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Understanding differentials

  1. Aug 7, 2013 #1
    Hello all,

    I have sort of a fundamental elementary calculus question. So, when trying to understand differentials, I always interpret is as change in the function corresponding to the change in the inputs. I always thought of these changes as "infinitesimal" changes and the differential for me was basically the change in function for an infinitesimal perturbation to its input. Is that interpretation correct?

    I was browsing the wikipedia entry here (https://en.wikipedia.org/wiki/Differential_of_a_function) and it says that the modern interpretation of a differential is seen as the function of two independent variables x and [itex]\Delta[/itex] x. So essentially the differential depends on where it is computed (the value of x) and how much the value is perturbed by i.e. [itex]\Delta[/itex] x.

    However, now the differential is not this "infinitesimal" quantity i.e. dy can be any real number depending on the product f'(x) * dx. Is that correct? Somehow I have thoroughly managed to confuse myself...

    Many thanks for any help you can give me.

    Luca
     
  2. jcsd
  3. Aug 7, 2013 #2

    Stephen Tashi

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    See the recent thread https://www.physicsforums.com/showthread.php?t=701423&highlight=calculus

    Discussions of differentials on the forum reveal that they are a "matter of opinion" issue, with people in physics and applied math taking one view and those who take a rigorous approach to math another.

    The practical bottom line is that in undergraduate calculus they are not defined with the same precision as things such as the definition of the limit of a function or the definition of a derivative. So if your course is "fair" you won't have to do any problems that require that you know a precise definition for them.
     
  4. Aug 7, 2013 #3
    Thanks for pointing me to that thread. Sorry I missed it but I am not more confused that ever... About 40 posts but still no rigorous definition of what dx and dy might be!
     
  5. Aug 7, 2013 #4
    In elementary calculus, you can imagine the differential as a "magical functional thing" that does stuff in calculus. It's far better that way for now.

    Do you just want to know a rigorous definition of a differential, or do you want to fully comprehend it? From experience, I know the latter kind of sucks when you have basically only elementary understanding of calculus. To fully comprehend what a differential is, starting with only a basic knowledge of calculus, it takes some effort.

    However, if you just want to know a rigorous definition of the differential in order to confirm that there is one, I'd be happy to give it to you. However, if you want to actually understand, giving you a rigorous definition of the differential would likely confuse you more.
     
  6. Aug 8, 2013 #5
    Hi Mandelbrot,

    Thanks for the reply. I have been looking at a lot of material on this since last night and you are right, it is probably quite difficult to understand this at my level. However, can you be kind enough to give me the definition anyway, so I know what to aim for at some point!

    Thanks,
    Luca
     
  7. Aug 8, 2013 #6
    Alright...

    Consider a smooth map ##\varphi:M\to N## between smooth manifolds ##M## and ##N##. For a point ##p\in M##, the differential of this map, ##d\varphi##, gives a linear map defined by $$d\varphi_p:T_pM\to T_{\varphi(p)}N.$$ Given a curve ##\gamma:[-1,1]\to M## such that ##\gamma(0)=p##, we define ##d\varphi_p(\gamma'(0))=(\varphi\circ\gamma)'(0)##.

    That's a general definition of a differential, but I think you'll be more satisfied with a special case:

    For a smooth function ##f:M\to\mathbb{R}##, the exterior derivative of ##f##, ##df##, is a smooth section of the cotangent bundle of ##M##, defining at each point ##p\in M## a map ##df_p:T_pM\to\mathbb{R}##.

    In this case, which you work with the most in elementary calculus, ##f## is something called a differential 0-form, and its differential ##df## is a differential 1-form. Differential forms generalize the idea of the differential for scalar functions, and provide a beautiful structure to the mathematics of calculus. Put simply for now, you can just think of differentials of real functions as linear functionals that take tangent vectors on the domain to real numbers.

    And, again, if none of that made sense to you, don't worry. I can almost guarantee you that your calculus teacher wouldn't understand it either if you presented it to him or her. It's pretty legit math, and definitely not something you'd be expected to grasp mathematically when you're just entering the realm of calculus.
     
    Last edited: Aug 8, 2013
  8. Aug 8, 2013 #7

    HallsofIvy

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    This use of "differential" is definitely NOT "fundamental elementary Calculus"!
     
  9. Aug 8, 2013 #8
    This. I feel like we should reemphasize this repeatedly.

    pamparana, you will most certainly not be expected to know all of this. This is very sophisticated and advanced mathematics. I'd guess that this is at least upper level undergraduate math. I'd highly suggest just knowing that the differential is a rigorous thing and leaving it alone for a couple years.
     
  10. Aug 8, 2013 #9

    micromass

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    Sorry, but I don't see why you need to bring manifolds in to this. You can really give a way easier definition.

    Let ##f:\mathbb{R}\rightarrow \mathbb{R}## be a differentiable function. We define ##df## as the function

    [tex]df: \mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}:(p,h)\rightarrow f^\prime(p)h[/tex]

    In particular, if ##f(x) = x##, then we define

    [tex]dx: \mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}: (x,h)\rightarrow h[/tex]

    In multiple variables, we get

    [tex]df: \mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}: (p,h)\rightarrow h_1\frac{\partial f}{\partial x_1}(p) + .... + h_n\frac{\partial f}{\partial x_n}(p)[/tex]

    The intuition behind this is not so difficult. Given a differentiable function ##f:\mathbb{R}\rightarrow \mathbb{R}## and a point ##p##, then it's value at the point ##p+h## is ##f(x+h)##. But we also have the tangent to ##f## at the point ##p##. The value at the point ##p+h## of the tangent line is given by ##f(p) + df(p,h)## If ##h## is small, then this is a reasonable approximation of ##f(p+h)##.

    I know you can get everything sound complicated with manifolds and advanced math. But the real trick is to let something sound easy and intuitive! The reason I love people like Feynman is not because they are insanely smart and they won a Nobel prize, but it is because they can give such a simple and intuitive explanations for rather complicated topics. That's what I respect most in a person.
     
    Last edited: Aug 8, 2013
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