
020812, 01:09 AM #1
Independent variables and partial differentiation
I hope this makes sense... I'm not familiar with the terminology.
There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
There is a set of points (a,b) in (x,y) defined by the following function:
Where c is an independent variable.
 What is db/dc? Is it zero? Is it undefined?
The points (a,b) are transformed to (a',b').
 Can we say from the above whether b' and c are independent?
 Is there a rigorous definition of what 'independent' means?
 Does it depend on the particular equation?
Say we use the above to derive an expression like this:
 Are b' and c independent variables in that expression?
 Can I simply treat b' as a constant when finding , or do I need to treat it as a function of c?

020812, 03:21 AM #2
I'm not sure whether I can answer this question in the abstract. So let me see if I can construct a simple example...
I'll use
There is a set of points (a,b) in (x,y) defined by the following function:
Where c is an independent variable.
What is db/dc? Is it zero? Is it undefined?
This suggests to me that db/dc=0.
The points (a,b) are transformed to (a',b').
so
 Can we say from the above whether b' and c are independent?
Is there a rigorous definition of what 'independent' means?
Say we use the above to derive an expression like this:
 Are b' and c independent variables in that expression?
 Can I simply treat b' as a constant when finding , or do I need to treat it as a function of c?

020812, 07:02 PM #3
Thanks James,
But the same logic suggests that dc/db=0, ie db/dc is infinite or something.
It seems to me that db/dc=0 would imply that b can't vary:
...but I'm not enough of a mathematician to be certain that this proof is valid.
Doesn't it mean that you can vary one variable without affecting the other?
Do you mean ? Using my example we have
No, because
In that function, it seems that b' is an dependent variable, depending on b and c. (Or are they three interdependent variables?)
But in the function:
Where b is not involved, it seems to me that b' is independent of c, because they can vary without affecting the other.
For any given value of c, b' can have any value (because b can have any value).
For any given value of b', c can have any value (because b can have any value).
Does that make sense, or am I right off the track?
Doesn't the notation mean that all variables other than c are to be held constant? That would include b', wouldn't it?
We can hold b constant or b' constant, but not both (because that would imply holding c constant).
b is not in the equation, but b' was earlier derived/defined as a function of b and c.
So, do we hold b' constant, because b' is in the equation and b isn't?
Or do we hold b constant, for some other reason?

020812, 07:29 PM #4
Consider a horizontal line, which is really what you're describing.
The value of Y (or in this case b) is both constant and independent of the value of X (or in this case c).
Or to put it another way, if you set the variable c to some constant, you make it independent of b.
I'm not sure it neccessarily follows that it's true of all independent variables, however, that the derivative is 0.

020812, 07:37 PM #5

020812, 08:01 PM #6

020812, 09:17 PM #7
What if y = z, and z is independent of x?
In general, if y=f(x,z), then
So, if y=f(x,z)=z, then
so, we need to know how z varies with x to find the total derivative of y with respect to x.
If z is independent of x, then it seems to me that we must have , and it then follows also that .

020912, 12:29 AM #8
I think that should be
So, if y=f(x,z)=z, then
so, we need to know how z varies with x to find the total derivative of y with respect to x.
Maybe I'm just not getting the concept of independence.
It seems to me that independence is not an absolute thing, but is relative to the equation.
eg:
y = x+z
implies y depends on independent variables x and z.
But rewriting it as:
x = y  z
implies that x depends on independent variables y and z.
Is it better in this case to say that x, y, and z are interdependent?Last edited by Pete; 020912 at 12:36 AM.

020912, 01:13 AM #9
 Posts
 3,741
Penrose states in his book "The Road to Reality", that for a pair of independent variables, say u and v, expressing that some quantity, say Z, is a function of u and v but independent of v is given by: dZ/dv = 0. It also says that for any value of u, Z is constant in v, and so Z depends only on u.
He also makes the point that this only holds locally, it might not be true for all Z.

020912, 01:43 AM #10

020912, 01:55 AM #11
 Posts
 3,741
Originally Posted by Pete
Ed: but the principle holds for ordinary differentiation: if Z is constant in v the "slope" dZ/dv is zero, and v is independent.Originally Posted by PeteLast edited by arfa brane; 020912 at 02:56 PM.

020912, 04:51 PM #12
If you fix c, you'll get a curve a=f(b,c) in general. So c cannot be taken as the only independent variable. Rather, we should consider b as a function of a and c. Differentiate the equation a=f(b,c) with respect to c to get
so
or in short ,
provided the latter term can be inverted.
In effect, it is the same as holding a fixed, and defining the implicit function b(c) by a=f(b,c).
 Not from the above, bu you can choose to consider b' and c as independent, because you have one equation and 3 variables.
 It means you can freely choose the values of the variables, and still have freedom to satisfy the given equation/relation.
 Yes, but in a weak way. If you have 1 equation involving 3 variables, you can choose 2 of them to be independent, unless some singular situations arise (look up implicit function theorem).
 You can choose so.
 Yes.

020912, 05:28 PM #13
Thanks temur.
What about the total derivative ?
Does it have a value?
Is it even meaningful?
 Not from the above, bu you can choose to consider b' and c as independent, because you have one equation and 3 variables.
 It means you can freely choose the values of the variables, and still have freedom to satisfy the given equation/relation.
 Yes, but in a weak way. If you have 1 equation involving 3 variables, you can choose 2 of them to be independent, unless some singular situations arise (look up implicit function theorem).
One more...
We have:
Is this correct:
Or this:
My understanding is that the use of the expression \frac{dx}{dy} implies that z is a function of y, so x is implicitly a function of only y, and that the first process is correct.
But I'm not certain, and there's disagreement about in another thread (it's a debate thread, so only Tach and I can post there).

020912, 05:53 PM #14
It is not meaningful. In order to define this, you need to specify how a depends on c. The "total" derivative is taken along a given curve. Without a curve (i.e., a dependence such as a=a(c)), there is no "total" derivative.
Again you need to know how z depends on y. Otherwise it has no meaning.

020912, 07:58 PM #15
Right, we can't even talk about unless z depends on y...
So, if:
Then is this correct?
Thanks againLast edited by Pete; 020912 at 08:11 PM.

020912, 08:30 PM #16

020912, 08:34 PM #17
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