L’Hospital’s Rule

Hello, my name is L’Hopital.

My name is L’Hospital!

And that first guy can’t spell.

To Tell the Truth 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

L’Hospital #1 L’Hospital #2 L’Hospital #3

Will the real L’Hospital please stand up!!!

Johann Bernoulli

L’Hospital #3

It’s true, L’Hospital’s Rule can be directly applied to the

limit forms and . 0

0

whatever

But the rule says nothing if doesn’t exist.

limx a

f x

g x

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.

If or ,

lim lim 0x I x I

x a x af x g x

limx I

x ag x

and , lim

x Ix a

f xL a number or

g x

then .

limx I

x a

f xL

g x

[Suppose that and for all x in I.]

0g x 0g x

This condition is typical, but not needed if we assume

exists.

limx I

x a

f x

g x

, , , , , , , ,I a b b a b b orb a c

This just takes care of one-sided limits and limits at infinity all at once.

The many different cases of L’Hospital’s Rule can all be proven using Cauchy’s Mean Value Theorem:

Let f and g be continuous on and

differentiable on . Then there is a c

in with

or .

,a b

,a b

,a b f c g b g a g c f b f a

; 0,

f c f b f ag c g a g b

g c g b g a

Extra letters and limits of rachieauxs seem to be French things. Maybe it’s something in the Perrier. Eau well!

Cauchy’s Mean Value Theorem can be proven from Rolle’s Theorem:

F x f x f a g b g a g x g a f b f a

Apply Rolle’s Theorem to the function

on the interval . ,a b

See Handout!

The case is covered in most textbooks, but the

case isn’t mentioned.whatever

0

0

From Cauchy, we get that

1

f x f a

f c f x f a g x g x

g ag c g x g ag x

a c x or x c a

;1

f x f b

f c f x f b g x g xb c x

g bg c g x g bg x

Case #1:

, lim , limx a x a

f xa g x exists

g x

So we get that .

lim limx x

f x f x

g x g x

If and are sufficiently large, then we

can make and arbitrarily close to

zero, since , and arbitrarily close

to .

x bx b

f b

g x

g b

g x lim

xg x

f c

g c

limx

f x

g x

1

f x f b

f c f x f b g x g x

g bg c g x g bg x

b c x a or a x c b

Case #2:

, lim , limx a x a

f xa g x exists

g x

So we get that .

lim limx a x a

f x f x

g x g x

If and are sufficiently close to a, then we

can make and arbitrarily close to

zero, since , and arbitrarily close

to .

x b

f b

g x

g b

g x lim

x ag x

f c

g c

limx a

f x

g x

Examples where L’Hospital’s Rule doesn’t apply:

1.

2 1sin

sin

0

f x xx

g x x

a

1 12 sin cos

cos

1sin

sin

xf x x xg x x

f x xx

g x x x

2.

2

sin

1

xf x

x

g xx

a

2sincos

sin

f x xx

g x x

f x x

g x x

See Handout!

See Handout!

Examples where L’Hospital’s Rule doesn’t apply(cont.):

3.

2

2

sinf x x x

g x x

a

2

2

sin 2 sin

2 2

sin

f x x x

g x x

f x x

g x x

4.

sinf x x x

g x x

a

1 cos

sin1

f xx

g x

f x x

g x x

See Handout!

See Handout!

Examples where L’Hospital’s Rule doesn’t apply(cont.):

5. sin

f x x

g x x x

a

1 cos 0

1sin

1

f xundefined if x

g x

f xxg x

x

An example where you can’t get away from the zeros of . g x

See Handout!

Surprising examples where L’Hospital’s Rule applies:

1.1

limx x

0

1

f x

g x

2.1tan

limx

x

x

2

1

1

0

2

f x

g x x

f x

g x x

Watch out for !!! sin

limx

x

x

See Handout!

See Handout!

More surprising examples where L’Hospital’s Rule applies:

3. If exists on and ,

then find .

