Results 1 to 7 of 7

Math Help - Stuck on a problem from Rudin Ch. 1 (16)

  1. #1
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98

    Stuck on a problem from Rudin Ch. 1 (16)

    I'm doing this problem as part of my effort to self-study my way through the first 7 chapters of Rudin.

    Suppose k  \ge 3, x, y \in \mathbb{R}^k, ||x-y|| = d > 0, and r > 0. Prove (a) that if 2r > d, there are infinitely many z \in \mathbb{R}^k such that ||z - x|| = ||z - y|| = r, (b) there is exactly one such z if 2r = d and (c) there is no such z if 2r < d.

    My progress:

    I proved (b) and (c) using the condition for attainment in Cauchy-Schwarz. Part (a) I proved after assuming that x = 0 and y = (y_1, 0, ..., 0). I'm probably just being stupid, but the problem I seem to be running into is that I'm having a hard time undoing my rotation and translation without going beyond what can be done with the material learned so far. Any help would be appreciated. If anyone could give a simple justification (i.e. one that doesn't use anything but definition of the Euclidean norm and the basic algebraic properties of the dot product) for assuming WLOG that x = 0 and y = (y_1, 0, ..., 0), this would certainly do the trick. Any other permissible solution would be fine too.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Guy View Post
    I'm doing this problem as part of my effort to self-study my way through the first 7 chapters of Rudin.

    Suppose k  \ge 3, x, y \in \mathbb{R}^k, ||x-y|| = d > 0, and r > 0. Prove (a) that if 2r > d, there are infinitely many z \in \mathbb{R}^k such that ||z - x|| = ||z - y|| = r, (b) there is exactly one such z if 2r = d and (c) there is no such z if 2r < d.

    My progress:

    I proved (b) and (c) using the condition for attainment in Cauchy-Schwarz. Part (a) I proved after assuming that x = 0 and y = (y_1, 0, ..., 0). I'm probably just being stupid, but the problem I seem to be running into is that I'm having a hard time undoing my rotation and translation without going beyond what can be done with the material learned so far. Any help would be appreciated. If anyone could give a simple justification (i.e. one that doesn't use anything but definition of the Euclidean norm and the basic algebraic properties of the dot product) for assuming WLOG that x = 0 and y = (y_1, 0, ..., 0), this would certainly do the trick. Any other permissible solution would be fine too.

    Thanks.
    Can't you just find one and extend the line from the origin it lies on outwards?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98
    I'm not sure I understand. I'm probably misunderstanding your post, but the set of z should be a circle (in R^3), so I don't see what extending a line comes in.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Guy View Post
    I'm not sure I understand. I'm probably misunderstanding your post, but the set of z should be a circle (in R^3), so I don't see what extending a line comes in.
    It can't be a sphere, then all the points would be a fixed distance from "one" point.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98
    The set of z such that ||z-x|| = r is clearly a sphere with in Euclidean space centered at x. Ditto for the set of z such that ||z - y|| = r. The set of z that satisfy both of these is an intersection of the two spheres, which is (I hope) a circle (for the case k = 3).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member nimon's Avatar
    Joined
    Sep 2009
    From
    Edinburgh, UK
    Posts
    64
    Hey,

    Can't you just take the points along the perpendicular bisector of y-x? By the triangle inequality, d = \|x-y \| \leq \|x-z\| + \|y-z\| = 2r for any point z equidistant from x,y. So you could never have 2r < d, and 2r = d precisely at the midpoint of y-x, and if we move from this midpoint along the perpendicular to y-x then we get an infinite number of points equidistant from x,y. That is, there are an infinite number of isosceles triangles whose base is y-x.

    That is not very mathematical, but it should be possible to translate it: the essential ingredient appears to be the triangle inequality.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98
    Ah yes, I should be able to, for any line that bisects and is orthogonal to y-x, find two points on it such that the equality holds for a given r. Since there are an infinite number of lines that satisfy those conditions, I'll have an infinite number of points for each r.

    Thanks for the help, much appreciated.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rudin's analysis
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: July 2nd 2011, 06:58 AM
  2. Theorem from Baby Rudin (Thm. 2.36)
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: January 13th 2011, 09:45 PM
  3. Baby Rudin problem 1.5
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: December 29th 2010, 07:11 AM
  4. Baby Rudin - ch 2, ex 27
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 21st 2010, 01:24 PM
  5. Rudin: chapter 6 exercise 13
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: May 11th 2010, 03:38 AM

Search Tags


/mathhelpforum @mathhelpforum