Question

Working on something

Answer 1

\[\large\rm a_o =\frac{1}{2L}\int\limits_{-L}^{L} f(x) dx \] L=0.5\[\large\rm =\frac{1}{2(\frac{1}{2})} \int\limits_{-0.5}^{0.5} f(x) ~dx ~ ~~\normalsize ~~~~~~~~ Substitute~ L=0.5\]\[\large \rm =\int\limits_{-0.5}^{0} -1~dx + \int\limits_{0}^{0.5} 1~ dx~ ~~~~ \normalsize ~Separates~ the ~integral \] \[\large \rm=[-x]^0_{-0.5} + [x]^{0.5}_0 ~~~~~~~~~~~~\normalsize Integrate \]\[\large\rm a_o=[0-0.5]+[0+0.5]=0~~~~\normalsize ~~~~plug ~in ~the~ upper~ and~ lower~ limit\]

Answer 2

\[\large\rm =\frac{1}{\frac{1}{2}} \int\limits_{-0.5}^{0.5} f(x) \cdot \cos( \frac{\pi n x}{\large\rm \frac{1}{2}}) ~dx~~~~~~\normalsize ~~Substitute~~ L=0.5\]\[\large\rm =2[\int\limits_{-0.5}^{0} - \cos( 2 \pi n x) ~dx+\int\limits_{0}^{0.5} \cos( 2 \pi n x ) ~dx] ~~~~~ \normalsize Separates ~the~ integral \]\[\large\rm =2([-\frac{1}{2 \pi n }\sin(2 \pi n x)]^0_{-0.5} + [-\frac{1}{2 \pi n}\sin(2 \pi nx)]^0_{0.5}) ~~~~~\normalsize Integrate \] \[\large\rm a_n=2(\frac{1}{2 \pi n})(0)= 0~~~~~~\normalsize Plug~ in~ upper~ and~ lower~ limits \]

Answer 3

\[\large\rm b_n = 2 [\int\limits_{-0.5}^{0}-\sin(2 \pi nx) +\int\limits_{0}^{0.5} \sin(2 \pi nx)]~~~~~\normalsize ~~~Substitute ~L=0.5\]\[\large\rm =2([\frac{1}{2\pi n}\cos (2 \pi nx)]^0_{-0.5} +[ -\frac{1}{2\pi n}\cos(2 \pi nx)]^0_{0.5} )~~~~~~~\normalsize ~~~Integrate \]\[\large\rm =\frac{1}{\pi n }[(1-\cos(\pi n)) +(-\cos(\pi n)-(-1))]~~~\normalsize ~~~Plug ~in~upper~and~lower~limits \]\[\large\rm b_n=\frac{1}{\pi n}(2-2\cos(\pi n))~~~~~~~\normalsize ~~~~~~Simplify \]

Answer 4

\[\large\rm b_n = 2 [\int\limits_{-0.5}^{0}-\sin(2 \pi nx) +\int\limits_{0}^{0.5} \sin(2 \pi nx)]~~~~~\normalsize ~~~Substitute ~L=0.5\]\[\large\rm =2([\frac{1}{2\pi n}\cos (2 \pi nx)]^0_{-0.5} +[ -\frac{1}{2\pi n}\cos(2 \pi nx)]^0_{0.5} )~~~~~~~\normalsize ~~~Integrate \]\[\large\rm =\frac{1}{\pi n }[(1-\cos(\pi n)) +(-\cos(\pi n)-(-1))]~~~\normalsize ~~~Plug ~in~upper~and~lower~limits \]\[\large\rm b_n=\frac{1}{\pi n}(2-2\cos(\pi n))~~~~~~~\normalsize ~~~~~~Simplify \]

Answer 5

\[\large\rm b_n = 2 [\int\limits_{-0.5}^{0}-\sin(2 \pi nx) +\int\limits_{0}^{0.5} \sin(2 \pi nx)]~~~~~\normalsize ~~~Substitute ~L=0.5\]\[\large\rm =2([\frac{1}{2\pi n}\cos (2 \pi nx)]^0_{-0.5} +[ -\frac{1}{2\pi n}\cos(2 \pi nx)]^0_{0.5} )~~~~~~~\normalsize ~~~Integrate \]\[\large\rm =\frac{1}{\pi n }[(1-\cos(\pi n)) +(-\cos(\pi n)-(-1))]~~~\normalsize ~~~Plug ~in~upper~and~lower~limits \]\[\large\rm b_n=\frac{1}{\pi n}(2-2\cos(\pi n))~~~~~~~\normalsize ~~~~~~Simplify \]\[\normalsize \rm Find~first~5~terms\]\[\large\rm b_1 = \frac{1}{\pi } (2+2)\cdot \sin(2\pi n)\]\[\large\rm b_2=\frac{1}{2\pi}{(2-2)} =0\]\[\large\rm b_3=\frac{1}{3\pi}{(2+2})\cdot sin (3\cdot 2 \pi n)\]\[\large\rm b_4=\frac{1}{4 \pi } (2-2)=0\]\[\large\rm b_5=\frac{1}{5 \pi }(2+2) \cdot \sin(5 \cdot 2 \pi n)\]\[\large\rm \sum_{n=1}^{\infty} 4(\frac{1}{(2n-1)\pi})\sin( 2 \pi x(2n-1))\]

Answer 6

\[\large\rm a_n =2 \int\limits_{-0.5}^{0.5} f(x) \cos ((2 n \pi x) dx\]\[\large\rm =2(2)\int\limits_{0}^{0.5} (-4x+1)\cos(2n \pi x) dx\]\[\large\rm =4[-4(\frac{x}{2 \pi n })\sin(2 \pi n x)+ \frac{1}{(2 \pi n)^2} \cos (2n \pi x)+\frac{1}{2 \pi n }\sin (2 \pi n x)]_0^{0.5}\]

Answer 7

\[\large\rm= 4[-\frac{4}{(2 \pi n )^2} \cos ( \pi n ) - (-4\frac{1}{(2 \pi n )^2} )]~~~~\normalsize Plug ~in~ upper~and~lower~limits\]\[\large\rm =4[-\frac{4}{(2 \pi n )^2} \cos ( \pi n ) + \frac{4}{(2\pi n)^2}] =\frac{16}{(2\pi n)^2} [1-\cos ( \pi n)]~~~Simplify \]\[a_n=\frac{16}{(2 \pi n )^2 }[1-(-1)^n]\]

Answer 8

\[\large\rm a_n =2 \int\limits_{-0.5}^{0.5} f(x) \cos ((2 n \pi x) dx\]\[\large\rm =2(2)\int\limits_{0}^{0.5} (-4x+1)\cos(2n \pi x) dx\]\[\rm =4[-4(\frac{x}{2 \pi n })\sin(2 \pi n x)+ \frac{1}{(2 \pi n)^2} \cos (2n \pi x)+\frac{1}{2 \pi n }\sin (2 \pi n x)]_0^{0.5} ~
~Integrate ~by~Parts\]