\magnification=\magstep1

$$
f(t) \quad = \quad 3 \ + \ \sin \left( {{2 \pi t}\over{5}} \right) 
$$

$$
\sin \left( {{2 \pi}\over{P}} t \right) \quad = \quad \sin ( \omega t )
$$


$$
\eqalign{
A_0 \quad &= \quad 3  \cr
B_1 \quad &= \quad 1  \cr
}
$$

$$
A_0 \quad = \quad {{1}\over{P}} \int_0^P f(t) dt \quad = \quad
           {{1}\over{5}} \int_0^5 \left[ 3 \ + 
                            \sin \left( {{2\pi t}\over{5}} \right) \right] dt
$$

$$
{{1}\over{5}} \int_0^5 \sin \left( {{2\pi t}\over{5}} \right)  dt 
   \quad = \quad 
     {{1}\over{5}} \left( {{5}\over{2\pi}} \right)
      \left[ {-\cos \left( {{2\pi t}\over{5}} \right) } \right]_0^5
$$


$$
{{1}\over{2 \pi}} 
      \left[ {-\cos (2\pi) - ( -\cos(0) ) } \right] 
           \quad = \quad
{{1}\over{2 \pi}} \left[ -1 + 1 \right]
           \quad = \quad
  0
$$

$$
A_0 \quad = \quad 3 \ + \ 0 \quad = \quad 3  
$$


$$
\eqalign{
A_1 \quad &= \quad {{2}\over{P}} \int_0^P f(t) \ 
                \cos \left( {{2\pi t}\over{P}} \right) dt 
        \quad = \quad
            \quad {{2}\over{5}} \int_0^P 
              \left[ 3 \ + \ \sin \left( {{2 \pi t}\over{5}} \right) \right] 
                \cos \left( {{2\pi t}\over{5}} \right) dt  \cr
B_1 \quad &= \quad {{2}\over{P}} \int_0^P f(t) \ 
                \sin \left( {{2\pi t}\over{P}} \right) dt 
        \quad = \quad
            \quad {{2}\over{5}} \int_0^P 
              \left[ 3 \ + \ \sin \left( {{2 \pi t}\over{5}} \right) \right] 
                \sin \left( {{2\pi t}\over{5}} \right) dt  \cr
}
$$

$$
A_1 \quad = \quad 
            \quad {{2}\over{5}} \int_0^P 
              3 \cos \left( {{2\pi t}\over{5}} \right) dt  
    \quad + 
            \quad {{2}\over{5}} \int_0^P 
               \sin \left( {{2\pi t}\over{5}} \right) 
               \cos \left( {{2\pi t}\over{5}} \right) dt  
$$

$$
A_1 \quad = \quad
            \quad {{6}\over{5}} 
              \left( {{5}\over{2 \pi}} \right) 
               \sin \left[ {{2\pi t}\over{5}} \right]_0^5
  \quad = \quad 
            \quad {{6}\over{2 \pi}} 
               \left[ \sin (2\pi) - \sin(0) \right]
  \quad = \quad 0
$$

$$
A_1 \quad = \quad 0
$$


$$
B_1 \quad = \quad 
            \quad {{2}\over{5}} \int_0^5 
              3 \sin \left( {{2\pi t}\over{5}} \right) dt  
    \quad + 
            \quad {{2}\over{5}} \int_0^5 
               \sin \left( {{2\pi t}\over{5}} \right) 
               \sin \left( {{2\pi t}\over{5}} \right) dt  
$$


$$
B_1 \quad = 
            \quad {{6}\over{5}} 
              \left( - {{5}\over{2 \pi}} \right) 
               \cos \left[ {{2\pi t}\over{5}} \right]_0^5
   \ + \quad
            {{2}\over{5}} \int_0^5 
                {{1}\over{2}}
               \left[ 1 - 
               \cos \left( {{4\pi t}\over{5}} \right) 
               \right] dt
$$

$$
B_1 \quad = 
            \quad {{6}\over{5}} 
              \left[ \cos(2 \pi) - \cos(0) \right] 
   \ + \quad
            {{1}\over{5}} \int_0^5 dt
   \ - \quad 
            {{1}\over{5}} \int_0^5 
               \cos \left( {{4\pi t}\over{5}} \right) dt
$$


$$
B_1 \quad = \quad
            \quad 0 \quad
   \ + \quad
            \quad 1 \quad
   \ - \quad 
            {{1}\over{5}} \left( {{5}\over{4 \pi}} \right) 
               \sin \left[ {{4\pi t}\over{5}} \right]_0^5
$$


$$
B_1 \quad = \quad
            \quad 0 \quad
   \ + \quad
            \quad 1 \quad
   \ - \quad 
            {{1}\over{5}} \left( {{5}\over{4 \pi}} \right) 
               \left[ \sin(4 \pi) - \sin(0) \right] 
$$


$$
B_1 \quad = \quad
            \quad 0 \quad
   \ + \quad
            \quad 1 \quad
   \ - \quad 
            0 
     \quad = \quad 1
$$

$$
f(t) \quad = \quad \biggl\lbrace {{1 \quad \quad 0 \leq t \leq 1 } \atop
                   {0 \quad \quad 1 \leq t \leq 2 }}
$$

