I am currently studying the textbook Introduction to Tensor Analysis and the Calculus of Moving Surfaces, by Pavel Grinfeld. On page 3, the author begins with the following motivating example:
What is the gradient of a function $F$ and a point $P$? You are familiar with two definitions, one geometric and one analytical. According to the geometric definition, the gradient $\nabla F$ of $F$ is the vector that points in the direction of the greatest increase of the function $F$, and its magnitude equals the greatest rate of increase. According to the analytical definition that requires the presence of a coordinate system, the gradient of $F$ is the triplet of numbers
$$\nabla F = \left( \dfrac{\partial{F}}{\partial{x}}, \dfrac{\partial{F}}{\partial{y}} \right) \tag{1.1}$$
Are the two definitions equivalent in some sense? If you believe that the connection is
$$\nabla F = \dfrac{\partial{F}}{\partial{x}}\mathbf{i} + \dfrac{\partial{F}}{\partial{y}} \mathbf{j}\tag{1.2}$$
where $\mathbf{i}$ and $\mathbf{j}$ are the coordinate basis, you are in for a surprise! Equation (1.2) can only be considered valid if it produces the same vector in all coordinate systems. You may not be familiar with the definition of a coordinate basis in general curvilinear coordinates, such as spherical coordinates. The appropriate definition will be given in Chap. 5. However, equation (1.2) yields different answers even for the two coordinate systems in Fig. 1.1.
For a more specific example, consider a temperature distribution $T$ in a two-dimensional rectangular room. Refer the interior of the room to a rectangular coordinate system $x$, $y$ where the coordinate lines are one meter apart. This coordinate system is illustrated on the left of Fig. 1.1. Express the temperature field in terms of these coordinates and construct the vector gradient $\nabla T$ according to equation (1.2).
Alternatively, refer the interior of the room to another rectangular system $x'$, $y'$, illustrated on the right of Fig. 1.1), whose coordinate lines are two meters apart. For example, at a point where $x = 2$, the new coordinate $x'$ equals $1$. Therefore, the new coordinates and the old coordinates are related by the identities
$$x = 2x' \ \ \ \ \ y = 2y' \tag{1.3}$$
Now repeat the construction of the gradient according to equation (1.2) in the new coordinate system: refer the temperature field to the new coordinates, resulting in the function $F'(x', y')$, calculate the partial derivatives and evaluate the expression in equation (1.2), except with “primed” elements:
$$(\nabla T)' = \dfrac{\partial{F'}}{\partial{x'}} \mathbf{i}' + \dfrac{\partial{F'}}{\partial{y'}} \mathbf{j}' \tag{1.4}$$
How does $\nabla T$ compare to $(\nabla T)'$? The magnitudes of the new coordinate vectors $\mathbf{i}'$ and $\mathbf{j}'$ are double those of the old coordinate vectors $\mathbf{i}$ and $\mathbf{j}$. What happens to the partial derivatives? Do they halve (this would be good) or do they double (this would be trouble)?
They double! This is because in the new coordinates, quantities change twice as fast. In evaluating the rate of change with respect to, say, $x$, one increments $x$ by a small amount $\Delta x$, such as $\Delta x = 10^{-3}$, and determines how much the function $F(x, y)$ changes in response to that small change in $x$. When one evaluates the partial derivative with respect to $x'$ in the new coordinate system, the same increment in the new variable $x'$ is, in physical terms, twice as large. It results in twice as large a change $\Delta F'$ in the function $F'(x', y')$. Therefore, $\Delta F' / \Delta x'$ is approximately twice as large as $\Delta F / \Delta x$ and we conclude that partial derivatives double:
$$\dfrac{\partial{F'(x', y')}}{\partial{x'}} = 2\dfrac{\partial{F(x, y)}}{\partial{x}} \tag{1.5}$$
The first point I'm seeking clarification on is a minor problem: (1.1) is actually a doublet, right? The author refers to it as a triplet.
