2 1 day 2 Step Functions Miraculous Staircase

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

“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.

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function: The greatest integer function is also called the floor function. The notation for the floor function is: Some usewas introduced or Thisbooks notation in. 1962 bynot Kenneth E. Iverson. We will use these notations. Recent by math standards!

The TI-89 command for the floor function is floor (x). Graph the floor function for and . Y= CATALOG F floor( 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 . If you have the older TI-89 you could try this: Go to Y= Highlight the function. 2 nd F 6 Style 2: Dot ENTER The open and closed circles do not show, but we. GRAPH can see the discontinuities.

Least Integer Function:

Least Integer Function:

Least Integer Function:

Least Integer Function:

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

Using the Sandwich theorem to find

If we graph , it appears that

If we graph , it appears that 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.

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

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

T P(x, y) 1 O Unit Circle A (1, 0)

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

By the sandwich theorem: p