If L is a number, then find .

f 0, limx

f x f x L

lim

xf x

limx

f x

{Hint: Apply L’Hospital’s Rule to , and

then observe that .}

x

x

e f xf x

e

f x f x f x f x

What’s if ? Is this a problem? limx

f x

f x x

See Handout!

More surprising examples where L’Hospital’s Rule applies:

4. If exists on and , and

exists as a number, then what must

be the value of L?

f 0, limx

f x L

limn

f n

{Hint: Apply L’Hospital’s Rule to , and use it to

determine . Determine the limit from

the fact that exists as a number .}

f x

x limn

f n

n

limn

f n

See Handout!

5. If exists on and , where A

is a number. Show that there is a sequence

with and .

f 0,

limx

f xA

x

nx

nx lim nnf x A

{Hint: The Mean Value Theorem implies that for each n

for some . So

.}

2n

f n f nf x

n

2 22

2n

f n f n f n f nf x

n n n

2nn x n

You might think that , by applying L’Hospital’s Rule in reverse, but consider .

limx

f x A

sinf x x

See Handout!

6. You can see that doesn’t exist. If we

write the limit as and try

L’Hospital’s Rule, we get

. What’s wrong?

sin

1lim

xx e

sin

2 sin 2lim

2 sin 2 xx

x x

x x e

sin

4coslim 0

2 4cos sin 2 xx

x

x x x e

See Handout!

2

2 sin 2 2 1 cos2

4cos

x x x

x

sin sin

2 sin

sin

2 sin 2 2 1 cos2 cos 2 sin 2

4cos cos 2 sin 2

cos 2 4cos sin 2

x x

x

x

x x e x x x x e

x x x x e

x x x x e

2

sinsin

2 sin 2 4cos

cos 2 4cos sin 22 sin 2x

x

x x x

x x x x ex x e

Hint:

7. Use L’Hospital’s Rule to evaluate

Where the numbers are arbitrary

real numbers.

1 2, , , na a a

1 2

0

1 1 1 1lim

nn

x

a x a x a x

x

See Handout!

Hint:

1

1 2

1 2

1

1 1 1

1 1 11

n

nn

n

in

ii

da x a x a x

dx

aa x a x a x

a x

n

The Nth Derivative Test

II. Suppose that exists in an open interval containing c.

III. Suppose that exists in an open interval containing c, and

exists.

I. Suppose that is continuous in an open interval containing c.f

f

f

f c

A common test for determining the nature of critical numbers in a first semester Calculus course is the 2nd Derivative Test. Here isa list of three common hypotheses from six Calculus textbooks:

The weakest hypothesis is III.

If and , then f has a local minimum at .

If and , then f has a local maximum at .

0f c 0f c x c

The following conclusions are common to all versions of the 2nd Derivative Test:

0f c 0f c x c

If and , then the 2nd Derivative Test fails. 0f c 0f c

Suppose that exists in an open interval containing c,

exists, and .

f

f c

0f c

0 0

0

0

lim lim

lim

lim

h h

h

h

f c h f c f c hf c

h h

f c h

hf c h

h

If , then the sign chart of looks like: 0f c f

f c

0

If , then the sign chart of looks like: 0f c f

f c

0

See Handout!

For , suppose that are continuous in an open interval containing and that , but .

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 Theoremalong with the following hypotheses in second semester Calculus or higher:

2n , , nf fx c 10, , 0nf c f c

0nf c

For , suppose that exist in an open interval

containing , , exists

and .

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 and its derivatives.

f

Beginning of the Off-the-rack Nth Derivative Test:

2n 1, , nf f

x c 10, , 0nf c f c nf c

0nf c

Now we’ll investigate four cases:

Case I: is odd and : n 0nf c

The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following:

2nf

2nf c

0 + ++ +Odd derivative:

3nf c

0 + +- -Even derivative:

f c

0 + ++ +Odd derivative:

Case II: is odd and : n 0nf c

The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following:

2nf

2nf c

0 - -- -Odd derivative:

3nf c

0 - -+ +Even derivative:

f c

0 - -- -Odd derivative:

Case III: is even and : n 0nf c

The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following:

2nf

2nf c

0 + ++ +Even derivative:

3nf c

0 + +- -Odd derivative:

f c

0 + +- -Odd derivative:

Case IV: is even and : n 0nf c

The weakest 2nd Derivative Test applied to along with the Mean Value Theorem yield the following:

2nf

2nf c

0 - -- -Even derivative:

3nf c

0 - -+ +Odd derivative:

f c

0 - -+ +Odd derivative:

If is odd, then f has neither a maximum or minimum at .

