Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The Greatest Integer Function - Problem 3

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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I want to graph another transformation of the greatest integer function. Let’s graph y equals ½, the greatest integer less than or equal to –x plus 1 minus 2. The parent function that’s being transformed here is, y equals the greatest integer less than or equal to x. And here’s a graph of that function.

Now my first step is usually to make a table of values for the parent function, and then to transform those values, and finally to graph the transformed values. Well, first let me make a substitution, I’ll call u, xu and then I’ll make a table for u and greatest integer of u.

Now the thing that’s unique about this function is that there’s a whole integer of values that will give you any integer output. For example, you can get 0 by plugging in 0, but you can also get it by plugging in ½ or .9 or .99. So any number from 0 to 1, not including 1 will give you an output of 0. Any number from 1 to 2 not including 2, will give you an output of 1, and so on.

I’ll start with these values and see where that gets me. Now my transformed function is ½ greatest integer of –x plus 1 minus 2. I’m going to make a substitution u equals –x plus 1. If u equals –x plus 1, then u minus 1 equals –x, and that means x equals –u plus 1.

To get my x values, I need to take these values, take their opposites and then add 1. So the opposites of these numbers plus 1. Notice we have to pay special attention to what happens to the endpoints of this interval; which endpoint is included, which endpoint is not included. When I take the opposite of 0 I get 0, then I add 1, and I get 1.

When I take the opposite of 1, I get -1 and adding 1, I get 0. What’s happened here is the 1’s become a 0 and the 0’s become 1. The bracket has to follow where the 0 went. The 0 went to 1. And the parenthesis has to follow where the 1 went, 1 became 0. So the kind of interval we’ve got has switched. We’ve got conclusion on the right side rather than the left side.

Let’s see if that’ll happen again, -u plus 1, -1 plus 1 is 0. Minus 2 plus 1 is -1. So the 2 became -1, my parenthesis goes here and the 1 became 0, so my bracket goes here. And this will continue. -2 plus 1 is -1, -3 plus 1 is -2, so the parenthesis follows the -2 and the bracket goes with the -1. And this will continue.

What happens with the y values? We multiply the greatest integer by ½ and subtract 2. So we take these, multiply by ½ and subtract 2. So 0 time ½ is 0, minus 2 is -2. 1 times ½ is ½ minus 2 is -1.5. 2 times ½ is 1 minus 2 is -1. Let’s plot these segments and see if we can continue the pattern.

From 0 to 1 I get -2. Now notice 0 is not included, 1 is included. 0, 1 is here, I get -2. 0 is not included, 1 is included. So this is what I’m going to get, my segment looking like that. And then from -1 to 0, I get -1.5. So in this interval I get -1.5. I include 0 but not -1. And then here, from -2 to -1, I get the value -1. I don’t want to include -2. There’ll be an open circle on the left. -2 to -1 I get -1. And this pattern just continues.

So I can go up to here. I just keep going up ½ and move the segment over one unit. Here and so on. And I can go down ½ and over and so on. That’s my graph of the transformed greatest integer function.

Notice a big difference here, first of all, this function’s decreasing. But second of all, the open circle is on the left rather than the right. How did that happen? It’s this –x. Somewhere in the transformation there is the reflection about the y axis and that puts the open circle from the right to the left.