Question

Помогите пожалуйста решить задачу ​

Answer 1

Объяснение:

$f(x)=x+1.\\f(x)=\frac{a_0}{2} +\sum \limits _{n=1}^\infty} (a_ncos(nx)+b_nsin(nx)).$

$a_0=\frac{1}{\pi } *\int\limits^\pi _{-\pi} {(x+1)} \, dx=\frac{1}{\pi } *(\frac{x^2}{2}+x)|_{-\pi }^\pi =\frac{1}{\pi }*((\frac{\pi ^2}{2} -\frac{(-\pi)^2 }{2} )+(\pi -(-\pi )))=\\ =\frac{1}{\pi } *((\frac{\pi ^2}{2}-\frac{\pi ^2}{2})+(\pi +\pi ))=\frac{1}{\pi }*(0+2\pi )=\frac{2\pi }{\pi } =2.\\ a_n=\frac{1}{\pi } *\int\limits^\pi _{-\pi } {(x+1)*cos(nx)} \, dx=\frac{1}{\pi } *(\int\limits^\pi _{-\pi } {(x*cos(nx))} \, dx +\int\limits^\pi _{-\pi } {cos(nx)} \, dx )=$

$=\frac{1}{\pi } *(\left | {{u=x\ \ \ \ v= cos(nx)} \atop {du=dx\ \ \ \ \ dv=\frac{1}{n}sinx }} \right.| +\int\limits^\pi _{-\pi } {cos(nx)} ) =\\=\frac{1}{\pi } *(x*cos(nx)\ |^\pi _{-\pi }-\int\limits^\pi _{-\pi } {cos(nx)} \, dx +\int\limits^\pi _{-\pi } {cos(nx)})=\\=\frac{1}{\pi }*(\pi *cos(\pi n)-(-\pi)*cos(-\pi n))=\frac{1}{\pi } *(\pi *cos(\pi n)+\pi *cos(\pi n))=\\ =\frac{2*\pi *cos(\pi n)}{\pi }=2cos(\pi n)=2*(-1)^n.$

$b_n=\frac{1}{\pi } *\int\limits^\pi _{-\pi } {(x+1)*sin( nx)} \, dx =\frac{1}{\pi }*(\int\limits^\pi _{-\pi } {(x*sin(nx)} \, dx +\int\limits^\pi _{-\pi } {sin(nx))} \, dx=\\ =\frac{1}{\pi } *(x*sin(nx)\ |^\pi _{-\pi } -\int\limits^\pi _{-\pi } {sin(nx)}+\int\limits^\pi _{-\pi } {sin(nx)} \, dx =\\ =\frac{1}{\pi } *(\pi *sin(\pi n)-(-\pi )*sin(-\pi n))=\frac{1}{\pi }*(\pi *0-\pi *0)=\frac{1}{\pi }*0=0.\ \ \Rightarrow$

$f(x)=\frac{2}{2} +\sum \limits _{n=1}^\infty} (2*(-1)^ncos(nx)+0*sin(nx)=1+\sum \limits _{n=1}^\infty} 2*(-1)^ncos(nx).$