# Is the gradient really just a first-order approximation?

• psholtz
In summary, the conversation discusses the use of differential forms and their relation to first-order approximations. The difference between dT and ∆T is also mentioned, with dT being a 1-form and ∆T being an actual difference between two points. The conversation also highlights the importance of using integrals to pass back into the world of "ordinary" mathematical objects.

#### psholtz

In physics texts, its customary to write (and even to define the gradient as) the following:

$$dT = (\nabla T) \cdot dl$$

Working in Cartesian coordinates, we can expand this as follows:

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

But the above equation is really just a "first-order" approximation for the differential dT, is it not? Working just in the differential dx, a more "complete" way to write this expression would be:

$$dT \approx \frac{\partial T}{\partial x} dx + \frac{1}{2} \frac{\partial ^2T}{\partial x^2} (dx)^2 + \frac{1}{6} \frac{\partial ^3T}{\partial x^3} (dx)^3 + ...$$

along with the additional, commensurate expansions in terms of the y and z variables as well.

So while we can define equations in terms of gradient, and work with gradients, etc, is it fair to say that all these gradient expressions are really just giving us answers which are first-order approximations? And that if we desired more precise answers, we would have to delve deeper into the higher-order terms?

Nope; $dT = (\nabla T) \cdot d\ell$ is an exactly correct equation of differential forms.

You're thinking about differential approximation, which says $\Delta T \approx (\nabla T) \cdot \Delta \ell$, or more explicitly,

$$T(\ell + v) - T(\ell) = \nabla_v T(\ell) + \frac{1}{2} \nabla_v \nabla_v T(\ell) + \cdots$$

You have to remember that differential forms are not numbers (or number-valued fields); they're differential forms. Their very definition is arranged so that these "first-order" equations are exactly correct. If you want to pass back into the world of "ordinary" mathematical objects, you have to use some sort of integral.

That said, people do often use expressions like dT when they really mean $\Delta T$.

They are first order approximations of ∆T, not of dT.

dT is exactly equal to $\partial_{x}T dx + \partial_{y}T dy + \partial_{z}T dz$

dx said:
They are first order approximations of ∆T, not of dT.

dT is exactly equal to $\partial_{x}T dx + \partial_{y}T dy + \partial_{z}T dz$

Where can I read up on the difference between dT and $$\Delta T$$?

I don't think I've ever (formally) encountered that distinction before..

So in my first equation above, what I should have written, should have been:

$$\Delta T \approx \frac{\partial T}{\partial x} \Delta x +\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2 + \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...$$

Yes?

psholtz said:
So in my first equation above, what I should have written, should have been:

$$\Delta T \approx \frac{\partial T}{\partial x} \Delta x +\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2 + \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...$$

Yes?

Yes.

∆T is an actual difference of T between two points. dT on the other hand is what is called a 1-form. You can read about 1-forms in books about differentiable manifolds and differential geometry.

## 1. What is the gradient?

The gradient is a mathematical concept used in vector calculus to describe the rate of change of a function at a specific point. It is represented by a vector that points in the direction of the steepest increase of the function, and its magnitude represents the steepness of the increase.

## 2. How is the gradient calculated?

The gradient is calculated by taking the partial derivatives of a multivariable function with respect to each of its independent variables and combining them into a vector. This vector represents the direction and magnitude of the steepest increase of the function at a specific point.

## 3. Is the gradient always a first-order approximation?

No, the gradient is not always a first-order approximation. It is a first-order approximation only when the function is differentiable at the point of interest. If the function is not differentiable, the gradient cannot accurately represent the rate of change at that point.

## 4. What does it mean for the gradient to be a first-order approximation?

When the gradient is a first-order approximation, it means that it is a close estimation of the actual rate of change of the function at a specific point. It is not an exact value, but it becomes more accurate as the distance from the point of interest decreases.

## 5. Are there any limitations to using the gradient as a first-order approximation?

Yes, there are limitations to using the gradient as a first-order approximation. It is only accurate near the point of interest and becomes less accurate as the distance from the point increases. Additionally, it cannot be used for functions that are not differentiable at the point of interest.