Part 1 of a step-by-step tutorial for balancing chemical equations, along with a worked example.

Instructor: John Appleseed

To be honest, I have no idea why I have a blog. I’ve only told a handful of people about it, really, and don’t seem to have much of a theme in terms of what I write. I’m not even putting much of my own thought into the posts. I mostly just post links to things that I think are important or of interest. But if I’m not publicizing the blog, who do I think will be interested in these links and such? One thought that I do like, is that this blog acts as a time capsule, and archive, of things that I’m interested in at this point in my life. In the future I can go back and see how my thoughts or interests may have changed. I think that is enough to give this blog a sense of purpose. So I’ll run with it.

The following is an unsorted, unexpanded list of ideas that I think of often enough, and would like to post about in the future. Note, I am nowhere near an expert on any of these subjects, and will never claim to be:
- Kan’ts idea that space and time are inherent ideas of ours, which our minds (insert word I can not, for the life of me, think of right now) it upon the world; how this is related to our current understanding of general relativity and evolutionary psychology.
- Infinite: does calculus (when applied to the world) really use infinity to account for ratios (I’m talking specifically about limits)? If not, does calculus really solve Zeno’s paradox? What would this mean for the universe: is time and space discreet, i.e. digital (see loop quantum gravity). This would certainly pose a problem for black holes and the singularity…
- Interpretations of quantum mechanics, specifically the difference between the Copenhagen vs. a hidden variable one, like Bohmian mechanics; also the EPR paradox and the Bell experiments.
- Dark matter, how it is defined as unobservable, how this relates the the philosophy of science (Popper, and logical positivism), and that other theories like modified gravity may be more scientific, and therefore ‘correct’. I guess I would also want to talk about string theory, Lee Smolin’s criticism of string theory, and the LHC. And what would a post about the philosophy of science be without Kuhn and pragmatism. Oh, then I’d also have to talk about Hume and his views on cause and effect, and the problem of induction.
- Determinism, free will, psychology and Dennett’s idea of ‘elbow room’. Pinker’s treatment in ‘The Blank Slate’.
- Consciousness, specifically Hofstradter’s strange loop, and Dennett’s multiple drafts. Maybe Penrose a bit, but I haven’t read ‘The Emeror’s New Mind’, so that might have to wait.
- Common errors of human psychology, and how knowing them and being aware of them can profoundly change the way we think about how we think. This might be the first post I make, since all I have to do really is look in my first year psych textbook and write out the list. Oh, also I should touch on change blindness, split brain experiments, and confabulations. Then maybe relate this back to consciousness and free will…
- Evolutionary psychology and the various theories that have been put forth towards explaining religious and superstitious beliefs (specifically Boyer’s treatment).
- Nietzsche’s idea of a God shaped hole, and how it relates to meanings of our lives, and ethics. Thus I would talk about: Sartre’s ‘existence precedes essence‘; Singer; Rorty; Rawls’ ‘Theory of Justice‘; and perhaps Sen (thanks Mike), but I haven’t read much on Capabilities Approach.
- That would transition into my vegetarianism, and how it’s is probably the thing I’m most proud of in my life (what good is learning if it doesn’t change me in a meaningful way?).
- Certainty, rhetoric, sophistry, Socrates…
- Religion, some of the philosophical ideas in favor of it (which, admittedly, will be very secularly weighted, like Tillich or Kierkegaard), the historical perspective of Jesus (Michael Grant), and the history of Christian thought (Saul; the jewish revolt; Platonism; Catholicism; Europe, Luther, and the plague; and eventually Darwin).
- Madness, mental health, and Richard Bentall.

And that’s all I can think of right now, in the middle of a café.

As we said last time, we have an idea for a necessary condition for finding local extrema subject to constraints of a certain form. To be explicit, we assume that is continuously differentiable on an open region , and we also assume that is a continuously differentiable vector-valued function on the same region (and that ). We use to define the region consisting of those points so that . Now if is a point with a neighborhood so that for all we have (or for all such ). And, finally, we assume that the determinant

is nonzero at . Then we have reason to believe that there will exist real numbers , one for each component of the constraint function, so that the equations

are satisfied at .

Well, first off we can solve the first equations and determine the right off. Rewrite them as

a system of equations in the unknowns . Since the matrix has a nonzero determinant (by assumption) we can solve this system uniquely to determine the . What’s left is to verify that this choice of the also satisfies the remaining equations.