The point that I'm uncomfortable with is the idea that $\Delta F' / \Delta x'$ is approximately twice as large as $\Delta F / \Delta x$. If we're increasing the denominator by $10^{-3}$, then why doesn't the numerator also increase by approximately the same amount, so that the result approximately cancels out to result in approximately the same value as before? I'm not satisfied that the author explains this clearly enough to make it evident.
I would greatly appreciate it if people would please take the time to clarify this.
EDIT
The first +50 bounty has expired, so I might as well post my thoughts so far, so that it might help others formulate an answer.
In trying to understand this, I had two streams of thought. The first stream of thought was focused on trying to understand (1.5), whilst the second stream of though was focused on trying to understand what I alluded to in my main post (which was the ideas behind the $\Delta F / \Delta x$ and $\Delta F' / \Delta x'$). It seems to me that an answer to the latter type of problem would naturally be suited to a real-analytic approach, so that's what I've been trying to do. I tried to understand the former by just using straightforward manipulations of the functions and their various implications. The main problem is that I don't have a lot of experience in real analysis, so I'm very unsure of my reasoning.
I am not claiming that this post is an answer to the questions I have; rather, it is an expansion of my main post. In this expansion, I present my thoughts in trying to understand this section of the textbook, as well as some new questions that arose in the process of my reasoning.
I am now going to post a second, larger bounty for this post. In addition to my original questions, I would appreciate it if people would review my reasoning here, and perhaps build upon it in their answer, and answer the questions that arose in this reasoning.
The first problem I see here is that we don't know the form of the function $F(x, y)$. It seems to me that this would have made things much easier. So we don't know whether, for instance, the function takes the form $F(x, y) = x + y$, and, therefore, $\nabla F = (1, 1)$, or whether the function takes some other form. I also wondered if we could use the antiderivative to recover the form of the function somehow, as is done to solve certain forms of differential equations, but I'm not sure whether that is possible/helpful in this situation. On the other hand, based on what I can infer from reasoning about this situation, it may not matter what form the function takes, or, rather, we may not need to know the precise form of the function, since we are told that the function form is such that $x = 2x', y = 2y'$ and $\dfrac{\partial{F'(x', y')}}{\partial{x'}} = 2\dfrac{\partial{F(x, y)}}{\partial{x}}$, which might already give us enough information to understand what's going on here without needing the precise form of the function.
The first part of my thoughts:
So, we have that the change of coordinates
$$x = 2x', \ \ \ y = 2y'$$
leads to the new function
$$F'(x', y') = F' \left( \dfrac{x}{2}, \dfrac{y}{2} \right) = \dfrac{1}{2} F' \left( x, y \right).$$
Side-note: This assumes that $F' \left( \dfrac{x}{2}, \dfrac{y}{2} \right) = \dfrac{1}{2} F' \left( x, y \right)$. This begs the question: If we have a function of the form $F'(x', y') = F'\left( \dfrac{x}{2}, \dfrac{y}{2} \right)$, where $x = 2x'$ and $y = 2y'$ are a change of variables, then what conditions must be satisfied for us to be able to say that $F'(x', y') = F'\left( \dfrac{x}{2}, \dfrac{y}{2} \right) = \dfrac{1}{2} F'\left( x, y \right)$? I ask that question here, but have, as of yet, received no answers. I proceed assuming that it is valid in this case.
EDIT: This property of functions seems to be referred to as "homogeneity" (see the aforementioned question for further details). So I proceed assuming that $F'$ is a homogeneous function.
