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L’Hospital’s Rule

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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. ^

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Question: Messieurs, can you tell us something about your famous rule? L’Hospital #1 L’Hospital #2 L’Hospital #3

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L’Hospital #1 If has the form or and doesn’t exist, then doesn’t exist.

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My rule can only be directly applied to the limit forms and. L’Hospital #2

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My rule can be directly applied to the limit forms and *, but if doesn’t exist, then my rule can’t be applied. L’Hospital #3 * can also be replaced with.

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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. But the rule says nothing if doesn’t exist.

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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, and, then. [Suppose that and for all x in I.] This condition is typical, but not needed if we assume exists. or b ac This just takes care of one-sided limits and limits at infinity all at once.

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

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Cauchy’s Mean Value Theorem can be proven from Rolle’s Theorem: Apply Rolle’s Theorem to the function on the interval. See Handout!

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The case is covered in most textbooks, but the case isn’t mentioned. From Cauchy, we get that

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Case #1: So we get that. If and are sufficiently large, then we can make and arbitrarily close to zero, since, and arbitrarily close to.

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Case #2: So we get that. If and are sufficiently close to a, then we can make and arbitrarily close to zero, since, and arbitrarily close to.

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Examples where L’Hospital’s Rule doesn’t apply: 1. 2. See Handout!

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Examples where L’Hospital’s Rule doesn’t apply(cont.): 3. 4. See Handout!

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Examples where L’Hospital’s Rule doesn’t apply(cont.): 5. An example where you can’t get away from the zeros of. See Handout!

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Surprising examples where L’Hospital’s Rule applies: 1. 2. Watch out for !!! See Handout!

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More surprising examples where L’Hospital’s Rule applies: 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? See Handout!

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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? {Hint: Apply L’Hospital’s Rule to, and use it to determine. Determine the limit from the fact that exists as a number.} See Handout!

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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!

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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!

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Hint:

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7. Use L’Hospital’s Rule to evaluate Where the numbers are arbitrary real numbers. See Handout! Hint:

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The N th Derivative Test

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

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If and, then f has a local minimum at. If and, then f has a local maximum at. The following conclusions are common to all versions of the 2 nd Derivative Test: If and, then the 2 nd Derivative Test fails.

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Suppose that exists in an open interval containing c, exists, and. If, then the sign chart of looks like: c 0 c 0 See Handout!

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For, suppose that are continuous in an open interval containing and that, but. If you ask a veteran Calculus student about the 2 nd Derivative Test, you’ll probably get a positive response, but if you ask about the N th Derivative Test, you’re likely to get a puzzled look. Typically, the N th Derivative Test is proved using Taylor’s Theorem along with the following hypotheses in second semester Calculus or higher:

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For, suppose that exist in an open interval containing,, exists and. We can prove an Off-the-rack N 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 N th Derivative Test: Now we’ll investigate four cases:

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Case I: is odd and : The weakest 2 nd Derivative Test applied to along with the Mean Value Theorem yield the following: c 0 + Odd derivative: c 0 + - Even derivative: c 0 + Odd derivative:

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Case II: is odd and : The weakest 2 nd Derivative Test applied to along with the Mean Value Theorem yield the following: c 0 - Odd derivative: c 0 - + Even derivative: c 0 - Odd derivative:

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Case III: is even and : The weakest 2 nd Derivative Test applied to along with the Mean Value Theorem yield the following: c 0 + Even derivative: c 0 + - Odd derivative: c 0 + - Odd derivative:

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Case IV: is even and : The weakest 2 nd Derivative Test applied to along with the Mean Value Theorem yield the following: c 0 - Even derivative: c 0 - + Odd derivative: c 0 - + Odd derivative:

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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 N th Derivative Test: For, suppose that exist in an open interval containing,, exists and. If is even and, then f has a local minimum at. If is even and, then f has a local maximum at.

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

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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 ? Consider the functions:

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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: c 0 + - In the case of, we can conclude that the sign chart for f” near x = c is as follows: c 0 - + The N th Derivative Test is fairly definitive. See Handout!

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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!

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If, then complete the graph of the function g on the interval that has the maximum length. See Handout!

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If, then complete the graph of the function g on the interval that has the minimum length. See Handout!

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Do the same under the assumptions: or See Handout!

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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!

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

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Fermat/Steiner Problems

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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 17 th 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.

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power supply 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. Steiner point or vertex

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

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

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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.

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Compare this with the soap film configuration using the frames.

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

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A four point example: We want to link the four points,,, and with in a minimal way. Steiner points

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There are two competing arrangements for the position of the Steiner points: horizontal or vertical.

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The length of the connection as a function of x with the parameters a and h in the vertical case is given by

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

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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.

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

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

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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.

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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:

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Not possible.

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Here’s the Phase Transition diagram in the ha parameter plane One vertical Steiner Point and two horizontal Steiner Points Two Steiner Points for the vertical and horizontal One horizontal and two vertical

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minimum

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Compare this with the soap film configuration using the frames.

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

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The Blancmange Function

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In the 19 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:

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Now extend it periodically to all the nonnegative real numbers to get the function. Here’s a portion of its graph:

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

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Here’s an approximate graph of the function f known as the Blancmange Function:

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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.

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Here are some selected plots of on the interval

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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

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Bounded above: Nondecreasing: See Handout!

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So for every x in. Use this to show that you can make. Show that the function f is continuous on. for every x in by choosing N large enough. See Handout!

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So if, then and.

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Since is a continuous function, there is a so that if, then Choose x in, let, and consider Finish the proof of the continuity of f. See Handout!

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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

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so we get that

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What does this imply about ?

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Since as, if exists, then, but

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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,

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So we get that Since g is periodic of period 1, we get that

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What does this imply about ? See Handout!

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Consider the sequence of points. Examine What does this imply about ?

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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.

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For example, let, and, then and

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For, if x is not a dyadic rational, then show that for every value of m,. See Handout!

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Prove that. First we’ll show that.

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Since is a linear function on the interval. For example, let, and, then and

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If exists, then and, but the Squeeze Theorem would imply something about. You can do a similar argument to show that. Using the previous results, show why doesn’t exist.

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Differentiability of Powers of the Popcorn Function This function was originally defined by the mathematician Johannes Thomae. Consider the function

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It’s called the popcorn function, ruler function, raindrop function,… Here is a portion of its graph: Popcorn/Raindrop Ruler

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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

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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.

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And if and is rational with, then, and. If and is irrational, then Let See Handout!

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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

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To accomplish this, let. For every, there is an irrational number with, but See Handout!

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Is the darn thing differentiable? Let’s look at the difference quotient for an irrational number a:

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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.

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What about powers of the popcorn function? is not differentiable at, but its square is.

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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.

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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. Liouville’s Theorem:

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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.

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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:

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is differentiable at for (1953)is differentiable at for (1993)is differentiable at for (1996)is differentiable at for (1987)

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