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Topic: Proof that mixed partials commute.
Replies: 20   Last Post: Nov 22, 2013 11:57 PM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Proof that mixed partials commute.
Posted: Nov 19, 2013 6:59 PM
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On 11/19/2013 11:17 AM, dullrich@sprynet.com wrote:
> On Mon, 18 Nov 2013 18:45:53 -0500, Hetware <hattons@speakyeasy.net>
> wrote:
>

>> In _Calculus and Analytic Geometry_ 2nd ed.(1953), Thomas provides:
>>
>> Theorem. If the function w=f(x,y) together with the partial derivatives
>> f_x, f_y, f_xy and f_yx are continuous, then f_xy = f_yx.
>>
>> Both Thomas and Anton (1980) provide rather long-winded proofs of this
>> theorem. These proofs involved geometric arguments, auxiliary
>> functions, the mean-value theorem, epsilon error variables, a
>> proliferation of symbols, and a generous helping of obscurity.
>>
>> Starting from the definition of partial differentiation, and using the
>> rules of limits, along with a modest amount of basic algebra, I came up
>> with this:
>>
>> f_x(x,y) = Limit[[f(x+Dx,y)-f(x,y)]/Dx, Dx->0]
>>
>> f_yx(x,y) = Limit[[f_x(x,y+Dy)-f_x(x,y)]/Dy, Dy->0]
>> = Limit[
>> [[f(x+Dx,y+Dy)-f(x,y+Dy)]-[f(x+Dx,y)-f(x,y)]]/DyDx
>> , {Dy->0, Dx->0}]
>>
>> f_xy(x,y) = Limit[
>> [[f(x+Dx,y+Dy)-f(x+Dx,y)]-[f(x,y+Dy)-f(x,y)]]/DxDy
>> , {Dx->0, Dy->0}]

>
> That's very bad notation. It's not one limit, it's the limit of
> a limit. Should be
>
> Limt(Limt(...)[x->c][y->c].
>
> And now the big question is why
>
> Limt(Limt(...)[x->c][y->c] = Limt(Limt(...)[y->c][x->c]


I guess I should have included the intermediate steps. I had intended
that the order of taking limits should be ambiguous.

>>
>> The only caveat is that the rules for limits, such as /the product of
>> limits is equal to the limit of the products/, are stated in terms of a
>> single variable. For example:
>>
>> Limit[F(t) G(t), t->c] = Limit[F(t), t->c] Limit[G(t), t->c]
>>
>> Whereas I am assuming
>>
>> Limit[F(x) G(y), {x->c, y->c}] = Limit[F(x), x->c] Limit[F(y), y->c].
>>
>> I argue as follows. The statement that x->c as y->c can be formalized by
>> treating x and y as functions of t such that
>>
>> Limit[x(t), t->c] = Limit[y(t), t->c] = c
>>
>> |F(x)-F(c)| < epsilon_F ==> |x-c| < delta_F exists
>>
>> |G(y)-G(c)| < epsilon_G ==> |y-c| < delta_G exists
>>
>> |x(t)-c| < epsilon_x ==> |t-c| < delta_x exists
>>
>> |y(t)-c| < epsilon_y ==> |t-c| < delta_y exists
>>
>> Now epsilon_F ==> delta_F can be used as delta_F = epsilon_x which
>> implies delta_x > |t-c| exists. So
>>
>> Limit[F(x(t)), t->c] = Limit[F(x), x->c], etc.
>>
>> It follows that
>>
>> Limit[F(x(t)) G(y(t)), t->c]
>> = Limit[F(x(t)), t->c] Limit[G(y(t)), t->c]
>> = Limit[F(x), x->c] Limit[G(y), y->c]
>>
>> Am I making sense here?

>
> Not as far as I can see.


Suppose that x(t)=at+c and y(t)=bt+c where a and b are arbitrarily
chosen real number constants where at least one is not zero. Is it true
that

Limit[F(x(t)) G(y(t)), t->c]
= Limit[F(x(t)), t->c] Limit[G(y(t)), t->c]
= Limit[F(x), x->c] Limit[G(y), y->c]

for all possible a and b? Assuming F and G are continuous in the
neighborhood of {x(c),y(c)}. Is there any case of Limit[F(x), x->c]
Limit[G(y), y->c] not covered by the set of all pairs {{a,b}}?



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