Taking the partial derivatives with respect to $x$, we have
$$\dfrac{\partial{F'(x', y')}}{\partial{x'}} = \dfrac{1}{2} \dfrac{\partial{F'\left( x, y \right)}}{\partial{x}'} \Rightarrow 2 \dfrac{\partial{F'(x', y')}}{\partial{x'}} = \dfrac{\partial{F'\left( x, y \right)}}{\partial{x}'}.$$
I wonder if $\dfrac{\partial{F'\left( x, y \right)}}{\partial{x}'} = \dfrac{\partial{F\left( x, y \right)}}{\partial{x}}$? Because then we would have that
$$\dfrac{\partial{F'(x', y')}}{\partial{x'}} = \dfrac{1}{2} \dfrac{\partial{F'\left( x, y \right)}}{\partial{x}'} \Rightarrow 2 \dfrac{\partial{F'(x', y')}}{\partial{x'}} = \dfrac{\partial{F'\left( x, y \right)}}{\partial{x}'} = \dfrac{\partial{F\left( x, y \right)}}{\partial{x}} \\ \therefore 2 \dfrac{\partial{F'(x', y')}}{\partial{x'}} = \dfrac{\partial{F\left( x, y \right)}}{\partial{x}},$$
as required.
The second part of my thoughts:
Using the definition of partial differentiation, I think we can proceed as follows:
$$\begin{align} \dfrac{\Delta F}{\Delta x} = \dfrac{\partial{F(x, y)}}{\partial{x}} &= \lim_{\Delta x \to 10^{-3}} \dfrac{F(x + \Delta x, y) - F(x, y)}{\Delta x} \\ &= \lim_{\Delta (2x') \to 10^{-3}} \dfrac{F(2x' + \Delta (2x'), 2y') - F(2x', 2y')}{\Delta (2x')} \\ &= \lim_{\Delta (x') \to \frac{10^{-3}}{2}} \dfrac{2[F(x' + \Delta (x'), y') - F(x', y')]}{2\Delta (x')} \ \ \text{(Assuming that $F$ is a **homogeneous function**.)} \\ &= \lim_{\Delta (x') \to \frac{10^{-3}}{2}} \dfrac{F(x' + \Delta (x'), y') - F(x', y')}{\Delta (x')} \end{align}$$
I'm not sure whether I made an error here or whether it's just a dead-end, but it seems to that proceeding in this way, using the definition of partial differentiation, is a good way to gain an understanding of what the author meant.
EDIT 2:
My new reasoning is as follows:
The denominator is $\Delta x$. And since $x = 2x' \Rightarrow x' = \dfrac{x}{2}$, we have that $\dfrac{\Delta F'}{\Delta x'} = \dfrac{\Delta F'}{\Delta \frac{x}{2}} = \dfrac{2\Delta F'}{ \Delta x}$. If $\Delta x = 10^{-3}$, then $\dfrac{2\Delta F'}{10^{-3}}$. So, by this, we can only say that $\dfrac{\Delta F'}{\Delta x'}$ is approximately double $\dfrac{\Delta F}{\Delta x}$ if $F' = F$, so that $\dfrac{\Delta F'}{\Delta x'} = \dfrac{2\Delta F'}{10^{-3}} = \dfrac{2\Delta F}{10^{-3}}$.
I find it difficult to deciver all your "thoughts", so I'll focus on the concrete question you ask. In the one dimensional case, what you are looking for is the chain rule. $\frac{\partial F(u(x))}{\partial x} = \frac{\partial F(u)}{\partial u}\frac{\partial u(x)}{\partial x}$. In this higher dimensional example, $F'(x',y') = F(u(x'),u(y'))$ where $x = u(x') = 2x'$ (similar for $y$). So $\frac{\partial F'(x', y')}{\partial x'} = \frac{\partial F(u(x'),u(y'))}{\partial x'} = \frac{\partial F(u(x'),u(y'))}{\partial (u(x'))}\frac{u(x')}{\partial x'} = \frac{\partial F(x, y)}{\partial x} \cdot 2$. The reason that the text uses approximate, is that when you calculate the approximate derivative $\frac{\Delta F}{\Delta x}$ for $\Delta x = 10^{-3}$ you will not get exactly 2.
I noticed in your thoughts that you asked whether $F(x/2, y/2) = F(x,y)/2$. This requiress some knowledge about the function $F$ which you do not have here.