cuti3panda Posted September 11, 2004 Share Posted September 11, 2004 -y^2-z^2=1.......from this equation, is it a circle or what?? help me anyone!! Link to comment Share on other sites More sharing options...

Dave Posted September 11, 2004 Share Posted September 11, 2004 Don't think so. It's almost the equation of a circle, but with a radius of -1 - heh. Link to comment Share on other sites More sharing options...

BrainMan Posted September 11, 2004 Share Posted September 11, 2004 err...nevermind. The answer will have to be complex... Link to comment Share on other sites More sharing options...

Dave Posted September 11, 2004 Share Posted September 11, 2004 Yup. I don't think that equation represents any special kind of geometric object, but I suppose I'm probably wrong Link to comment Share on other sites More sharing options...

pulkit Posted September 11, 2004 Share Posted September 11, 2004 No real numbers can satisfy that equation. Please specify the domain. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 14, 2004 Share Posted September 14, 2004 Maybe a complex circle ? It is rather hard to imagine such higher dimensional circles and it depends on what you call a circle too. In any case there are no real solutions and many many complex ones. You could for exemple assume that y is real and z is complex and let y parametrize z. You will find that the solutions of your equation are situated on the imaginary axis covering almost all except for the part in between -i and i. You could decompose y and z in imaginary and real parts and obtain 2 (non linear) equations with 4 variables. The thing is you need 4 dimensions to show the collection of all points satisfying your equation. Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted September 14, 2004 Share Posted September 14, 2004 I thought maths for higher dimensions was well defined, although I myself have no experience in this field, but have heard of it on several occasions. Could this not be a 4-d surface/volume ? Link to comment Share on other sites More sharing options...

Dave Posted September 14, 2004 Share Posted September 14, 2004 I thought maths for higher dimensions was well defined, although I myself have no experience in this field, but have heard of it on several occasions. Well, sure. Something like [math]\mathbb{R}^{n}[/math] is defined mathematically, it's just your physical interpretation of it. What exactly is as 4-dimensional cylinder? A cylinder is well defined in [math]\mathbb{R}^3[/math], but that may not necessarily extend to higher dimensions. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 15, 2004 Share Posted September 15, 2004 I thought maths for higher dimensions was well defined' date=' although I myself have no experience in this field, but have heard of it on several occasions. Could this not be a 4-d surface/volume ?[/quote'] Yeah. Take for instance the [math]\sum_{i=1}^n x_i^2 \leq 1[/math] as an equation in [math]\mathbb{R}^n[/math], it is the equation of a ball, but well pretty hard to make a picture of an n-dimensional ball right ? [math]\max_{i} |x_i| \leq 1[/math] would be the n-dimensional hypercube, but same difficulty to imagine what it would look like. Maths is well defined also in infinitely many dimensions, but the problem is just to imagine what the geometrical structure looks like. Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted September 15, 2004 Share Posted September 15, 2004 I have studied about balls in a course on metric spaces. I understand that they are (a)in general very hard to visualise (b) not always circular even in 2/3 dimensions. What I really want to know is wether in maths there is defined an extension into n dimensions for most geometrical entities. I have heard about n dimensional spheres, the notion of distance here, is it the same definition as we have in 2/3 dimensions ? , or like metric spaces it is also customized ? Link to comment Share on other sites More sharing options...

matt grime Posted September 15, 2004 Share Posted September 15, 2004 the distance between two points x and y in R^n is [MATH](\sum_1^n(x_i-y_i)^2)^{1/2}[/MATH] for x=(x_1,x_2,..x_n) etc why would you need to picture any of these things in your head though? it doesn't help particularly. Link to comment Share on other sites More sharing options...

Dave Posted September 15, 2004 Share Posted September 15, 2004 And they don't really serve any use in a geometrical sense. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 16, 2004 Share Posted September 16, 2004 the distance between two points x and y in R^n is [MATH](\sum_1^n(x_i-y_i)^2)^{1/2}[/MATH] for x=(x_1' date='x_2,..x_n) etc why would you need to picture any of these things in your head though? it doesn't help particularly.[/quote'] Or any other distance, many can be chosen. Like [MATH](\sum_1^n |x_i-y_i|[/MATH] or even [MATH](\sum_1^n |x_i-y_i|^p)^{\frac{1}{p}}[/MATH] or distance based on infinity norm, discrete metric etc..... the whole point of using metric spaces theory or even banach space theory (or hilbert spaces) is that you actually have a geometry in some abstract space that you can use (though not visualise). There is little point in trying to visualise such structures and if you perse want some 3d or 2d image suffices to get the idea of how stuff works.... Mandrake Link to comment Share on other sites More sharing options...

