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

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David C. Ullrich

Posts: 3,085
Registered: 12/13/04
Re: Proof that mixed partials commute.
Posted: Nov 20, 2013 11:07 AM
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On Wed, 20 Nov 2013 06:55:30 -0500, Hetware <hattons@speakyeasy.net>
wrote:

>On 11/20/2013 3:22 AM, Robin Chapman wrote:
>> On 19/11/2013 23:59, Hetware wrote:
>>>>
>>>> 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.

>>
>> That's the nub of the matter. Iterated limits need not commute.
>> One has to show that in this case they do. Putting in deliberate
>> ambiguities in your notation sounds a really bad idea.
>>
>> Of course there are examples where mixed partials are different,
>> so your original argument can't have been valid, since it didn't
>> use the necessary hypotheses about continuity of partials etc.
>>

>
>But I added my reason for assuming the limits commute. I expressed a
>function of two independent variables as the function of a single
>variable and appealed to the limit rules for a function of a single
>variable to the result.
>
>The question is whether that reasoning is valid.


It can't be valid, since it "proves" something false!
Mixed partials are the same _under_ certain hypotheses.
Your proof, if valid, would show that they commute
_wiithout_ those hypotheses. And that's not true.

Many theorems in analysis amount to showing
that some particular two limits commuute.





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