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Selected Fourier series animations

Nasser M. Abbasi

October 4, 2019   Compiled on October 4, 2019 at 8:19pm



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Fourier series over original domain

Fourier series over periodic extended domain




1

        (
        { x  − 2 < x < 0
f (x) = (                   − 2 <  x < 2
          1  0 < x <  2

                         n                   n
f (x) ∼ 2 ∑ ∞n=1− 2(−1+2(−21))cos( πnx) + 1−3(−1)-sin (πnx )
                    n π         2         nπ       2




2

f (x) = − x   −  π < x < π

f (x) ∼ 2 ∑ ∞  (−1)nsin(nx )
            n=1  n




3

        (
        { x   |x| < π2
f (x) = (                  − π < x <  π
          0  otherwise

        ∑        (   (   )           (  ))
f (x) ∼   ∞n=1 π2n2 sin  n2π  − 12n πcos  nπ2   sin (nx )




4

        (
        { − 1 − π ≤ x ≤  0
f (x) = (  1   0 < x ≤ π     −  π < x < π

        4∑  ∞    1
f (x ) ∼ π   n=1 2n−1 sin((2n − 1 )x )




5

        ({           1
f (x) =   1    |x| < 2π      −  π < x < π
        ( 0  1π < |x| < π
             2

        1   ∑ ∞  -2     nπ
f (x ) ∼ 2 +   n=1πn sin(2 )cos(nx )




6

f (x) = x2    − π < x < π

        1  2    ∑ ∞   1-     n
f (x) ∼ 3π  + 4   n=1 n2 (− 1) cos(nx )




7

        (
        { sin x   0 < x ≤ π
f (x) =                        − π < x <  π
        (  0    − π ≤ x < 0

f (x) ∼ 1 + 1 sin (x) + 2 ∑ ∞  ---1--2 cos(2nx )
        π   2          π   n=11−(2n)




8

        (
        {  x2   0 < x ≤  π
f (x) = (    2                −  π < x < π
          − x  − π ≤ x <  0

                 (                              )
f (x) ∼ 2π2 ∑ ∞    1-(− 1)n+1 − --23-(1 − (− 1)n)  sin (nx)
              n=1  nπ           (nπ)




9

            1
f (x) = x + 4x2    − π <  x < π

                         (               )
f (x) ∼ π2 + ∑ ∞   (− 1)n  cos(n2x)−  2sin(nx)-
        12     n=1          n        n




10

f (x) = ex   −  π < x < π

               (                                      )
f (x) ∼ 2sinh(π) 1 + ∑ ∞   (−1)n2-(acos (nx) − n sin (nx))
           π    2     n=1 1+n




11

        (
        {1   |x | < π
f (x ) = (          2   −  π < x < π
         0   True

            ∑
f (x ) ∼ 12 +   ∞n=1n2π sin(n2π)cos(nx )




12

         3
f (x ) = x    − π <  x < π

        ∑          n    2 2
f (x ) ∼   ∞n=1− 2(−1)(n−36n-π-)sin (nx )




13

f (x ) = sin(x )   − 1 < x < 1

                     n
f (x ) ∼ ∑ ∞n=1− 2nπ2(−21)-sin (1)sin (nπx)
                n π −1