By John Montroll

Step by step directions and over 1,000 transparent diagrams convey starting and skilled paperfolders the best way to create 27 remarkable polyhedra from one sheet of paper. Graded in response to hassle, the tasks variety from an easy dice, tetrahedron and octahedron to a hard rhombic dodecahedron, sunken icosahedron, and an antidiamond with pentagonal base.

It's easily remarkable to determine a fancy form as a dodecahedron(12 sided polyhedron) or an icosahedron (20 sided polyhedron) shape on your fingers after a large amount of twiddling and scratching your head thinking about the right way to get to the following step...

Scanning caliber is nice. All textual content (and the photographs of course!) is clear

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**Extra info for A Plethora of Polyhedra in Origami**

**Example text**

2B R The Fourier inversion formula for f holds ( f is integrable) and it holds everywhere ( f is continuous). In particular, the nth Fourier coefficient of is equal to f − The formal Fourier series of n . 2B is therefore 1 2B f n∈Z n n e−2iπ 2B ν . e. 9), that is is almost everywhere equal to its Fourier series. This proves (†). 9), a normally convergent series of bounded and continuous functions). 1 Fourier Theory in L 1 1 f˜(t) = 2B 43 f n∈Z = R n∈Z n 2B n R h(ν) e2iπ ν(t− 2B ) dν n n 1 e−2iπ ν 2B f 2B 2B h(ν) e2iπ νt dν.

13 Let f : R → C be an integrable function such that for some α > 1, f (t) = O 1 1 + |t|α (|t| → ∞), and f (ν) = O 1 1 + |ν|α (|ν| → ∞), then the strong Poisson summation formula holds for all T ∈ R+ \{0}. 21. ) with rapid convergence, or to obtain some remarkable formulas. Here is a typical example. Let a > 0. The ft of the function f (t) = e−2πa|t| is f (ν) = π(a 2a+ν 2 ) . 1 Fourier Theory in L 1 41 is a continuous function with bounded variation, we have by the strong Poisson summation formula e−2πa|n| = n∈Z The left-hand side is equal to as 1 πa +2 a n≥1 π(a 2 +n 2 ) .

1 Fourier Theory in L 1 39 Proof The ft g of the (integrable) function g is g(ν) e−2iπ νt dt . g(t) = R By the inversion formula, the last integral is equal to g(−t). Since g is assumed continuous, this equality holds everywhere, and in particular for t = nT . By the weak version of the Poisson sum formula, T g(nT ) = T g(− nT ) is the nth Fourier coefficient of n∈Z g(ν + n/T ). Therefore if (b) is true, then (a) is necessarily true. Conversely, if (a) is true, then the sequence {T g(− nT )}n∈Z is the sequence of Fourier coefficients of two functions, the constant function equal to T g(0), and n∈Z g(ν + n/T ), and therefore the two functions must be equal almost everywhere.