2.1 day 2: Step Functions “Miraculous Staircase” Loretto Chapel, Santa Fe, NM Two 360 o turns...

2.1 day 2: Step Functions

“Miraculous Staircase”Loretto Chapel, Santa Fe, NM

Two 360o turns without support!

Greg Kelly, Hanford High School, Richland, Washington

Photo by Vickie Kelly, 2003

“Step functions” are sometimes used to describe real-life situations.

Our book refers to one such function: int( )y x

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.

int( )x

Greatest Integer Function:

greatest integer that is xy

x y

0 00.5 0

0.75 01 1



x y

0 00.5 0

0.75 01 1

1.5 12 2

This notation was introduced in 1962 by Kenneth E. Iverson.

Recent by math standards!



The greatest integer function is also called the floor function.

The notation for the floor function is:

y x

We will not use these notations.

Some books use or . y x y x

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 .8 8x 4 4y

Y= floor x

CATALOG F floor(

Go to Y=

Highlight the function.

2nd F6 Style 2: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 .8 8x 4 4y

If you have the older TI-89 you could try this:

Least Integer Function:

least integer that is xy

x y

0 00.5 1

0.75 11 1

x y

0 00.5 1

0.75 11 1

1.5 22 2



The least integer function is also called the ceiling function.

The notation for the ceiling function is:

y x



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 0

sinlimx

x

x

If we graph , it appears thatsin x

yx

0

sinlim 1x

x

x

If we graph , it appears thatsin x

yx

0

sinlim 1x

x

x

We might try to prove this using the sandwich theorem as follows:

sin 1 and sin 1x x

0 0 0

1 sin 1 lim lim lim

x x x

x

x x x

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.

(1,0)

1

Unit Circle

cos

sin

P(x,y)

Note: The following proof assumes positive values of . You could do a similar proof for negative values.

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

tan1

AT

tanAT

1, tan

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP Area sector AOP

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP Area sector AOP Area OAT

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2


(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2

Area sector AOP

2

2r

2

2


(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2

2

11 tan

2


11 sin

2

2

11 tan

2

sin tan multiply by two

sinsin

cos

11

sin cos

divide by sin

sin1 cos

Take the reciprocals, which reverses the inequalities.

sincos 1

Switch ends.

11 sin

2

2

11 tan

2

sin tan

sinsin

cos

11

sin cos

sin1 cos

sincos 1

0 0 0

sinlim cos lim lim1

0

sin1 lim 1

By the sandwich theorem:

0

sinlim 1

p

2.1 day 2: Step Functions “Miraculous Staircase” Loretto Chapel, Santa Fe, NM Two 360 o turns...

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