limits

[peterkirby]

Can somebody help me with this pleasa - How can I use mathcad to determine the limit of Z=f(x,y) as (x,y) approaches (0,0)? It's easy for y=f(x) but I can't find a way of dealing with a second variable in a limit. Thanks.

[StuartBruff]

On 5/30/2009 6:08:59 PM, peterkirby wrote:
= Can somebody help me with this pleasa - How can I use mathcad to determine the limit of Z=f(x,y) as (x,y) approaches (0,0)? It's easy for y=f(x) but I can't find a way of dealing with a second variable in a limit.

It appears to be as simple as employing the limit operator twice.

Stuart

[TomGutman]

That is fine, provided you know that the limit exists, and you have uniform convergence. As your examples show, where the limit does not exist the iterated limit can exist, and can have different values for the two orders. But even if the two iterated limits have the same value, that does not guarantee that the double limit exists.
__________________
� � � � Tom Gutman

[StuartBruff]

On 5/30/2009 9:46:09 PM, Tom_Gutman wrote:
== That is fine, provided you know that the limit exists, and you have uniform convergence. As your examples show, where the limit does not exist the iterated limit can exist, and can have different values for the two orders. But even if the two iterated limits have the same value, that does not guarantee that the double limit exists.

That's simply caveat emptor. I took the request as a 'how do I in principle'.

However, what I'm not sure about is how to specify a path of convergence, eg along x=y.

Stuart

[AlvaroDíaz]

On 5/30/2009 10:14:50 PM, stuartafbruff wrote:
>However, what I'm not sure about is how to specify a path of convergence, eg along x=y.

See attachad.

Regards. Alvaro.

[ptc-1368288]

On 5/30/2009 10:58:16 PM, adiaz wrote:
>On 5/30/2009 10:14:50 PM, stuartafbruff
>wrote:
>However, what I'm not sure about
>is how to specify a path of convergence,
>eg along x=y.

See attached.

Regards.
>Alvaro.
_______________________

"Usually general conditions of existence are established with continuity testing with partial derivatives. This is: if partial derivatives can be iterated, then limits can. This because if partial derivatives are continuous (or exist, depend what hard do you want to set theorems hypothesis), then the function is continuous at this point, and then limits can be iterated."

Very correct for the mental.
All that is cast in the theorems "Uniqueness & Existence".

jmG

[TomGutman]

A path is defined by a pair of functions of some parameter t. One can get the limit along the path by taking the limit as t approaches the end point. This is now a single variable limit. Typically if looking for the limit at (0,0) on will choose a path such that x(0)=y(0)=0, and take the limit of f(x(t),y(t)) as t→0.

In principle all paths have to result in the same limit for the two variable limit to exist. In practice, for "reasonable" paths and functions, on can use the fact that at a small enough scale everything is a straight line and look only at radial paths, x(t)=t·cos(θ), y(t)=t·sin(θ) for some θ. If the resulting limit is independent of θ, it is a good indication that the desired limit exists.
__________________
� � � � Tom Gutman

[AlvaroDíaz]

On 5/31/2009 12:52:15 AM, Tom_Gutman wrote:
>A path is defined by a pair of functions of some parameter t.

This is only the parametric representation of a path. There are others.

>One can ...

That's true also for path in the form y=f(x) or x=f(y). For the lines along the origin are more eassy writing y=mx.

>In principle ...

This not true. See attachad.

The best technique to demonstrate that a double limit exist (and, in this case, the two iterating limits are equals) is ... using the limit definition.

Regards. Alvaro.

[PhilipOakley]

On 5/31/2009 4:19:20 AM, adiaz wrote:
>The best technique to
>Regards. Alvaro.

Surely a "good" technique...

Best is so one-sided...
😉

Definately some educational stuff. It shows how hard some of this can be. Many thanks.

Philip Oakley

[TomGutman]

Yes, it's not enough that all the lines converge at zero, they must do so uniformly.
__________________
� � � � Tom Gutman

[ptc-1368288]

On 5/31/2009 9:51:19 PM, Tom_Gutman wrote:
>Yes, it's not enough that all
>the lines converge at zero,
>they must do so uniformly.
>__________________
>� � � � Tom Gutman
______________________________

That was the point I was going to make or missed to make. When the Cauchy "Uniqueness & Existence" fail at the limit = 0, The study switches from analytical elements [Deriv ...] to "elemental Delta". Such DE's fall in the category "Singular Points". They present themselves under the "Homographic form" and are solved from the "Characteristic Equation" . Some of their solution are known as: knots, saddle, focus, center. The solution(s) are known a "Integral Curves".

