Logarithmic approximation of ∑ 0 N − 1 1 2 i + 1 Can anyone...

sedeln5w

sedeln5w

Answered

2022-06-27

Logarithmic approximation of 0 N 1 1 2 i + 1
Can anyone confirm that it's possible to approximate the sum 0 N 1 1 2 i + 1 with the log N 2 ? And why?
It's clearly visible that the sum has a logarithmic growth over i but it's unclear to me how to prove it.

Answer & Explanation

kuncwadi17

kuncwadi17

Expert

2022-06-28Added 16 answers

i = 0 N 1 1 2 i + 1 = 1 2 i = 0 N 1 1 i + 1 / 2 = 1 2 [ Ψ ( 1 2 + N ) Ψ ( 1 2 ) ] = 1 2 Ψ ( N + 1 2 ) + 1 2 γ + ln ( 2 )
where Ψ ( z ) is the digamma function and γ = 0.577215664901533 is the Euler-Mascheroni constant . Ψ ( 1 2 ) = γ 2 ln ( 2 )
Also, Ψ ( z ) ln ( z ) 1 2 z   when   | z | 1
i = 0 N 1 1 2 i + 1 1 2 [ ln ( N + 1 2 ) 1 2 ( N + 1 / 2 ) ] + 1 2 γ + ln ( 2 ) , N 1
i = 0 N 1 1 2 i + 1 1 2 ln ( N + 1 2 ) 1 2 ( 2 N + 1 ) + 1 2 γ + ln ( 2 ) , N 1
polivijuye

polivijuye

Expert

2022-06-29Added 16 answers

This is a Riemann sum for the integral from 0 to N of 1 / ( 2 x + 1 ), so it approximates ( 1 / 2 ) ln ( 2 N + 1 ), which is approximately ( 1 / 2 ) ln N

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