Jump to content
Science Forums

Recommended Posts

Posted

With the usual norm, which induces the usual or euclidean metric, Pythagoras' theorem is valid. With the metric induced by another norm it might not be.

Posted

Qfwfq: Thanks. As far as I think I understand it, with the Euclidean norm the diagonal is incommensurable with the sides on account of the root of 2. With the 1-norm does the diagonal equal twice the side? It seems to me if the square is considered in cells the diagonal is equal to one side, I dont know if this has a 'norm' name(?) I dont understand what's meant by the infinity-norm or how this shows the square to be a circle. Zero-norm and any others, I'm equally lost with.

Posted

I've never looked up the definitions of zero-norm, 1-norm infinity-norm.

 

For an application from V to R to be a norm it's enough for it to satisfy the 3 properties.

Posted

V is the vector space, R is the real field and the 3 properties may be summed up as follows:

 

|x| = 0 if x is the null vector otherwise |x| > 0 (positive definite norm)

 

for any scalar a, |a x| = |a||x|

 

|x| + |y| > |x + y|

 

I hope that's plain enough English! :hihi: I'm not sure what kind of textbook you have been following.

Posted

Qfwfq: The problem is that I know nothing about maths and I'm not using any textbook. Rather than trying to take in the full generalities, I'd like to be able to relate the norms to each other in the specific context of the diagonal of the square.

Vectors are magnitudes with fixed direction(?), so in this case the diagonal and the side are defined as distinct vectors? Is the "vector space" the locality of potential vectors ie the square? What is the "real field" in this situation? I'm not familiar with the vertical lines enclosing the x and y or the meaning of "scalar" in this situation, so I dont understand the relationships required by the 3 properties.

Posted

How did you happen on to the problem?

 

A norm is a generalized notion of the length (or magintude) of a vector, defined as an element of a linear vector space, in terms of its components. Its value is a real number.

 

You seem to be concerned with geometrical vectors (albeit with other norms) so your scarlars are real numers. Associating the sides of a square with two perpendicular vectors, the diagonal is their vector sum, x + y = d, and this would remain true even for an "odd" norm (supposing the scalar product isn't changed and hence the definition of perpendicular). What changes is the length of each of these three vectors.

 

I found what's meant by p-norm in Mathworld with the special case of infinity but no specification about p=0, which runs into difficulties. With the 1-norm, the length of d is the sum of the lengths of x and y. I don't see how infinity norm would give the square being a circle though.

Posted

As Qfwfq said, a norm is the generalized notion of length/size for a vector. Generalized is the key here, meaning firstly that the particular norm being used may be nothing like what you are familiar with. Secondly, the vectors are generalized vectors that may or may not have anything to do with numbers.

 

The 3 properties (actually I'd prefer the term "requirements") a norm must have is what links them to the idea of "size" (i will use norm and size interchangeably):

 

As stated by Qfwfq, the first property says that a zero vector should have zero size.

 

The second property involves scalars, which are things that "expand" or "contract" vectors whatever that means. It says that if you apply the scalar to the vector before finding its size, the answer should be the same as if you found the size of the vector first and then applied the scalar to it.

 

The third property is the "triangle property" meaning the sum of the sizes of the two sides of any triangle should be greater than or equal to the third side.

 

Notice there is no requirement that the pythagorean rule be satisfied for triangles.

 

I hope that sheds some light on what the meaning of the requirements of norms.

 

Qfwfq: I dont understand what's meant by the infinity-norm or how this shows the square to be a circle. Zero-norm and any others, I'm equally lost with.

 

THe infinity norm is |x|_inf = max(|x_i|).

x is any old vector. x_i are the components of that vector. In terms of Euclidean spaces, say R^2 (the number plane) x_i would be a real number.

The inf norm of any vector is equal to the largest component of that vector.

 

The definition of a unit circle is that the size of the vectors from the center to the circle must always be one. I.e. the norms of these vectors are always one.

 

With respect to the infinity norm, for a vector to lie on the "circle", we need |x|_inf = 1. THis means at least one of the two components must be one, and the other number no greater than one. For example x = (0.2, 1). This happens to trace out a square.

 

Does that answer it?

Posted

Qfwfq and Woog:

Thanks for the replies. I think I'm now okay with the properties/requirements.

I'm still fogged by the circles and squares with infinity-norm. Do you know where I can view some diagramatic representations that might make it clearer to me?

 

Qfwfq: This was just something that came from thinking about chess queens finding a8 and h8 equally distant from a1.

Posted

I found a page claiming that the infinity-norm is the same as my chess board cell-norm. If that's the case, I guess I can see what they mean, as all straight lines joining two sides and passing through the centre are equally long, the model describes a circle. However, the original cell model gives the diagonal as equal to the side whereas the conceived circle would give the diagonal as four times the side divided by pi, no?

Posted
Qfwfq: This was just something that came from thinking about chess queens finding a8 and h8 equally distant from a1.
Wow! :hihi:

 

That explains it to the tee!

Posted

Sorry, I hadn't seen your last post.

 

However, the original cell model gives the diagonal as equal to the side whereas the conceived circle would give the diagonal as four times the side divided by pi, no?
No, if the two sides are y=(0, 1) and x=(1, 0) the diagonal is d=(1, 1) and these three all have the same infinity-norm. So do all vectors that take you to points on the other two sides of the square which can each be described by:

ax + y and respectively ay + x, with -1 < a < 1.

 

With Woog's interpretation I now see what's meant about the circle. All I lack is a clear cut definition of 0-norm; I suppose the exponent is 1 but what about one or more components being 0? Is the summation defined extended only to nonzero ones?

Posted

Qfwfq: I understand that's the case of it as a square but as, with this model, it's a square with the properties of a circle why cant it be treated mathematically as a circle when using this model?

Posted

Oooooh, :hihi: I'll have to figure that sentence out!

 

Let's see... perhaps we could say that using a norm that isn't the ordinary one favoured by Euclid, the same set of points will be a square from one point of view and a circle from another.

Posted

I'm assuming norms are optional, if so I dont see a reason why choice would be restricted to the norm level. It doesn't matter either way to me as I'm only looking at these norms for fun, so if there's a conventional restriction on how to treat the infinity-norm diagonal, I'll ignore it and treat both the square and circle as valid.

 

A question raised by the manifold logical interpretations of something so apparently simple, is as to whether or not mathematical logic is a suitable tool for attempting confirmation of complex propositions?

Posted
A question raised by the manifold logical interpretations of something so apparently simple, is as to whether or not mathematical logic is a suitable tool for attempting confirmation of complex propositions?
Tricky point that so many people get stuck on. Math doesn't care about reality, you can define math that describes one thing or that describes another. It's a matter of picking the right definitions and axioms. There has been a discussion along these lines here.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...