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What is the total variation of sin(x) on [a,b]?

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data
    For a given function ##g:[a,b]→ℝ, 0 < a < b##, compute its total variation [itex]
    \underset{[a,b]}{\mathrm{Var}}
    (g)[/itex] where ##g(x) = \sin(x), x\in[a,b].##
    2. Relevant equations


    3. The attempt at a solution
    I know that between odd multiples of ##\frac{\pi}{2}##, ##\sin(x)## is monotone, so the interval ##[a,b]## needs to be broken up accordingly. If ##a## and ##b## are both in the same monotone interval, we simply have ##\sin(b)-\sin(a)##. If they're split over a turn, it becomes ##|1-\sin(a)| + |\sin(b)-1|##. I've so far been unable to come up with a closed form way of describing that for arbitrary ##a## and ##b## stretching over an area greater than ##\pi##, though.
     
  2. jcsd
  3. Apr 13, 2016 #2

    micromass

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    Please define the variation. Besides its formal definition, is there an easier way to compute the variation for special (differentiable) functions?
     
  4. Apr 13, 2016 #3
    ##\underset{[a,b]}{\text{Var}}(g) = \sup\{v(g,P):P\text{ is a partition of }[a,b]\}## and ##v(g,P)=\sum_{i=1}^n|g(x_i)-g(x_{i-1})|, P=\{x_0,x_1,...,x_n\}##
    I'm aware of no special rules for finding variation aside from when a function is monotonic. In that case, ##\underset{[a,b]}{\mathrm{Var}}(g) = |g(b)-g(a)|##
     
  5. Apr 13, 2016 #4

    micromass

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    Did you not see how to compute the variation using an integral?
     
  6. Apr 13, 2016 #5
    I'm not aware of any way to compute the variation using an integral. ##\int_a^b|\sin(x)|dx## gives the correct answers for multiples of ##\frac{\pi}{2}## but not for say ##\frac{3\pi}{4}## to ##\frac{5\pi}{4}##.
     
  7. Apr 13, 2016 #6

    Ray Vickson

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    On line I have seen expressions like
    [tex] \text{Var}_{[a,b]}(f) = \int_a^b |f'(x)| \, dx,[/tex]
    at least for functions that are in ##C^1([a,b])##. In your case that would reduce to
    [tex] \int_a^b |\cos(x)| \, dx [/tex]
    Does that give you correct results?
     
  8. Apr 13, 2016 #7

    micromass

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    Indeed, that is what I was refering to. You can easily see it heuristically from:

    [tex]\text{Var}_a^b(f) =\text{sup} \sum |f(x_{j+1}) - f(x_j)|= \text{sup} \sum \left|\frac{f(x_{j+1}) - f(x_j)}{x_{j+1} - x_j}\right|\Delta x_j = \int_a^b |f'(x)|dx[/tex]

    While perhaps not a completely rigorous proof, it definitely gives an indication of where this formula comes from.
     
  9. Apr 13, 2016 #8
    I was completely unaware of that. It seems to work perfectly, thank you.
     
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