0
$\begingroup$

I have a function, e.g. $f(x,y)=2x^2+xy+y^2$. I can figure out how much of the change in $f$ with respect to $t$ is because of changes in either $x$ or $y$ instantaneously with the chain rule.

$$\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt}$$

So at $t=0$ suppose $x'(0)=5$, $y'(0)=3$, $x(0)=10$, $y(0)=20$, we get:

$$\frac{df}{dt} = 60*5 + 50*3 = 300 + 150 = 450$$

This has a nice interpretation that $300/450 = 2/3$ of the change in $f$ at $t=0$ can be attributed to $x$

However, I would like to figure out how much of the actual change in $f$ is attributable to $x$. Say, $\Delta t = 5$, $x_0=10$, $y_0=20$, $x_1=30$, $y_1=50$.

Does this even make sense? I can consider the hypothetical situation where $y$ didn't change, and compute how much $f$ would increase as we changed $x$. But, really, $y$ is changing as $x$ is changing, so this may not make sense, in light of the term $xy$. Can someone help me make sense out of this concept of "attributing" change to the individual variables, and provide some formulae for computing such a thing?

$\endgroup$
0
$\begingroup$

This is my opinion:

I think you can attribute changes to specific variables almost only when the change is linear. One can study the case of $f(x,y)=xy$ and see that the change of the function from the origin can't be attributed to one of the variables or some of the change...

That's why maybe in the differential this is possible, there, the function is linear and you can tell which part of the change came from different variables.

In the end of the day you're looking into a formula of the form:

Change in f(x,y) = [part attributed to x] * change in x + [ same in y] * change in y

But that has meaning only if the [part attributed to x] doesn't depend on y and vice versa, meaning only if $$ \frac{\partial^2 f }{\partial x \partial y} = 0 $$ I am not aware of a field in analysis that deels specifically with these kind of functions, but if there is one, maybe they do have a special term for what you are describing.

To summarize, I think that for this to have meaning your function needs to obey a certain equation that most functions do not obey but people that explore that area maybe do have a term for this.

$\endgroup$
  • $\begingroup$ but even if the change isn't linear, it's "locally linear", so maybe we could add up all of the tiny derivative changes as we go. So $\int_t \frac{df}{dx}\frac{dx}{dt}$ could be the total change in $f$ attributable to $x$ over our time interval? Does that mesh with your thoughts here? $\endgroup$ – Scott May 27 '17 at 21:23
  • $\begingroup$ That isn't well defines because it depends on the contour you're taking between the two points $\endgroup$ – Ofek Gillon May 28 '17 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.