Independent variables and partial differentiation

**Pete** · 02-08-12, 01:09 AM

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'):
$x' = g_1(x, y) \\ y' = g_2(x, y)$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
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:
$a' = h(a', b', c)$
- Are b' and c independent variables in that expression?
- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$ , or do I need to treat it as a function of c?

**James R** · 02-08-12, 03:21 AM

I'm not sure whether I can answer this question in the abstract. So let me see if I can construct a simple example...

Originally Posted by Pete

There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
$x' = g_1(x, y) \\ y' = g_2(x, y)$

I'll use

$x' = x + y\\ y'=x-y$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
Where c is an independent variable.

I'll take

$a=e^{bc}$

What is db/dc? Is it zero? Is it undefined?

You said c is an independent variable, by which I assume you mean c is independent of b. That is, we can vary b as much as we want to without affecting c, and vice versa.

This suggests to me that db/dc=0.

The points (a,b) are transformed to (a',b').

Using my example

$a=e^{bc} \\ x' = x + y\\ y'=x-y$

so

$a'=a + b = e^{bc} + b\\ b'=a-b = e^{bc} - b$

- Can we say from the above whether b' and c are independent?

It seems clear to me from the example that b' depends on c.

Is there a rigorous definition of what 'independent' means?

Doesn't it mean that you can vary one variable without affecting the other?

Say we use the above to derive an expression like this:
$a' = h(a', b', c)$

Do you mean $a'=h(a,b',c)$ ? Using my example we have

$a'= e^{bc} + b = b' + 2b = b' + 2\frac{\ln a}{c}$

- Are b' and c independent variables in that expression?

No, because $b'=e^{bc}-b$

- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$ , or do I need to treat it as a function of c?

Does't the notation $\frac{\partial a'}{\partial c}$ mean that all variables other than c are to be held constant? That would include b', wouldn't it?

**Pete** · 02-08-12, 07:02 PM

Thanks James,

Originally Posted by James R

You said c is an independent variable, by which I assume you mean c is independent of b. That is, we can vary b as much as we want to without affecting c, and vice versa.

This suggests to me that db/dc=0.

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:
$\begin{align} \frac{db}{dc} &= 0 \\ \int \frac{db}{dc} dc &= \int 0 dc \\ b + C_1 &= C_2 \\ b &= C \end{align}$
...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?

That's my understanding, but I was wondering if there was any formal definition.

Do you mean $a'=h(a,b',c)$ ? Using my example we have

$a'= e^{bc} + b = b' + 2b = b' + 2\frac{\ln a}{c}$

In the specific case this question is based on, the expression was $a'=h(a',b',c)$ , but the key point is that the expression involves b' and c, and doesn't involve b, so your example is good.

No, because $b'=e^{bc}-b$

This is where I get confused.
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:
$a'= b' + 2\frac{\ln a}{c}$
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 $\frac{\partial a'}{\partial c}$ mean that all variables other than c are to be held constant? That would include b', wouldn't it?

I think so... but I'm getting confused by b.
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?

**Trippy** · 02-08-12, 07:29 PM

Originally Posted by Pete

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:
$\begin{align} \frac{db}{dc} &= 0 \\ \int \frac{db}{dc} dc &= \int 0 dc \\ b + C_1 &= C_2 \\ b &= C \end{align}$
...but I'm not enough of a mathematician to be certain that this proof is valid.

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.

**Pete** · 02-08-12, 07:37 PM

Originally Posted by Trippy

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 necessarily follows that it's true of all independent variables, however, that the derivative is 0.

So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?

**Trippy** · 02-08-12, 08:01 PM

Originally Posted by Pete

So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?

I would have said so, yes.

If you consider

Y=0
Y=6
Y= $\frac{\sqrt{2}}{\pi}$
and Y=999999999999999999999999999999999999999999999999 999999999999999

In all cases $\frac{dy}{dx}=0$

**James R** · 02-08-12, 09:17 PM

Originally Posted by Pete

So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?

What if y = z, and z is independent of x?

In general, if y=f(x,z), then

$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{\partial z}{\partial x}$

So, if y=f(x,z)=z, then

$\frac{dy}{dx} = 0 + 1\times \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}$

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 $\frac{\partial z}{\partial x} = 0$ , and it then follows also that $\frac{dy}{dx}=0$ .

**Pete** · 02-09-12, 12:29 AM

Originally Posted by James R

What if y = z, and z is independent of x?

In general, if y=f(x,z), then
$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{\partial z}{\partial x}$

I think that should be
$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{dz}{dx}$

So, if y=f(x,z)=z, then

$\frac{dy}{dx} = 0 + 1\times \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}$

so, we need to know how z varies with x to find the total derivative of y with respect to x.

Right... but "how z varies with x" implies some relationship between z and x. It doesn't make sense if z and and are independent, because when you vary x, z can do anything.

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?

