Transcript A Functional Bestiary (Powerpoint)

A Functional Bestiary
Dan Kennedy
Baylor School
Chattanooga, TN
From WikipediA, the
free
encyclopedia:
A bestiary, or Bestiarum vocabulum is a
compendium of beasts. Bestiaries were
made popular in the Middle Ages in
illustrated volumes that described various
animals, birds and even rocks. The natural
history and illustration of each beast was
usually accompanied by a moral lesson.
Most functions we encounter in the real
world are fairly tame and domesticated.
They are continuous.
They are differentiable.
They are smooth.
Have
a nice
day!
Exponential
Rational
Logistic
Trig
Log
Polynomials
Functions
Functions
Functions
Functions
Algebraic
Absolute
Functions
Value
Piecewise-Defined
Greatest Integer
Loosely speaking, a function is
continuous at a point if it can be graphed
through that point with an unbroken
curve.
Removable
Discontinuity
Continuous
Discontinuity
atatxx==0at
0 x=1
Loosely speaking, a function is
differentiable at a point if its graph
resembles a non-vertical line when you
zoom in closely enough at that point.
Points of non-differentiability include:
corners (e.g., absolute value);
points of verticality (e.g., cube root);
cusps (e.g., cube root of absolute value)
And, of course, points of discontinuity
are also points of non-differentiability.
So now let us look at a few
discontinuous functions from our
functional bestiary.
Usually we see functions that are
discontinuous at a single point.
They make their point…
but they are not very beastly.
This function does not have a limit at 0,
so it cannot be continuous there.
It does have a right-hand limit and a
left-hand limit, but the two are not
equal.
This is the wimpy way to foil continuity
at a point.
A true beast would fail even to have a
right-hand limit or a left-hand limit!
Here’s a good one:
 1
sin  
f ( x)    x 
 0

if x  0
if x  0
The right-hand and left-hand limits at 0
diverge by oscillation.
In the limit, this function pretty much
smears itself against the entire
interval [-1, 1] on the y-axis.
Since a limit at 0 must be unique, the
points cannot all be limits.
So, we call them cluster points.
How about a function with domain all real
numbers that is continuous nowhere?
Here’s one:
1 if x is rational
f ( x)  
0 if x is irrational
This is called a salt-and-pepper function.
It cannot, of
course, be drawn
accurately.
The real definition of continuity…
A function f is continuous at a if
lim f ( x)  f (a ).
x a
Note, however, that the function
1 if x is rational
f ( x)  
0 if x is irrational
has no limit at any a.
Now that’s a beast!
How about a function with domain all
real numbers that is continuous only
at x = 0?
You can construct one by making a
slight alteration to the salt-and-pepper
function
 x1 if x is rational
f ( x)  
0 if x is irrational
lim f ( x)  f (0)  0.
x 0
How about a function with domain
all real numbers that is continuous
everywhere except at the integers?
That’s not too hard.
The greatest integer function is
such a beast.
We have a beast that is continuous
everywhere except at the integers.
How about a function that is
discontinuous everywhere except at the
integers?
Pass the salt and pepper…
sin( x) if x is rational
f ( x)  
 0 if x is irrational
–3
–2
–1
1
2
3
lim f ( x)  sin(k )  0.
x k
It is now time to introduce my favorite
beast of discontinuity:
a function that is continuous at every
irrational number…
but discontinuous at every rational
number!
I am not making this up.
For x  [0,1) :

 1 if x  0

f ( x)   0 if x is irrational
 1
m
if x 
in lowest terms

n
 n
Extend f to a function of period 1 on
the whole real line.
1
0
1
You want proof?
A careful proof involves epsilons and
deltas, but here’s the gist.
The function is discontinuous at every
rational for the same reason as the
various salt-and-pepper functions.
No matter how small 1/n is, it is too far
away from 0 to share a 0 limit with its
irrational neighbors.
(m / n, 1/ n)
m/n
For any irrational number I, the
continuity requirement boils down to
lim f ( x)  0.
x I
For irrational x, the function values are
already at 0.
For rational x, the key is that only a
finite number of function values can be
very far from zero. Consider the picture
again…
No matter how close the blue line is to
zero, only a finite number of dots are
above it.
1
0
1
So you can always find a small enough
neighborhood around I so that none of
the points are above it!
1
0
I
1
So now let us consider the beasts of nondifferentiability.
The interesting ones are the continuous
functions that fail to be differentiable.
Continuous but not differentiable:
corners (e.g., absolute value,
x
2
);
points of verticality (e.g., cube root);
cusps (e.g., cube root of absolute value)
A warm-up:
Find a function that is continuous for all
real numbers but non-differentiable at
every integer.
f ( x)  sin( x)
One of the strangest beasts in function
history is Karl Weierstrass’s function
that is continuous everywhere but
differentiable nowhere!
Since his time, simpler functions with
this property have been constructed.
Also, we now know that “most” functions
that are continuous everywhere are, in
fact, differentiable nowhere!
Define G(x) to be the distance from x to
the nearest integer to x.
1/2
-2
-1
1
-1/2
2
y  G ( x)
1
y  G (2 x)
2
1
y  G (4 x)
4
Note that each of these functions has
half the amplitude and half the period of
its predecessor.
Now add them all up:
1
1
f ( x)  G ( x)  G (2 x)  G (4 x) 
2
4

