According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994): sum_{i = 1}^{n}\log_2(i)=nlog_2(n)-2^{log_2(n)) + 1 Now, I was wondering if someone knows what the following may equal? sum_(i = 1)^(n)i log_2(i)=?

grippeb9

grippeb9

Answered question

2022-08-11

Summation with Ceilinged Logarithmic Function
According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
i = 1 n log 2 ( i ) = n log 2 ( n ) 2 log 2 ( n ) + 1
Now, I was wondering if someone knows what the following may equal?
i = 1 n i log 2 ( i ) = ?

Answer & Explanation

yassou1v

yassou1v

Beginner2022-08-12Added 14 answers

Use sum by parts:
1 k n x k Δ y k = x n + 1 y n + 1 x 1 y 1 1 k n y k + 1 Δ x k
In your case Δ y k = log 2 k and x k = k. You'll get a sum similar to the original, you should be able to solve for that one.
rivasguss9

rivasguss9

Beginner2022-08-13Added 4 answers

Concretely,
1 n i log a ( i ) d i = n . n ( log a ( n ) 1 ) 1 ( log a ( 1 ) 1 ) 1 n i ( log a ( i ) 1 ) d i 1 n i log a ( i ) d i = n 2 ( log a ( n ) 1 ) + 1 [ i 2 ( 2 log a ( i ) 3 ) 4 ] 1 n 1 n i log a ( i ) d i = n 2 ( log a ( n ) 1 ) + 1 [ n 2 ( 2 log a ( n ) 3 ) 4 1 ( 2 log a ( 1 ) 3 ) 4 ] 1 n i log a ( i ) d i = n 2 ( log a ( n ) 1 ) + 1 [ n 2 ( 2 log a ( n ) 3 ) 4 + 3 4 ] 1 n i log a ( i ) d i = n 2 log a ( n ) n 2 + 1 4 n 2 log a ( n ) 2 + 3 n 2 4 1 n i log a ( i ) d i = n 2 log a ( n ) 2 n 2 4 + 1 4 Therefore,  i = 1 n i log a ( i ) n 2 log a ( n ) 2 n 2 4 + 1 4

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