficulties is to use differential notation. Sławomir Biały (talk) 14:25, 6 April 2012 (UTC)

Another approach would be to stop using variables altogether, and write everything in terms of function composition, inverse functions, and so forth (which I think is like what FbN is suggesting you do.) Sławomir Biały (talk) 14:31, 6 April 2012 (UTC)

I would write

$z\left(x,{\frac {dx}{du}}\right)={\frac {d}{du}}y(x)$.

or

$z\left(x,x\prime \right)={\frac {d}{du}}y(x)$.

Dmcq (talk) 14:50, 6 April 2012 (UTC)

You contrast

$z(x)={\frac {d}{dx}}y(x)$

versus

$z(x)={\frac {d}{du}}y(x){\text{ , }}x=x(u)$.

This comes up a lot in economics. In the former equation, x is viewed as the ultimately exogenous variable, while in the latter x is itself determined by a more fundamental variable u. It seems common to me to write the first equation as shown, since dy/dx is a function of x (hence z is a function of x), but to rewrite the left side of the second equation as g(u) since (a) it's a different function and so should not be given the same name z and (b) g is the chain rule product (dy/dx) as a function of x(u) times (dx/du) as a function of u -- so g is actually a function of u, not of x. Duoduoduo (talk) 18:03, 6 April 2012 (UTC)

Ordinarily, I would entirely agree with Duoduoduo. They are not the same functions, and it is dangerous to suggest that they are. However, perhaps if I provide some additional context, my motives for giving them similar names should become clear.

(Some of) the functions I am looking at evaluate the kinetic energy of parametric motions, with some parameter, which may or may not be the argument of the function, being regarded as time. The function mapping the argument of the function to the parameter regarded as time is always monotonically increasing and continuous. I want to distinguish different parametrizations of the (otherwise) same motion. By different parametrizations, I mean both (a) what variable the motion is parametrized in (the argument of the function) and (b) what parameter is being regarded as time.

So, is there a convention for doing this? --Leon (talk) 11:30, 7 April 2012 (UTC)

One way is to write a little subscript giving the parameter. You can see something like this for instance in Planck's law where the spectral radiance is expressed for various different parameters like $B_{\nu }(T)$ or $B_{\lambda }(T)$ Dmcq (talk) 12:08, 7 April 2012 (UTC)

There are two ways looking at this. One is the language of functions and function values, and the other is the language of variables and equations. The notation f'(x) refers to functions: f and f', and to the function value f'(x). The Leibniz notation dy/dx refers to variables: x,y,dx,dy. Above you have mixed the two languages and so you are in trouble. The endless discussion about the interpretation of differentials is waste of time. Take it formally and axiomatically. If x is a variable, then so is the differential dx. The rules are:

1. d(x+y)=dx+dy
2. d(xy)=dxy+xdy (where dxy always means (dx)y=dx·y)
3. dx=0 iff x is constant
4. dx is constant iff x is an independent variable, (and so ddx=d²x=0).

Now you can differentiate!

d(x²)=d(xx)=dxx+xdx=2xdx

Another example.

x=at²/2+bt+c

dx=dat²/2+atdt+dbt+bdt+dc.

d²x=d²at²/2+2datdt+adt²+atd²t+d²bt+2dbdt+bd²t+d²c.

If a,b,c are constants, and t is independent, you have

d²x=adt²

Your two examples are: zdx=dy and wdu=dy. Bo Jacoby (talk) 07:14, 7 April 2012 (UTC).

Leon says: The function mapping the argument of the function to the parameter regarded as time is always monotonically increasing and continuous. I want to distinguish different parametrizations of the (otherwise) same motion. In your equations

$z(x)={\frac {d}{dx}}y(x)$

$z(x)={\frac {d}{du}}y(x){\text{ , }}x=x(u)$

you could use one combined notation: $y(x(u))$ where u is time. Then you could write dy/dx for the one, and dy/du, equivalently (d/y/dx)(dx/du), for the other. Duoduoduo (talk) 16:11, 7 April 2012 (UTC)