1
  n G  2n x 
n 1 2
1
 n G (2 n x) 
2
This function is continuous everywhere
and differentiable nowhere.
Thanks to Tom Vogel’s Gallery of Calculus Pathologies!
Some miscellaneous functions from the
bestiary…
f ( x)  x  x  41
2
This is a pretty normal-looking quadratic
function until you carefully consider a
table of values…
…which we do in the next slide.
x
y
x
y
x
y
x
y
0
41
10
151
20
461
30
971
1
43
11
173
21
503
31
1033
2
47
12
197
22
547
32
1097
3
53
13
223
23
593
33
1163
4
61
14
251
24
641
34
1231
5
71
15
281
25
691
35
1301
6
83
16
313
26
743
36
1373
7
97
17
347
27
797
37
1447
8
113
18
383
28
853
38
1523
9
131
19
421
29
911
39
1601
All these function values are primes!
f (40)  40  40  41  40(41)  41  41
2
2
Here’s a harmless-looking function that
caused trouble on the AB Calculus AP
examination one year:
f ( x)  x  x
4
2
For one point, students had to give its
domain.
x  x  x  x  1
4
2
2
2
(,  1]  [1, ) {0}
Sadly, some students simplified the
function a little too well:
x  x  x  x  1  x x  1
4
2
2
2
Now…is 0 still in the domain?
(,  1]  [1, ) {0}
2
George Rosenstein showed me this one.
How about a continuous function with
domain all real numbers that has range
all real numbers and a zero derivative
almost everywhere?
For comparison, the greatest integer
function has a zero derivative almost
everywhere…
but, of course, it does not have range all
real numbers.
We will define this beast on [0, 1] and
then extend it to a function on the
whole real line.
We start with
f1 ( x) :
1
1/2
1/3
2/3
1
1
f132((xx))
1/2
1/3
2/3
1
The function we want is simply
lim f n ( x).
n 
It has range [0, 1].
It is constant on intervals of total
measure
1
1  1 
 2   4  
3
 9   27 
1/ 3

1 2 / 3
1 2 4
  

3 9 27
1
Finally, extend the function to make a
continuous “ladder” that has domain all
real numbers.
The function will have range all real
numbers, and its derivative will be
constant except on a set of measure
zero…
…by George!
We’ll finish today’s look at the
Bestiary with a pair of beasts that
every young calculus student should
know.
First, consider these two limits:
1
limsin   does not exist.
x 0
 x
1
lim x sin    0  (bounded )  0
x 0
 x

1
 x sin   if x  0
f ( x)  
 x

0
if x  0

This function is continuous at x = 0.
It does not have a derivative at x = 0.
It does not exhibit “local linearity” at
the origin.
 2 1
 x sin   if x  0
f ( x)  
x

0
if x  0

This function has a derivative at x = 0.
You can show it by using the definition
of the derivative at x = 0.
It also exhibits local linearity at the
origin.
Here’s the derivative at x = 0:
1
h sin    0
h



f (0)  lim
h 0
h
1
 lim h sin  
h 0
h
 0  (bounded)
0
2
Here’s the derivative for x ≠ 0:
1 2
1  1 


f ( x)  2 x  sin    x  cos      2 
 x
 x  x 
1
1
 2 x sin    cos  
 x
 x
Note that this function does not have
a limit as x approaches 0.
So f is differentiable everywhere, but
the derivative is not continuous at 0.
Perhaps more significantly, this function
has a derivative at 0, but you can not
find it by comparing the derivative on
one side of 0 to the derivative on the
other side of 0.
This is often how students show that
split functions are differentiable.
Luckily for them, that works for nice
functions.
But not all the functions in the
FUNCTIONAL BESTIARY!
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