From the sign patterns in the previous four cases, we can now state an Off-the-rack Nth Derivative Test:

For , suppose that exist in an open interval

containing , , exists

and .

2n 1, , nf f

x c 10, , 0nf c f c nf c

0nf c

If is even and , then f has a local minimum at .

nx c

n 0nf c x c

If is even and , then f has a local maximum at .

n 0nf c x c

Determine what’s going on at zero for the following functions:

3 2

1 cos6 2

x xf x x

4 2

1 cos24 2

x xf x x

1

x

f x g t dt

2 1sin ; 0

0 ; 0tt t t

g t t

Suppose that f has derivatives of all orders in an open interval

containing and they’re all equal to zero at . Can we

conclude anything about the nature of f at ?

x c

x c

Consider the functions:

2

1

; 0

0 ; 0

xe xf x x

x c

2

1

; 0

0 ; 0

xe xf x x

4

x

f x g t dt

2

1

; 0

0 ; 0

xe xg x x

In the case of n being odd, can we conclude anything about the graph of f at x = c?

In the case of , we can conclude that the sign chart for f” near x = c is as follows:

0nf c

f c

0 + +- -

In the case of , we can conclude that the sign chart for f” near x = c is as follows:

0nf c

f c

0 - -+ +

The Nth Derivative Test is fairly definitive.

See Handout!

Suppose that g has a (piecewise) continuous derivative on the interval and on . By considering the formula for the length of the graph of g on the interval ,

0,2 2 3g x

2

2

0

1 g x dx

0,2 0,2

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 , then complete the graph of the function g on the

interval that has the maximum length.

0 10g

0,2

See Handout!

If , then complete the graph of the function g on the

interval that has the minimum length.

0 10g

0,2

See Handout!

Do the same under the assumptions:

3 2g x

or

3 2g x

See Handout!

Suppose that f and g have (piecewise) continuous derivatives,

, , and , then use

the surface area of revolution about the y-axis formula

2 1g t 3 2f t 8 10f t

2 2

2dx dy

x dtdt dt

to find a decent upper bound on the surface area of revolution about the y-axis of the curve

;1 5

x f t

t

y g t

See Handout!

to find the minimal surface area of revolution about the y-axis of the curve

;1 5

x f t

t

y g t

See Handout!

Fermat/Steiner Problems

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.

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.” 

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.

0,h

,0a ,0a

x

power supply

factory factory

A practical example:

Two factories are located at the coordinates and with their power supply located at ,

. Find x so that the total length of power line from the power supply to the factories is a minimum.

,0a ,0a 0,h

, 0a h

Steiner point or vertex

The length of the power line as a function of x with the parameters a and h is given by

2 2; , 2 ; 0L x a h x a h x x h

2 2

2 2

2 2

2 2

2 2 2 2

2 2 2 2

2; , 1

2

3

2

33 3

2

xL x a h

x a

x x a

x a

x a

x x a x a

a ax x

x x a x a

Here are the possible sign charts for L’ depending on the values of the parameters a and h.

0 h3

a

0

3

ah

0 ,3

ah

0

3

ah

0 h3

a

0

3

ah

So in the case of the Steiner point should

be located units above the factories. If ,

then the power lines should go directly from the factories to the power supply without a Steiner point.

3

ah

3

a

3

ah

0,h

,0a ,0a

3

a 0,h

,0a ,0a

3

ah 3

ah

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 h

,a h ,a h

x

A four point example:

We want to link the four points , ,

, and with in a minimal way.

