There is something special about functions that fix certain elements. Today I was asked to show that if, a function from [0,1] to [0,1] is such that for every x there is a y, with x not equal to y and f(x) = y, then the function cannot be continuous. Anyone have any ideas?
Here is a way to visualize this for those who aren't as mathematically inclined:
One morning, at sunrise, a monk leaves his monastery to walk to a mountaintop along a very narrow path. He walks at varying speeds and takes several breaks along the way to eat and rest. The next morning, again starting at sunrise, he begins to walk back, along the exact same path. Again, his speed varies and he stops at some points, though obviously his speed going downhill is faster than what it was going uphill. Show that there is a point in the path that the monk will occupy both days at the exact same time.
I know two very different ways to answer this question, depending on which of the two above formulations is preferred. However, I can't really explain why they are in a sense, equivalent. I haven't thought about it much, though I've decided I'd like to know.
So, can anyone answer either question, or explain why they are equivalent?
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Thread: A math question
12-02-2011 06:58 PM #1
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- Sep 2011
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A math question
12-11-2011 09:25 PM #2
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- Apr 2008
Imagine that the man is walking back on the second day and that his twin brother starts walking up the mountain at the exact same pace that the monk did on the first day. It is clear that they must meet at some point (because position as a function of time is continuous in this example), where they meet is the place the monk was at the same time on both days.
For the first part, it is enough to note that the line must intersect g(x) = x, because f(0)>0, and f(1)<1, so g-f must have a root in the interval.
If you consider in the second, f(x) = position with respect to time, then for day 2 it is g(x). Then, clearly f(0) < g(0), since it is the starting position for each day, while f(1)>g(1), since he ends at the mountaintop on the first day but at home on the second. then since f,g are continuous, so is f-g, and it is clear that (f-g)(0) is positive, while (f-g)(1) is negative, so that there must be a point in the interval where f-g =0, so f(x) = g(x) at that point. This doesn't make them equivalent though, just shows how the theorem can be applied(and gives a decent idea of how to prove it).