1 / 30

MY FAVORITE FUNCTIONS

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.

moses-barton
Download Presentation

MY FAVORITE FUNCTIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MY FAVORITE FUNCTIONS OR Continuous from what to WHAT?!?

  2. Section 1.Accordions y = sin x _ x 1 y = sin

  3. f(x) = x sin . 1 x f is continuous at 0 even though there are nearly vertical slopes.

  4. 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

  5. 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

  6. 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

  7. 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∞

  8. 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.

  9. 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.

  10. 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:

  11. 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

  12. 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.

  13. 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.

  14. Section 4.Advancing dimension. 2 NN 2 (2m-1) <m,n> n-1

  15. Theorem. There is a continuous function from [0,1] onto the square. We’ll cheat and do it with the triangle.

  16. Section 5.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.

  17. Our aim is to consider another object which has an addition And also “looks like” P. THEOREM5. The set P of positive irrationals has an addition. Continued fraction

  18. Let G(x) denote the greatest integer ≤ x. Let a0 = 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).

  19. 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,…>.

  20. We let <2> 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+

  21. Prove <1> = 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,….>

  22. Lemma. 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.

  23. However, strange things happen:

  24. problems 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.

  25. Problems 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

More Related