,a h ,a h

,a h , 0a h

Steiner points

,a h

,a h

x

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

2 2; , 2 4 ; 0VL x a h x h x a x h

2 2

2 2

2 2

2 22

2 22 2

4; , 2

2 4

4 4 16

2 4

V

h xL x a h

h x a

h x a h x

h x a

h x a h x

h x a h x h x a

22

2 22 2

2 22 2

2 22 2

4 3

2 4

4 3 3

2 4

4 3 3 3 3

2 4

a h x

h x a h x h x a

a h x a h x

h x a h x h x a

x h a x h a

h x a h x h x a

2 22 2

123 3

2 4

a ax h x h

h x a h x h x a

Here are the possible sign charts for LV’, depending on the values of the parameters a and h.

0,3

ah ,

3

ah

0

3

ah

3

ah

U

0 h3

ah

0

3

ah

3

ah

U

0 h3

ah

U

3

ah

So in the case of the two vertical Steiner

points should be located units above and

below the origin. If then there should

only be one Steiner point at the origin.

3

ah

3

ah

3

ah

From the symmetry of the problem, we can quickly get the results for the horizontal case by switching a and h.

2 2; , 2 4 ; 0HL x a h x a x h x a

2 22 2

123 3

; ,2 4

H

h hx a x a

L x a ha x h a x a x h

Here are the possible sign charts for LH’, depending on the values of the parameters a and h.

0,3

ha ,

3

ha

0

3

ha

3

ha

U

0 a3

ha

0

3

ha

3

ha

U

0 a3

ha

U

3

ha

So in the case of the two horizontal Steiner

points should be located units left and

right of the origin. If then there should

only be one Steiner point at the origin.

3

ha

3

ha

3

ha

Here are the possible relationships between a and h in both the horizontal and vertical arrangements:

3

ah

3

ah

3

ha

3

ha

Vertical:

Horizontal:

There are four combinations of the inequalities:

1. , 0 33 3 3

a h hh a a h

3a h

3

ha

2. , 3 03 3 3 3

a h h hh a a and a h a

3a h

3

ha

3. , 3 33 3 3

a h hh a a and a h a h

3a h

3

ha

4. , 0 33 3 3

a h hh a h a

3a h

3

ha

Not possible.

Here’s the Phase Transition diagram in the ha parameter plane

One ve

rtica

l Ste

iner

Point a

nd two h

orizonta

l

Stein

er P

oints

Two Steiner Points fo

r the

vertical a

nd horizontal

One horizontal and two vertical

3a h

3

ha

2 20; , 4VL a h h a

; , 2 33

H

hL a a h a h

; , 2 33

V

aL h a h h a

; , 2 33

H

hL a a h a h

; , 2 33

V

aL h a h h a

2 20; , 4HL a h h a

3a h

3

ha

minimum

a h

minimum

minimum

minimum

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 with a single corner at ,

0,112x 1

21 2h x x

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:

0,1

0

2

2

2 4 8

2 4 8

n

n

n

g xf x

g x g x g xg x

Now we can define the continuous, nondifferentiable function on the interval ,

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 of the component functions

increases with n.

0,1 2

2

n

n

g x

Here are some selected plots of on the

interval

2

2

n

n

g x

0,1

0n 1n

2n

0

2

2

n

n

n

g xf x

0,1

0

2lim

2

N

nN

nNn

f x

g xf x

0

2

2

nN

N n

n

g xf x

0

2lim

2

N

nN

nNn

f x

g x

Show that the formula for the function f actually makes sense.

, which means that for each x in

, . If you can show that

is bounded from above

must exist as a number.

for each x

and is nondecreasing in N, then

0 0 0

2 1 1 12

12 2 2 12

nN N

N n n n

n n n

g xf x

Bounded above:

Nondecreasing:

1

1 1

20

2

N

N N N

g xf x f x

See Handout!

So for every x

in . Use this to show that you can make

. 0,1 lim NNf x f x

1 1

2 1

2 2

n

N n n

n N n N

g xf x f x

0,1

3Nf x f x

0,1

Show that the function f is continuous on

.

for every x in

by choosing N large enough. See Handout!