To take care of this, we’ll write , so we can write the point as and particularly . Now we can invoke the implicit function theorem! We find an -dimensional neighborhood of and a unique continuously differentiable function so that and for all . Without loss of generality, we can choose so that , where is the neighborhood from the assumptions above.

This is the parameterization we discussed last time, and we can now substitute these functions into the function . That is, we can define

Or if we define we can say this more succinctly as and .

Anyhow, now all of these are identically zero on as a consequence of the implicit function theorem, and so each partial derivative is identically zero as well. But since the are composite functions we can also use the chain rule to evaluate these partial derivatives. We find

Similarly, since has a local minimum (as a function of the ) at we must find its partial derivatives zero at that point. That is

Now let’s take the previous equation involving , evaluate it at , multiply it by , sum over , and add it to this latest equation. We find

Now, the expression in brackets is zero because that’s actually how we defined the way back at the start of the proof! And then what remains is exactly the equations we need to complete the proof.

We can consider the problem of maximizing or minimizing a function, as we have been, but insisting that our solution satisfy some constraint.

For instance, we might have a function to maximize, but we’re only concerned with unit-length vectors on . More generally, we’ll be concerned with constraints imposed by setting a function equal to zero. In the example, we might set . If we want to impose more conditions, we can make a vector-valued function with as many components as constraint functions we want to set equal to zero.

Now, we might be able to parameterize the collection of points satisfying . In the example, we could use the usual parameterization of the sphere by latitude and longitude, writing

where I’ve used the physicists’ convention on the variables instead of the common one in multivariable calculus classes. Then we could plug these expressions for , , and into our function , and get a composite function of the variables and , which we can then attack with the tools from the last couple days, being careful about when we can and can’t trust Cauchy’s invariant rule, since the second differential can transform oddly.

Besides even that care that must be taken, it may not even be possible to parameterize the surface, or it may be extremely difficult. At least we do know that such a parameterization will often exist. Indeed, the implicit function theorem tells us that if we have continuously differentiable constraint functions whose zeroes describe a collection of points in an -dimensional space , and the determinant

is nonzero at some point satisfying , then we can “solve” these equations for the first variables as functions of the last . This gives us exactly such a parameterization, and in principle we could use it. But the calculations get amazingly painful.

Instead, we want to think about this problem another way. We want to consider a point is a point satisfying which has a neighborhood so that for all satisfying we have . This does not say that there are no nearby points to where takes on larger values, but it does say that to reach any such point we must leave the region described by .

Now, let’s think about this sort of heuristically. As we look in various directions from , some of them are tangent to the region described by . These are the directions satisfying — where to first order the value of is not changing in the direction of . I say that in none of these directions can change (again, to first order) either. For if it did, either would increase in that direction or not. If it did, then we could find a path in the region where along which was increasing, contradicting our assertion that we’d have to leave the region for this to happen. But if decreased to first order, then it would increase to first order in the opposite direction, and we’d have the same problem. That is, we must have whenever . And so we find that

The kernel of consists of all vectors orthogonal to the gradient vector , and the line it spans is the orthogonal complement to the kernel. Similarly, the kernel of consists of all vectors orthogonal to each of the gradient vectors , and is thus the orthogonal complement to the entire subspace they span. The kernel of is contained in the kernel of , and orthogonal complements are order-reversing, which means that must lie within the span of the . That is, there must be real numbers so that

or, passing back to differentials

So in the presence of constraints we replace the condition by this one. We call the “Lagrange multipliers”, and for every one of these variables we add to the system of equations, we also add the constraint equation , so we should still get an isolated collection of points.

Now, we reached this conclusion by a rather handwavy argument about being able to find increasing directions and so on within the region . This line of reasoning could possibly be firmed up, but we’ll find our proof next time in a slightly different approach.

So let’s say we’ve got a critical point of a multivariable function . That is, a point where the differential vanishes. We want something like the second derivative test that might tell us more about the behavior of the function near that point, and to identify (some) local maxima and minima. We’ll assume here that is twice continuously differentiable in some region around .

The analogue of the second derivative for multivariable functions is the second differential . This function assigns to every point a bilinear function of two displacement vectors and , and it measures the rate at which the directional derivative in the direction of is changing as we move in the direction of . That is,

If we choose coordinates on given by an orthonormal basis , we can write the second differential in terms of coordinates

This matrix is often called the “Hessian” of at the point .

As I said above, this is a bilinear form. Further, Clairaut’s theorem tells us that it’s a symmetric form. Then the spectral theorem tells us that we can find an orthonormal basis with respect to which the Hessian is actually diagonal, and the diagonal entries are the eigenvalues of the matrix.

