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

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SUMMARY

The total variation of the function \( g(x) = \sin(x) \) on the interval \([a,b]\) can be computed by breaking the interval into monotonic segments, specifically around odd multiples of \( \frac{\pi}{2} \). If both \( a \) and \( b \) lie within the same monotonic segment, the variation is simply \( \sin(b) - \sin(a) \). For intervals that span multiple segments, the total variation can be expressed as \( |1 - \sin(a)| + |\sin(b) - 1| \). Additionally, for differentiable functions in \( C^1([a,b]) \), the total variation can be computed using the integral \( \int_a^b |\cos(x)| \, dx \).

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  • Understanding of total variation in real analysis
  • Knowledge of the sine function and its properties
  • Familiarity with the concept of monotonicity
  • Basic calculus, specifically integration and differentiation
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Ainlaen
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Homework Statement


For a given function ##g:[a,b]→ℝ, 0 < a < b##, compute its total variation <br /> \underset{[a,b]}{\mathrm{Var}}<br /> (g) where ##g(x) = \sin(x), x\in[a,b].##

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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.
 
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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?
 
micromass said:
Please define the variation. Besides its formal definition, is there an easier way to compute the variation for special (differentiable) functions?
##\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)|##
 
Ainlaen said:
##\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_1,x_2,...,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)|##

Did you not see how to compute the variation using an integral?
 
micromass said:
Did you not see how to compute the variation using an integral?
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}##.
 
Ainlaen said:
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}##.

On line I have seen expressions like
\text{Var}_{[a,b]}(f) = \int_a^b |f&#039;(x)| \, dx,
at least for functions that are in ##C^1([a,b])##. In your case that would reduce to
\int_a^b |\cos(x)| \, dx
Does that give you correct results?
 
Ray Vickson said:
On line I have seen expressions like
\text{Var}_{[a,b]}(f) = \int_a^b |f&#039;(x)| \, dx,
at least for functions that are in ##C^1([a,b])##. In your case that would reduce to
\int_a^b |\cos(x)| \, dx
Does that give you correct results?

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

\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&#039;(x)|dx

While perhaps not a completely rigorous proof, it definitely gives an indication of where this formula comes from.
 
Ray Vickson said:
On line I have seen expressions like
\text{Var}_{[a,b]}(f) = \int_a^b |f&#039;(x)| \, dx,
at least for functions that are in ##C^1([a,b])##. In your case that would reduce to
\int_a^b |\cos(x)| \, dx
Does that give you correct results?
I was completely unaware of that. It seems to work perfectly, thank you.
 

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