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

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Homework Statement


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]

Homework 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]


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]
 
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