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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 spellers.

L’Hospital #3

L’Hospital #2

Question: Messieurs, can you tell us

something about your famous

rule?

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

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

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.

Let f and g be continuous on and which contains a or has a as an endpoint.

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!

Apply Rolle’s Theorem to the function which contains a or has a as an endpoint.

on the interval .

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

See Handout!

The case is covered in most textbooks, but the which contains a or has a as an endpoint.

case isn’t mentioned.

From Cauchy, we get that

Case #1: which contains a or has a as an endpoint.

If and are sufficiently large, then we

can make and arbitrarily close to

zero, since , and arbitrarily close

to .

So we get that .

Case #2: which contains a or has a as an endpoint.

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 .

1. which contains a or has a as an endpoint.

2.

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

See Handout!

See Handout!

3. which contains a or has a as an endpoint.

4.

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

See Handout!

See Handout!

5. which contains a or has a as an endpoint.

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

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

See Handout!

1. which contains a or has a as an endpoint.

2.

Watch out for !!!

Surprising examples where L’Hospital’s Rule applies:

See Handout!

See Handout!

3. which contains a or has a as an endpoint. 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!

4. which contains a or has a as an endpoint. 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!

5. which contains a or has a as an endpoint. 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!

6. which contains a or has a as an endpoint. 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!

Hint: which contains a or has a as an endpoint.

Hint: which contains a or has a as an endpoint.

7. Use L’Hospital’s Rule to evaluate

Where the numbers are arbitrary

real numbers.

See Handout!

The N which contains a or has a as an endpoint.th 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 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.

The following conclusions are common to all versions of the in a

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.

0 in a

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:

If you ask a veteran Calculus student about the 2 in a nd 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

.

We can prove an Off-the-rack N in a th 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:

Case I: in a 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:

Case II: in a 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:

Case III: in a 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:

Case IV: in a 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:

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

.

Consider the functions: 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 ?

- - functions:

+ +

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.

Suppose that g has a functions:(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!

1 functions:

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

interval that has the maximum length.

See Handout!

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

interval that has the minimum length.

See Handout!

Suppose that f and g have functions:(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!

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

See Handout!

Fermat/Steiner Problems y-axis of the curve

Fermat’s problem: y-axis of the curve 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: y-axis of the curve

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

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

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

So in the case of the Steiner point should values of the parameters a and h.

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.

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

See Handout!

A four point example: minimization to the Calculus predictions.

We want to link the four points , ,

, and with in a minimal way.

Steiner points

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

Here are the possible sign charts for L parameters a and h in the vertical case is given byV’, depending on the values of the parameters a and h.

So in the case of the two vertical Steiner parameters a and h in the vertical case is given by

points should be located units above and

below the origin. If then there should

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.

Here are the possible sign charts for L results for the horizontal case by switching a and h.H’, depending on the values of the parameters a and h.

So in the case of the two horizontal Steiner results for the horizontal case by switching a and h.

points should be located units left and

right of the origin. If then there should

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:

Horizontal:

There are four combinations of the inequalities:

Not possible. the horizontal and vertical arrangements:

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

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

See Handout!

The Blancmange Function Nature’s minimization to the Calculus predictions.

In the 19 Nature’s minimization to the Calculus predictions.th 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:

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

Here’s a portion of its graph:

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.

Here are some selected plots of 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.on the

interval

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.

Show that the function f is continuous on sense.

.

So for every x

in .Use this to show that you can make

for every x in

.

by choosing N large enough.

See Handout!

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.

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

, but

Consider the sequence of points . then

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 then

Since g is periodic of period 1, we get that

What does this imply about ? then

See Handout!

Try similar thinking to show that then 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

.

For example, let then , and , then and

For example, let , and , then and

For example, let , and , then and

First we’ll show that .

Since .is a linear function on the interval .

For example, let , and , then and

For example, let ., and , then and

For example, let ., and , then and

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.

Differentiability of Powers of the Popcorn Function .

Consider the function

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

Remember, to show that Function 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

: .

To show that the Popcorn Function is discontinuous at a Function

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

Is the darn thing differentiable? Function

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

Any irrational number has an infinity of rational Function

approximations which satisfy .

Hurwitz’s Theorem:

So Hurwitz’s Theorem implies that

So it’s not differentiable anywhere.

What about powers of the popcorn function? Function

is not differentiable at , but its square is.

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.

Liouville’s Theorem: number a:

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,

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 .

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:

is differentiable at for (1953) point of

is differentiable at for (1993)

is differentiable at for (1996)

is differentiable at for (1987)

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