Here is a wok sheet in french, of the 4 categories.

The 5 th category not presented in the work sheet is the case from the originator in this thread, dy/dx = P(x,y)/Q(x,y). Read more: Bernstein p. 669, 670.

jmG

[StuartBruff]

On 5/31/2009 12:52:15 AM, Tom_Gutman wrote:
== A path is defined by a pair of functions of some parameter t. One can get the limit along the path by taking the limit as t approaches the end point. This is now a single variable limit. Typically if looking for the limit at (0,0) on will choose a path such that x(0)=y(0)=0, and take the limit of f(x(t),y(t)) as t→0.

OK, thanks Alvaro & Tom. I thought that might be the solution but wondered whether it might not be "The Solution", ie that there might have been someway of elegantly specifying the trajectory in the limit definition rather than the function specification.

Stuart

[AlvaroDíaz]

On 5/31/2009 7:46:18 AM, stuartafbruff wrote:
>I thought that might be the solution but wondered whether it might not be "The Solution", ie that there might have been someway of elegantly specifying the trajectory in the limit definition rather than the function specification.

You can't (Sorry for that!). In the image, formal definitions for Disk and Limit (in R^2). Last, is the geometrical interpretation. There are no paths in the definition, are domains (plane domains for R^2 limits, volume domains for R^3 limits, etc).



Paths (or trajectorys, which is similar, but not the same) appears only for convenience, but ONLY with a linear path can't demonstate that a limit exist. Even with the inclusion of "all" paths to the limit point (I believe, but not sure). This for the complicated structure of the topolgy of R^2 domains (fundamental, the presence of holes, related with the concept of homomorphic functions in complex domain). This definition ensure that there are no holes in the Domain where the limit is taking, but a more exact definition must to restrict the Disk to the intersection of the disk with the function domain ... and this can have holes (and not only punctual ones).

Regards. Alvaro.

[AlvaroDíaz]

This can go in Puzzles section:

Calculate lim 1/x for x->0.

Regards. Alvaro.

[AlvaroDíaz]

On 5/31/2009 11:25:57 AM, adiaz wrote:
>This can go in Puzzles section:
>
>Calculate lim 1/x for x->0.
>

PS: The correct answer is not Infinity. (And it is not a joke).

Alvaro.

[RichardJ]

The correct answer isn't -infinity either 🙂

I think the general idea with a puzzle is that it actually has an answer (other than the answer being "there isn't any answer").

Richard

[AlvaroDíaz]

On 5/31/2009 6:07:54 PM, rijackson wrote:
>The correct answer isn't -infinity either 🙂

It isn't.

>I think the general idea with a puzzle is that it actually has an answer (other than the answer being "there isn't any answer").

That's true. The answer 'there are not anwer' is good for equations like 2+2=5. But this is not the case. Be leave, there are an answer, more correct than lim 1/x = infinity for x->0, which have a 'general failure'.

Alvaro.

[AlvaroDíaz]

A tip: lim 1/x = infinity for x->0 it's true only under a very hard (and unusual in CAS systems) assumption.

So, question is: which is the correct value for this limit.

Regards. Alvaro.

[ptc-1368288]

On 5/31/2009 6:07:54 PM, rijackson wrote:
>The correct answer isn't
>-infinity either 🙂
>
>I think the general idea with
>a puzzle is that it actually
>has an answer (other than the
>answer being "there isn't any
>answer").
>
>Richard
______________________________

There is no puzzle, the "algebraic" limit of 1/x, i.e: in popular words "mathematical algebraic domain", the word is "Domain" limit 1/x x--> 0 = � infty, and that's it. Now, if you switch domain like deriv, integral ... the limit may be different, but 1/x does not really fall in the Lbesgue measurable domain either because the integrand does not oscillates.

For the puzzling case 1/x, the "symbolic algebraic rule" is a bit different than the college algebra. For appropriate numerical evaluation of 1/x, 1 is not treated as a number, rather as a function. And that fact is the basis of L'Hospital rules, for instance resulting in exact pure numbers 0^0 = 1 and 0^1 = 0 as exact algebraic an and exact numerical values. Again, these two are not "convention", they are exact numbers.

