- #1
EvLer
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How can I show that function defined as
f(n) = f(n-1) + 1/n is O(logn) ?
Do I need to use induction here?
Thanks in advance.
f(n) = f(n-1) + 1/n is O(logn) ?
Do I need to use induction here?
Thanks in advance.
Firstly, we assume that f(n) = O(log n) is true for some constant k. Then, we use induction to prove that f(n+1) = O(log(n+1)) by showing that the inequality f(n+1) ≤ k * log(n+1) holds true for all n ≥ 1. This can be done by rewriting f(n+1) in terms of f(n) and using the induction hypothesis.
The intuition behind using induction is that we can break down the proof into smaller steps, starting with the base case (n=1) and then showing that if the statement holds for n, it also holds for n+1. This allows us to build up the proof for all values of n, and ultimately prove that f(n) = O(log n) for all n ≥ 1.
Sure, let's say we have the function f(n) = n^2 + log n. We want to prove that f(n) = O(log n) using induction. Firstly, we assume that f(n) = O(log n) is true for some constant k. Then, we show that f(n+1) ≤ k * log(n+1) by rewriting f(n+1) as (n+1)^2 + log(n+1) and using the induction hypothesis for f(n). This will prove that f(n) = O(log n) for all n ≥ 1.
No, there are other methods such as direct proof, proof by contradiction, and proof by contrapositive. However, induction is a commonly used method for proving statements about recursive functions, and it can be particularly helpful for proving f(n) = O(log n).
Yes, induction can be used to prove f(n) = O(log n) for non-recursive functions as long as the function can be broken down into smaller steps that can be proven using induction. For example, if we have a function that involves a summation, we can use induction to prove that the summation is bounded by a logarithmic function.