Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 96194 by mathmax by abdo last updated on 30/May/20

let f(x) =ln(cosx) developp f at fourier serie

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{cosx}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Answered by john santu last updated on 30/May/20

ln(cos x) = ln (((e^(ix) +e^(−ix) )/2))  = ln(e^(ix) +e^(−ix) ) −ln(2)  = ln(e^(−ix) (1+e^(2ix) ))−ln(2)  =−ix−ln(2)+ln(1+e^(2ix) )   =−ix−ln(2)+Σ_(p = 1) ^∞ (−1)^(p−1)  ((e^(−2ixp) /p))

$$\mathrm{ln}\left(\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{ln}\:\left(\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right) \\ $$$$=\:\mathrm{ln}\left({e}^{{ix}} +{e}^{−{ix}} \right)\:−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=\:\mathrm{ln}\left({e}^{−{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)\right)−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=−{ix}−\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)\: \\ $$$$=−{ix}−\mathrm{ln}\left(\mathrm{2}\right)+\underset{\mathrm{p}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{p}−\mathrm{1}} \:\left(\frac{{e}^{−\mathrm{2}{ixp}} }{{p}}\right) \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 30/May/20

we have ln(cosx) =ln(((e^(ix)  +e^(−ix) )/2)) =ln(e^(ix)  +e^(−ix)  −ln(2)  =ln(e^(ix) (1+e^(−2ix) ))−ln2 =ix +ln(1+e^(−2ix) )−ln2  =ix +Σ_(n=1) ^∞  (((−1)^(n−1) )/n) (e^(−2ix) )^n  −ln2  =ix +Σ_(n=1) ^∞  (((−1)^(n−1) )/n) (cos(2nx)−isin(2nx))−ln2  =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) cos(2nx)−ln2 +i(x−Σ_(n=1) ^∞  (((−1)^(n−1) )/n)sin(2nx))  but ln(cosx)∈ R ⇒ln(cosx) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n)cos(2nx)−ln2  and x−Σ_(n=1) ^∞  (((−1)^(n−1) )/n)sin(2nx) =0

$$\mathrm{we}\:\mathrm{have}\:\mathrm{ln}\left(\mathrm{cosx}\right)\:=\mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{ix}} \:+\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2}}\right)\:=\mathrm{ln}\left(\mathrm{e}^{\mathrm{ix}} \:+\mathrm{e}^{−\mathrm{ix}} \:−\mathrm{ln}\left(\mathrm{2}\right)\right. \\ $$$$=\mathrm{ln}\left(\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}+\mathrm{e}^{−\mathrm{2ix}} \right)\right)−\mathrm{ln2}\:=\mathrm{ix}\:+\mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2ix}} \right)−\mathrm{ln2} \\ $$$$=\mathrm{ix}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\left(\mathrm{e}^{−\mathrm{2ix}} \right)^{\mathrm{n}} \:−\mathrm{ln2} \\ $$$$=\mathrm{ix}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\left(\mathrm{cos}\left(\mathrm{2nx}\right)−\mathrm{isin}\left(\mathrm{2nx}\right)\right)−\mathrm{ln2} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\mathrm{cos}\left(\mathrm{2nx}\right)−\mathrm{ln2}\:+\mathrm{i}\left(\mathrm{x}−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{2nx}\right)\right) \\ $$$$\mathrm{but}\:\mathrm{ln}\left(\mathrm{cosx}\right)\in\:\mathrm{R}\:\Rightarrow\mathrm{ln}\left(\mathrm{cosx}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{cos}\left(\mathrm{2nx}\right)−\mathrm{ln2} \\ $$$$\mathrm{and}\:\mathrm{x}−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{2nx}\right)\:=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com