L’Hospital’s Rule
To Tell the Truth
Hello, my name is
L’Hopital.
My name is
L’Hospital!
And that first
guy can’t spell.
I’m the real
^
L’Hopital!
And
they’re both bad
spellers.
Question: Messieurs, can you tell us
something about your famous
rule?
L’Hospital #1
L’Hospital #2
L’Hospital #3
Will the real L’Hospital please stand up!!!
L’Hospital #3
Johann Bernoulli
It’s true, L’Hospital’s Rule can be directly applied to the
0
whatever
limit forms
and
.
0

f  x 
But the rule says nothing if lim
doesn’t exist.
x a g   x 
L’Hospital #1
L’Hospital #3
L’Hospital
#2
Here’s a correct statement of L’Hospital’s Rule:
Let a be a real number or  and I an open interval
which contains a or has a as an endpoint.
I  just
a, b  takes
,  b, a  ,care
,  
, b  , or limits and limits at
This
one-sided
 b,  of
b
a
c
infinity
all at
g  x   0 and g   x   0 for all x in I.]
[Suppose
thatonce.
f  x 
This condition is typical, but not needed if we assume lim
If lim f  x   lim g  x   0 or lim g  x    , xa g   x 
xI
x a
x a
x a
xI
xI
exists. xI
f  x 
 L  a number or    ,
and lim
x a g   x 
xI
f  x
 L.
then lim
x a g  x 
xI
The many different cases of L’Hospital’s Rule can all
be proven using Cauchy’s Mean Value Theorem:
Extra letters and limits of rachieauxs seem
to be French things. Maybe it’s something
in the Perrier. Eau well!
Let f and g be continuous on  a, b and
differentiable on  a, b  . Then there is a c
in  a, b  with f   c   g b   g  a   g  c   f b   f  a 
f  c  f b   f  a 

; g   c   0, g  a   g  b  .
or
g c  g b   g  a 
Cauchy’s Mean Value Theorem can be proven from
Rolle’s Theorem:
Apply Rolle’s Theorem to the function
F  x    f  x   f  a   g  b   g  a    g  x   g  a   f b   f  a 
on the interval  a, b .
See Handout!
0
The
case is covered in most textbooks, but the
0
whatever
case isn’t mentioned.

From Cauchy, we get that
f  x f a

f c  f  x   f  a  g  x  g  x 


g a
g c  g  x   g  a 
1
g  x
a  c  x or
xca
f  x 
g  x   , lim
exists
Case #1: a   , lim
x a
x a g   x 
f  x  f b 

f  c  f  x   f b  g  x  g  x 


g b
g c  g  x   g b 
1
g  x
; bc x
If x and b  x  b  are sufficiently large, then we
f b
g b
can make
and   arbitrarily close to
g  x
g  x
f c 
g  x    , and
zero, since lim
arbitrarily close
x 
g c 

f  x
to lim
.
x  g   x 
f  x
f  x 
 lim
So we get that lim
.
x  g  x 
x  g   x 
f  x 
g  x   , lim
exists
Case #2: a   , lim
x a
x a g   x 
f  x  f b 

f  c  f  x   f b  g  x  g  x 


g b
g c  g  x   g b 
1
g  x
b  c  x  a or
a xcb
If x and b are sufficiently close to a, then we
f b
g b
can make
and   arbitrarily close to
g  x
g  x
f c 
g  x    , and
zero, since lim
arbitrarily
close
xa
g c 

f  x
to lim
.
x a g   x 
f  x
f  x 
 lim
So we get that lim
.
x a g  x 
x a g   x 
Examples where L’Hospital’s Rule doesn’t apply:
1
f  x   x sin  
 x
g  x   sin x
2
1.
a0
See Handout!
2.
sin x
f  x  2
x
1
g  x 
x
a  
See Handout!

