Closed Shimuuar closed 5 years ago
Testing recurrence Γ(x+1) = x·Γ(x) shows that error is significantly larger than estimate for small x. Error was estimated as follows: δ(logΓ(x+1) - logΓ(x) - log x) =< δ(logΓ(x+1)) + δ(logΓ(x)) + δ(log x) where δ(f(x)) is estimated as ε·f(x).
Γ(x+1) = x·Γ(x)
δ(logΓ(x+1) - logΓ(x) - log x) =< δ(logΓ(x+1)) + δ(logΓ(x)) + δ(log x)
δ(f(x))
ε·f(x)
Fixed by 9045155eb23223deb7e349ca87c6acccccfa830f
Testing recurrence
Γ(x+1) = x·Γ(x)
shows that error is significantly larger than estimate for small x. Error was estimated as follows:δ(logΓ(x+1) - logΓ(x) - log x) =< δ(logΓ(x+1)) + δ(logΓ(x)) + δ(log x)
whereδ(f(x))
is estimated asε·f(x)
.