$$
A_0 \quad = \quad {{1}\over{P}} \int_0^P f(t) dt \quad = \quad
           {{1}\over{2}} \int_0^1 ( 1 ) dt 
            \quad + \quad
           {{1}\over{2}} \int_1^2 ( 0 ) dt
$$


$$
A_1 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \cos \left( {{2 \pi}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \cos (\pi t) dt 
            \quad + \quad
           \int_1^2  0 \cos (\pi t) dt
$$


$$
B_1 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \sin \left( {{2 \pi}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \sin (\pi t) dt 
            \quad + \quad
           \int_1^2  0 \sin (\pi t) dt
$$

$$
A_1 \quad = \quad {{1}\over{\pi}} \sin \bigl[ \pi t \bigr]_0^1 
           \quad = \quad
                  {{1}\over{\pi}} (0 - 0) \quad = \quad 0
$$


$$
B_1 \quad = \quad -{{1}\over{\pi}} \cos \bigl[ \pi t \bigr]_0^1 
           \quad = \quad
                  -{{1}\over{\pi}} (-1 - (1) ) \quad = \quad 
                  -{{1}\over{\pi}} (-2) \quad = \quad {{2}\over{\pi}}
$$



$$
A_2 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \cos \left( {{2 \pi \cdot 2}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \cos (2 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \cos (2 \pi t) dt
$$


$$
B_2 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \sin \left( {{2 \pi \cdot 2}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \sin (2 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \sin (2 \pi t) dt
$$


$$
A_3 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \cos \left( {{2 \pi \cdot 3}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \cos (3 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \cos (3 \pi t) dt
$$


$$
B_3 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \sin \left( {{2 \pi \cdot 3}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \sin (3 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \sin (3 \pi t) dt
$$



$$
A_2 \quad = \quad {{1}\over{2 \pi}} \sin \bigl[ 2 \pi t \bigr]_0^1 
           \quad = \quad
                  {{1}\over{2 \pi}} (0 - 0) \quad = \quad 0
$$


$$
B_2 \quad = \quad -{{1}\over{2 \pi}} \cos \bigl[ 2 \pi t \bigr]_0^1 
           \quad = \quad
                  -{{1}\over{2 \pi}} (1 - 1 ) \quad = \quad  0
$$



$$
A_3 \quad = \quad {{1}\over{3 \pi}} \sin \bigl[ 3 \pi t \bigr]_0^1 
           \quad = \quad
                  {{1}\over{3 \pi}} (0 - 0) \quad = \quad 0
$$


$$
B_3 \quad = \quad -{{1}\over{3 \pi}} \cos \bigl[ 3 \pi t \bigr]_0^1 
           \quad = \quad
                  -{{1}\over{3 \pi}} (-1 - (1) ) \quad = \quad  {{2}\over{3 \pi}}
$$

$$
f(t) \quad \sim \quad 
       {{1}\over{2}} 
         \quad + \quad
       {{2}\over{\pi}} \sin(\pi t)
         \quad + \quad
       {{2}\over{3 \pi}} \sin(3 \pi t)
$$



$$
A_4 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \cos \left( {{2 \pi \cdot 4}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \cos (4 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \cos (4 \pi t) dt
$$


$$
B_4 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \sin \left( {{2 \pi \cdot 4}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \sin (4 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \sin (4 \pi t) dt
$$


$$
A_5 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \cos \left( {{2 \pi \cdot 5}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \cos (5 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \cos (5 \pi t) dt
$$


$$
B_5 \quad = \quad {{2}\over{P}} \int_0^P f(t) 
                    \sin \left( {{2 \pi \cdot 5}\over{P}} t \right) dt \quad = \quad
    		  \int_0^1 1 \sin (5 \pi t) dt 
            \quad + \quad
           \int_1^2  0 \sin (5 \pi t) dt
$$






$$
A_4 \quad = \quad {{1}\over{4 \pi}} \sin \bigl[ 4 \pi t \bigr]_0^1 
           \quad = \quad
                  {{1}\over{4 \pi}} (0 - 0) \quad = \quad 0
$$


$$
B_4 \quad = \quad -{{1}\over{4 \pi}} \cos \bigl[ 4 \pi t \bigr]_0^1 
           \quad = \quad
                  -{{1}\over{4 \pi}} (1 - 1 ) \quad = \quad  0
$$



$$
A_5 \quad = \quad {{1}\over{5 \pi}} \sin \bigl[ 5 \pi t \bigr]_0^1 
           \quad = \quad
                  {{1}\over{5 \pi}} (0 - 0) \quad = \quad 0
$$


$$
B_5 \quad = \quad -{{1}\over{5 \pi}} \cos \bigl[ 5 \pi t \bigr]_0^1 
           \quad = \quad
                  -{{1}\over{5 \pi}} (-1 - (1) ) \quad = \quad  {{2}\over{5 \pi}}
$$


$$
f(t) \quad \sim \quad 
       {{1}\over{2}} 
         \quad + \quad
       {{2}\over{\pi}} \sin(\pi t)
         \quad + \quad
       {{2}\over{3 \pi}} \sin(3 \pi t)
         \quad + \quad
       {{2}\over{5 \pi}} \sin(5 \pi t)
$$


\bye
