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The question is from the following problem:

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If $f$ is the function whose graph is indicated in the figure above, then the least upper bound (supremum) of $\big\{\sum_{k=1}^n|f(x_k)-f(x_{k-1})|:0=x_0 appears to be

$A. 2\quad B. 7\quad C. 12\quad D. 16\quad E. 21$

I don't know what the set above means. And I am curious about the background of the set in real analysis.

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    This question seems to be designed to test whether the client has studied an anal$y$sis cou$r$se that cove$r$s the notion of bounded va$r$iation. And we see that Jack hasn't, so the question worked.2011-06-29

3 Answers 3

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If the supremum is finite, the function is called of bounded variation.

Function of bounded variation appear in some context. A simple example I like is the following: a function is of bounded variation if and only if it is the sum of an increasing and a decreasing function.

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You take a partition of $\{0,12\}$ just like you would do when you compute a riemann sum, and then calculate the sum of all $f(x_k)-f(x_{k-1})$, and this will be an element of the given set.

edit : sorry I was wrong about the terms cancelling

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    You are missing the absolute values. Nothing cancels out since all terms of the sum are positive.2011-06-29
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Total variation sums up how much a function bobs up and down. Yours does this 16 units. Therefore choose D.

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    He didn't ask for the answer.2011-06-29