So let’s go back and assume we’re working with such a basis. This means that our second partial derivatives are particularly simple. We find that for we have

and for , the second partial derivative is an eigenvalue

which we can assume (without loss of generality) are nondecreasing. That is, .

Now, if all of these eigenvalues are positive at a critical point , then the Hessian is positive-definite. That is, given any direction we have . On the other hand, if all of the eigenvalues are negative, the Hessian is negative definite; given any direction we have . In the former case, we’ll find that has a local minimum in a neighborhood of , and in the latter case we’ll find that has a local maximum there. If some eigenvalues are negative and others are positive, then the function has a mixed behavior at we’ll call a “saddle” (sketch the graph of near to see why). And if any eigenvalues are zero, all sorts of weird things can happen, though at least if we can find one positive and one negative eigenvalue we know that the critical point can’t be a local extremum.

We remember that the determinant of a diagonal matrix is the product of its eigenvalues, so if the determinant of the Hessian is nonzero then either we have a local maximum, we have a local minimum, or we have some form of well-behaved saddle. These behaviors we call “generic” critical points, since if we “wiggle” the function a bit (while maintaining a critical point at ) the Hessian determinant will stay nonzero. If the Hessian determinant is zero, wiggling the function a little will make it nonzero, and so this sort of critical point is not generic. This is the sort of unstable situation analogous to a failure of the second derivative test. Unfortunately, the analogy doesn’t extent, in that the sign of the Hessian determinant isn’t instantly meaningful. In two dimensions a positive determinant means both eigenvalues have the same sign — denoting a local maximum or a local minimum — while a negative determinant denotes eigenvalues of different signs — denoting a saddle. This much is included in multivariable calculus courses, although usually without a clear explanation why it works.

So, given a direction vector so that , then since is in , there will be some neighborhood of so that for all . In particular, there will be some range of so that . For any such point we can use Taylor’s theorem with to tell us that

for some . And from this we see that for every so that . A similar argument shows that if then for any near in the direction of .

Now if the Hessian is positive-definite then every direction from gives us , and so every point near satisfies . If the Hessian is negative-definite, then every point near satisfies . And if the Hessian has both positive and negative eigenvalues then within any neighborhood we can find some directions in which and some in which .

This theorem is relatively simple. “Differentiation and integration re inverse processes. Integrals are antiderivatives. The derivative of an integral is the function.” These are just a few of the ways to state the theorem. Antiderivatives, the concept the theorem revolves about, are relatively algorithmic and easy to calculate (first apply the rules, then u-substitution, then integration by parts, then partials, etc.).

I’m going to take a closer look at the theorem. Derivatives and integrals are both defined in terms of limits, but they are otherwise unrelated. A derivative is the slope of the tangent line at a point; an integral is an area. It is not obvious that the derivative (d/dx) of an integral (from a to x of f(t) dt) is the original function f(x). 

Let’s consider the following exercise. Suppose I am given the graph of a function g’(x) (x-axis labeled, y-axis also labeled) and I would like to sketch an antiderivative given that point (a,b) is on the antiderivative g(x). In practice, this is a simple routine exercise; we determine where the antiderivative g(x) is concave upwards, concave downwards, increasing, and decreasing based on the given graph; then we sketch. However, I am more interested in the amount of increase and decrease; differential calculus does not provide us with this information. 

Suppose I start with the given point (a,b) and assume that g(x) is increasing and concave upwards until a point (c,d). Let me divide the subinterval [a,c] into n subintervals, where n is a very large number. I note values of g’(x) at the midpoints of the subintervals (midpoints to get a goodapproximation of the entire subinterval) and use this as the slope for the subinterval on the graph of g(x). This results in a series of tiny lines that closely approximate the function g(x). Intuitively, it should be obvious that as n becomes closer to infinity, the lines become better and better approximations. That is, with infinity subintervals, we can produce an exact graph of g(x).

This exercise may seem pointless; after all, a sketch of g(x) without amounts of increase/decrease is useful for most purposes. However, I will examine this a bit closer. With infinity subintervals, I am looking at all real values along [a,c]. In addition, each value can represent the distance of a line segment from the x-axis to  the graph of g’(x). Intuitively, infinity line segments can make up an area…. suggesting an integral.

I have intuitively proven that the amount of increase from a point (a,b) on the graph of g(x) is the integral from a to c of g’(x), one version of the Fundamental Theorem of Calculus. While this is not rigorous, it is an interesting viewpoint on the theorem. The regular proof of this theorem would involve the Mean Value Theorem and subintervals as well, but it would not appeal to the intuitive mind.

