My Math Forum visual space with more than three dimensions

 Topology Topology Math Forum

May 13th, 2015, 02:35 PM   #21
Senior Member

Joined: Dec 2014
From: Brazil

Posts: 203
Thanks: 1

Quote:
 Originally Posted by v8archie Well, my guess might be wrong (see here), but yours is definitely not correct either. What's a 'viewable space' then?

In portuguese "viewable" means "visualizavel", when a space had more than 3 dimensions, can't be viewable, understand now?

about the link than you give, a got a prove, with think solve the question.

We can write U1+U2+⋯ as the sum of two subspaces, for example U1+(U2+U3+⋯), and then:
dim(U1+U2+⋯)=dim(U1)+dim(U2+U3+⋯)−dim(U1∩( U2+U3+⋯)).
So then applying the same argument to the term dim(U2+U3+⋯) recursively you can then eventually arrive at an equation consisting of the sum of the dimensions of each subspace minus a bunch of nasty terms involving intersections similar to dim(U1 ∩(U2+U3+⋯))..
In our example, this term uderline, being the dimension of minor, in case, U1,
then, 3 dimensions, the other term will be 3n dimensions, so, the sum will be 3(n-1). You can get in adress linear algebra - The dimension of the sum of subspaces $(U_1,\ldots,U_n)$ - Mathematics Stack Exchange.

seems the same results, than I show in my post.

Lucio

Last edited by lemgruber; May 13th, 2015 at 02:38 PM.

 May 14th, 2015, 01:23 PM #22 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 (xo,yo,zo)= (0,0,0) be the center of sphere with rn it's ray of n-esima sphere: Change the equation of sphere, (x1 - x0)² + (y1 - y0)² + (z1 - z0)²= r²1 ∈ R³1 (x2 - x0)² + (y2 - y0)² + (z2 - z0)²= r²2 ∈ R³2 . . . . . . . . . . . . . . . (xn - x0)² + (yn - y0)² + (zn - z0)²= r²n ∈ R³n to coordinates who made a sphere in subspace vector, we got: (ρ1 sin φ cos θ)² + (ρ1 sin φ sin θ)² + (ρ1 cos φ)²= r²1 ∈ R³1 (ρ2 sin φ cos θ)² + (ρ2 sin φ sin θ)² + (ρ2 cos φ)²= r²2 ∈ R³2 . . . . . . . . . . . . (ρn sin φ cos θ)² + (ρn sin φ sin θ)² + (ρn cos φ)²= r²n ∈ R³n where R³1 ⊂ R³2 ⊂.........⊂ R³n where R³n are subspaces vectors and n ∈ N ( natural set numbers) ρ1 ≠ ρ2 ≠.............≠ ρn where n ∈ N ( natural set numbers) Dimension of Sum of subspaces vector: Proposition- Be V a vector space of dimension finite, if U1 and U2 are vectors subspace of V, then, Dim(U1 + U2)= Dim(U1) + Dim(U2) - Dim(U1 ∩ U2). in case Un, where n ∈ N ( natural set numbers), we got: dim(U1+U2+...Un)=dim(U1)+dim(U2+U3+....+Un) - dim(U1 ∩(U2+U3+....+Un)). see:http://math.stackexchange.com/questi...ldots-u-n?rq=1 In case of subspaces of spheres, we got: Dim (R³1 + R³2 +.....+ R³n) = Dim (R³1) + Dim(R³2 + .... +R³n) - Dim( R³1 ∩ (R³2 + ... +R³n) = 3 + 3(n - 1) - 3 = 3(n - 1) Dim (R³1 + R³2 + .... + R³n) = 3(n - 1) Dimensions this sum make a space vector compound of n -1 spheres concentric, that can be drawn on a graph, a graph 3(n - 1) dimensional. This, at least to me, a new thing. Apologize delay to make right form, but now seems ready. First I understand this with faith, after I could demonstrate mathematically. First I receive the idea, and I like thanks to being who make this possible. Lucio Last edited by lemgruber; May 14th, 2015 at 02:13 PM.
 May 14th, 2015, 05:55 PM #23 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 In time: ρ1=r1 ; ρ2=r2 ;....; ρn=rn and ρ1<ρ2<......<ρn
 May 15th, 2015, 09:52 AM #24 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 The utility for this, are inumerous, although the geometry of n-dimensional space, was much developed, the utilization of figures geometrics depicts on space n dimensional, let the understand about the geometry of fenomenous much more easy and fast. Lucio
 May 15th, 2015, 11:21 AM #25 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The utility is 0 until someone can understand you.
 May 15th, 2015, 11:46 AM #26 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 In economy, for example, the utility function operates in n dimensional space, but we can't draw the geometric figures, start of four dimensions, with this tool, became possible viewable the geometric figures and understand in another way, more intuitive what be said in language not geometric. The utility function don't have a spheric equation, but can be transformed in equations of sphere or became a subspace vector trough another transformation like a example in this post. Please, if someone understand, what I try to say, help me. Say something! Lucio Last edited by lemgruber; May 15th, 2015 at 11:57 AM.
May 15th, 2015, 12:06 PM   #27
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
 Originally Posted by lemgruber In economy, for example, the utility function operates in n dimensional space, but we can't draw the geometric figures
I don't understand why this would be a problem. I mean, there are ways to visualize things in more than three dimensions (see, for example, literally any complex analysis course), but it's not a limitation to fail to draw something.

 May 15th, 2015, 12:18 PM #28 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 I agree Greathouse, but when we draw geometry in graphic, the understand gain in clarity, again, I don't say what not exist ways to interpret things without draw figures, but it's much better when we do. At least I think so. Lucio
May 19th, 2015, 02:42 PM   #29
Senior Member

Joined: Dec 2014
From: Brazil

Posts: 203
Thanks: 1

Quote:
 Originally Posted by lemgruber In time: ρ1=r1 ; ρ2=r2 ;....; ρn=rn and ρ1<ρ2<......<ρn

Just possible if, ρ1≠r1 ; ρ2≠r2 ;....; ρn≠rn

 Tags dimensions, space, viewable, visual, whith

,

### r³2#%>-.*

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ProJO Calculus 4 February 10th, 2011 04:59 PM jamin1001 New Users 0 October 23rd, 2007 01:20 AM ^e^ Algebra 1 February 14th, 2007 10:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top