• n \mapsto a^{-n} is negligible for any a \geq 2: •
Step: This is an
exponential decay function where a is a constant greater than or equal to 2. As n \to \infty, a^{-n} \to 0 very quickly, making it negligible. • f(n) = 3^{-\sqrt{n}} is negligible: •
Step: This function has exponential decay with a base of 3, but the exponent grows slower than n (only at \sqrt{n}). As n \to \infty, 3^{-\sqrt{n}} \to 0, so it’s still negligible but decays slower than 3^{-n}. • f(n) = n^{-\log n} is negligible: •
Step: In this case, n^{-\log n} represents a polynomial decay, with the exponent growing negatively due to \log n. Since the decay rate increases with n, the function goes to 0 faster than polynomial functions like n^{-k} for any constant k, making it negligible. • f(n) = (\log n)^{-\log n} is negligible: •
Step: This function decays as the
logarithm of n raised to a negative exponent -\log n, which leads to a fast approach to 0 as n \to \infty. The decay here is faster than inverse logarithmic or polynomial rates, making it negligible. • f(n) = 2^{-c \log n} is not negligible, for positive c: •
Step: We can rewrite this as f(n) = n^{-c}, which is a polynomial decay rather than an exponential one. Since c is positive, f(n) \to 0 as n \to \infty, but it doesn’t decay as quickly as true exponential functions with respect to n, making it non-negligible. Assume n > 0, we take the limit as n \to \infty:
Negligible: • f(n) = \frac{1}{x^{n/2}}: •
Step: This function decays exponentially with base x raised to the power of -\frac{n}{2}. As n \to \infty, x^{-\frac{n}{2}} \to 0 quickly, making it negligible. • f(n) = \frac{1}{x^{\log{(n^k)}}} for k \geq 1: •
Step: We can simplify x^{-\log(n^k)} as n^{-k \log x}, which decays faster than any polynomial. As n \to \infty, the function approaches zero and is considered negligible for any k \geq 1 and x > 1. • f(n) = \frac{1}{x^{(\log n)^k}} for k \geq 1: •
Step: The decay is determined by the base x raised to the power of -(\log n)^k. Since (\log n)^k grows with n, this function approaches zero faster than polynomial decay, making it negligible.= • f(n) = \frac{1}{x^{\sqrt{n}}}: •
Step: Here, f(n) decays exponentially with a base of x raised to -\sqrt{n}. As n \to \infty, f(n) \to 0 quickly, so it’s considered negligible. • f(n) = \frac{1}{x^{n(\log n)}}: •
Step: With an exponential base and exponent n(\log n), this function would approach zero very rapidly, suggesting negligibility.
Non-negligible: • f(n) = \frac{1}{n^{1/n}}: •
Step: Since n^{1/n} \to 1 as n \to \infty, this function decays very slowly, failing to approach zero quickly enough to be considered negligible. ==See also==