Задача 71700 ...
Условие
Решение
[m]a_{o}=\frac{1}{π} ∫_{-π} ^{π}f(x)dx=\frac{1}{π} ∫_{-π} ^{0}0dx+\frac{1}{π} ∫_{0} ^{π}(x+2)dx=\frac{1}{π}(\frac{x^2}{2}+2x)|_{0} ^{π}=\frac{π}{2}+2=\frac{π+4}{2}[/m]
[m]a_{n}=\frac{1}{π} ∫_{-π} ^{π}f(x) \cdot cos\frac{nπx}{π}dx=\frac{1}{π} ∫_{-π} ^{0}0\cdot cos nx dx+\frac{1}{π} ∫_{0} ^{π}(x+2)\cdot cos nx dx=[/m]
Интегрирование по частям
u=x+2
dv=cosnx dx
du=dx
v=[m]\frac{1}{n} sinnx[/m]
⇒
[m]a_{n}=\frac{1}{π} ∫_{0} ^{π}(x+2)\cdot cos nx dx=\frac{1}{π}\cdot (\frac{x+2}{n} sinnx[)_{0} ^{π}-\frac{1}{π}\cdot ∫ _{0} ^{π}\frac{1}{n} sinnxdx=\frac{1}{πn}(π+2)\underbrace{sinπn}_{0}-\frac{1}{πn}(0+2)\underbrace{sin0}_{0} -\frac{1}{πn}\cdot ∫ _{0} ^{π}sinnxdx=[/m]
[m]-\frac{1}{πn}∫ _{0} ^{π}sinnxdx=-\frac{1}{πn^2}(-cosnx)| _{0} ^{π}=\frac{1}{πn^2}(cosπn-cos0)=\frac{1}{πn^2}((-1)^{n}-1)[/m]
при n=2k в скобках 0
При n=2k-1
[m]\frac{1}{π(2k-1)^2}(-1-1)=-\frac{2}{π(2k-1)^2}[/m]
[m]b_{n}=\frac{1}{π} ∫_{-π} ^{0}0\cdot sin nx dx+\frac{1}{π} ∫_{0} ^{π}((x+2)\cdot sin nx dx=[/m]
Интегрирование по частям
u=x+2
dv=sinnx dx
du=dx
v=[m]\frac{1}{n} (-cosnx)[/m]
⇒
[m]b_{n}=\frac{1}{π} \cdot (∫_{0} ^{π}((x+2)\cdot sin nx dx=\frac{1}{π}\cdot \frac{x+2}{n} (-cosnx[)_{0} ^{π}-\frac{1}{π}\cdot∫ _{0} ^{π}(-\frac{1}{n} cosnx)dx=-\frac{1}{πn}(π+2)\underbrace{cosπn}_{(-1)^{n}}+\frac{1}{πn}(0+2)\underbrace{cos0}_{1} -\frac{1}{πn}\cdot ∫ _{0} ^{π}sinnxdx=[/m]
[m]=-(-1)^{n}\frac{π+2}{πn}+\frac{2}{πn}-\frac{1}{πn^2}(-cosnx)| _{0} ^{π}=\frac{2}{πn}-(-1)^{n}\frac{π+2}{πn}+\frac{1}{πn^2}(-1)^{n}-1)=[/m]
[m]b_{n}=[/m]
⇒
[m]f(x)=\frac{a_{o}}{2}+ ∑^{n= ∞} _{n=1} (a_{n}cosπnx+b_{n}sinπnx)[/m]
[m]f(x)=\frac{\frac{π+4}{2}}{2}+ ∑^{n= ∞} _{k=1} (-\frac{2}{π(2k-1)^2}cosπ(2k-1)x+∑^{n= ∞} _{n=1}(\frac{2}{πn}-(-1)^{n}\frac{π+2}{πn}+\frac{1}{πn^2}(-1)^{n}-1)sinπnx[/m]