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MY FAVORITE FUNCTIONS. OR Continuous from what to WHAT?!?. Section 1. Accordions. y = sin x. _ x. 1. y = sin. f(x) = x sin. 1. x. f is continuous at 0 even though there are nearly vertical slopes. g(x) = x sin. 2. 1. _ x. g’(0) = lim. g(h). ___ h. = 0. h  0.

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My favorite functions

MY FAVORITE FUNCTIONS

OR

Continuous from what to WHAT?!?


Section 1 accordions
Section 1.Accordions

y = sin x

_

x

1

y = sin


My favorite functions

f(x) = x sin

.

1

x

f is continuous at 0 even though there are nearly vertical slopes.


G x x sin
g(x) = x sin

2

1

_

x

g’(0) = lim

g(h)

___

h

= 0

h  0

So g has a derivative at 0.

However, there are sequences <a > and <b > such that

lim b - a = 0, but lim

n

n

g(b ) - g(a )

___________

n

n

= ∞.

n

n

n∞

b - a

n∞

n

n


Section 2 le blancmange function
Section 2. Le Blancmange function

Given x, 0 ≤ x ≤ 1, and non-negative integer n, let k be the greatest non-negative integer such that a = 2-nk ≤ x and

b = 2-n(k+1). Define

(x,n)

(x,n)

fn : [0,1][0,1] by

fn(x) = min{x-a , b - x}.

(x,n)

(x,n)

k = 0


F x f x
f(x) = ∑f (x)

¥

≤ ∑2

¥

-n

n

n=1

n=1

Example:

7

7 ;

1 ;

1 .

__

16

__

16

__

16

7

__

16

__

16

7

__

16

7 ;

__

16

7

__

16

f ( ) =

f ( ) =

f( ) =

f ( ) =

1

2

0


My favorite functions

Lemma2: Suppose a function h : RR is differentiable at x. If an and if bn are such that n, an ≤ x ≤ bn, then h'(x) =

h(bn) - h(an)

lim

__________

bn - an

n∞


Section 3 stretching zero to one
Section 3.Stretching zero to one.

Cantor's Middle Third SetC is a subset of [0,1] formed inductively

by deleting middle third open intervals.

Say (1/3,2/3) in step one.


My favorite functions

In step two, remove the middle-thirds of the

remaining two intervals of step one, they are

(1/9,2/9) and (7/9,8/9).

In step three, remove the middle thirds of the

remaining four intervals.

and so on for infinitely many steps.


My favorite functions

What we get is C,Cantor’s

Middle Thirds Set.

C is very “thin” and a

“spread out” set whose

measure is 0 (since the sum of the lengths of intervals remived from [0,1] is 1.

As [0,1] is thick, the following is surprising:


Theorem there is a continuous function from c onto 0 1
THEOREM. There is a continuous function from C onto [0,1].

The points of C are the points equal to the sums of

infinite series of form

∑2s(n)3

¥

-n

where s(n) {0,1}.

n=1


My favorite functions

F( ∑2s(n)3 ) = ∑s(n)2

¥

-n

¥

-n

n=1

n=1

defines a continuous surjective function

whose domain is C and whose range is [0,1].

Example. The two geometric series

show F(1/3)=F(2/3)=1/2.


My favorite functions

Picturing the proof.

Stretch the two halves of step 1 until they join at 1/2.

Now stretch the two halves of each pair of step 2

Until they join at 1/4 and 3/4…

Each point is moved to the sum of an infinite series.


Section 4 advancing dimension
Section 4.Advancing dimension.

2

NN

2 (2m-1) <m,n>

n-1


Theorem there is a continuous function from 0 1 onto the square
Theorem. There is a continuous function from [0,1] onto the square.

We’ll cheat and do it with the triangle.


Section 5 an addition for the irrationals
Section 5. square. An addition for the irrationals

By an addition for those objects X[0,∞) we mean a continuous function s : X´XX (write x+y instead of s(<x,y>)) such that for x+y the following three rules hold:

(1). x+y = y+x (the commutative law) and

(2). (x+y)+z = x+(y+z) (the associative law).

With sets like Q, the set of positive rationals, the

addition inherited from the reals R works, but

with the set P of positive irratonals it does not work:

(3+√2 )+(3-√2) = 6.


My favorite functions

Our aim is to consider another object which has an addition square.

And also “looks like” P.

THEOREM5.

The set P of positive irrationals has an addition.

Continued fraction


My favorite functions

Let G(x) denote the greatest integer ≤ x. Let a square. 0 = G(x).

Given an irrational x, the sequence <an> is computed as follows:

If a0,,,an have been found as below, let an+1 = G(1/r).


My favorite functions

Continuing in this fashion we get a sequence which converges to x. Often the result is denoted by

However, here we denote it by CF(x) = < a0, a1, a2,a3,…>.


My favorite functions

We let < to x. Often the result is denoted by2> denote the constant < 2, 2, 2, 2,…>.

Note <2> = CF(1+ √2) since

2

__x

1

or x - 2x - 1 = 0.

Hint: A quick way to prove the above is to solve for x in x = 2+


Prove 1 cf and 1 2 cf
Prove < to x. Often the result is denoted by1> = CF( ) and <1,2> = CF( )

We add two continued fractions “pointwise,” so

<1>+<1,2> =<2,3> or <2,3,2,3,2,3,2,3,…>.

  • Here are the first few terms for , <3,7,15,1,…>.

  • No wonder your grade school teacher told you

  • = 3 + . The first four terms of CF(), <3,7,15,1>

    approximate  to 5 decimals.

1

__

7

Here are the first few terms fo e,

<2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,….>


My favorite functions

Lemma to x. Often the result is denoted by. Two irrationals x and y are "close" as real numbers iff the "first few" partial continued fractions of CF(x) and CF(y) are identical.

For example <2,2,2,2,2,1,1,1,…> and <2> are close, but <2,2,2,2,2,2,2,2,2,91,5,5,…> and <2> are closer.

Here is the “addition:” We define xy = z if

CF(z) = CF(x) + CF(y).

Then the lemma shows  is continuous.


However strange things happen
However, strange things happen: to x. Often the result is denoted by


Problems
problems to x. Often the result is denoted by

1. How many derivatives has

1

g(x) = x sin

2

_

x

2. Prove that each number in [0,2] is the

sum of two members of the Cantor set.


Problems1
Problems to x. Often the result is denoted by

3. Prove there is no distance non-increasing function

whose domain is a closed interval in N and whose

range is the unit square [0,1] ´ [0,1].

4. Determine