1-developp-at-fourier-serie-f-x-ln-sinx-2-developp-at-fourier-serie-g-x-ln-cosx-sinx-3-developp-at-fourier-seri-e-h-x-ln-cosx-2sinx-

Question Number 97230 by mathmax by abdo last updated on 07/Jun/20

1) developp at fourier serie f (x) = \ln (sinx)

2) developp at fourier serie g (x) = \ln (cosx + sinx)

3) developp at fourier seri e h (x) = \ln (cosx + 2 sinx)

Answered by mathmax by abdo last updated on 07/Jun/20

1) we have f^{^{'}} (x) = \frac{cosx}{sinx} = \frac{e^{ix} + e^{- ix}}{\frac{e^{ix} - e^{- ix}}{i}} = i \frac{e^{ix} + e^{- ix}}{e^{x} - e^{- ix}} changement e^{ix} = z give

f^{^{'}} (x) = i \frac{z + z^{- 1}}{z - z^{- 1}} = i \frac{z^{2} + 1}{z^{2} - 1} = - i \frac{z^{2} + 1}{1 - z^{2}} = - i (z^{2} + 1) \sum_{n = 0}^{\infty} z^{2 n}

= - i \sum_{n = 0}^{\infty} (z^{2 n + 2} + z^{2 n}) = - i \sum_{n = 0}^{\infty} e^{i (2 n + 2) x} - i \sum_{n = 0}^{\infty} e^{i 2 nx}

= - i {\sum_{n = 0}^{\infty} (\cos (2 n + 2) x + isin (2 n + 2) x) + \sum_{n = 0}^{\infty} (\cos (2 nx) + isin (2 nx))}

= - i (\sum_{n = 0}^{\infty} \cos (2 n + 2) x + \sum_{n = 0}^{\infty} \cos (2 nx)) + \sum_{n = 0}^{\infty} \sin (2 n + 2) x + \sum_{n = 0}^{\infty} \sin (2 nx)

but f^{^{'}} (x) is real \Rightarrow f^{^{'}} (x) = \sum_{n = 0}^{\infty} \sin (2 n + 2) x + \sum_{n = 0}^{\infty} \sin (2 nx)

= \sum_{n = 1}^{\infty} \sin (2 nx) + \sum_{n = 1}^{\infty} \sin (2 nx) = 2 \sum_{n = 1}^{\infty} \sin (2 nx) \Rightarrow

f (x) = 2 \sum_{n = 1}^{\infty} (- \frac{1}{2 n}) \cos (2 nx) + C = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} + C

f (\frac{π}{2}) = 0 = - \sum_{n = 1}^{\infty} \frac{\cos (n π)}{n} + C = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + C = \ln 2 + C \Rightarrow

C = - \ln 2 \Rightarrow f (x) = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} - \ln (2)

Commented by mathmax by abdo last updated on 07/Jun/20

2) g (x) = \ln (cosx + sinx) \Rightarrow g (x) = \ln (\sqrt{2} \sin (x + \frac{π}{4}))

= \frac{1}{2} \ln (2) + \ln (x + \frac{π}{4}) = \frac{1}{2} \ln (2) - \sum_{n = 1}^{\infty} \frac{\cos (2 n (x + \frac{π}{4}))}{n} - \ln (2)

= - \frac{\ln (2)}{2} - \sum_{n = 1}^{\infty} \frac{\cos (2 nx + n \frac{π}{2})}{n}

Commented by mathmax by abdo last updated on 07/Jun/20

3) h (x) = \ln (cosx + 2 sinx) we have cosx + 2 sinx = \sqrt{5} (\frac{1}{\sqrt{5}} cosx + \frac{2}{\sqrt{5}} sinx)

let \sin α = \frac{1}{\sqrt{5}} and \cos α = \frac{2}{\sqrt{5}} \Rightarrow \tan α = \frac{1}{2} \Rightarrow α = \arctan (\frac{1}{2}) \Rightarrow

cosx + 2 sinx = \sqrt{5} \sin (x + α) \Rightarrow h (x) = \frac{1}{2} \ln 5 + \ln (\sin (x + α))

= \frac{\ln 5}{2} - \sum_{n = 1}^{\infty} \frac{\cos (2 n (x + α))}{n} - \ln (2)

= \frac{\ln 5}{2} - \ln (2) - \sum_{n = 1}^{\infty} \frac{\cos (2 nx + 2 n \arctan (\frac{1}{2}))}{n}

Answered by mathmax by abdo last updated on 07/Jun/20

another method for f (x) = \ln (sinx) we have

f (x) = \ln (\frac{e^{ix} - e^{- ix}}{2 i}) = \ln (e^{ix} - e^{- ix}) - \ln (2 i)

= \ln (e^{ix}) + \ln (1 - e^{- 2 ix}) - \ln (2 i) = ix - \ln 2 - lni + \ln (1 - e^{- 2 ix})

= ix - \ln (2) - \frac{i π}{2} - \sum_{n = 1}^{\infty} \frac{e^{- 2 inx}}{n}

= i (x - \frac{π}{2}) - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} + i \sum_{n = 1}^{\infty} \frac{\sin (2 nx)}{n} - \ln (2)

f (x) \in R \Rightarrow f (x) = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} - \ln (2) also we get

\sum_{n = 1}^{\infty} \frac{\sin (2 nx)}{n} = \frac{π}{2} - x