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The question is from the following problem:

enter image description here

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

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

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

  • 2
    It looks like you're calculating what real analysis texts call the [total variation](http://en.wikipedia.org/wiki/Total_variation) of $f$ on $[0, 12]$.2011-06-29
  • 1
    There is a nice animation on [Wikipedia](http://en.wikipedia.org/wiki/Total_variation) that helps explain total variation.2011-06-29
  • 0
    This question seems to be designed to test whether the client has studied an analysis course that covers the notion of bounded variation. And we see that Jack hasn't, so the question worked.2011-06-29

3 Answers 3

3

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.

0

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

  • 0
    You are missing the absolute values. Nothing cancels out since all terms of the sum are positive.2011-06-29
-1

Total variation sums up how much a function bobs up and down. Yours does this 16 units. Therefore choose D.

  • 0
    He didn't ask for the answer.2011-06-29