I don't get why this troll physics is wrong.

[DaveC426913]

Note that the limit of these jagged paths is the true diagonal; they converge uniformly to the line in the euclidean norm.

But they don't converge on the euclidean norm.

No matter how small you make the increment, the "taxi-cab diagonal" is always exactly 2.

 

[D H]

It does have a length of 2 -- using the taxicab, or L₁, norm, that is. What this troll physics shows is that the circumference of a circle using the taxicab norm is c=4d. That does not mean that pi is 4. It just means that different norms will yield different answers for length.

Well, the "diagonal" of a square is a straight line between two opposing corners, so it will not have a length of 2.

That diagonal line does have a length of 2 using the L₁ norm.

The L_p norm of a vector x=[x₁,x₂,...,x_n] is defined as

||{\mathbf x}||_p \equiv \left(\sqrt{\sum_i |x_i|^p}\right)^{1/p}

For p≥1, this definition

• Is scalable, ||ax||_p = |a| ||x||_p,
• Is positive definite, ||x||_p>0 for all x but x=0, in which case ||x||_p=0, and
• Obeys the triangle inequality, ||x+y||_p≤||x||_p+||y||_p

In other words, it is a norm.

The L₁ norm of this vector is

||\mathbf x}||_1 = \sum_i |x_i|

This is also called the taxicab norm or Manhattan norm. Think of how a taxicab must drive in New York City. Taxicabs cannot follow the route taken by a crow flying from point A to point B.

 

[disregardthat]

But they don't converge on the euclidean norm.

No matter how small you make the increment, the "taxi-cab diagonal" is always exactly 2.

All in the euclidean norm (nevermind the taxi-cab-norm for a moment):
The lengths does not, but the curves do. The limit curve is the circle, or in your example, the diagonal. The limit path does not resemble the elements of the sequence. It is not "infinitely jagged" or anything of the sort. That is why things like these cause so much confusion. You have no guarantee for that the lengths will converge. For this the curves will need to satisfy stronger conditions, e.g. being differentiable and having uniformly convergent derivative-curves.

In the taxi-cab-norm:
The troll and your example illustrates a perfectly proper way of measuring the perimeter of the circle and the diagonal in this norm, and the limiting length of the curves does converge what you intend to measure. We can say that pi = 4 in the taxi-cab-norm (though pi conventionally refers to the limit in the euclidean norm under the conventional definition of the length of a curve), and the diagonal has length 4.

Bear in mind that in this context the jags only illustrate which norm are being used, and are not the approximating curves themselves.

 

[Dadface]

I just don't get this .Is it being claimed that by drawing a stepped line of length 4 around a circle of diameter 1,the said line touching the circle at different places,then the perimeter of the circle is equal to the perimeter of the line or that the perimeter of the line ,which is independant of the number of steps used,converges to the perimeter of the circle?Are there different ways of defining length or perimeter and that in a taxi cab norm definition pi=4?Where does that leave things such as Coulombs law?
As far as I understand it the stepped line has nothing to do with the dimensions of the circle other than it surrounds it and touches it at different points.I also think that a simple problem is being overcomplicated.I am probably misreading or misunderstanding so can someone please clarify.

 

[D H]

Pi is not 4. It is defined in terms of the Euclidean norm in Euclidean space. Use some other norm and you will get a different ratio of circumference to diameter. The ratio might not even be constant, as is the case in many non-Euclidean geometries. That doesn't mean pi has changed to some other constant or isn't even a constant. Pi is 3.141592653589793..., period. All that getting a different, and possibly non-constant, ratio of circumference to diameter means just means you are using a different norm than the Euclidean norm or a different geometry than Euclidean geometry. The resolution of this troll physics is that it isn't using the Euclidean norm.

 

[disregardthat]

Are there different ways of defining length or perimeter and that in a taxi cab norm definition pi=4?

Yes, the length of a curve can be defined relative to different norms, but normally the euclidean norm is used. If we say that pi is the ratio between the perimeter and diameter of a circle relative to the taxi-cab norm then we will get that pi = 4. Pi is however defined relative to the euclidean norm, not the taxi-cab norm, so the conclusion is wrong.

 

[G037H3]

Because Manhattan/taxicab geometry doesn't apply to Euclidean geometry.