**arfa brane** · 02-09-12, 01:13 AM

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.

**Pete** · 02-09-12, 01:43 AM

Originally Posted by arfa brane

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

...so Z is really only a function of u?

... is given by: dZ/dv = 0.

Are you sure it didn't say $\partial Z/\partial v = 0$ ?

It also says that for any value of u, Z is constant in v, and so Z depends only on u.

That looks very much like a description of $\partial Z/\partial v = 0$ .

**arfa brane** · 02-09-12, 01:55 AM

Originally Posted by Pete

Are you sure it didn't say $\partial Z/\partial v = 0$ ?

Yes, sorry, it actually concerns the Cauchy-Riemann equations. Partial derivatives are of one variable by definition.

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 Pete

so Z is really only a function of u?

For a function of two variables, Z(u,v), there is no problem with Z being constant in v but still a function of u and v. The v could be contour lines of constant height (latitude) above the equator of a sphere, say, so they have gradients = 0.

**temur** · 02-09-12, 04:51 PM

Originally Posted by Pete

There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
$x' = g_1(x, y) \\ y' = g_2(x, y)$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
Where c is an independent variable.
- What is db/dc? Is it zero? Is it undefined?

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

$ 0=\frac{\partial f}{\partial b} \frac{\partial b}{\partial c} + \frac{\partial f}{\partial c}, $
so
$ \frac{\partial b}{\partial c} = - \frac{\partial f}{\partial c} \left(\frac{\partial f}{\partial b}\right)^{-1}, $ or in short $b_c = - f_c/f_b$ ,
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).

Originally Posted by Pete

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?

- 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).

Originally Posted by Pete

Say we use the above to derive an expression like this:
$a' = h(a', b', c)$
- Are b' and c independent variables in that expression?
- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$ , or do I need to treat it as a function of c?

- You can choose so.
- Yes.

**Pete** · 02-09-12, 05:28 PM

Originally Posted by temur

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

$ 0=\frac{\partial f}{\partial b} \frac{\partial b}{\partial c} + \frac{\partial f}{\partial c}, $
so
$ \frac{\partial b}{\partial c} = - \frac{\partial f}{\partial c} \left(\frac{\partial f}{\partial b}\right)^{-1}, $ or in short $b_c = - f_c/f_b$ ,
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).

Thanks temur.
What about the total derivative $\frac{db}{dc}$ ?
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).

Thanks, I'll exercise due care

One more...
We have:
$\begin{align} x &= f(y,z) \\ \frac{dx}{dy} &= 0 \end{align}$

Is this correct:
$\begin{align} \int \frac{dx}{dy} dy &= \int 0 dy \\ x + C_1 &= C_2 \\ x &= C \end{align}$

Or this:
$\begin{align} \int \frac{dx}{dy} dy &= \int 0 dy \\ x + g(z) &= C_2 \end{align}$

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).

**temur** · 02-09-12, 05:53 PM

Originally Posted by Pete

What about the total derivative $\frac{db}{dc}$ ?
Does it have a value?
Is it even meaningful?

It is not meaningful. In order to define this, you need to specify how a depends on c. The "total" derivative $\frac{db}{dc}$ is taken along a given curve. Without a curve (i.e., a dependence such as a=a(c)), there is no "total" derivative.

Originally Posted by Pete

We have:
$\begin{align} x &= f(y,z) \\ \frac{dx}{dy} &= 0 \end{align}$

Is this correct:
$\begin{align} \int \frac{dx}{dy} dy &= \int 0 dy \\ x + C_1 &= C_2 \\ x &= C \end{align}$

Or this:
$\begin{align} \int \frac{dx}{dy} dy &= \int 0 dy \\ x + g(z) &= C_2 \end{align}$

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).

Again you need to know how z depends on y. Otherwise it has no meaning.

**Pete** · 02-09-12, 07:58 PM

Right, we can't even talk about $\frac{dx}{dy}$ unless z depends on y...

So, if:
$\begin{align} x &= f(y,z) \\ z &= g(y) \\ \frac{dx}{dy} &= 0 \end{align}$

Then is this correct?
$\begin{align} \int \frac{dx}{dy} dy &= \int 0 dy \\ x + C_1 &= C_2 \\ x &= C \end{align}$

Thanks again

**temur** · 02-09-12, 08:30 PM

Originally Posted by Pete

Right, we can't even talk about $\frac{dx}{dy}$ unless z depends on y...

So, if:
$\begin{align} x &= f(y,z) \\ z &= g(y) \\ \frac{dx}{dy} &= 0 \end{align}$

These conditions imply only that the function f(y,z) is constant along the curve z=g(y). For example, take g(y)=0. Then any of f(y,z)=1, f(y,z)=zy, f(y,z)=sin(z) satisfy the equation $\frac{dx}{dy} = f_y+f_zg_y=0$ .

**Pete** · 02-09-12, 08:34 PM

*click*

Thanks!

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