Below I have attached an illustrative picture. The point is (0,0), g(x) is the purple function, and g’(x) is the blue function. You can observe that my above predictions are true (it is easiest to view this if the graph is printed and you draw in the lines, the subintervals, and the area).

Pascal, Blaise     

I cannot forgive Descartes. In all his philosophy he would have been quite willing to dispense with God. But he had to make Him give a fillip to set the world in motion; beyond this, he has no further need of God.

– Pensees # 77

(19 June 1623, at Clermont, France – 19 August 1662, Paris, France)

Invented a computing machine (age 18), helped invent calculus of probabilities, invented Pascal’s Triangle   

Converted to Christ, 23 November 1654

Pascal, Blaise     

All great amusements are dangerous to the Christian life; but among all those which the world has invented there is none more to be feared than the theatre. – Pensees # 11

(19 June 1623, at Clermont, France – 19 August 1662, Paris, France)

Invented a computing machine (age 18), helped invent calculus of probabilities, invented Pascal’s Triangle   

Converted to Christ, 23 November 1654

Just like in one variable, we’re interested in local maxima and minima of a function , where is an open region in . Again, we say that has a local minimum at a point if there is some neighborhood of so that for all . A maximum is similarly defined, except that we require in the neighborhood. As I alluded to recently, we can bring Fermat’s theorem to bear to determine a necessary condition.

Specifically, if we have coordinates on given by a basis , we can regard as a function of the variables . We can fix of these variables for and let vary in a neighborhood of . If has a local extremum at , then in particular it has a local extremum along this coordinate line at . And so we can use Fermat’s theorem to draw conclusions about the derivative of this restricted function at , which of course is the partial derivative .

So what can we say? For each variable , the partial derivative either does not exist or is equal to zero at . And because the differential subsumes the partial derivatives, if any of them fail to exist the differential must fail to exist as well. On the other hand, if they all exist they’re all zero, and so as well. Incidentally, we can again make the connection to the usual coverage in a multivariable calculus course by remembering that the gradient is the vector that corresponds to the linear functional of the differential . So at a local extremum we must have .

As was the case with Fermat’s theorem, this provides a necessary, but not a sufficient condition to have a local extremum. Anything that can go wrong in one dimension can be copied here. For instance, we could define . Then we find , which is zero at . But any neighborhood of this point will contain points and for small enough , and we see that , so the origin cannot be a local extremum.

But weirder things can happen. We might ask that have a local minimum at along any line, like we tried with directional derivatives. But even this can go wrong. If we define

we can calculate

which again is zero at . Along any slanted line through the origin we find

and so the second derivative is always positive at the origin, except along the -axis. For the vertical line, we find

so along all of these lines we have a local minimum at the origin by the second derivative test. And along the -axis, we have , which has the origin as a local minimum.

Unfortunately, it’s still not a local minimum in the plane, since any neighborhood of the origin must contain points of the form for small enough . For these points we find

and so cannot have a local minimum at the origin.

What we’ll do is content ourselves with this analogue and extension of Fermat’s theorem as a necessary condition, and then develop tools that can distinguish the common behaviors near such critical points, analogous to the second derivative test.

Chapter 4 Assignment # 6

Quiz Wednesday Cahpter 4

I was battling through Walter Rudin’s Principles of Mathematical Analysis the other day when I came across the following theorem:

Theorem Suppose f is a continuous real function on a compact metric space X, and

Then there exist points such that and .

And then the proof is one line; it is a corollary of two other theorems:

1. If f is a continuous mapping of a compact metric space X into , then is closed and bounded. Thus, f is bounded.

2. Let E be a nonempty set of real numbers which is bounded above. Let y = sup E. Then . Hence if E is closed.

I was reminded of the same, albeit more specific theorem, in Spivak’s Calculus.

Theorem If f is continuous on [a, b], then there is a number y in [a, b] such that for all x in [a,b].

And then the proof

Proof f is bounded above on [a,b], which means that the set

is bounded. This set is not empty, so it has a least upper bound . Since for x in [a, b] it suffices to show that for some y in [a, b].

Suppose instead that for all y in [a,b]. Then the function g defined by

is continuous on [a, b] since the denominator is never 0. On the other hand, is the least upper bound of A; this means that

for every there is x in [a, b] with .

This, in turn, means that,

for every there is x in [a, b] with

But this means that g is not bounded on [a,b], contradicting continuous functions are bounded.

