Can someone explain how Big-Oh works with Summations?

Question

I know this isn\'t strictly a programming question, but it is a computer science question so I\'m hoping someone can help me.

I\'ve been working on my Algor

Answer 1

• Σ (i=1 to n) i² = n(n+1)(2n+1)/6, which is O(n³).

• Note that (n!)² = (1 n) (2(n-1)) (3(n-2))...((n-1)2) (n 1)
  = Π (i=1 to n) i (n+1-i)
  >= Π (i=1 to n) n
      [E.g., because for each i=1 to n, (i-1)(n-i) >= 0. Compare with Graham/Knuth/Patashnik, section 4.4]
  = nⁿ.
  Thus, n! >= n^n/2, and therefore
  Σ (i=1 to n) log i = log Π (i=1 to n) i = log n! >= log n^n/2 = (n/2) log n, which is Ω(n log n).

Answer 2

http://en.wikipedia.org/wiki/Big_O_notation

N repetitions of g(m)=O(f(m)) is O(N*f(m)) for any f(N).

Sum of i=1..N of i*g(i) is O(N*f(N)) if g(n)=O(f(n)) and f is monotonic.

Definition: g(n)=O(f(n)) if some c,m exist so for all n>m, g(n)<=c*f(n)

The sum is for i=1..N of i*f(i).

If f is monotonic in i this means every term is <= if(N) <= Nf(N). So the sum is less than N*c*f(N) so the sum is O(N*f(N)) (witnessed by the same c,m that makes g(n)=O(f(n)))

Of course, log_2(x) and x^2 are both monotonic.

Answer 3

Probably, your CPU will multiply 32-bit integers in constant time. But big-Oh doesn't care about "less than four billion", so maybe you have to look at multiplication algorithms?

According to Wolfram, the "traditional" multiplication algorithm is O(n²). Although n in this case is the number of digits, and thus really log(n) in the actual number. So I should be able to square the integers 1..n in time O(n.log(n)). Summing is O(log(n²)), which is obviously O(log(n)), for an overall complexity of O(n.log(n)).

So I can quite understand your confusion.

Answer 4

The simplest approach that jumps out to me is a proof by induction.

For the first one, essentially you need to show that

sum (i=1 to n) i^2 < k*n^3, k > 2,n > 0

If we use the generalized principle of induction and take a base case of n=1 and k=2.

we get 1<2*1.

Now of course take the inductive hypothesis, then we know that

sum(i=1 to n) i^2<k *n^3, with a bit of fun math we get to

sum(i=1 to n) i^2+(n+1)^2 < k *n^3+(n+1)^2.

• Now show k * n^3+(n+1)^2 < k *(n+1)^3

• k *n^3+n^2+2n+1 < k *n^3+k *(3n^2+3n+1)

• k *n^3 < k *n^3+(3k-1) *n^2+(3k-2) *n+k-1

Therefore, our original result is correct.

For the second proof you need to show that sum(i=1 to n) log_2(i) >= k*n*log(n), which I'll leave as an exercise for the reader ;).

The main step though is sum(i=1 to n) log_2(i)+log_2(n+1)>=k*n*log(n)+k*log(n+1), for some k, so clearly k is 1.

Answer 5

My guess is that what the question statement means is if you're summing the results of some calculation for which the running time is proportional to i² in the first case, and proportional to log₂i in the second case. In both cases, the running time of the overall summation is "dominated" by the larger values of N within the summation, and thus the overall big-O evaluation of both will be N*O(f) where f is the function you're summing.