# Asymptotic growth rate of functions

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Thread starter 11 months ago
#1
My first question is, are n^2 and n^3 in different complexity classes?
Secondly, does 3^n have a higher growth rate than 2'n log n?
Similarly is n^2 log n bigger in terms if growth rate than n^3 log n?
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11 months ago
#2
(Original post by loginrunner)
My first question is, are n^2 and n^3 in different complexity classes?
Secondly, does 3^n have a higher growth rate than 2'n log n?
Similarly is n^2 log n bigger in terms if growth rate than n^3 log n?
1) one is quadratic, the other cubic. Both are polynomial.
2) is this (2^n) log(n), or is the log in the exponent?
3) similar to 1. The cubic must be bigger?
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Thread starter 11 months ago
#3
(Original post by mqb2766)
1) one is quadratic, the other cubic. Both are polynomial.
2) is this (2^n) log(n), or is the log in the exponent?
3) similar to 1. The cubic must be bigger?
For point 2 the log is not in the exponent. (2^n)log(n)
I know that n log n grows slower than n, so I'm wondering if 2^n log n grows slower than 3^n, even though 3^n is so much bigger than 2^n that it's hard to see on a graph if 2^n log n will ever overtake it
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11 months ago
#4
(Original post by loginrunner)
For point 2 the log is not in the exponent. (2^n)log(n)
I know that n log n grows slower than n, so I'm wondering if 2^n log n grows slower than 3^n, even though 3^n is so much bigger than 2^n that it's hard to see on a graph if 2^n log n will ever overtake it
The 3^n must be larger. Can you think how to relate them?
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Thread starter 11 months ago
#5
(Original post by mqb2766)
The 3^n must be larger. Can you think how to relate them?
I guess 2^n < (2^n) log(n) < 3^n?
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11 months ago
#6
(Original post by loginrunner)
I guess 2^n < (2^n) log(n) < 3^n?
Sort of, Id do
3^n = 1.5^n*2^n > n2^n > log(n)2^n
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Thread starter 11 months ago
#7
(Original post by mqb2766)
Sort of, Id do
3^n = 1.5^n*2^n > n2^n > log(n)2^n
That makes a lot of sense! Thanks
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