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I'm not sure if this is a duplicate (it might even just be silly), but why isn't the differential of some function $f\left(x_1,x_2\right)$, \begin{align}df&=\frac{\partial f}{\partial x_1}\:dx_1+\frac{\partial f}{\partial x_2}\:dx_2,\end{align} defined as \begin{align}df=\frac{\partial f}{\partial x_1}\:dx_1+\frac{\partial f}{\partial x_2}\:dx_2+\frac{\partial^2f}{\partial x_1^2}\:dx_1^2+\frac{\partial^2 f}{\partial x_1\partial x_2}\:dx_1\:dx_2+\frac{\partial^2 f}{\partial x_2\partial x_1}\:dx_2\:dx_1+\frac{\partial^2 f}{\partial x_2^2}\:dx_2^2,\end{align} or with even more terms, so as to resemble a Taylor series? I'm curious about general functions $f\left(x_1,x_2,\cdots,x_n\right)$ too, I just don't think I'd have enough room to write out a three variable "quadratic" term, etc.

Has this concept ever been explored before?

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  • $\begingroup$ I would assume it's at least partly related to the fact that we like the definition of the gradient $$\left(\dfrac{\partial}{\partial x_{1}},\dfrac{\partial}{\partial x_{2}},\ldots,\dfrac{\partial}{\partial x_{n}}\right),$$ and we can express the "usual definition" in terms of this gradient function, which in turn gives a nice interpretation. An additional reason might be that the rule you're proposing seems to come out of nowhere - it's not a very natural generalization of the chain rule from single-variable calculus. Finally, a very important reason is that we like derivatives to be linear. $\endgroup$
    – Will R
    Oct 9, 2015 at 23:15
  • $\begingroup$ In fact, this last reason is so important that I'm tempted to post it as an answer: part of the point of the derivative is that it gives you the best linear approximation to the given function near a given point. $\endgroup$
    – Will R
    Oct 9, 2015 at 23:23
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    $\begingroup$ @WillR I beat you to it. :) $\endgroup$
    – amd
    Oct 9, 2015 at 23:24
  • $\begingroup$ @WillR I understand the whole "linear approximation" scheme. Not from a differential geometry or topological perspective as I haven't taken any of those advanced courses yet, but what if I'm looking for a non-linear approximation with higher order terms? What would it look like and how does one use it? (Assuming we can?) $\endgroup$
    – bjd2385
    Oct 9, 2015 at 23:27
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    $\begingroup$ You might want to take a look at en.wikipedia.org/wiki/Differential_form $\endgroup$
    – user9464
    Oct 9, 2015 at 23:39

2 Answers 2

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The differential is specifically defined to be a linear map that approximates the change in $f$ near a given point, so second-degree and higher terms in $dx$ and $dy$ are ignored. There is, however, a multi-variable version of Taylor’s formula that extends this approximation to higher-order terms, just as you suspect.

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  • $\begingroup$ What does the approximation look like? I've seen the multivariable Taylor formula, I'm just curious about the differentials.. like suppose I wanted to use more terms than just \begin{align}\frac{\partial f}{\partial x}\:dx+\frac{\partial f}{\partial y}\:dy?\end{align} $\endgroup$
    – bjd2385
    Oct 9, 2015 at 23:29
  • $\begingroup$ It’s pretty much what you’ve got, though the coefficients of some of the terms are different. E.g., expanding $f(x,y)$ about the origin, you’ll have $f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+\frac12x^2f_{xx}(0,0)+xyf_{xy}(0,0)+\frac12 y^2 f_{yy}(0,0)+O(||\langle x,y\rangle||^3)$. You get higher-order terms by differentiating further and evaluating at $\langle 0,0\rangle$. For more than two dimensions, for $f(\mathbf u+\mathbf v)$ you’ll have an nth-degree polynomial in the coordinates of $\mathbf v$. $\endgroup$
    – amd
    Oct 10, 2015 at 0:07
  • $\begingroup$ The differentials, $dx$ and $dy$ of the coordinates are basically just the components of the delta from a given point, so you’d plug them in for $x$ and $y$ in the above formula. I.e., if $\mathbf h=\langle dx,dy\rangle$, then you have $f(\mathbf u+\mathbf h)=f(\mathbf u)+f_x(\mathbf u)dx+f_y(\mathbf u)dy+\frac12f_{xx}(\mathbf u) dx^2+f_{xy}(\mathbf u){dx}{dy}+\frac12f_{yy}(\mathbf u)dy^2+\ldots$ $\endgroup$
    – amd
    Oct 10, 2015 at 0:16
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According to Wikipedia:

The differential change in $f$ is caused by the differential change in its variables.

When we write the linear terms out, that is what it means.

However, since the differential changes are small, the quadratic terms do not matter.

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