1
 1 
 f   x  2 x sin  x   cos  x  
 
 




g
x
cos
x




 f  x

x
1




 x sin  
x
 g  x  sin x

 f   x  2sin x


 cos x 
 
x
 g  x



 f  x   sin x

 g  x

x


Examples where L’Hospital’s Rule doesn’t apply(cont.):
3.
f  x   x sin 2 x
g  x   x2
a  
See Handout!
4.
f  x   x  sin x
g  x  x
a  
See Handout!
 f   x  sin 2 x sin 2 x 


 

g
x
2
2
x
  



2
 f  x   sin x

 g  x

x


 f  x 

 1  cos x 
 
 g  x



 f  x   1  sin x 
 g  x

x


Examples where L’Hospital’s Rule doesn’t apply(cont.):
 f  x 

 undefined if 1  cos x  0 
 
5.
g  x
g  x   x  sin x 


 f  x

1
a  



 g  x  1  sin x

See Handout!


x
f  x  x
An example where you can’t get away from
the zeros of g   x  .
Surprising examples where L’Hospital’s Rule applies:
 f   x  0 
 

 g   x  1 
See Handout!
 f  x 
1 

 
2
1
tan x  g  x  1  x 
2. lim


x
x
 f   x   0 


g
x
2
x


See Handout! 


1
1. lim
x  x
sin x
Watch out for lim
!!!
x 
x
More surprising examples where L’Hospital’s Rule
applies:
3. If f  exists on  0,  and lim  f  x   f   x   L ,
then find lim f  x .
x
x 
f  x  .
If L is a number, then find lim
x 
See Handout!
ex f  x 
{Hint: Apply L’Hospital’s Rule to f  x  
, and
x
e
then observe that f   x    f  x   f   x   f  x  .}
What’s lim f   x  if f  x   x ? Is this a problem?
x 
More surprising examples where L’Hospital’s Rule
applies:
4. If f  exists on  0,  and lim f   x   L , and
x 
lim f  n  exists as a number, then what must
n 
be the value of L?
f  x
{Hint: Apply L’Hospital’s Rule to
, and use it to
x
f  n
determine lim
. Determine the limit from
n
n
f  n  exists as a number .}
the fact that lim
n 
See Handout!
f  x
 A , where A
5. If f  exists on  0,  and lim
x
x
is a number. Show that there is a sequence  xn 
with xn   and lim f   xn   A.
n
See Handout!
{Hint: The Mean Value Theorem implies that for each n
f  2n   f  n 
for some n  xn  2n . So
f   xn  
n
f  2n   f  n 
f  2n  f  n 
f   xn  
 2

.}
n
2n
n
You might think that lim f   x   A , by applying
x 
L’Hospital’s Rule in reverse, but consider f  x   sin x .
6. You can see that lim
x 
1
e
sin x
doesn’t exist. If we
2 x  sin 2 x
write the limit as lim
and try
x   2 x  sin 2 x  esin x
L’Hospital’s Rule, we get
4cos x
lim
 0. What’s wrong?
sin
x
x   2 x  4cos x  sin 2 x  e
See Handout!
Hint:
 2 x  sin 2 x   2 1  cos 2 x 
 4cos x
2
 2 x  sin 2 x  esin x    2 1  cos 2 x   cos x  2 x  sin 2 x   esin x
  4cos 2 x  cos x  2 x  sin 2 x   esin x
 cos x  2 x  4cos x  sin 2 x  esin x
 2 x  sin 2 x  
4cos 2 x
sin x
cos
x
2
x

4cos
x

sin
2
x
e



 2 x  sin 2 x  esin x 
7. Use L’Hospital’s Rule to evaluate
n
lim
1  a1x 1  a2 x  1  an x   1
x 0
x
Where the numbers a1 , a2 ,
, an are arbitrary
real numbers.
See Handout!
Hint:
d n
1  a1 x 1  a2 x 

dx 

1  a1 x 1  a2 x 
1  an x  
1  an x 
1
n

n
i 1
n
ai
1  ai x 
The Nth Derivative Test
A common test for determining the nature of critical numbers in a
first semester Calculus course is the 2nd Derivative Test. Here is
a list of three common hypotheses from six Calculus textbooks:
I. Suppose that f  is continuous in an open interval containing c.
II. Suppose that f  exists in an open interval containing c.
III. Suppose that f  exists in an open interval containing c, and
f   c  exists.
The weakest hypothesis is III.
The following conclusions are common to all versions of the
2nd Derivative Test:
If f   c   0 and f   c   0 , then f has a local maximum at
x  c.
If f   c   0 and f   c   0 , then f has a local minimum at
x  c.
If f   c   0 and f   c   0 , then the 2nd Derivative Test fails.
Suppose that f  exists in an open interval containing c,
f   c  exists, and f   c   0 .
f c  h  f c 
f c  h 
f   c   lim
 lim
h0
h0
h
h
f c  h

hlim
0 
h
See Handout!

