Uniqueness of the Differential
Okay, for the moment let’s pick an orthonormal basis for our vector space . This gives us coordinates on the Euclidean space of points. It also gives us the dual basis of the dual space . This lets us write any linear functional as a unique linear combination . The component measures how much weight we give to the distance a vector extends in the direction.
Now if we look at a particular point we can put it into our differential and leave the second (vector) slot blank: . We will also write this simply as , and apply it to a vector by setting the vector just to its right: . Now is a linear functional, and we can regard as a function from our space of points to the dual of the space of displacements. We can thus write it out uniquely in components , where each is a function of the point , but not of the displacement .
We want to analyze these components. I assert that these are just the partial derivatives in terms of the orthonormal basis we’ve chosen: . In particular, I’m asserting that if the differential exists, then the partial derivatives exist as well.
By the definition of the differential, for every there is a so that if , then
Now we can write out in components
Next for a specific index we can pick for some value . Then , , and for all the other indices . Putting all these and the component representation of into the definition of the differential we find
Dividing through by we find
And this is exactly what we need to find that exists and equals .
Therefore if the function has a differential at the point , then it has all partial derivatives there, and these uniquely determine the differential at that point.
[…] of showing that the differential of a function at a point — if it exists at all — is unique (and thus we can say “the” differential), we showed that given an orthonormal basis we […]
Pingback by Differentiability Implies Continuity « The Unapologetic Mathematician | September 30, 2009 |
[…] haven’t yet seen any conditions that tell us that any such function exists. We know from the uniqueness proof that if it does exist, then given an orthonormal basis we have all partial derivatives, and […]
Pingback by An Existence Condition for the Differential « The Unapologetic Mathematician | October 1, 2009 |
minor typo: epsilon-eta confusion at beginning. A akways wonder if this sort of thign is worth pointing out
Comment by Avery Andrews | October 1, 2009 |
You may as well, since it’s easy enough to correct. Usually that sort of thing happens when I go back to review my earlier pieces, use old notation, then later in the process of writing decide to change it. The editor for WordPress isn’t the best, but I have to use it if I’m going to get any sort of preview.
Comment by John Armstrong | October 1, 2009 |
[…] showed that these partial derivatives are the components of the differential (when it exists), and so there should be some connection between the two […]
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[…] Chain Rule Since the components of the differential are given by partial derivatives, and partial derivatives (like all single-variable derivatives) […]
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[…] and is given by the product of these matrices, and the entries of the resulting matrix must (by uniqueness) be the partial derivatives of the composite […]
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[…] by uniqueness we can read off the partial derivatives of in terms of and […]
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[…] First let’s look at the second-order differential of a real-valued function of variables . We’ll use the as a basis for the space of differentials, which allows us to write out the components of the differential: […]
Pingback by Higher-Order Differentials « The Unapologetic Mathematician | October 16, 2009 |
[…] the partial derivative either does not exist or is equal to zero at . And because the differential subsumes the partial derivatives, if any of them fail to exist the differential must fail to exist as well. On the other hand, if […]
Pingback by Local Extrema in Multiple Variables « The Unapologetic Mathematician | November 23, 2009 |