Question Number 189865 by uchihayahia last updated on 23/Mar/23
![i know b_n =0, but a_0 =(4/3)F_0 (according to solution) my answer is a_0 =(8/3)F_0 where did i did wrong how to find answer (Fourier transformation) like picture below (in the comment? f(t)= { ((((3F_0 )/t_0 )t,0≤t≤(t_0 /3))),((F_0 , (t_0 /3)≤t≤((2t_0 )/3))),((((-3F_0 )/t_0 )t+3F_0 ,((2t_0 )/3)≤t≤t_0 )) :} a_n =(2/t_0 )∫_0 ^t_0 f(t) cos(nωt)dt a_n =(2/t_0 )∫_0 ^t_0 f(t) cos(nωt)dt a_0 =(2/t_0 )∫_0 ^t_0 f(t) dt ∫_0 ^((4t_0 )/3) f(t) dt =∫_0 ^(t_0 /3) ((3F_0 )/t_0 )tdt+∫_(t_0 /3) ^((2t_0 )/3) F_0 dt+∫_((2t_0 )/3) ^t_0 ((-3F_0 )/t_0 )t+3F_0 dt =[((3F_0 )/t_0 )t]_0 ^(t_0 /3) +[F_0 t]_(t_0 /3) ^((2t_0 )/3) +[((-3F_0 )/t_0 )t+3F_0 t]_((2t_0 )/3) ^t_0 =F_0 +(1/3)F_0 t_0 −F_0 +F_0 t_0 =(4/3) F_0 t_0 so a_0 =(2/t_0 )×(4/3)F_0 t_0 =(8/3)F_0](https://www.tinkutara.com/question/Q189865.png)
$$ \\ $$$$\:{i}\:{know}\:{b}_{{n}} =\mathrm{0},\:{but}\:{a}_{\mathrm{0}} =\frac{\mathrm{4}}{\mathrm{3}}{F}_{\mathrm{0}} \left({according}\:\right. \\ $$$$\left.\:{to}\:{solution}\right)\:{my}\:{answer}\:{is}\:{a}_{\mathrm{0}} =\frac{\mathrm{8}}{\mathrm{3}}{F}_{\mathrm{0}} \\ $$$$\:{where}\:{did}\:{i}\:{did}\:{wrong}\:{how}\:{to} \\ $$$${find}\:{answer}\:\left({Fourier}\:{transformation}\right) \\ $$$$\:{like}\:{picture}\:{below}\:\left({in}\:{the}\:{comment}?\right. \\ $$$$\:{f}\left({t}\right)=\begin{cases}{\frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t},\mathrm{0}\leqslant{t}\leqslant\frac{{t}_{\mathrm{0}} }{\mathrm{3}}}\\{{F}_{\mathrm{0}} ,\:\frac{{t}_{\mathrm{0}} }{\mathrm{3}}\leqslant{t}\leqslant\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}}\\{\frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} ,\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}\leqslant{t}\leqslant{t}_{\mathrm{0}} }\end{cases} \\ $$$$\:{a}_{{n}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{cos}\left({n}\omega{t}\right){dt}\: \\ $$$$\:{a}_{{n}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{cos}\left({n}\omega{t}\right){dt}\: \\ $$$$\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }\int_{\mathrm{0}} ^{{t}_{\mathrm{0}} } {f}\left({t}\right)\:{dt}\: \\ $$$$\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}{t}_{\mathrm{0}} }{\mathrm{3}}} {f}\left({t}\right)\:{dt}\:=\int_{\mathrm{0}} ^{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} \frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{tdt}+\int_{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} ^{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} {F}_{\mathrm{0}} {dt}+\int_{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} ^{{t}_{\mathrm{0}} } \frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} {dt} \\ $$$$\:\:=\left[\frac{\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}\right]_{\mathrm{0}} ^{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} +\left[{F}_{\mathrm{0}} {t}\right]_{\frac{{t}_{\mathrm{0}} }{\mathrm{3}}} ^{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} +\left[\frac{-\mathrm{3}{F}_{\mathrm{0}} }{{t}_{\mathrm{0}} }{t}+\mathrm{3}{F}_{\mathrm{0}} {t}\right]_{\frac{\mathrm{2}{t}_{\mathrm{0}} }{\mathrm{3}}} ^{{t}_{\mathrm{0}} } \\ $$$$\:={F}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{3}}{F}_{\mathrm{0}} {t}_{\mathrm{0}} −{F}_{\mathrm{0}} +{F}_{\mathrm{0}} {t}_{\mathrm{0}} \\ $$$$\:=\frac{\mathrm{4}}{\mathrm{3}}\:{F}_{\mathrm{0}} {t}_{\mathrm{0}} \\ $$$$\:{so}\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{{t}_{\mathrm{0}} }×\frac{\mathrm{4}}{\mathrm{3}}{F}_{\mathrm{0}} {t}_{\mathrm{0}} =\frac{\mathrm{8}}{\mathrm{3}}{F}_{\mathrm{0}} \\ $$$$ \\ $$$$ \\ $$
Commented by uchihayahia last updated on 23/Mar/23
$$ \\ $$
Commented by uchihayahia last updated on 23/Mar/23