 lim f   c  h 
h0
h
If f   c   0 , then the sign chart of f  looks like:
0
f
c
If f   c   0 , then the sign chart of f  looks like:
0
c
f
If you ask a veteran Calculus student about the 2nd Derivative
Test, you’ll probably get a positive response, but if you ask about
the Nth Derivative Test, you’re likely to get a puzzled look.
Typically, the Nth Derivative Test is proved using Taylor’s Theorem
along with the following hypotheses in second semester Calculus
or higher:
 n

f
,
,
f
n

2
For
, suppose that
are continuous in an open
n 1
interval containing x  c and that f   c   0, , f    c   0, but
n
f    c   0.
We can prove an Off-the-rack Nth Derivative Test (without using
Taylor’s Theorem) and with weaker hypotheses, i. e. first semester
Calculus style.
First, let’s find some general hypotheses on f and its derivatives.
Beginning of the Off-the-rack Nth Derivative Test:
For n  2 , suppose that f ,
containing x  c, f   c   0,
, f  n 1 exist in an open interval
,f
 n 1

f
c

0
,
 
n
c
exists
 n
and f  c   0 .
Now we’ll investigate four cases:
Case I:
n is odd and f  n  c   0 :
 n2
The weakest 2nd Derivative Test applied to f
along with the
Mean Value Theorem yield the following:
++ 0 ++
Odd derivative:
c
--
f
n2
0 ++
Even derivative:
c
f
 n 3
++ 0 ++
Odd derivative:
c
f
Case II: n is odd and f  n   c   0 :
 n2
The weakest 2nd Derivative Test applied to f
along with the
Mean Value Theorem yield the following:
--
0 --
Odd derivative:
c
f
n2
++ 0 -Even derivative:
c
--
f
 n 3
0 --
Odd derivative:
c
f
Case III: n is even and f  n   c   0 :
 n2
The weakest 2nd Derivative Test applied to f
along with the
Mean Value Theorem yield the following:
++ 0 ++
Even derivative:
c
--
f
0 ++
Odd derivative:
c
--
n2
f
 n 3
0 ++
Odd derivative:
c
f
Case IV: n is even and f  n   c   0 :
 n2
The weakest 2nd Derivative Test applied to f
along with the
Mean Value Theorem yield the following:
--
0 --
Even derivative:
c
f
n2
++ 0 -Odd derivative:
c
f
 n 3
++ 0 -Odd derivative:
c
f
From the sign patterns in the previous four cases, we can now
state an Off-the-rack Nth Derivative Test:
For n  2 , suppose that f ,
containing x  c, f   c   0,
, f  n 1 exist in an open interval
,f
 n 1
c  0 , f 
n
 c  exists
n
and f    c   0 .
If n is odd, then f has neither a maximum or minimum at
x  c.
n
If n is even and f    c   0 , then f has a local minimum at
x  c.
 n
n
f
If
is even and
 c   0 , then f has a local maximum at
x  c.
Determine what’s going on at zero for the following
functions:
x3 x 2
f  x     1  cos x
6 2
x4 x2
f  x      1  cos x
24 2
x
f  x 

1
t 2 sin  1t   t ; t  0
g  t  dt g  t   
0
;t  0

Suppose that f has derivatives of all orders in an open interval
containing x  c and they’re all equal to zero at x  c . Can we
conclude anything about the nature of f at x  c ?
Consider the functions:
  x12
e ; x  0
f  x  
 0 ; x  0
  x12
 e ; x  0
f  x  
 0 ; x  0
x
f  x 

