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2.1 day 2: Step Functions “Miraculous Staircase” Loretto Chapel, Santa Fe, NM Two 360 o turns without support! Greg Kelly, Hanford High School, Richland,

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Presentation on theme: "2.1 day 2: Step Functions “Miraculous Staircase” Loretto Chapel, Santa Fe, NM Two 360 o turns without support! Greg Kelly, Hanford High School, Richland,"— Presentation transcript:

1 2.1 day 2: Step Functions “Miraculous Staircase” Loretto Chapel, Santa Fe, NM Two 360 o turns without support! Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2003

2 “Step functions” are sometimes used to describe real-life situations. Our book refers to one such function: This is the Greatest Integer Function. The TI-89 contains the command, but it is important that you understand the function rather than just entering it in your calculator.

3 Greatest Integer Function:

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7 This notation was introduced in 1962 by Kenneth E. Iverson. Recent by math standards! Greatest Integer Function: The greatest integer function is also called the floor function. The notation for the floor function is: We will not use these notations. Some books use or.

8 The older TI-89 calculator “connects the dots” which covers up the discontinuities. (The Titanium Edition does not do this.) The TI-89 command for the floor function is floor (x). Graph the floor function for and. Y= CATALOG F floor(

9 Go toY= Highlight the function. 2nd F6 Style2:Dot ENTER GRAPH The open and closed circles do not show, but we can see the discontinuities. The TI-89 command for the floor function is floor (x). Graph the floor function for and. If you have the older TI-89 you could try this:

10 Least Integer Function:

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14 The least integer function is also called the ceiling function. The notation for the ceiling function is: Least Integer Function: The TI-89 command for the ceiling function is ceiling (x). Don’t worry, there are not wall functions, front door functions, fireplace functions!

15 Using the Sandwich theorem to find

16 If we graph, it appears that

17 We might try to prove this using the sandwich theorem as follows: Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match. We will have to be more creative. Just see if you can follow this proof. Don’t worry that you wouldn’t have thought of it. Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match.

18 (1,0) 1 Unit Circle P(x,y) Note: The following proof assumes positive values of. You could do a similar proof for negative values.

19 (1,0) 1 Unit Circle P(x,y) T AO

20 (1,0) 1 Unit Circle P(x,y) T AO

21 (1,0) 1 Unit Circle P(x,y) T AO

22 (1,0) 1 Unit Circle P(x,y) T AO

23 (1,0) 1 Unit Circle P(x,y) T AO

24 (1,0) 1 Unit Circle P(x,y) T AO

25 (1,0) 1 Unit Circle P(x,y) T AO

26 (1,0) 1 Unit Circle P(x,y) T AO

27 multiply by two divide by Take the reciprocals, which reverses the inequalities. Switch ends.

28 By the sandwich theorem: 


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