Sum of order of O(1)+O(2)+ .... +O(n)

Question

What does the sum O(1)+O(2)+ .... +O(n) evaluate to?

I have seen its solution somewhere it was written:

O(n(n+1) / 2) = O(n^2)

but I am not satisfied with it because O(1) = O(2) = constant, so according to me it must evaluate to O(n) only. I have also seen this in Cormen:

Σ(i=1 to n) O(i)

In above expression there is only a single anonymous function. This function is not the same as O(1) + O(2) + ... + O(n) which doesn't really have a clean interpretation.

This question appears to be off-topic because it is about math — Sean Owen, Sep 12 '13 at 07:15
If it's [Big O notation](http://en.wikipedia.org/wiki/Big_O_notation) as used to calculate [time complexity](http://en.wikipedia.org/wiki/Time_complexity), then this question might be better on http://programmers.stackexchange.com/ — Some programmer dude, Sep 12 '13 at 07:19
This question appears to be off-topic because it is more conceptual, and probably a better fit for http://programmers.stackexchange.com/ — Some programmer dude, Sep 12 '13 at 07:20
@JoachimPileborg Time complexity isn't the only thing that can be dealt with using Big O notation. — , Sep 12 '13 at 07:55
This question makes no sense. Big-O expressions O(...) are not values that can be summed in this way. — Chris Martin, Jan 23 '16 at 10:11

score 4 · Answer 1 · edited May 23 '17 at 10:27

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The question seems to be perfectly on-topic since there is a tag asymptotic_complexity...

According to CLRS, p. 49,

"The number of anonymous functions in an expression is understood to be equal to the number of times the asymptotic notation appears. For example, in the expression

sum(O(i), i=1..n)

there is only a single anonymous function (a function of i). This expression is thus not the same as O(1) + O(2) + ... + O(n), which doesn't really have a clean interpretation"

Actually, in your formula, the constants behind the "O" notation may be all different, and their growth may change the asymptotic behaviour of the whole sum. Don't write this!

To answer your question more completely, in sum(O(i), i=1..n), you can use the fact that (see GKP p. 450 for the following)

O(f(n)g(n)) = f(n) O(g(n))

Thus, O(i) = i O(1), this time with the same O(1) in your formula. Therefore,

sum(O(i), i=1..n) = sum(i, i=1, n) O(1)

=n(n+1)/2 O(1) = O(n^2)

This way you can eliminate your sum without trouble.

edited May 23 '17 at 10:27

Community

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answered Sep 12 '13 at 07:40

"the constants begind the "O" notation may be all different,"; `O(.)` yields a set, not a value. The 'constants' don't exist per se. – Sep 12 '13 at 07:57
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the actual confusion stems from the fact that in expanding the sum, `n ` is no longer treated as a parameter denoting the size of the input but as a cardinal constant. so what you actually have is a countably infinite family of expansions, one for each input size, which does not make for a good tool to describe complexity. however, you can still operate formally on the definitions behind big-Oh notation. – collapsar Sep 12 '13 at 07:58
@Rhymoid. O(n) is a family of functions, but f \in O(n) really means there is a **constant** M such that f(n) <= M g(n). Well, at least it's what all authors say, including the two books I point to in the answer, but also [Knuth TAOCP](https://en.wikipedia.org/wiki/The_Art_of_Computer_Programming) and Aho, Hopcroft and Ullman. – Sep 12 '13 at 08:02
That constant is existentially bound in the definition of the set. The argument of `O(.)` is actually a function, and another function can be in that set. Instead of `O(n)`, one should actually write `O(n :> n)` (`n` maps to `n`), but for convenience, this binding is implicit. Strictly speaking, if you'd write `O(n)` with `n` already bound to a value, you have a subset of `O(1)` (more formally, `O(x :> 1)`). I've seen this explained imprecisely (or even incorrectly) many times, often to the point of confusion of my fellow students. I don't think TAOCP tells it differently, just less precise. – Sep 12 '13 at 08:45
In summary: there are many interpretations of `O(.)`, some precise, some brief. This question calls for a very precise definition of `O(.)`, one that TAOCP does not give (or you paraphrased it incorrectly). (Also, my interpretation is heavily influenced by functional programming.) – Sep 12 '13 at 08:50

collapsar · Answer 2 · 2013-09-12T07:49:02.457

Go by the definition of the Big-Oh notation.

You have n functions f_i, each of which satisfies

a_i * i <= f_i(x) <= b_i * i; x > X_i

for some positive constants a_i, b_i and sufficiently large X_i. let X = max_i ( X_i ) and sum over the n inequalities to get

sum_i=1..n ( a_i * i ) <= sum_i=1..n ( f_i(x) ) <= sum_i=1..n ( b_i * i )

noting that

sum_i=1..n ( min(a_i) * i ) <= sum_i=1..n ( a_i * i )
sum_i=1..n ( b_i * i )      <= sum_i=1..n ( max(b_i) * i )

arriving at

    min(a_i) * 0.5*(n(n-1))  <= sum_i=1..n ( f_i(x) ) <= max(b_i) * 0.5*(n(n-1))
<=>       A       * (n(n-1)) <= sum_i=1..n ( f_i(x) ) <=        B      *(n(n-1))

thus the sum of the functions you started with is indeed O(n^2).

score 0 · Answer 3 · answered Sep 12 '13 at 07:22

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sure, O(1) and O(2) is a constant, so we can forget about them. But is O(n/2) constant, I guess no. So try (for your homework) count just second half of the elements:

O(n/2) + O(n/2+1) + ... O(n) = ??

you will come up with O(n^2) :)

answered Sep 12 '13 at 07:22

sodik

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score 0 · Answer 4 · answered Sep 12 '13 at 13:34

There is a trick to find the formula, really not that hard.

You have: 1 + 2 + 3 + 4 + .... + n

If you count the list twice (one time ordered from low to high one time from high to low you get twice the result)

((1 + 2 + 3 + 4 + ... + n) + (n + (n - 1) + (n - 2) ... + 2 + 1)) / 2

This is the same as

((1 + n) + (2 + n - 1) + (3 + n - 2) + ... + (n + 1)) / 2

And this is:

((n + 1) * n) / 2

or as we right in o notion O(n²).

Sum of order of O(1)+O(2)+ .... +O(n)

4 Answers4