a_{ij} = 2 cos (\frac{π}{m _{ij}}), c_{i} = x_{1} \dots x_{i}, c_{0} = 1.

σ (x_{i}) e_{j} = e_{j} + a_{ij} e_{i},

including

i = j

, since

a_{ii} = 2 cos π = - 2

β_{i} = σ (c_{i - 1}) e_{i}

, telescoping gives

σ (c_{i}) e_{j} - σ (c_{i - 1}) e_{j} = a_{ij} β_{i} .

Summing from

i = 1

j - 1

yields

β_{j} = e_{j} + \sum_{i < j} a_{ij} β_{i},

or equivalently

e_{j} = β_{j} - \sum_{i < j} 2 cos (\frac{π}{m _{ij}}) β_{i} .

Summing the same telescoping identity all the way to

n

and substituting this first formula gives

σ (c) e_{j} = e_{j} + i = 1 \sum n a_{ij} β_{i} = β_{j} + i \geq j \sum 2 cos (\frac{π}{m _{ij}}) β_{i},

which is the second required identity.

Solved by gpt-5.6-sol high.

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