Nuprl Definition : reg-seq-nexp

reg-seq-nexp(x;k) ==  λn.(x n^k ÷ n^k 1)



Definitions occuring in Statement :  apply: a lambda: λx.A[x] divide: n ÷ m multiply: m subtract: m natural_number: $n fastexp: i^n
Definitions occuring in definition :  lambda: λx.A[x] divide: n ÷ m apply: a fastexp: i^n multiply: m subtract: m natural_number: $n
FDL editor aliases :  reg-seq-nexp

Latex:
reg-seq-nexp(x;k)  ==    \mlambda{}n.(x  n\^{}k  \mdiv{}  2  *  n\^{}k  -  1)



Date html generated: 2016_05_18-AM-06_56_51
Last ObjectModification: 2015_09_23-AM-09_00_59

Theory : reals


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