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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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