 

[D H]

Exactly. While both Euclidean geometry and taxicab geometry "fit" on R², taxicab geometry violates some key characteristics of Euclidean geometry, one of them being the Pythagorean theorem. An AMS article on taxicab geometry: http://www.ams.org/samplings/feature-column/fcarc-taxi

 

[Dadface]

Hello DH and Jarle,thanks for clarifying.Of course I know that pi is 3.142 etc but I was mainly wondering why a definition in terms of taxicab geometry is made.Perhaps the definition can have useful applications.There is more than one way to answer many question and I think that in this thread the question has been answered with and without reference to taxicab geometry.
DH,I really like the look of the content in the link you posted above and I have put it on my list of "things to read".

 

[D H]

I was mainly wondering why a definition in terms of taxicab geometry is made.Perhaps the definition can have useful applications.

In mathematics there are many definitions of 'distance' other than Euclidean distance. Some have names because they are rather useful in some circumstances. A couple of examples in addition to the taxicab norm discussed in this thread:

• Chebyshev distance, or L_∞ norm, ||{\boldsymbol x}||_{\infty} = \max_i |x_i|.
  This is at the opposite end of the L_p famility of norms from the taxicab norm. Suppose you and a coworker are each asked to come up with a simple approximation of some complicated function. Your coworker uses a root mean square approach to minimize the error while you seek to minimize the worst-case error. In many cases, the function that minimizes the absolute error is deemed to be a better fit than a least squares "best fit".
• Mahalanobis distance, ||{\boldsymbol x}||_S=\sqrt{{\boldsymbol x}^T S^{-1} {\boldsymbol x}}.
  Suppose you are told that a good-sized meteorite is bearing down on your city. Odds of living if you are within 100 meters of the impact are rather low. Outside a kilometer away, not much of an impact at all. The impact is estimated to occur within an ellipse 50 km long by 1 km wide. You live 10 km from the center of the ellipse. Should you get out of Dodge? The answer depends on direction as well as Euclidean distance. If that 10 km is along the long axis of the ellipse, going somewhere else is a rather good idea. If that 10 km is along the short axis of the ellipse you might as well stay home. The odds of getting in an accident due to all the idiots on the road trying to escape the meteorite are a lot higher than you getting hit by the meteorite in this case. The Mahalanobis distance is a much more meaningful metric in this case than is Euclidean distance.

 

[Edgardo]

An article on the diagonal paradox which has been mentioned before can be found on http://mathworld.wolfram.com/DiagonalParadox.html" .
Another interesting paradox involving sine functions converging to zero is described on http://www128.pair.com/r3d4k7/Mathematicae4.html" .

 

[OmCheeto]

An article on the diagonal paradox which has been mentioned before can be found on http://mathworld.wolfram.com/DiagonalParadox.html" .
Another interesting paradox involving sine functions converging to zero is described on http://www128.pair.com/r3d4k7/Mathematicae4.html" .

The diagonal paradox also references the http://mathworld.wolfram.com/CoastlineParadox.html" , which, from the link listed earlier:

Could you be more specific? Just like an integral is the riemann sum of n number of rectangles as n goes to infinity. I assume that is the same reasoning being used here.

EDIT: http://www.axiomaticdoubt.com/?p=504

Makes sense.

Makes me wonder if it was correct to correct the author:

The limit of f as n approaches infinity is not a circle, it’s some right angled fractal beast [edit: I've been corrected, the limit is a circle, but the length of the curve is defined in terms of the derivatives, which means that it is not defined on the corners], π does not equal 4, proceed to infinity with caution.

It strikes me as a fractal beast. At least as far as I understand fractals. You can hypothetically take the limit to infinity, but the circle would then be infinitely jagged. Which in my mind, must violate some rule of integration.

Help! It's been 30 years since I've studied Calculus!

 

[disregardthat]

It strikes me as a fractal beast. At least as far as I understand fractals. You can hypothetically take the limit to infinity, but the circle would then be infinitely jagged. Which in my mind, must violate some rule of integration.

Help! It's been 30 years since I've studied Calculus!

