Big-O notation confusion

Question

I have come across some of the difficulties during doing this question. The question is: sort the following functions by order of growth from slowest to fastest:

7n^3 − 10n, 4n^2, n, n^8621909, 3n, 2^(log log n), n log n, 6n log n, n!, 1.1^n

And my answer to the question is

n,3n
nlogn, 6nlogn
4n^2(which equals to n^2)
7n^3 – 10n(which equals to n^3)
n^ 8621909
2^loglogn
1.1^n(exponent 2^0.1376n)
n!

Just wondering: can I assume that 2^(loglogn) has the same growth as 2^n? Should I take 1.1^n as a constant?

A bit off-topic but ... if had investment giving you 10% annually, for sure you would notice how `1.1^n` is different from a constant! In 7.27 years you would have doubled your initial investment! — eddie, Aug 21 '16 at 14:30

A. Sarid · Accepted Answer · 2016-09-26T20:45:12.420

9

Just wondering can i assume that 2^(loglogn) has the same growth as 2^n?

No. Assuming that the logs are in base 2 then 2^log(n) mathematically equals to n, so 2^(log(log(n)) = log(n). And of course it does not have the same growth as 2^n.

Should I take 1.1^n as a constant?

No. 1.1^n is still an exponent by n that cannot be ignored - of course it is not a constant.

The correct order is:

2^loglogn = log(n)
n,3n
nlogn, 6nlogn
4n^2
7n^3 – 10n
n^8621909
1.1^n
n!

I suggest to take a look at the Formal definition of the Big-O notation. But, for simplicity, the top of the list goes slower to infinity than the lower functions.
If for example, you will put two of those function on a graph you will see that one function will eventually pass the other one and will go faster to infinity.

Take a look at this comparing n^2 with 2^n. You will notice that 2^n is passing n^2 and going faster to infinity.
You might also want to check the graph in this page.

edited Sep 26 '16 at 20:45

answered Aug 21 '16 at 13:54

A. Sarid

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I think the final three in your series could use an explanation (i.e. why `n^8621909`, `1.1^n`, and `n!` appear in this order). – Tim Biegeleisen Aug 21 '16 at 14:07
"2^log(x) mathematically equals `n`". No. `2^(log base 2 of x)` is equal to `n`. However, `O(2^log(x)) = x` – intboolstring Aug 21 '16 at 18:30
@intboolstring thanks for this correction to make it precise. Edited my answer accordingly. – A. Sarid Aug 21 '16 at 18:43

score 4 · Answer 2 · answered Aug 21 '16 at 13:49

4

2^log(x) = x, so 2^log(log(n)) = log(n), which grows far more slower than 2ⁿ.
1.1ⁿ is definitely not a constant. 1.1ⁿ tends to infinity, and a constant obviously does not.

answered Aug 21 '16 at 13:49

Ami Tavory

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The OP needs to order the functions from slowest to fastest growing. Can you do this? – Tim Biegeleisen Aug 21 '16 at 13:55

Big-O notation confusion

2 Answers2