Answer:
a. The sequence x_n = n^-2 converges quadratically to zero as n approaches ∞.
b. The sequence x_n = 2^-n converges linearly to zero as n approaches ∞.
c. The sequence x_n = 2^-2n converges quadratically to zero as n approaches ∞.
d. The sequence x_n = 2^-a_n with a_0 = a_1 = 1 and a_n+1 = a_n+a_n-1 for n≥2 converges superlinearly to zero as n approaches ∞.
Explanation:
a. The sequence {x_n} = {n^(-2)} converges quadratically to zero as n approaches ∞.
To see this, we can calculate the limit of the ratio between consecutive terms:
lim_{n -> ∞} (x_{n+1} / x_n) = lim_{n -> ∞} ((n+1)^(-2) / n^(-2)) = lim_{n -> ∞} ((n / (n+1))^2) = 1
Since the limit is equal to 1, we know that the convergence rate of {x_n} is at least quadratic. This can also be seen from the fact that the error in the nth term is O(n^(-3)), which is smaller than the error in a linearly converging sequence (which is O(n^(-1))).
b. The sequence {x_n} = {2^(-n)} converges linearly to zero as n approaches ∞.
To see this, we can calculate the limit of the ratio between consecutive terms:
lim_{n -> ∞} (x_{n+1} / x_n) = lim_{n -> ∞} (2^(-(n+1)) / 2^(-n)) = lim_{n -> ∞} (1/2) = 1/2
Since the limit is strictly less than 1, we know that the convergence rate of {x_n} is linear.
c. The sequence {x_n} = {2^(-2n)} converges quadratically to zero as n approaches ∞.
To see this, we can calculate the limit of the ratio between consecutive terms:
lim_{n -> ∞} (x_{n+1} / x_n) = lim_{n -> ∞} (2^(-(2(n+1))) / 2^(-2n)) = lim_{n -> ∞} (1/4) = 1/4
Since the limit is equal to 1/4, we know that the convergence rate of {x_n} is at least quadratic. This can also be seen from the fact that the error in the nth term is O(n^(-2)), which is smaller than the error in a linearly converging sequence (which is O(n^(-1))).
d. The sequence {x_n} = {2^(-a_n)}, where a_0 = a_1 = 1 and a_{n+1} = a_n + a_{n-1} for n >= 2, converges superlinearly to zero as n approaches ∞.
To see this, we can use the fact that a_n grows exponentially with n. We can prove by induction that a_n >= phi^n, where phi is the golden ratio. Since the terms of {x_n} are decreasing and bounded below by 0, they must converge to a limit L. Taking the limit of the ratio of consecutive terms, we get:
lim_{n -> ∞} (x_{n+1} / x_n) = lim_{n -> ∞} (2^(-a_{n+1}) / 2^(-a_n)) = lim_{n -> ∞} (2^(-a_n - a_{n-1})) = 0
This shows that the convergence rate of {x_n} is superlinear.