The limit path is the circle. The curves converge uniformly to the circle, and not some "fractal beast". When you "zoom in" on the limit path you will never see jags, it has a constant curvature. It is not "infinitely jagged", whatever it means. But in order to be certain of that length is preserved under the limit you will need the curves to be differentiable, and the differentiated curves will have to be continuous and converge uniformly as well. This is why the length of the sine curves (being differentiable with continuous derivatives) in the other link does not converge to the length of the unit interval [0,1]; the derivatives does not have a limit. But the curves do converge to the unit interval.

What rules of integration do you think it violates?

 

[OmCheeto]

The limit path is the circle. The curves converge uniformly to the circle, and not some "fractal beast". When you "zoom in" on the limit path you will never see jags, it has a constant curvature. So it is not infinitely jagged. But in order to be certain of that length is preserved under the limit you will need them to be differentiable, and the differentiated curves will have to converge uniformly as well. This is why the length of the sine curves (being differentiable) in the other link does not converge to the length of the unit interval [0,1]; the derivatives does not have a limit. But the curves do converge to the unit interval.

I disagree. You can go all the way to infinity, and it will still be jagged. Otherwise, pi would equal 4.

And can someone point me to the definition of "limit path". I'm not familiar with the term.

 

[disregardthat]

I disagree. You can go all the way to infinity, and it will still be jagged. Otherwise, pi would equal 4.

And can someone point me to the definition of "limit path". I'm not familiar with the term.

This is wrong, the limit of the curves is the circle, but the lengths does not converge to the length of the limit curve. It's that simple.

A sequence of functions (f_n) converge uniformly to the function f if \sup||f_n(x)-f(x)|| \to 0 as n \to \infty, where the supremum is taken for x over the domain. In this case the relevant functions would be parametrizations of the curves approximating the circle.

Explain what you mean by infinitely jagged.

 

[OmCheeto]

This is wrong, the limit of the curves is the circle, but the lengths does not converge to the length of the limit curve. It's that simple.

A sequence of functions (f_n) converge uniformly to the function f if \sup||f_n(x)-f(x)|| \to 0 as n \to \infty, where the supremum is taken for x over the domain. In this case the relevant functions would be parametrizations of the curves approximating the circle.

Explain what you mean by infinitely jagged.

I think you are wrong. I see no curves. The function represented in reducing the square to a circle consists of an infinite number of non-continuous functions. Isn't there some rule about the function needing to be continuous?

But perhaps I don't speak maths well enough to explain myself.

ps. I went to Wolframs and could find no definition of "limit path". It also shows up only 6 times when googling: calculus "limit path" wolfram.
(One of which being your usage.)

Are you sure you used the correct term? Or is it an ellipsis?

 

[disregardthat]

I think you are wrong. I see no curves. The function represented in reducing the square to a circle consists of an infinite number of non-continuous functions. Isn't there some rule about the function needing to be continuous?

But perhaps I don't speak maths well enough to explain myself.

ps. I went to Wolframs and could find no definition of "limit path". It also shows up only 6 times when googling: calculus "limit path" wolfram.
(One of which being your usage.)

Are you sure you used the correct term? Or it an ellipsis?

It's a limiting function under the uniform norm which happens to be a path/curve, therefore I call it a limit path/curve, or limiting path/curve if you will. I have explained what I meant, it should provide no further confusion.

See http://en.wikipedia.org/wiki/Uniform_convergence

Of course there are curves. The jagged curves around the circle are approximating the circle. There are no non-continuous functions here.

You may believe I am wrong, but I can't see any evidence from your side demonstrating it.

 

[OmCheeto]

It's a limiting function which happens to be a path, therefore I call it a limit path, or limiting path if you will. I have explained what I meant, it should provide no further confusion.

See http://en.wikipedia.org/wiki/Uniform_convergence

Of course there are curves. The jagged curves around the circle are approximating the circle. There are no non-continuous functions here.

You may believe I am wrong, but I can't see any evidence from your side demonstrating it.

The function of the jaggedness of the functions defining the broken square approaching the circle is: f(x)=4*2^x-4

As x approaches infinity, the jaggedness approaches infinity even faster.

If that even makes sense.

 

[disregardthat]

The function of the jaggedness of the functions defining the broken square approaching the circle is: f(x)=4*2^x-4

As x approaches infinity, the jaggedness approaches infinity even faster.

If that even makes sense.

That does not mean the limiting function has an infinite number of jags. That is faulty logic. It just means that the number of jags are increasing without bound for the functions in the sequence. The limiting function need not resemble these.