4
g  t  dt
  x12
e ; x  0
g  x  
 0 ; x  0
In the case of n being odd, can we conclude anything
about the graph of f at x = c? See Handout!
In the case of f  c   0 , we can conclude that the sign
chart for f” near x = c is as follows:
 n
--
0 ++
c
f 
In the case of f  c   0, we can conclude that the sign
chart for f” near x = c is as follows:
 n
++ 0 -c
f 
The Nth Derivative Test is fairly definitive.
Suppose that g has a (piecewise) continuous derivative on the
interval 0,2 and 2  g   x   3 on 0,2 . By considering the
formula for the length of the graph of g on the interval 0,2,
2

1   g   x   dx
2
0
Determine the maximum possible length of the graph of g on the
interval 0,2 .
Determine the minimum possible length of the graph of g on the
interval 0,2 .
See Handout!
1
If g  0   10, then complete the graph of the function g on the
interval 0,2 that has the maximum length. See Handout!
If g  0   10, then complete the graph of the function g on the
interval 0,2 that has the minimum length.
See Handout!
Do the same under the assumptions:
3  g   x   2
or
3  g   x   2
See Handout!
Suppose that f and g have (piecewise) continuous derivatives,
3  f   t   2 ,2  g   t   1, and 8  f  t   10 , then use
the surface area of revolution about the y-axis formula

2
2
 dx   dy 
2 x      dt
dt   dt 



to find a decent upper bound on the surface area of revolution
about the y-axis of the curve  x  f t

See Handout!

;1  t  5

y  g t


to find the minimal surface area of revolution about the y-axis of
the curve
 x  f t 

;1  t  5

y  g t


See Handout!
Fermat/Steiner Problems
Fermat’s problem: Given three points in the plane,
find a fourth such that the sum of its distances to the
three given ones is a minimum.”
Euclidean Steiner tree problem: Given N points in the
plane, it is required to connect them by lines of minimal
total length in such a way that any two points may be
interconnected by line segments either directly or via
other points and line segments.
The Euclidean Steiner tree problem is solved by finding a minimal length
tree that spans a set of vertices in the plane while allowing for the
addition of auxiliary vertices (Steiner vertices). The Euclidean Steiner
tree problem has long roots that date back to the 17th century when the
famous scientist Pierre Fermat proposed the following problem: Find in
the plane a point, the sum of whose distances from three given points is
minimal.
A practical example:
Two factories are located at the coordinates  a,0 
and  a,0  with their power supply located at  0,h ,
a, h  0. Find x so that the total length of power
line from the power supply to the factories is a
minimum.
 0,h  power supply
Steiner point
or vertex
x
factory
 a,0 
factory
 a,0 
The length of the power line as a function of x
with the parameters a and h is given by
L  x; a, h   2 x 2  a 2   h  x  ; 0  x  h
2x
L  x; a, h  
1
x2  a2


2 x  x2  a2
x2  a2
3x 2  a 2
 2x 
x a
2
2

x a
2
2
a 
a 

3 x 
 x 

3
3




2
2
2
2
2x  x  a
x a


Here are the possible sign charts for L’ depending on the
values of the parameters a and h.


0
a
h
3
a
0
h
3



a
h
3
a
h,
3
0

0
0


 0
h
a
3
a
h
3
a
So in the case of h 
the Steiner point should
3
a
a
be located
units above the factories. If h  ,
3
3
then the power lines should go directly from the
factories to the power supply without a Steiner
point.
 0,h 
a
a
h
h
3
3
a
 0,h 
3
 a,0 
 a,0   a,0 
 a,0 
Compare this with the soap film
configuration using the frames.
Vary the height of this
suction cup and
compare Nature’s
minimization to the
Calculus predictions.
See Handout!
A four point example:
We want to link the four points  a, h  ,  a, h ,
 a, h  , and  a, h  with a, h  0 in a minimal way.
  a, h 
 a, h 
Steiner points
x
x
  a,  h 
 a,  h 
There are two competing arrangements for the position
of the Steiner points: horizontal or vertical.
The length of the connection as a function of x
with the parameters a and h in the vertical case is
given by
LV  x; a, h   2 x  4
LV  x; a, h   2 