Manifold Posted September 17, 2004 Share Posted September 17, 2004 -y^2-z^2=1.......from this equation, is it a circle or what?? help me anyone!! This seems to be a complex elliptical cylinder with respect to y- and z-axes...or simply an empty set. Link to comment Share on other sites More sharing options...

bloodhound Posted September 17, 2004 Share Posted September 17, 2004 but to visualise [math]\mathbb{C}^2[/math] you need 4 orthogonal axes. which is impossible for a human mind to comprehend. i dont know if a 3 axes projection of it can be made Link to comment Share on other sites More sharing options...

Manifold Posted September 17, 2004 Share Posted September 17, 2004 you're right... well...I told about the axes because they are defined...but this doesn't mean by far that they exist...it all just loses sense here...that's why it's an empty set. Link to comment Share on other sites More sharing options...

pulkit Posted September 18, 2004 Share Posted September 18, 2004 When you say elliptical cylider, yet you know that its a 4-dimensional figure, exactly which property/properties of the ellipse or cylider do you generalise to the fourth dimension. It seems easy to carry forward spheres into n dimensions, but how do you carry forward cylinders and ellipses ? Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 20, 2004 Share Posted September 20, 2004 you're right...well...I told about the axes because they are defined...but this doesn't mean by far that they exist...it all just loses sense here...that's why it's an empty set. There are at least infinitely many solutions. Try fixing y to a real number and solving for z complex. You will see that almost all the complex axis is a solution (parametrized by y) except for a small part in the center of it. Mandrake Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 20, 2004 Share Posted September 20, 2004 When you say elliptical cylider, yet you know that its a 4-dimensional figure, exactly which property/properties of the ellipse or cylider do you generalise to the fourth dimension. It seems easy to carry forward spheres into n dimensions, but how do you carry forward cylinders and ellipses ? It is the equation that does the trick. [math]x^2 + y^2 = 1[/math] is a nice circle right ? [math]x^2 + y^2 + z^2= 1[/math] would be a ball So well [math]\sum_{i=1}^n x_i^2= 1[/math] would be the n-dimensional ball (hyperball if you like). Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted September 20, 2004 Share Posted September 20, 2004 As I said I have no problems with carrying forward spheres, now that I think of it even ellipsoids. But how do you carry forward a cylinder. I am unaware of the explicit form of the equation of a cylinder in cartesian coordinates (i only have a slight idea - working with cylindrical coordinates), could someone write that down please ? Link to comment Share on other sites More sharing options...

Dapthar Posted September 20, 2004 Share Posted September 20, 2004 I am unaware of the explicit form of the equation of a cylinder in cartesian coordinates (i only have a slight idea - working with cylindrical coordinates), could someone write that down please ? Equations: [math](x-x_0)^2+(y-y_0)^2=r^2[/math] [math](x-x_0)^2+(z-z_0)^2=r^2[/math] [math](z-z_0)^2+(y-y_0)^2=r^2[/math] The "centers" are at the points with the subscripts. There is the more general equation of a cylinder that is not parallel to one of the axes, however, it isn't particularly enlightening. The intuitive notion is that one picks two axes, and simply describes a circle in that coordinate system. Then, when one moves to the 3-D Cartesian plane, all the cross sections from the axes that was excluded from the initial equation are circles, thus one gets a cylinder. Alternatively, one could simply convert the equation of a cylinder in cylindrical coordinates to Cartesian coordinates. Link to comment Share on other sites More sharing options...

pulkit Posted September 20, 2004 Share Posted September 20, 2004 COnverting into cartesian is what I also mentioned. But that won't be too clever a method when your cylinder isn't aligned to the axis. About the equations you gave, I supose in order to specify finite cylinders, you just specify a range on third coordinate. Link to comment Share on other sites More sharing options...

Dave Posted September 20, 2004 Share Posted September 20, 2004 Well, your standard cylindrical change of co-ordinates is: [math]\begin{cases} x = r \cos\theta\\ y = r \sin\theta\\ z = z \end{cases} [/math] Then if you wanted, for example, to find the volume of a basic cylinder of height h and radius r_{0}, the set representing this would be: [math]\left\{ (r,\theta,z) \; \vert \; r \in [0, r_{0}], \theta \in [0,2\pi], z \in [0, h] \right\}[/math] From there, it's pretty straightforward to set up a triple integral to find your volume. Of course, this is complete overkill for a simple cylinder, but demonstrates the method pretty well. Link to comment Share on other sites More sharing options...

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