Consider a similar variant along these lines: You iteratively pick some arbitrary number from the stack of rational numbers. Each pick leaves a new disjoint interval in the rational number line. As you continue, the number of disjoint intervals increase. Still, the "limit" of this process will exhaust the rational numbers since they are countable (given that you actually pick from a list of the rational numbers), leaving no such intervals.

 

[D H]

This is wrong, the limit of the curves is the circle, but the lengths does not converge to the length of the limit curve. It's that simple.

No, its not that simple. The limit of the steps is a curve that converges uniformly to the circle but is nowhere differentiable. In short, it is not the circle.

 

[OmCheeto]

That does not mean the limiting function has an infinite number of jags. That is faulty logic. It just means that the number of jags are increasing without bound for the functions in the sequence. The limiting function need not resemble these.

Consider a similar variant along these lines: You iteratively pick some arbitrary number from the stack of rational numbers. Each pick leaves a new disjoint interval in the rational number line. As you continue, the number of disjoint intervals increase. Still, the "limit" of this process will exhaust the rational numbers since they are countable, leaving no such intervals.

Exhaust?

Please.

I may be daft, but I'm not stupid.

Where is the wizard?

 

[disregardthat]

Exhaust?

Please.

I may be daft, but I'm not stupid.

Where is the wizard?

Don't you know the word?
Definitions of exhaust on the Web:
consume: use up (resources or materials); "this car consumes a lot of gas"; "We exhausted our savings"; "They run through 20 bottles of wine a week"
run down: deplete; "exhaust one's savings"; "We quickly played out our strength"
use up the whole supply of; "We have exhausted the food supplies".

Please explain what genuine questions you have if any.

 

[disregardthat]

No, its not that simple. The limit of the steps is a curve that converges uniformly to the circle but is nowhere differentiable. In short, it is not the circle.

If you are right then that is if you insist on the induced parametrization of the circle. I may have been sloppy and written path when I meant curve and vice versa. Sometimes it matter. Only the range (curve) of this limit function is relevant though, and that is the circle. Why would you say "it is not the circle" when you say it converges uniformly towards the circle? The circle does not depend on any parametrization and does not have a preferred (differentiable) one.

As far as I know the length of a curve is defined independently of any particular parametrization, and is only applicable if a differentiable parametrization exists. Hence the length of the limit curve is not necessarily calculated using the induced parametrization. If I have understood you right.

 

[OmCheeto]

Don't you know the word?
Definitions of exhaust on the Web:
consume: use up (resources or materials); "this car consumes a lot of gas"; "We exhausted our savings"; "They run through 20 bottles of wine a week"
run down: deplete; "exhaust one's savings"; "We quickly played out our strength"
use up the whole supply of; "We have exhausted the food supplies".

Please explain what genuine questions you have if any.

Since there are an infinite number of rational numbers, how long do we have to wait before they are exhausted? Or is that the next question? "Prove that there are an infinite number of rational numbers."

I must say, is General Math always this lively?

 

[disregardthat]

Since there are an infinite number of rational numbers, how long do we have to wait before they are exhausted? Or is that the next question? "Prove that there are an infinite number of rational numbers."

I must say, is General Math always this lively?

Note that is said the limit of this process. A limit is never "reached". I don't see what you are getting at.

 

[alphachapmtl]

http://i254.photobucket.com/albums/hh116/balthamossa2b/1290457745312.jpg

Can someone explain the flaw in this logic?

In the same way, a square diagonal will lead to 1+1=sqrt(2).

 

[qwerty4056]

The correct way to approximate pi would be to take the diagonals of the "bumps", which are illustrated in the attached image (View attachment Pi aproximation.bmp). Now you add up all the diagonals' lengths and then you get some number under pi.

 

[1MileCrash]

Maybe I'm not seeing it, but I picture repeating this process "infinitely" just yields a diamond or sideways square, never a circle.

 

[lurflurf]

The main point (what makes circumference difficult at times) is that circumference is a local property. That is a small (in some metric) deviation can cause a large difference in circumference. Our jagged circle thing is almost a circle in some sense, but it has a very different circumference.

 

[coolul007]

An interesting book containing "proofs" that are wrong is:

The Mathematical Recreations of Lewis Carroll: Pillow Problems and a Tangled Tale

He is a master at them.