Question Number 190623 by mnjuly1970 last updated on 07/Apr/23
$$ \\ $$$$\:\:\:\:\:\:\mathrm{calculate}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{{n}}\:\mathrm{cos}\:\left(\frac{\:{n}\pi}{\mathrm{3}}\:\right)\:=? \\ $$$$ \\ $$
Answered by mahdipoor last updated on 07/Apr/23
![get f(x)=x^2 in [−3,3] a_0 =(1/3)∫_(−3) ^( 3) x^2 .dx=18 a_n =(1/3)∫_(−3) ^( 3) x^2 .cos(((nπx)/3))= =(1/3)[((x^2 sin(((nπx)/3)))/((nπ)/3))+((2xcos(((nπx)/3)))/((nπ)/3))−((2sin(((nπx)/3)))/((nπ)/3))]_(−3) ^( 3) =(2/(nπ))[9sin(nπ)+6cos(nπ)−2sin(nπ)]=((12)/(nπ))(−1)^n b_n =(1/3)∫_(−3) ^( 3) x^2 .sin(((nπx)/3))=0 ⇒⇒ x^2 =((18)/2)+Σ_(n=1) ^∞ (((12(−1)^n )/(nπ))cos(((nπx)/3)))⇒x=1⇒ 1−((18)/2)=−((12)/π)Σ_(n=1) ^∞ ((((−1)^(n−1) )/n)cos(((nπ)/3)))⇒ Σ_(n=1) ^∞ ((((−1)^(n−1) )/n)cos(((nπ)/3)))=((2π)/3)](https://www.tinkutara.com/question/Q190632.png)
$${get}\:{f}\left({x}\right)={x}^{\mathrm{2}} \:\:{in}\:\left[−\mathrm{3},\mathrm{3}\right]\:\:\: \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{3}}\int_{−\mathrm{3}} ^{\:\mathrm{3}} \:{x}^{\mathrm{2}} .{dx}=\mathrm{18} \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\int_{−\mathrm{3}} ^{\:\mathrm{3}} {x}^{\mathrm{2}} .{cos}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{{x}^{\mathrm{2}} {sin}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)}{\frac{{n}\pi}{\mathrm{3}}}+\frac{\mathrm{2}{xcos}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)}{\frac{{n}\pi}{\mathrm{3}}}−\frac{\mathrm{2}{sin}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)}{\frac{{n}\pi}{\mathrm{3}}}\right]_{−\mathrm{3}} ^{\:\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{{n}\pi}\left[\mathrm{9}{sin}\left({n}\pi\right)+\mathrm{6}{cos}\left({n}\pi\right)−\mathrm{2}{sin}\left({n}\pi\right)\right]=\frac{\mathrm{12}}{{n}\pi}\left(−\mathrm{1}\right)^{{n}} \\ $$$${b}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\int_{−\mathrm{3}} ^{\:\mathrm{3}} {x}^{\mathrm{2}} .{sin}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{18}}{\mathrm{2}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{12}\left(−\mathrm{1}\right)^{{n}} }{{n}\pi}{cos}\left(\frac{{n}\pi{x}}{\mathrm{3}}\right)\right)\Rightarrow{x}=\mathrm{1}\Rightarrow \\ $$$$\mathrm{1}−\frac{\mathrm{18}}{\mathrm{2}}=−\frac{\mathrm{12}}{\pi}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\right)\Rightarrow \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 07/Apr/23
$$\:\:{answer}\::\:\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$
Answered by Peace last updated on 08/Apr/23
$$={Re}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({e}^{{i}\frac{\pi}{\mathrm{3}}} \right)^{{n}} }{{n}} \\ $$$$={Reln}\left(\mathrm{1}+{e}^{{i}\frac{\pi}{\mathrm{3}}} \right) \\ $$$$={ln}\left(\mid\mathrm{1}+{e}^{{i}\frac{\pi}{\mathrm{3}}} \mid\right)={Re}\left(\mid{e}^{{i}\frac{\pi}{\mathrm{6}}} \mid\mid\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{6}}\right)\mid\right) \\ $$$$={ln}\left(\sqrt{\mathrm{3}}\right)=\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 08/Apr/23
$$\:\:{so}\:{nice}\:{sir}\:\:{thx}\:{alot}=. \\ $$
Commented by mehdee42 last updated on 08/Apr/23
$${very}\:{good} \\ $$