I need to find the total variation of $V_g([0,2\pi])$, when $g(x)=cos(x)$
According to the formula:
$V_g([a,b])=g(b)-g(a)$
$V_g([0,2\pi])=cos(2\pi)-cos(0)=1-1=0$
But this answer is shown as wrong. What is wrong there?
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mirik
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Your formula is wrong. – xbh Sep 20 '18 at 05:57
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This formula $V_g([a,b])=g(b)−g(a)$ ? – mirik Sep 20 '18 at 06:28
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The formula you have is valid for monotonic increasing functions: if you check your source it will probably state this as such. – Tom Collinge Sep 20 '18 at 06:31
1 Answers
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Your formula is wrong and so is the anwser!. Since $\cos 0=1$ and $\cos \pi =-1$ there is no way the total variation can be $1$ (because $|\cos 0-\cos \pi|=2$). The correct formula is $V_g=\int_0^{\pi} |g'(x)|\, dx$ which is $4.$
Kavi Rama Murthy
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I fixed the original question. In my question I had $cos(2\pi)$ not $cos(\pi)$, so your answer can also be erroneous. – mirik Sep 20 '18 at 06:25
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My answer is correct. The total variation of $f$ on $[a,b]$ is $f(b)-f(a)$ if $f$ is an increasing function. Since $\cos x$ is not increasing on $[0,2\pi]$ you cannot use this formula. – Kavi Rama Murthy Sep 20 '18 at 06:28
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When I said that the total variation cannot be $1$ I used the fact that $|f(y)-f(x)| \leq V_g$ for any two points $x$ and $y$. I chose $x=0$ and $y=\pi$ which are points in the given interval. What I can conclude from this is $V_g \geq 2$. – Kavi Rama Murthy Sep 20 '18 at 06:32