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Question Number 144702 by mathmax by abdo last updated on 28/Jun/21

let g(x)=log(cosx +2sinx) developp f at fourier serie

$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\left(\mathrm{cosx}\:+\mathrm{2sinx}\right) \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 28/Jun/21

we have cosx +2sinx)=(√5)((1/( (√5)))cosx +(2/( (√5)))sinx) let cosα=(1/( (√5))) and sinα=(2/( (√5))) ⇒tanα=2 ⇒α=arctan2 ⇒ cosx+2sinx=(√5)cos(x−α) ⇒g(x)=(1/2)log5 +log(cos(x−α)) the problem is turned to developp ϕ(t)=log(cost) we have ϕ(t)=log(((e^(it) +e^(−t) )/2))=log(e^(it) +e^(−it) )−log(2) =it +log(1+e^(−2it) )−log2 we[have log^′ (1+u)=(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ log(1+u)=Σ_(n=0) ^∞ (((−1)^n u^(n+1) )/(n+1))=Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒log(1+e^(−2it) )=Σ_(n=1) ^∞ (−1)^(n−1) (e^(−2int) /n) ⇒ϕ(t)=−log2 +it+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)(cos(2nt)+isin(2nt)) =−log2+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)cos(2nt)+i(t+Σ(...)sint) ϕ(t)real ⇒ϕ(t)=−log2 +Σ_(n=1) ^∞ (((−1)^(n−1) )/n)cos(2nt) ⇒ log(cosx+2sinx)=(1/2)log5 +ϕ(x−α) =(1/2)log5 −log2 +Σ_(n=1) ^∞ (((−1)^(n−1) )/n)cos(2n(x−α)) =(1/2)log5−log2+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)cos(2nx−2narctan(2))

$$\left.\mathrm{we}\:\mathrm{have}\:\mathrm{cosx}\:+\mathrm{2sinx}\right)=\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\mathrm{cosx}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\mathrm{sinx}\right) \\ $$$$\mathrm{let}\:\mathrm{cos}\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\mathrm{and}\:\mathrm{sin}\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{tan}\alpha=\mathrm{2}\:\Rightarrow\alpha=\mathrm{arctan2}\:\Rightarrow \\ $$$$\mathrm{cosx}+\mathrm{2sinx}=\sqrt{\mathrm{5}}\mathrm{cos}\left(\mathrm{x}−\alpha\right)\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log5}\:+\mathrm{log}\left(\mathrm{cos}\left(\mathrm{x}−\alpha\right)\right) \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{turned}\:\mathrm{to}\:\mathrm{developp}\:\varphi\left(\mathrm{t}\right)=\mathrm{log}\left(\mathrm{cost}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\varphi\left(\mathrm{t}\right)=\mathrm{log}\left(\frac{\mathrm{e}^{\mathrm{it}} +\mathrm{e}^{−\mathrm{t}} }{\mathrm{2}}\right)=\mathrm{log}\left(\mathrm{e}^{\mathrm{it}} \:+\mathrm{e}^{−\mathrm{it}} \right)−\mathrm{log}\left(\mathrm{2}\right) \\ $$$$=\mathrm{it}\:+\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2it}} \right)−\mathrm{log2} \\ $$$$\mathrm{we}\left[\mathrm{have}\:\:\mathrm{log}^{'} \left(\mathrm{1}+\mathrm{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow\right. \\ $$$$\mathrm{log}\left(\mathrm{1}+\mathrm{u}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2it}} \right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{e}^{−\mathrm{2int}} }{\mathrm{n}} \\ $$$$\Rightarrow\varphi\left(\mathrm{t}\right)=−\mathrm{log2}\:+\mathrm{it}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\left(\mathrm{cos}\left(\mathrm{2nt}\right)+\mathrm{isin}\left(\mathrm{2nt}\right)\right) \\ $$$$=−\mathrm{log2}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{cos}\left(\mathrm{2nt}\right)+\mathrm{i}\left(\mathrm{t}+\Sigma\left(...\right)\mathrm{sint}\right) \\ $$$$\varphi\left(\mathrm{t}\right)\mathrm{real}\:\Rightarrow\varphi\left(\mathrm{t}\right)=−\mathrm{log2}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{cos}\left(\mathrm{2nt}\right)\:\Rightarrow \\ $$$$\mathrm{log}\left(\mathrm{cosx}+\mathrm{2sinx}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log5}\:+\varphi\left(\mathrm{x}−\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log5}\:−\mathrm{log2}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{cos}\left(\mathrm{2n}\left(\mathrm{x}−\alpha\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log5}−\mathrm{log2}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{cos}\left(\mathrm{2nx}−\mathrm{2narctan}\left(\mathrm{2}\right)\right) \\ $$

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