Refresher:



jmG

[AlvaroDíaz]

On 5/31/2009 11:36:40 PM, jmG wrote:
>For the puzzling case 1/x, the "symbolic algebraic rule" is a bit different than the college algebra.

That's the point in the question.

>... resulting in exact pure numbers 0^0 = 1 and 0^1 = 0 ...

Nope. By limit definitions, taking |x - a| < &e (and not <= ) point a (equal zero in this case) is excluding. So, never can take x = a (numeric or symbolic evaluation).

Regards. Alvaro.

[ptc-1368288]

On 5/31/2009 11:44:52 PM, adiaz wrote:
>On 5/31/2009 11:36:40 PM, jmG wrote:
>>For the puzzling case 1/x, the "symbolic algebraic rule" is a bit different than the college algebra.
>
>That's the point in the question.
>
>>... resulting in exact pure numbers 0^0 = 1 and 0^1 = 0 ...
>
>Nope. By limit definitions, taking |x -
>a| < &e (and not <= ) point a (equal
>zero in this case) is excluding. So,
>never can take x = a (numeric or
>symbolic evaluation).
>
>Regards. Alvaro.
________________________

"Nope. By limit definitions"

Yes, and by limit definitions.

The work sheets(s), in fact several work sheets have posted in this collab. Google for. If you don't find, I will post personally, because it will raise again useless argumentation from the wrong start in maths... I mean from "profane maths". You missed understanding or catching that the college algebra is not necessarily executable algebra, like going from Montreal to Rome on the thumb with a pack sack. In maths, there are lots of tickets and tokens between domains. Even with zillions of tokens valid for the Canso Causeway, Air Canada won't deliver a ticket for Vaparaiso or the MoukMouk Islands.

jmG

[GuyB]

On 5/31/2009 11:25:57 AM, adiaz wrote:
>This can go in Puzzles
>section:
>
>Calculate lim 1/x for x->0.
>
>Regards. Alvaro.

I'm confused as to why you even posed the question after your very clear graphic showing the definition of a true R^2 limit.

According to the graphic that you yourself posted, a true limit at (x,y) exists only when all the function values in a small circle around (x,y) differ from the limit by a very small number (call it epsilon).

Furthermore, no matter how small epsilon is, you can find a radius small enough that all the function values within it fall within epsilon of each other.

When these conditions are met (and only when these conditions are met) you can define a limit at (x,y).

By contrast, 1/z violates all aspects of that definition. In a small (complex-valued) circle about (0,0) the function values explode all over the complex plane! That's hardly a small deviation of the function value!

- Guy

[ptc-1368288]

You are right Guy,

A bit of Cauchy analysis would help vis the complex domain. You have best pointed the source of insanity in this "limit". Simply deceptive or deceptively simple ?

Jean

[AlvaroDíaz]

On 6/3/2009 12:12:57 PM, jmG wrote:
>You are right Guy,
>
>A bit of Cauchy analysis would help vis the complex domain.

Nice idea. Also a Lorentz serie could help.

>You have best pointed the source of insanity in this "limit".

Yes, it's insanity, more properly for puzzles forum section.

Regards. Alvaro.

[ptc-1368288]

On 6/3/2009 3:23:28 PM, adiaz wrote:
>On 6/3/2009 12:12:57 PM, jmG wrote:
... Also a Lorentz series could help.<<br> ....
>
>Regards. Alvaro.
_____________________

Did you mean "Laurent series" ?

jmG

[AlvaroDíaz]

On 6/3/2009 7:27:53 PM, jmG wrote:
>Did you mean "Laurent series" ?

Yes, poor Lorentz. I'm sorry.

Regards. Alvaro.

[AlvaroDíaz]

On 6/3/2009 11:08:13 AM, GuyBeadie wrote:
>On 5/31/2009 11:25:57 AM, adiaz wrote:
>>This can go in Puzzles
>>section:
>>
>>Calculate lim 1/x for x->0.
>>
>>Regards. Alvaro.
>

>I'm confused ...

My original intention was to show that CAS usual make the unusal assumption that values are reals taking limits. I think that this is a very hard assumption, and must to be more explicit in the help pages. Also, this happen (sometimes) with integrals. When I post the question I know that the better answer for the reals was unsigned infinity (even I despiste about the fact that implementation violates the principle of unitiy of the limit), and know that CAS have only two reals ininitys. Also, the mathematica implementation it's old and know this.

