递推式T(n)=T(n-1)+T(n-2)+n∈Θ(2ⁿ)的证明咨询

Great start with the telescoping series approach for your upper bound recurrence! Let's finish this up step by step, then tie it together with the lower bound to complete the tight Θ(2ⁿ) proof using the squeeze theorem.

First, let's re-align your expanded equations to make the telescoping subtraction clearer (matching like exponents of 2):

S(n) = 2^{n-1} + 2^{n-2}·2 + 2^{n-3}·3 + … + 2²(n-2) + 2(n-1) + n
2S(n) = 2ⁿ + 2^{n-1}·2 + 2^{n-2}·3 + … + 2³(n-2) + 2²(n-1) + 2n

Subtract the first equation from the second—this collapses most terms into a clean geometric series:

2S(n) - S(n) = S(n) = 2ⁿ + (2^{n-1}·2 - 2^{n-1}) + (2^{n-2}·3 - 2^{n-2}·2) + ... + (2²(n-1) - 2(n-1)) + (2n - n)

Simplify each parenthetical term: every coefficient difference is 1, leaving pure powers of 2. The only exception is the final linear term, which doesn't have a matching higher-exponent pair in the original S(n), so we subtract it as -n:

S(n) = (2ⁿ + 2^{n-1} + 2^{n-2} + ... + 2^1) - n

Use the geometric series sum formula: the sum of 2^1 to 2ⁿ is 2^{n+1} - 2 (derived from sum_{k=1}^n 2^k = 2(2ⁿ - 1)/(2-1)). Plug this in to get the closed-form:

S(n) = 2^{n+1} - 2 - n = 2·2ⁿ - (n + 2)

Proving S(n) ∈ Θ(2ⁿ) #

To confirm this upper bound is tight (i.e., it's not asymptotically larger than 2ⁿ), we need to show both upper and lower bounds relative to 2ⁿ:

• Upper Bound: For all n ≥ 2, n + 2 ≤ 2ⁿ (easily proven by induction: base case n=2 gives 4=4; if true for n=k, then 2^{k+1}=2·2ᵏ ≥ 2(k+2) ≥ k+3 for k≥2). Substitute this into our closed-form:

  S(n) = 2·2ⁿ - (n+2) ≤ 2·2ⁿ
  

  Choose c₂=2 and n₀=2—this holds for all n ≥ n₀.

• Lower Bound: Pick c₁=1. For n ≥ 2, 2ⁿ ≥ n + 2 (same induction as above). Substitute:

  S(n) = 2·2ⁿ - (n+2) ≥ 2·2ⁿ - 2ⁿ = 2ⁿ = c₁·2ⁿ
  

  This holds for all n ≥ 2.

With both constants c₁,c₂ > 0 and a threshold n₀ where c₁·2ⁿ ≤ S(n) ≤ c₂·2ⁿ, we've proven S(n) ∈ Θ(2ⁿ).

For the squeeze theorem, we need a lower-bound recurrence L(n) such that L(n) ∈ Θ(2ⁿ) and T(n) ≥ L(n) for all n ≥ n₀.

Assuming your original T(n) follows a recurrence like T(n) = 2T(n-1) + f(n) where f(n) ≥ 0 (since you chose S(n)=2S(n-1)+n as an upper bound, we know f(n) ≤ n), we can use the simplest valid lower bound:

L(n) = 2L(n-1)

With a positive initial condition (e.g., L(0)=1), the closed-form is L(n)=2ⁿ, which obviously belongs to Θ(2ⁿ).

By induction, we can confirm T(n) ≥ L(n):

1. Base Case: Assume T(0) ≥ L(0) = 1 (valid for any positive initial condition of T(n)).
2. Inductive Step: If T(k) ≥ L(k) = 2ᵏ, then T(k+1) = 2T(k) + f(k+1) ≥ 2·2ᵏ = 2^{k+1} = L(k+1).

We now have all the pieces:

• L(n) ≤ T(n) ≤ S(n) for all n ≥ 2 (the maximum threshold from our upper/lower bound proofs)
• Both L(n) and S(n) belong to Θ(2ⁿ)

By the squeeze theorem for asymptotic complexity, this directly implies T(n) ∈ Θ(2ⁿ).

内容的提问来源于stack exchange，提问作者Alex B.