2
h  x
2
 a2
; 0 xh
4h  x
h  x
2
 a2
2
h

x

a
 4h  x


2
2
h

x

a


2
4  h  x   4a  16  h  x 

 2  h  x 2  a 2  4  h  x    h  x 2  a 2


2
2
2
2
2

4 a  3 h  x  



 2  h  x 2  a 2  4  h  x    h  x 2  a 2



2

4  a  3  h  x    a  3  h  x  
 h  x   a  4  h  x 

2

2

h  x

2
 a2

4  3 x  3h  a    3 x  3h  a 




 2  h  x 2  a 2  4  h  x    h  x 2  a 2


 
a   
a 
12  x   h 
  x   h 

3   
3 



 2  h  x 2  a 2  4  h  x    h  x 2  a 2


Here are the possible sign charts for LV’, depending on
the values of the parameters a and h.





a
0, h 
3


0

a
h
3
0
 0

0
h
a
h
3

 U 
a
h,
3


 U 

a
h
3
 U 
h
a
h
3
a
h
3
a
h
3
a
h
3
a
So in the case of h 
the two vertical Steiner
3
a
points should be located h 
units above and
3
a
below the origin. If h 
then there should
3
only be one Steiner point at the origin.
From the symmetry of the problem, we can quickly
get the results for the horizontal case by switching
a and h.
LH  x; a, h   2 x  4
2
a

x

h


2
; 0 xa
 
h   
h 
12  x   a 
  x   a 

3   
3 


LH  x; a, h  
 2  a  x 2  h 2  4  a  x    a  x 2  h 2


Here are the possible sign charts for LH’, depending on
the values of the parameters a and h.



a


0


h
0, a 
3


h
a
3
0
 0

0
h
a,
3



 U 
h
a
3
 U 
h
a
3
 U 
a
h
a
3
h
a
3
h
a
3
h
a
3
h
So in the case of a 
the two horizontal Steiner
3
h
points should be located a 
units left and
3
h
right of the origin. If a 
then there should
3
only be one Steiner point at the origin.
Here are the possible relationships between a and h in
both the horizontal and vertical arrangements:
Vertical:
a
h
3
a
h
3
Horizontal:
h
a
3
h
a
3
There are four combinations of the inequalities:
a
h
h
1. h 
,a 
0
 a  3h
3
3
3
a  3h
h
a
3
a
h
h
h
2. h 
,a 
a
and a  3h  0  a 
3
3
3
3
a  3h
h
a
3
a
h
h
3. h 
,a 
a
and a  3h  a  3h
3
3
3
a  3h
h
a
3
a
h
h
4. h 
,a 
 0  3h  a 
3
3
3
a  3h
Not possible.
h
a
3
Here’s the Phase Transition diagram in the
ha parameter plane
a  3h
h
a
3
a  3h
LV  0; a, h   4 h  a
2

minimum
2
h


LH  a 
; a, h   2 a  3h
3



ah
minimum
minimum




minimum

a


LV  h 
; a, h   2 h  3a
3


h


LH  a 
; a, h   2 a  3h
3


a

h

3
a


LV  h 
; a, h   2 h  3a
3


LH  0; a, h   4 h 2  a 2
Compare this with the soap film
configuration using the frames.
Vary the distance
between pairs of suction
cups and compare
Nature’s minimization to
the Calculus predictions.
See Handout!
The Blancmange Function
In the 19th Century, mathematicians gave examples
of functions which were continuous everywhere on
there domain, but differentiable nowhere on their
domain.
One such example is constructed as
follows:
Start with a function defined on the interval 0,1 with
a single corner at x  12, h  x   1  2 x  12
Here’s its graph:
Now extend it periodically to all the nonnegative real
numbers to get the function g  x  .
Here’s a portion of its graph:
Now we can define the continuous, nondifferentiable
function on the interval 0,1,
f  x 


