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L’Hospital’s Rule. I’m the real L’Hopital! And they’re both bad spellers. ^. To Tell the Truth. My name is L’Hospital! And that first guy can’t spell. Hello, my name is L’Hopital. L’Hospital #1. L’Hospital #3. L’Hospital #2. Question: Messieurs, can you tell us

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slide2
I’m the real L’Hopital! And they’re both bad spellers.

^

To Tell the Truth

My name is L’Hospital!

And that first guy can’t spell.

Hello, my name is L’Hopital.

slide3
L’Hospital #1

L’Hospital #3

L’Hospital #2

Question: Messieurs, can you tell us

something about your famous

rule?

slide4
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

limit forms and .

But the rule says nothing if doesn’t exist.

L’Hospital #1

L’Hospital #3

L’Hospital #2

slide5
Let a be a real number or and I an open interval which contains a or has a as an endpoint.

or

b

a

c

[Suppose that and for all x in I.]

This condition is typical, but not needed if we assume

exists.

If or ,

and ,

then .

Here’s a correct statement of L’Hospital’s Rule:

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

slide6
Let f and g be continuous on and

differentiable on . Then there is a c

in with

or .

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!

slide7
Apply Rolle’s Theorem to the function

on the interval .

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

See Handout!

slide8
The case is covered in most textbooks, but the

case isn’t mentioned.

From Cauchy, we get that

slide9
Case #1:

If and are sufficiently large, then we

can make and arbitrarily close to

zero, since , and arbitrarily close

to .

So we get that .

slide10
Case #2:

If and are sufficiently close to a, then we

can make and arbitrarily close to

zero, since , and arbitrarily close

to .

So we get that .

slide11
1.

2.

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

See Handout!

See Handout!

slide12
3.

4.

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

See Handout!

See Handout!

slide13
5.

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

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

See Handout!

slide14
1.

2.

Watch out for !!!

Surprising examples where L’Hospital’s Rule applies:

See Handout!

See Handout!

slide15
3. If exists on and ,

then find .

If L is a number, then find .

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

then observe that .}

What’s if ? Is this a problem?

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

See Handout!

slide16
4. If exists on and , and

exists as a number, then what must

be the value of L?

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

determine . Determine the limit from

the fact that exists as a number .}

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

See Handout!

slide17
5. If exists on and , where A

is a number. Show that there is a sequence

with and .

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

for some . So

.}

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

See Handout!

slide18
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?

See Handout!

slide20
Hint:

7. Use L’Hospital’s Rule to evaluate

Where the numbers are arbitrary

real numbers.

See Handout!

slide22
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 is continuous in an open interval containing c.

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

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

exists.

The weakest hypothesis is III.

slide23
The following conclusions are common to all versions of the

2nd Derivative Test:

If and , then f has a local maximum at

.

If and , then f has a local minimum at

.

If and , then the 2nd Derivative Test fails.

slide24
0

0

c

c

Suppose that exists in an open interval containing c,

exists, and .

See Handout!

If , then the sign chart of looks like:

If , then the sign chart of looks like:

slide25
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:

For , suppose thatare continuous in an open

interval containing and that , but

.

slide26
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.

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

For , suppose that exist in an open interval

containing , , exists

and .

Now we’ll investigate four cases:

slide27
Case I: is odd and :

- -

+ +

0

c

+ +

+ +

0

c

+ +

+ +

0

c

The weakest 2nd Derivative Test applied to along with the

Mean Value Theorem yield the following:

Odd derivative:

Even derivative:

Odd derivative:

slide28
Case II: is odd and :

- -

+ +

0

c

- -

- -

0

c

- -

- -

0

c

The weakest 2nd Derivative Test applied to along with the

Mean Value Theorem yield the following:

Odd derivative:

Even derivative:

Odd derivative:

slide29
Case III: is even and :

+ +

+ +

0

c

- -

+ +

0

c

The weakest 2nd Derivative Test applied to along with the

Mean Value Theorem yield the following:

Even derivative:

- -

+ +

0

Odd derivative:

c

Odd derivative:

slide30
Case IV: is even and :

- -

- -

0

c

+ +

- -

0

c

The weakest 2nd Derivative Test applied to along with the

Mean Value Theorem yield the following:

Even derivative:

- -

+ +

0

Odd derivative:

c

Odd derivative:

slide31
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 .

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

.

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

.

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

.

slide33
Consider the functions:

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 ?

slide36
- -

+ +

0

c

- -

+ +

0

c

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

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

The Nth Derivative Test is fairly definitive.

slide37
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 ,

Determine the maximum possible length of the graph of g on the

interval .

