Re: Integration of 0

On Wed, 10 Aug 2005 17:12:10 GMT, "n3wb1e" <n3wb1e@xxxxxxxxxxxxxx>
wrote:

>
>"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> ha scritto nel messaggio
>news:r24kf1hdmg2f2l1ds98j86k65ggdge9bi2@xxxxxxxxxx
>> On Wed, 10 Aug 2005 11:30:17 GMT, "n3wb1e" <n3wb1e@xxxxxxxxxxxxxx>
>> wrote:
>>
>>>
>>>"agentmath" <Agent.Math@xxxxxxxxx> ha scritto nel messaggio
>>>news:1123617224.977190.237230@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>>>> In indefinite integraion,
>>>>
>>>> (*) integral 0 dx = constant.
>>>>
>>>> But by the property of integration,
>>>>
>>>> (**) integral 0 dx = integral (0 product 1) dx = 0 product (integral 1
>>>> dx) = 0.
>>>>
>>>> Why are (*) and (**) different?
>>>>
>>>
>>>
>>>Even though the totality of books you may read says that, the homogeneity
>>>property for indefinite integral DOES NOT HOLD for 0.
>>
>> That's not really the answer. The homogeneity property does hold
>> if you look at things right - the actual problem is something
>> else.
>>
>> There _is_ a problem with what you see in books, which has
>> already been pointed out. One way to put it is to say it's
>> an abuse of the equals sign, since the thing on the right
>> is not just one thing. Another way to look at it:
>>
>> If we want to talk about _the_ indefinite integral of a
>> function then the integral of a function cannot be a function,
>> it has to be an equivalence class of functions. Given a
>> function f, let's say [f] is the class of all functions
>> of the form g = f + c for some constant c. Say V is the
>> space of all these equivalence classes [f].
>>
>> Now the integral of a function should really be regarded
>> as an element of V. (Why? If we say int x dx = x^2/2 + c
>> we seem to be saying that x^2/2 = x^2/2 + 1, which is
>> wrong. But if we say that int x dx = [x^2/2] there's
>> no problem, because [x^2/2] _is_ precisely equal to
>> [x^2/2 + 1].)
>>
>> And V is a vector space in a natural/standard way
>> (standard because V is just the quotient space of
>> the space of all functions modulo the subspace of
>> constant functions); if f and g are functions and
>> a is a scalar we define
>>
>> (i) [f] + [g] = [f+g]
>> (ii) a[f] = [af].
>>
>> (You need to check that these are "well-defined";
>> for (ii) that amounts to showing that if [f] = [g]
>> then [af] = [ag], which is easy to see. Note for
>> example that we _cannot_ define the product of
>> two elements of V by
>>
>> (iii) [f][g] = [fg];
>>
>> why not?)
>>
>> And now the problem goes away - homogeneity
>> _does_ hold, even for the integral of 0 times
>> the scalar 0. Because 0[0] = [0]; both sides
>> are exactly the class of all constant functions.
>>
>>
>> ************************
>>
>> David C. Ullrich
>
>
>
>No, for 0 it is senseless.

Maybe you should point out an _error_ above.

>The reason why is that all books tell us wich is the "*" operation in
>$k*f(x) dx, but not in k*$f(x)dx. Some analysts repeat again and again the
>flaw that consists in seeing a SET as a number (or a function).
>we cannot deal with a set like it was a function.

Precisely. If you'd read what I wrote you'd see this was
exactly my point. An indefinite integral is really a
certain equivalence class (in particular a SET). Those
eqiuivalence classes are elements of a certain vector
space, and when you use the properly-defined operations
in that vector space the apparent problem with homogeneity
goes away.

>The same happens when you try to integrate
>$1/x dx
>by parts, obtaining 0=1 (where is the flaw?).

I can't tell you where the flaw is in an argument you
haven't shown me. But presumably if you redo the
argument as above then instead of 0 = 1 you get
[0] = [1], and there is no problem there, in fact
[0] _does_ equal [1].

>For all k in R-{0}, the sets {k*F(x)+c} and k#{F(x)+c} are the same set
>(where # is a product for each element of the set).
>For 0 that property fails, that is {0}<>R.

I can't quite make sense of this, because you haven't explained
what your notation # means - saying "# is a product for each
element of the set" is not quite a clear definition.

But I suspect that the confusion is here: If f is a function
and k is a scalar then the definition of k[f] is

(ii) k[f] = [kf].

The definition is _not_

(ii') {k(f+c) : c in R}.

If you think the definition is (ii') then of course
you're right, homogeneity fails. But the definition is
(ii), and with (ii) there's no problem with
homogeneity. I _think_ that your point is that (ii) and
(ii') are the same unless k = 0, in which case they
are not the same. That's correct, but it has no
relevance here, because the definition is (ii), not
(ii').

(What do I mean when I say the definition _is_ (ii)?
Two things: that's the definition that I _gave_, and
it's also the _standard_ definition for scalar multilcation
in a quotient space. With the definition corresponding
to (ii') there's not just a problem with integration,
with the definition as in (ii') the definition of a
quotient space V/W (for W a subspace of the vector space
V) simply doesn't work.)

>See for example:
>http://www.cs.unict.it/~emmanuele/calcolointegrale-am1-0203.pdf
>page 4.
>
>
>


************************

David C. Ullrich
.

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