Greatest Integer Function
Date: 12/04/2002 at 08:19:48 From: Teresa Leonard Subject: Greatest Integer function, Step Function in trigonometry I am a mother who is trying to home school my high school senior through trigonometry. We are studying special functions and I am stumped with how I can explain Greatest Integer Function to him. How can I explain to him how to graph f(x)=[2x]? and h(x)=[x]/x? I have never taken trigonometry and am limited in resources. If you can, please advise. Thank you, Teresa Leonard
Date: 12/04/2002 at 13:05:16 From: Doctor Peterson Subject: Re: Greatest Interger function, Step Function in trigonometry Hi, Teresa. Here are a few resources that can help - these sites may be useful in other cases as well. First, you can search our site (using the search form at the bottom of many pages, or the Search Dr. Math button) for your key word or phrase, in this case "greatest integer". In this case, I didn't find any pages that really explain the function. Another useful source, listed in our FAQ, is Eric Weisstein's World of Mathematics. This can be very advanced, so you have to be willing to ignore anything you don't understand; but often at least the start of an article gives a good definition. Here is the page to look at, which I found by searching for the same phrase: Greatest Integer Function http://mathworld.wolfram.com/GreatestIntegerFunction.html Now you'll see that the entry consists only of a cross reference to "Floor Function." Read that article, without worrying about any parts you don't follow; then you can come back to our site and search for that alternative name. I found this, which tells at least a little about it: Ceiling and Floor Function Notation http://mathforum.org/library/drmath/view/61697.html These sources will have told you what the function is, why the [x] notation is a little out of date, and what its graph looks like: [x] + *===o | + *===o | +---+---*===o---+---+- | *===o | *===o + Between any pair of integers, [x] is constant; for an integer, it is equal to x. So it starts on the line y=x at any integer, and goes flat over to the next integer, then hops up again. Now, let's look at how to graph your functions. f(x)=[2x] It will be very useful to think about how functions are transformed when you multiply or add on the "inside" or the "outside". This answer discusses that: Graph with f(x) http://mathforum.org/library/drmath/view/54509.html If you know the graph of [x], you can use it to make the graph of [2x]. Notice that the latter graph, when x=5, will have the same y-coordinate as the former graph when x=10, since both are . In general, the value for any x is the value of [x] twice as far from the origin. So we are squeezing the graph horizontally to half its size. Since we started with a function whose graph looks like this, [x] + *===o | + *===o | +---+---*===o---+---+- | *===o | *===o + the new equation will look like this: [2x] + *=o | + *=o | +---+---*=o-+---+---+- | *=o | *=o + You can also graph this directly. Think about what the function means: for any number from 0 to 1 (excluding 1), [x] is 0, the greatest integer that is not greater than x. From 1 to 2, it is 1, and so on. Now, when will 2x be between 0 and 1? When x is between 0 and 1/2. So in that interval, [2x] is 0. Continue the same way, and you will see the graph above. Now for the other function: h(x)=[x]/x Here we have to use the direct method, since we are not just stretching or compressing [x]. We know that [x] has the same value for all x between any pair of consecutive integers. So we can make a chart: domain [x] [x]/x -------+-----+------ -2<=x<-1| -2 | -2/x -1<=x<0 | -1 | -1/x 0<=x<1 | 0 | 0 1<=x<2 | 1 | 1/x 2<=x<3 | 2 | 2/x Just graph each simplified function over its domain, and you get a graph like this: o |+ / / | * * + *___ | o +---+---o===o---+ | + | + This is hard for me to draw, but you should at least get the idea that it will be a smooth curve in each segment, and will be 1 at each integer, except that at 0 it is not defined (0/0). If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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