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Question about Sawtoothwave Function

  1. Dec 29, 2011 #1
    1. The problem statement, all variables and given/known data

    For a sawtooth wave function with a period of P where the first branch of the function is given for [a, a+P) we can say that :

    f(x)=some expression in x for [a, a+P)

    f(x+nP)=f(x)

    2. Relevant equations
    3. The attempt at a solution

    Why is 'a+P' not included in the x-interval? Wouldn't f(x) equal 1 if x was 1? What would happen if 'a+P' was included in the interval?

    Thank You.
     
  2. jcsd
  3. Dec 29, 2011 #2

    micromass

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    We have that f(a+P)=f(a). So knowing the value of f(a) means that we also know the value of f(a+P). So there is no need to include a+P.
     
  4. Dec 29, 2011 #3
    The graph in my textbook has a straight line branch between x=0 and x=1 that repeats itself indefinitely. The highest Y equals 1. For [0,1) the output from f is given as f(x)=x.

    So if x is 0.5 then f(x) is 0.5. What is f(x) if x=1?
     
  5. Dec 29, 2011 #4

    SammyS

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    If f(0) = 0, and the function is periodic with period 1, then f(1) = 0 also.
     
  6. Dec 29, 2011 #5
    From the book:

    "f(x+1)=f(x) for [0,1)

    For example, f(1.5)=f(0.5+1)=f(0.5)=0.5"

    Period here is 1. y=1 when x=1 on the graph. Does it mean that x=1 is undefined or something?
     
  7. Dec 29, 2011 #6

    micromass

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    f(1)=f(0)=0.
     
  8. Dec 29, 2011 #7

    SammyS

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    It may look like y=1 when x=1, however y only gets very close to 1 when x gets very close to 1.

    f(0.999995) = 0.999995 , f(0.99999999993) = 0.99999999993 , f(1) = 0
     
  9. Dec 29, 2011 #8
    f(1)=0 if you go by definition, f(1)=1 if you go by the picture of the graph. Too confusing.
     
  10. Dec 29, 2011 #9
    So what would happen if I were to include 1 in the interval?
     
  11. Dec 29, 2011 #10

    micromass

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    The picture of the graph is wrong. On several points. For example, that line straight down doesn't happen, you just have a discontinuity


    You can't. If you include 1, then f(1) would have multiple definitions, this is not good.
     
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