What is ln(((i/2-1/2)!)/(sqrt(pi)(i/2)!)) ?

The solution to ln(((i/2-1/2)!)/(sqrt(pi)(i/2)!)) is ln((i-1)/2 !)-1/2 ln(pi)-ln(i/2 !)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln(((i/2-1/2)!)/(sqrt(pi)(i/2)!))

simplify

ln (\frac{( \frac{i}{2} - \frac{1}{2} ) !}{π ( \frac{i}{2} ) !})

ln (\frac{i - 1}{2}!) - \frac{1}{2} ln (π) - ln (\frac{i}{2}!)

Solution steps

ln (\frac{( \frac{i}{2} - \frac{1}{2} ) !}{π ( \frac{i}{2} ) !})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{( \frac{i}{2} - \frac{1}{2} ) !}{π ( \frac{i}{2} ) !}) = ln ((\frac{i}{2} - \frac{1}{2})!) - ln (π (\frac{i}{2})!)

= ln ((- \frac{1}{2} + \frac{i}{2})!) - ln (π \frac{i}{2}!)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (π \frac{i}{2}!) = ln (π) + ln (\frac{i}{2}!)

= ln ((- \frac{1}{2} + \frac{i}{2})!) - (ln (π) + ln (\frac{i}{2}!))

ln (π) = \frac{1}{2} ln (π)

= ln ((- \frac{1}{2} + \frac{i}{2})!) - (\frac{1}{2} ln (π) + ln (\frac{i}{2}!))

- (\frac{1}{2} ln (π) + ln (\frac{i}{2}!)) : - \frac{1}{2} ln (π) - ln (\frac{i}{2}!)

= ln ((\frac{i}{2} - \frac{1}{2})!) - \frac{1}{2} ln (π) - ln (\frac{i}{2}!)

Combine the fractions

\frac{i}{2} - \frac{1}{2} : \frac{i - 1}{2}

= ln (\frac{- 1 + i}{2}!) - \frac{1}{2} ln (π) - ln (\frac{i}{2}!)

= ln (\frac{i - 1}{2}!) - \frac{1}{2} ln (π) - ln (\frac{i}{2}!)

ln (a) - 5 ln (b)

ln (3) + ln (8)

- ln (\frac{3}{2})

ln (3.9 \cdot 1 0^{- 13})

ln (\frac{T ^{2}}{T ^{1}})

What is ln(((i/2-1/2)!)/(sqrt(pi)(i/2)!)) ?
The solution to ln(((i/2-1/2)!)/(sqrt(pi)(i/2)!)) is ln((i-1)/2 !)-1/2 ln(pi)-ln(i/2 !)