n 0
g  2n x 
2n
g  2 x  g  4 x  g 8x 
 g  x 



2
4
8
Here’s an approximate graph of the function f
known as the Blancmange Function:
It is an example of a fractal, in that it is infinitesimally
fractured, and self-similar. No matter how much you
zoom in on a point on the graph, the graph never
flattens out into an approximate non-vertical line
segment through the point.
The number of points of nondifferentiability in the
interval 0,1 of the component functions
increases with n.
g  2n x 
2n
n
g
2
Here are some selected plots of  x  on the
2n
interval 0,1
n0
n 1
n2
Show that the formula for the function f actually
makes sense.
f  x 


n 0
g  2n x 
2
n
, which means that for each x in
0,1 , f  x   Nlim
 
N
g  2n x 
2
n 0
n
. If you can show that
fN  x
for each x f N  x  

N
n 0
g  2n x 
2
n
is bounded from above

N
and is nondecreasing in N, then lim
N 
must exist as a number.
n 0
g  2n x 
2
fN  x
n
Bounded above:
fN  x 

N
n 0
g  2n x 
2
n

N

n 0
1

n
2


n 0
1
1

2
n
2 1 1
2
Nondecreasing:
f N 1  x   f N  x  
g  2 N 1 x 
2
N 1
See Handout!
0
Show that the function f is continuous on
fN  x
0,1 . f  x   Nlim

So f  x   f N  x  


n  N 1
g  2n x 
2
n


1

n
2
n  N 1
for every x
in 0,1 . Use this to show that you can make
f  x  fN  x 

3
for every x in 0,1
.
by choosing N large enough.
See Handout!


1
1 2 N 1
1

 N
n
1 2
2
n  N 1
1
2
1 
3
3
N
So if N  log 2   , then 2  and N  .
2
3

 
Since f N  x  

N
n 0
g  2n x 
2
n
is a continuous function,
there is a   0 so that if x  y   , then
fN  x  fN  y  

3
Choose x in 0,1 , let   0 , and consider
f  x  f  y  f  x  fN  x  fN  x  fN  y   fN  y   f  y 
 f  x  fN  x  fN  x  fN  y   fN  y   f  y 
Finish the proof of the continuity of f.
See Handout!
Show that the function f is nondifferentiable on 0,1 .
1
Consider the sequence of points xm  m .
2
f  xm   f  0 
 2m .
Show that
xm  0
 n 1 
 g2  m 
2 

0
n
2
f  xm   f  0  n0

1
xm  0
0
m
2
 n 1 
p
g
2


g
2


For values of n greater than or equal to m, 
m 
2 


for some positive whole number p, but g  whole number   0
so we get that
f  xm   f  0 

xm  0
m 1

n 0
g  2nm 
2nm
 1  m1  1  m2  1 
 2  g  m   2  g  m1   2  g  m2  
2 
2 
2 
m
1 1
1 1
2 m 2m1 2m2 2
1
 2 g 
2
1
1
1
m 1
m2
 2  2  m  2  2  m1  2  2  m2 
2
2
2
 2  2  2   2  2m
m
m times
f  xm   f  0 
 2m
xm  0
What does this imply about f   0  ?
1
 22
2
1
Since xm  m  0 as m   , if f   0  exists, then
2
f  xm   f  0 
f   0   lim
, but
m
xm  0
f  xm   f  0 
lim
 lim 2m  
m
m
xm  0
1
Consider the sequence of points xm  1  m .
2
f  xm   f 1
Examine
.
xm  1
 n 
1 
 g  2  1  m  
 2 

0
n
2
f  xm   f 1 n0

1
xm  1
1 m 1
2

For values of n greater than or equal to m,
 n 
1 
g  2  1  m    g  k  2 p 
 2 

for some positive whole numbers p and k, but as before,
g  whole number   0
So we get that
f  xm   f 1

xm  1
n 0
m 1

 2m  1 
g  mn 
 2

2n m
 m 
1  m1 
1 
   2  g 1  m   2  g  2  m1  
2 
 2 


 m 1 
 2  g  2  
2 

Since g is periodic of period 1, we get that
 m 
1  m1 
1  m2 
1 
   2  g 1  m   2  g 1  m1   2  g 1  m2  
 2 
 2 
 2 

