1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum of singular 1-cubes = boundary of a singular 2-cube?

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm doing question 23 in Chapter 4 of Spivak's Calculus on Manifolds. The question asks,
    For R > O, and n an integer, define the singular l-cube,
    [itex] c_{R,n} :[0,1] \rightarrow \mathbb {R}^2 - 0 [/itex] by [itex] c_{R,n} (t) = (Rcos2\pi nt, Rsin2\pi nt).[/itex] Show that there is a singular 2-cube [itex] c:[0,1]\rightarrow \mathbb {R}^2 - 0 [/itex] such that [itex] c_{R_1,n} - c_{R_2,n} = \partial c [/itex]

    2. Relevant equations
    [itex] \partial c = \sum_{i =1}^{n} \sum_{\alpha = 0,1} (-1)^{i+ \alpha} c_{i, \alpha} [/itex]
    [itex] c_{i, \alpha} = c(I^n_{i,\alpha})[/itex]


    3. The attempt at a solution
    What I'm confused about is probably the notation of
    [itex] c_{R_2,n} [/itex] and it's relation with [itex] c_{(i, \alpha)}[/itex], are they the same thing since
    [itex]c_{R,n} (t) = (Rcos2\pi nt, Rsin2\pi nt)[/itex]
    , which I find difficult to connect with the
    [itex](i,\alpha)[/itex]
    -face of c and I.
    To solve the question you use [itex] \partial [/itex] c, to get the (i, n) - face of c, from which c can be obtained. The solution is apparently
    [itex]c(t_1,t_2) = (t_1R_1 + (1-t_1)R_2)(cos2\pi nt_2, sin2\pi nt_1) [/itex]
     
    Last edited: Aug 17, 2014
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Sum of singular 1-cubes = boundary of a singular 2-cube?
  1. Finding singularities (Replies: 0)

Loading...