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

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1. Apr 13, 2016

### Ainlaen

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 $\underset{[a,b]}{\mathrm{Var}} (g)$ 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. Apr 13, 2016

### micromass

Please define the variation. Besides its formal definition, is there an easier way to compute the variation for special (differentiable) functions?

3. Apr 13, 2016

### Ainlaen

$\underset{[a,b]}{\text{Var}}(g) = \sup\{v(g,P):P\text{ is a partition of }[a,b]\}$ 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, $\underset{[a,b]}{\mathrm{Var}}(g) = |g(b)-g(a)|$

4. Apr 13, 2016

### micromass

Did you not see how to compute the variation using an integral?

5. Apr 13, 2016

### Ainlaen

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 $\frac{5\pi}{4}$.

6. Apr 13, 2016

### Ray Vickson

On line I have seen expressions like
$$\text{Var}_{[a,b]}(f) = \int_a^b |f'(x)| \, dx,$$
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?

7. Apr 13, 2016

### micromass

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

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

While perhaps not a completely rigorous proof, it definitely gives an indication of where this formula comes from.

8. Apr 13, 2016

### Ainlaen

I was completely unaware of that. It seems to work perfectly, thank you.