S-expression: Difference between revisions
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...which simplifies to: (''k'' + 2)/''k'' = (''k'' + 1)/(''k'' - 1). | ...which simplifies to: (''k'' + 2)/''k'' = (''k'' + 1)/(''k'' - 1). | ||
This means that if we temper: <math> {\rm S}k \cdot {\rm S}(k+1) \large = \frac{k/(k-1)}{(k+1)/k} \cdot \frac{(k+1)/k}{(k+2)/(k+1)} = \frac{k/(k | This means that if we temper: <math> {\rm S}k \cdot {\rm S}(k+1) \large = \frac{k/(k-1)}{(k+1)/k} \cdot \frac{(k+1)/k}{(k+2)/(k+1)} = \frac{k/(k-1)}{(k+2)/(k+1)}</math> | ||
...then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k'' + 1) individually unless other considerations seem to force your hand. | ...then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k'' + 1) individually unless other considerations seem to force your hand. | ||