1

1

11 12

12 212

N

n N

n N

So if , then and . 2

3logN

32N

1

2 3N

Since is a continuous function,

there is a so that if , then

0

2

2

nN

N n

n

g xf x

0 x y

3N Nf x f y

0,1 0

N N N N

N N N N

f x f y f x f x f x f y f y f y

f x f x f x f y f y f y

Choose x in , let , and consider

Finish the proof of the continuity of f.See Handout!

0,1

1

2m mx

02

0m

m

f x fm

x

0

122 0

2010 02

nm

nm n

mm

g

f x f

x

12 22

n pm

g g

0g whole number

Show that the function f is nondifferentiable on .

Consider the sequence of points .

Show that .

For values of n greater than or equal to m,

for some positive whole number p, but

so we get that

1

0

1 21 2

20

0 2

1 1 1 12 2 2 2

2 2 2 2

n mmm

n mm n

m m mm m m

gf x f

x

g g g g

1

2m 1

1

2m 2

1

2m

1

2

1 21 2

1 1 1 12 2 2 2 2 2 2 2

2 2 2 22 2 2 2 2

m m mm m m

m times

m

What does this imply about ? 0f

02

0m

m

f x fm

x

10

2m mx m 0f

00 lim

0m

mm

f x ff

x

0lim lim 2

0m

m mm

f x fm

x

Since as , if exists, then

, but

112m m

x

11

m

m

f x f

x

0

12 1

20

2111 1 12

nm

nm n

mm

g

f x f

x

12 1 2

2n p

mg g k

0g whole number

Consider the sequence of points .

Examine .

For values of n greater than or equal to m,

for some positive whole numbers p and k, but as before,

So we get that

1

0

11

2 11 2

1 2

1 1 12 1 2 2 2 2

2 2 2

m

m m nm

n mm n

m m mm m

gf x f

x

g g g

Since g is periodic of period 1, we get that

1 21 2

1 1 1 12 1 2 1 2 1 2 1

2 2 2 2m m m

m m mg g g g

1

2

112m

1

112m

2

112m

1 21 2

11

1 1 1 12 1 2 1 2 1 2 1

2 2 2 2

1 1 12 2 2 1 2 2 2 1 2 2 2 1

2 2 2

2 2 2 2 2 2

m m mm m m

m mm m

m times

g g g g

m

What does this imply about ? 1f

See Handout!

1 1

2 2m mx

12

12

m

m

f x f

x

1

2f

Consider the sequence of points .

Examine

What does this imply about ?

f x

2k

px

0,1

1

2 2m m

p px

Try similar thinking to show that doesn’t exist for any

where p and k are whole numbers.

These x’s are called dyadic rational numbers. If x is in

and is not a dyadic rational, then for a fixed value of m,

x falls between two adjacent dyadic rationals, .

2m m

px 1

2m m

py

mx

Let and , for each whole number m to get two sequences and so that and

.

my lim limm mm mx y x

m mx x y

1

3x 3m 3

2

8x 3

3

8y For example, let , and , then and

1

3x 5m 5

11

32x 5

12

32y For example, let , and , then and

1

3x 4m

4

5

16x 4

6

16y For example, let , and , then and

For , if x is not a dyadic rational, then

show that for every value of m, .

0

2

2

nm

m n

n

g xf x

1 2m mf x f x

1

1

0 0

1

2 2

2

m mn n

m m

n n

m

f x f x g x g x

g x

See Handout!

1

m mm

m m

f y f x f x f xf x

y x x x

1m

mm

f y f xf x

y x

Prove that .

First we’ll show that .

0 0

1

0

1 1

12 222 2

121

2 2 221 1

2 2 2 22 2

122

1 12 2

2 2

nnm

n nm n n

mm

n nnm m

n n n nn n m

m m

nm mm

n nn m

m m

pg g x

f y f xpy x x

pg g x g x

p px x

pf f x g x

p px x

1mf x

1,

2 2m m

p p

Since is a linear function on the interval .

