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Clairauts “equality of mixed partial derivatives” theorem

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davidbenari
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Jul15-14, 01:22 PM
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I know how to prove this via limits and I'm okay with that.

What I want to understand is the interpretation of the theorem and specifically a visualisation of why what the theorem states must be the case.

My guess is that this theorem is saying that change is symmetrical. But I don't know if this is only true for second derivatives.

If you don't know this theorem by its name the theorem basically says this:

∂/∂y(∂f/∂x)=∂/∂x(∂f/∂y)

Also, I would like to know if you consider my focus on visualisation to be not worthwhile and that I should instead just trust this theorem.

I thank you in advance.
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UltrafastPED
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Jul15-14, 07:28 PM
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Clairaut's theorem (1743) is valid when the first partial derivatives are continuous. If you are a mathematics student then it is worth while understanding why this is so.

More information is available here:
https://en.wikipedia.org/wiki/Symmet...nd_derivatives

and
http://calculus.subwiki.org/wiki/Cla...mixed_partials
davidbenari
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Jul15-14, 08:03 PM
P: 22
I am a physics student. I have a question about this "if you are a mathematics student then it is worthwhile" stuff. Would a great physicist, say like Feynman, know why this is so? Even if your answer is speculative, what do you think?

AlephZero
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Jul15-14, 08:20 PM
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Clairauts “equality of mixed partial derivatives” theorem

If the math is a model of some physical situation, you can often argue that derivatives are continuous, etc, on physical grounds. Usually, a math model only hits the "exceptional" conditions that mathematicians like to understand completely, if it's a poor model of the physics.

IMO it is always worthwhile (not to say essential) to know something about the conditions that make your math valid. But as a physicist or engineer you don't necessarily need to know the most general set of conditions that make it valid, or be able to prove why it is valid.

Of course you can never know "too much" math, but in real life, whether you learn more math or more physics is a time management problem.

(Full disclosure: I've seen both sides of this first hand - I have a math degree, and spent most of my life working on engineering problems).


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