 1 
 2  g 1   
 2 
1
1 1 1
1

m
m 2
2
2
1
2
1  m1
2
 m 
1  m1 
1  m2 
1 
   2  g 1  m   2  g 1  m1   2  g 1  m2  
 2 
 2 
 2 

 m 
1 
1 


m 1 
  2   2  2 1  m    2   2  2 1  m1   
 2 
 2 





   2  2  2  2  2   2m
m times


What does this imply about f  1 ?
See Handout!
 1 
 2  g 1   
 2 

 1  
 2   2  2 1    
 2  

1 1
Consider the sequence of points xm   m .
2 2
1
f  xm   f  
2
Examine
1
xm 
2
1

What does this imply about f    ?
2
Try similar thinking to show that f   x  doesn’t exist for any
p
x  k where p and k are whole numbers.
2
These x’s are called dyadic rational numbers. If x is in 0,1
and is not a dyadic rational, then for a fixed value of m,
p
p 1
x falls between two adjacent dyadic rationals, m  x  m .
2
2
p
p 1
Let xm  m and ym  m , for each whole number m to get
2
2
two sequences  xm  and  ym  so that lim xm  lim ym  x and
m
xm  x  ym .
m
3
1
2
For example, let x  , and m  3, then x3  and y3 
8
3
8
6
1
5
For example, let x  , and m  4 , then x4 
and y4 
16
3
16
For example, let
11
12
1
, and m  5 , then x5 
and y5 
x
32
32
3
For f m  x  

m
n 0
g  2n x 
2
n
, if x is not a dyadic rational, then
show that for every value of m, f m1  x   f m  x   2 .
f m1  x   f m  x  
m 1

g   2n x  
n 0
 g   2m1 x 
See Handout!

m
n 0
g   2n x 
f  ym   f  x 
f  xm   f  x 
 f m1  x  
Prove that
.
ym  x
xm  x
f  ym   f  x 
First we’ll show that
 f m1  x  .
ym  x
 n p 1
n
 g2 

m 
g
2
x

2 


n
n
2
2
f  ym   f  x  n  0
n 0

p 1
ym  x
x
m
2
 n p 1
n
g
2


g
2
x 
n
m 1


m 
g
2
x

2 



p 1
n p 1
n
n 0
nm 2 
2n  m  2n x

2
x
m
2
2
 p 1
f m1  m   f m1  x  
n
g
2
x

2 



p 1
n p 1
n
n

m

x
2


2
x
m
m
2
2





 p p  1
Since f m1  x  is a linear function on the interval  m , m  .
2 
2
3
1
2
y

For example, let x  , and m  3, then x3  and 3
8
3
8
f2  x 
6
1
5
For example, let x  , and m  4 , then x4 
and y4 
16
3
16
f3  x 
For example, let
11
12
1
, and m  5 , then x5 
and y5 
x
32
32
3
f4  x 
 p 1
f m1  m   f m1  x  
n
g
2
x

2 



p 1
n p 1
n
nm 2 

x

2
x
m
m
2
2

g  2n x 
 f m1  x  
n p 1
n
nm 2 

2
x
m
2


0
 f m1  x 
You can do a similar argument to show that
f  xm   f  x 
 f m1  x  .
xm  x
f  xm   f  x 
 f   x  and
If f   x  exists, then mlim

xm  x
f  ym   f  x 
lim
 f   x  , but the Squeeze Theorem would
m
ym  x
f m  x  .
imply something about mlim

Using the previous results, show why f   x  doesn’t exist.
Differentiability of Powers of the Popcorn Function
Consider the function
0 ; x is irrational

f  x  1
m
 n ; x is a reduced rational with x  n
This function was originally defined by the mathematician Johannes Thomae.
It’s called the popcorn function, ruler function, raindrop function,…
Here is a portion of its graph:
Popcorn/Raindrop
Ruler
Outline of the Proof of the continuity of the Popcorn Function
at the irrationals and the discontinuity at the rationals.
Let’s begin with a basic fact about rationals and irrationals. Both
types of numbers are dense in the real numbers: meaning that
every interval of real numbers contains both rational and
irrational numbers.
Since the Popcorn Function value at any irrational number is
zero, to show that the Popcorn Function is continuous at an
irrational number, x0 , we just have to show that
lim f  x   0   f  x0  
xx0
Remember, to show that lim f  x   L for any function f, we
x  x0
have to show that for every   0 , there is a   0 so that if
0  x  x0   , then f  x   L   .
So let’s begin the proof by letting   0 . Now we will choose
1
n so that   . Now consider the finitely many rational
n
numbers in 0,1 whose denominator is less than or equal to
n
:
r1, r2 ,
0
r1
, rk  .
r2
x0
rk  2
rk 1
1
rk
Let   minimum  x0  r1 , x0  r2 ,

