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Old 2017-12-16, 15:54   #23
Alberico Lepore
 
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I thought

6^X

where X is

N^(1/X)=1
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Old 2017-12-16, 16:13   #24
Alberico Lepore
 
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I am not practical in mathematics.
It should be about 6 ^ [log_2 (log_10 (N))]
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Old 2017-12-16, 16:59   #25
Dr Sardonicus
 
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Let's see here: Subject says O([log_9(N)]^3)

Post #5 says
Quote:
sorry it's 3 ^ [log_9 (N)] = sqrt (N).
but I will try again
Post #23 says
Quote:
I thought

6^X

where X is

N^(1/X)=1

I am not practical in mathematics.
Apparently not. if N > 1, and N^(1/X) = 1, then X = ln(N)/(2*Pi*I*k) for a non-zero integer k, where Pi is the circle number and I^2 = -1. So exp(ln(6)*X) is a non-real complex number of absolute value 1.

Post #24 says
Quote:
It should be about 6 ^ [log_2 (log_10 (N))]
So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.

I can hardly wait to see what's next

Last fiddled with by Dr Sardonicus on 2017-12-16 at 17:04
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Old 2017-12-16, 17:04   #26
Alberico Lepore
 
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Quote:
Originally Posted by Dr Sardonicus View Post
Let's see here: Subject says O([log_9(N)]^3)

Post #5 says

Post #23 says

Apparently not. if N > 1, and N^(1/X) = 1, then X = ln(N)/(2*Pi*I*k) for a non-zero integer k, where Pi is the circle number and I^2 = -1. So exp(ln(6)*X) is a non-real complex number of absolute value 1.

Post #24 says

So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.

I can hardly wait to see what's next
thank you
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Old 2017-12-17, 18:44   #27
CRGreathouse
 
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Quote:
Originally Posted by Dr Sardonicus View Post
So now you're claiming, if I did the algebra correctly,

O(log(N)^(log(6)/log(2))).

Since log(6)/log(2) < 3, that's smaller than your original claim.
I get the same.
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