So, the trick in the question is making a well assumption about the domain of the function 1/x, and work in it. Because the default domain in CAS systems are complexes, this is then natural domain for this question.

> a true limit at (x,y) exists only ...

This is the well english words explanation about a limit.

Central point in your post is:

>By contrast, 1/z violates all aspects of that definition. In a small (complex-valued) circle about (0,0) the function values explode all over the complex plane! That's hardly a small deviation of the function value!

Explode in the interpretation of the complex plane that ... this is a plane. In the complex plane the better geometrical way to understand the infinities (for me) are the elements from the proyective geometry, who says that there are a lot of "improper" point (points at the infinity), one for each "direction" in the plane, and all fails into the "improper" rect line, the only one line in the complex plane that have all points in the infinity. Is in this geometry that I says that the limit of 1/z as z->0 is ∞ - i∞.

But there are other geometrical representations for the complexes. For instance, the Riemman sphere. With this we can destroy the explosion.

To do this, we must (i guess that it is a must) add the "point at the infinity". The beatifull improper line now degenerates in only one point. When you proyect the plane points into an sphere, the proyection of the improper rect line fails (with all the improper points representing plane directions) in the north pole.

So, the north pole kills a lot of information, for instance, all the directions go to same point (all the paths go to Rome).

In the Riemman representation, so, the lim 1/z is preciselly the north pole of the sphere, actually, a well defined point (for the sphere sense of well defintion).

Also, for continiuty questions in the 'sphere-well-defined-questions' is usual to add the north pole (called now "the point at the infinity") to the complexes defining C union {∞}. Maybe for analysis ∞ remains only as a formal symbol, but in the geometrical sense (the sphere sense of the complex numbers) actually it's only a only one well defined and accesible point (hard to arrive, but exist. Ask to Fridtjof Nansen if it isn't).

Resuming: there are presented this answers to the question lim f(x) as x->0 with f(x):=1/x

1. If Domain(f)=(-∞,0), ans = -∞
2. If Domain(f)=(0,+∞), ans = +∞
3. If Domain(f)=(-∞,+∞), ans = unsigned ∞
4. If Domain(f)=Complexes,
A. Riemann Representation: ans = ∞ (representing the north pole)
B. Polar Form: ans = +∞e^iθ with 0<=θ<2π
C. Cartesian Form: ans = +∞ - i∞

I'm think that only the result 4.A. is which it is generating controversing.

Regards. Alvaro.

P.S.: Notice that the anser undefined for the result 3. kills information. For example: a*∞=∞ for a real not null and a+∞=∞ for a finite, but a*undefined=undefined and a+undefined=undefined. Also, Riemman representation do the same with the directions in the plane. There are an information theory question here: 1/+∞, 1/∞ and 1/∞ converges all to 0: right zero, left zero and 'unsigned' zero. 'Obviously', zero have not sign (or have?).

[ptc-1368288]

Alvaro,

It looks like you are out of the track completely. CAS do execute the l'Hospital rule(s), basically in terms of "algebraic numbers", i.e: in term of algebraic functions. That CAS do assume something else than algebraic number = NO. Only in the case root and you very well know about "roots of unity" as a special case.

jmG

[TomGutman]

>>By contrast, 1/z violates all aspects of that definition. In a small (complex-valued) circle about (0,0) the function values explode all over the complex plane! That's hardly a small deviation of the function value!<<

Not quite. Which is why there is an issue at all, and some sort of meaning to the limit. While the function does go in all directions, it does not go all over the complex plane. Rather it avoids a disk centered on zero. And that disk grows without bound as we look at smaller and smaller disks.

Thus there is no limit in the sense of a value that is approached. There is an asymptotic behaviour that can be described. This behaviour is unbounded growth, and is conventially described as being infinity.

But contrary to Alvaro's continued assertions the real and imaginary parts of 1/z do not have any such behaviour, and it is quite wrong to describe them as being infinite.
__________________
� � � � Tom Gutman

[StuartBruff]

On 5/31/2009 7:46:18 AM, stuartafbruff wrote:
== OK, thanks Alvaro & Tom. I thought that might be the solution but wondered whether it might not be "The Solution", ie that there might have been someway of elegantly specifying the trajectory in the limit definition rather than the function specification.

Forgot to attach the attached worksheet to the above message ...

Stuart