, x0  rk
x0
rk  2 rk 1
r2
0
r1
If 0  x  x0   and x is irrational, then

1
rk
f  x   0  ____________  ________
j
And if 0  x  x0   and x is rational with x 
, then
m
m___n , and f  x   0  __________  _________ .
See Handout!
To show that the Popcorn Function is discontinuous at a
m
, we have to show that for some   0
n
there is no   0 with the property that if 0  x  x0   , then
rational number, x0 
1
f  x     . In other words, we have to show that for every
n
  0 , there is at least one x with 0  x  x0   , but
1
f  x   
n
1
To accomplish this, let  
. For every,   0 there is an
2n
irrational number x with 0  x  x0   , but
1
f  x    __________  _________
n
See Handout!
Is the darn thing differentiable?
Let’s look at the difference quotient for an irrational number a:
0 ; x is irrational
f  x  f a f  x  1

 n
m
; x is a reduced rational,x 
xa
xa m
n
 n a
Hurwitz’s Theorem:
Any irrational number a has an infinity of rational
approximations
m
m
1
which satisfy a  
.
2
n
n
5n
So Hurwitz’s Theorem implies that
 0 ; x is irrational
f  x  f a  1
 n  5n ; x is a reduced rational,x  m
xa
 12
n
 5n
So it’s not differentiable anywhere.
What about powers of the popcorn function?
f  x   x is not differentiable at x  0 , but its square is.
2
3
Let’s look at the difference quotient for an irrational number a:
0 ; x is irrational
f  x  f a f  x  1

  n2
m
; x is a reduced rational,x 
xa
xa m
n
 n a
2
2
2
So Hurwitz’s Theorem implies that
 0 ; x is irrational
f  x  f a  1
 n2  5 ; x is a reduced rational,x  m
xa
 12
n
 5n
2
2
So it square is not differentiable anywhere.
Liouville’s Theorem:
An algebraic number, a , of degree k , has the property that
for each positive number c , there are only finitely many reduced
m
c
m
rationals
with a    for   k  1 .
n n
n
m
So for c  0 there are only finitely many rationals
with
n
a
m
c
  ,then for the remaining rationals in the reduced
n n
m
c
interval we’d have a    . This means that in this case,
n n
Let’s look at the difference quotient of the popcorn function raised
to the power  at an irrational algebraic number a of degree k:
0 ; x is irrational
f   x  f  a
f   x 

 c
m
xa
xa
   ; x is a reduced rational, x 

n
 n
If     k  1 then
 c
f   x   f   a      ;  
 n
. In other words, the
xa
 c ;  
popcorn Function raised to the  power is differentiable at all
algebraic irrationals of degree less than or equal to   1 . For
3
f
 x  is differentiable at the second degree algebraic
example,
irrational 2 , but not necessarily at the third degree irrational 3 2 .
So eventually, every algebraic irrational number will be a point of
differentiability of some power of the popcorn function. Furthermore,
f
is differentiable at every irrational algebraic number for   2 ,
using the Thue-Siegel-Roth Theorem for which Klaus Roth received
a Fields Medal in 1958. What about the transcendental numbers?
Some transcendental numbers are not points of differentiability for
any power of the popcorn function,the Liouville transcendentals.
Other transcendental numbers are eventually points of
differentiability for some power of the popcorn function:
f  is differentiable at  for   42 (1953)
f  is differentiable at  for   9 (1993)
f

is differentiable at
 2 for   6 (1996)
f  is differentiable at ln 2 for   4 (1987)