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Fourier of Half-Wave Rectified

  1. Feb 14, 2013 #1
    1. The problem statement, all variables and given/known data

    half_wave.jpg

    How did [itex]\frac{1}{2}x[/itex] come from at k=1?

    2. Relevant equations


    3. The attempt at a solution

    because k=1 will make the first term at denominator 2(k-1) = [itex]\frac{0}{0}[/itex]
     
  2. jcsd
  3. Feb 14, 2013 #2

    SammyS

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    Yes. The [itex]\ \frac{1}{2}x\ [/itex] comes from the fact that the first term has the form 0/0 as k → 1.

    What is [itex]\displaystyle \ \lim_{t\to0}\frac{\sin(t)}{t}\ ?[/itex]
     
  4. Feb 14, 2013 #3
    Use L' Hopital's Rule --> [itex]\displaystyle \ \lim_{t\to0}cos(t) = 1 [/itex]

    thank you SammyS
     
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