Determine the minimum possible length of the graph of g on the

interval .

See Handout!

slide39
If , then complete the graph of the function g on the

interval that has the maximum length.

See Handout!

slide40
If , then complete the graph of the function g on the

interval that has the minimum length.

See Handout!

slide42
Suppose that f and g have (piecewise)continuous derivatives,

, , and , then use

the surface area of revolution about the y-axis formula

to find a decent upper bound on the surface area of revolution

about the y-axis of the curve

See Handout!

slide45
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.

slide46
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.

power supply

Steiner point or vertex

factory

factory

slide49
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.

slide51
Vary the height of this suction cup and compare Nature’s minimization to the Calculus predictions.

See Handout!

slide52
A four point example:

We want to link the four points , ,

, and with in a minimal way.

Steiner points

slide53
There are two competing arrangements for the position of the Steiner points: horizontal or vertical.
slide54
The length of the connection as a function of x with the parameters a and h in the vertical case is given by
slide58
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.

slide59
From the symmetry of the problem, we can quickly get the results for the horizontal case by switching a and h.
slide61
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.

slide62
Here are the possible relationships between a and h in both the horizontal and vertical arrangements:

Vertical:

Horizontal:

There are four combinations of the inequalities:

slide67
One vertical Steiner Point and two horizontal Steiner Points

Two Steiner Points for the vertical and horizontal

One horizontal and two vertical

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

slide68
minimum

minimum

minimum

minimum

slide70
Vary the distance between pairs of suction cups and compare Nature’s minimization to the Calculus predictions.

See Handout!

slide72
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 ,

Here’s its graph:

slide73
Now extend it periodically to all the nonnegative real numbers to get the function .

Here’s a portion of its graph:

slide76
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.

slide79
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

for each x

and is nondecreasing in N, then

must exist as a number.

slide80
Bounded above:

Nondecreasing:

See Handout!

slide81
Show that the function f is continuous on

.

So for every x

in .Use this to show that you can make

for every x in

.

by choosing N large enough.

See Handout!

slide83
Since is a continuous function,

there is a so that if , then

Choose x in , let , and consider

See Handout!

Finish the proof of the continuity of f.

slide84
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

slide88
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,

slide89
So we get that

Since g is periodic of period 1, we get that

slide92
Consider the sequence of points .

Examine

What does this imply about ?

slide93
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, .

Let and , for each whole number m to get

two sequences and so that and

.

slide94
For example, let , and , then and

For example, let , and , then and

For example, let , and , then and

slide95
For , if x is not a dyadic rational, then

show that for every value of m, .

See Handout!

slide96
Prove that .

First we’ll show that .

slide98
Since is a linear function on the interval .

For example, let , and , then and

slide102
You can do a similar argument to show that

.

Ifexists, then and

, but the Squeeze Theorem would

imply something about .

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

slide103
Differentiability of Powers of the Popcorn Function

Consider the function

This function was originally defined by the mathematician Johannes Thomae.

slide104
It’s called the popcorn function, ruler function, raindrop function,…

Here is a portion of its graph:

Popcorn/Raindrop

Ruler

slide105
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 numberis zero, to show that the Popcorn Function is continuous at an irrational number, , we just have to show that

slide106
Remember, to show that for any function f, we

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

, then .

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

: .

slide107
Let

Ifand is irrational, then

And if and is rational with , then

, and .

See Handout!

slide108
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

slide109
To accomplish this, let . For every, there is an

irrational number with , but

See Handout!

slide110
Is the darn thing differentiable?

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

slide111
Any irrational number has an infinity of rational

approximations which satisfy .

Hurwitz’s Theorem:

So Hurwitz’s Theorem implies that

So it’s not differentiable anywhere.

slide112
What about powers of the popcorn function?

is not differentiable at , but its square is.

slide113
Let’s look at the difference quotient for an irrational number a:

So Hurwitz’s Theorem implies that

So it square is not differentiable anywhere.

slide114
Liouville’s Theorem:

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

for each positive number , there are only finitely many reduced

rationals withfor .

So for there are only finitely many rationalswith

,then for the remaining rationals in the reduced

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

slide115
Let’s look at the difference quotient of the popcorn function raised

to the power at an irrational algebraic number a of degree k:

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 .

slide116
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:

slide117
is differentiable at for (1953)

is differentiable at for (1993)

is differentiable at for (1996)

is differentiable at for (1987)

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