2f x

1

3x 3m 3

2

8x 3

3

8y For example, let , and , then and

3f x

1

3x 4m

4

5

16x 4

6

16y For example, let , and , then and

4f x

1

3x 5m 5

11

32x 5

12

32y For example, let , and , then and

1 1

1

0

1

122

1 12 2

2 2

2

12 2

2

nm mm

n nn m

m m

n

mn n

n mm

m

pf f x g x

p px x

g xf x

px

f x

If exists, then and

, but the Squeeze Theorem would

imply something about .

You can do a similar argument to show that

.

1m

mm

f x f xf x

x x

f x

lim m

mm

f y f xf x

y x

lim m

mm

f x f xf x

x x

lim mmf x

f x

Using the previous results, show why doesn’t exist.

0 ;

1;

x is irrationalf x m

x is a reduced rational with xn n

Differentiability of Powers of the Popcorn Function

This function was originally defined by the mathematician Johannes Thomae.

Consider the function

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, , we just have to show that 0x

0

0lim 0x x

f x f x

So let’s begin the proof by letting . Now we will choose so that . Now consider the finitely many rational

numbers in whose denominator is less than or equal to : .

Remember, to show that for any function f, we

have to show that for every , there is a so that if

, then .

0

limx x

f x L

0 0

00 x x f x L

0

n

1

n

0,1

n

1 2, , , kr r r

0 11r

2r 0x 2kr 1kr

kr

And if and is rational with , then

, and .

If and is irrational, then

Let 1 2, , ,0 0 0 kminimum x r x r x r

00 x x

x

0 ____________ ________f x

00 x x

x

jx

m

m___n 0 __________ _________f x

0 11r

2r 0x 2kr 1kr

kr

See Handout!

To show that the Popcorn Function is discontinuous at a

rational number, , we have to show that for some

there is no with the property that if , then

. In other words, we have to show that for every

, there is at least one with , but

0

mx

n 0

0 00 x x

1f x

n

0 x 00 x x

1f x

n

1

2n 0

x 00 x x

1__________ _________f x

n

To accomplish this, let . For every, there is an

irrational number with , but

See Handout!

Is the darn thing differentiable?

Let’s look at the difference quotient for an irrational number a:

1

0 ;

;n

mn

x is irrationalf x f a f x

mx is a reduced rational,xx a x a

a n

Any irrational number has an infinity of rational

approximations which satisfy .

a

m

n 2

1

5

ma

n n

Hurwitz’s Theorem:

2

1

15

0 ;

5 ;n

n

x is irrationalf x f a

mn x is a reduced rational,xx a

n

So Hurwitz’s Theorem implies that

So it’s not differentiable anywhere.

What about powers of the popcorn function?

23f x x is not differentiable at , but its square is.0x

Let’s look at the difference quotient for an irrational number a:

2

2 2 2

1

0 ;

;n

mn

x is irrationalf x f a f x

mx is a reduced rational,xx a x a

a n

So Hurwitz’s Theorem implies that

2

2

2 21

15

0 ;

5 ;n

n

x is irrationalf x f a

mx is a reduced rational,xx a

n

So it square is not differentiable anywhere.

So for there are only finitely many rationals with

,then for the remaining rationals in the reduced

interval we’d have . This means that in this case,

An algebraic number, , of degree , has the property that

for each positive number , there are only finitely many reduced

rationals with for .

ka

cm

n

m ca

n n 1k

Liouville’s Theorem:

0c m

n

m ca

n n

m ca

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 irrationalf x f a f x

c mx a x a x is a reduced rational, x

n n

If then . In other words, the

popcorn Function raised to the power is differentiable at all

algebraic irrationals of degree less than or equal to . For

example, is differentiable at the second degree algebraic

irrational , but not necessarily at the third degree irrational .

1k ;

;

cf x f a

nx a

c

1

2

3f x

3 2

So eventually, every algebraic irrational number will be a point of

differentiability of some power of the popcorn function. Furthermore,

is differentiable at every irrational algebraic number for ,

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 2

f is differentiable at for (1953) 42

f is differentiable at for (1993) 9

f is differentiable at for (1996)2 6

f is differentiable at for (1987)ln 2 4