Phew. The more complicated proof from real analysis is significantly shorter than the one from calculus, even though the one from analysis is actually strictly stronger. I could do a similar writeup for the calculus proof of the fundamental theorem of algebra vs. the complex analytic one using Liouville’s theorem, but the calculus proof would be like 10 hours of typing.

Okay, today we’re going to prove the implicit function theorem. We’re going to think of our function as taking an -dimensional vector and a -dimensional vector and giving back an -dimensional vector . In essence, what we want to do is see how this output vector must change as we change , and then undo that by making a corresponding change in . And to do that, we need to know how changing the output changes , at least in a neighborhood of . That is, we’ve got to invert a function, and we’ll need to use the inverse function theorem.

But we’re not going to apply it directly as the above heuristic suggests. Instead, we’re going to “puff up” the function into a bigger function that will give us some room to maneuver. For we define

just copying over our original function. Then we continue by defining for

That is, the new component functions are just the coordinate functions . We can easily calculate the Jacobian matrix

where is the zero matrix and is the identity matrix. From here it’s straightforward to find the Jacobian determinant

which is exactly the determinant we assert to be nonzero at . We also easily see that .

And so the inverse function theorem tells us that there are neighborhoods of and of so that is injective on and , and that there is a continuously differentiable inverse function so that for all . We want to study this inverse function to recover our implicit function from it.

First off, we can write for two functions: which takes -dimensional vector values, and which takes -dimensional vector values. Our inverse relation tells us that

But since is injective from onto , we can write any point as , and in this case we must have by the definition of . That is, we have

And so we see that , where is the -dimensional vector so that . We thus have for every .

Now define be the collection of vectors so that , and for each such define , so . As a slice of the open set in the product topology on , the set is open in . Further, is continuously differentiable on since is continuously differentiable on , and the components of are taken directly from those of . Finally, is in since , and by assumption. This also shows that .

The only thing left is to show that is uniquely defined. But there can only be one such function, by the injectivity of . If there were another such function then we’d have , and thus , or for every .

Today I learned, or rather re-learned, that math is fun. It’s especially fun when a whole department of motivated teachers purposes to just dive in, take risks, and try new things. And it’s even more fun when they start archiving their learning for all to see in a Glendale math wiki and other collaborative sharing spaces.  I have a secret desire to be back in Gr 9 or 10 with these fine folks as my teachers.

Will Richardson has long been a maven of mine and, as you may know, is a link in my personal learning network. Seriously, when I look at my personal philosophy of education, Will’s influence is all over the place…and I appreciate his input. He’s the original guru of wikis, blogs and podcasts in education and now his ideas are more and more common in schools. I’m very glad to say that my school is fast becoming one of these places and our math folks are leading the way. 

When we use wikis and other collaborative spaces we are doing ‘real’ work, says Will.  And I agree.

Grateful for:

• Melinda, Chantelle, Gary, and Rebecca
• our entire math department
• birthday dinners and cakes
• a box of teen-friendly clothes, mostly new, that a teacher brought in so that I could set up a situation where a couple of teens could ‘help’ me sort them. By the time we’d finished, most of the clothes were gone and claimed by my helpers. What a blessing.

Curious about:

• an entire calculus course in less than 20 minutes…it actually makes sense
• the 17-33 principle as it relates to looking after a person with dementia…apparently this is the age bracket in which many life decisions are made, therefore these years are easier to remember and talk about when dementia sets in. News flash to me, but it’s surely a practical tip.

Consider the integral .

Here’s a too clever solution: Since we have

Using the Weierstrass substitution here would be a less clever, more general way to solve this integral. Remember we set yielding , , and . Applying these here we find

We complete the square: and go on.

We make the substitution , and find

I’ll leave it to you to reconcile the two answers.

Let’s consider the function . The collection of points so that defines a curve in the plane: the unit circle. Unfortunately, this relation is not a function. Neither is defined as a function of , nor is defined as a function of by this curve. However, if we consider a point on the curve (that is, with ), then near this point we usually do have a graph of as a function of (except for a few isolated points). That is, as we move near the value then we have to adjust to maintain the relation . There is some function defined “implicitly” in a neighborhood of satisfying the relation .

We want to generalize this situation. Given a system of functions of variables

we consider the collection of points in -dimensional space satisfying .

If this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) dimensional. Indeed, we could use Gauss-Jordan elimination to put the system into reduced row echelon form, and (usually) find the resulting matrix starting with an identity matrix, like

This makes finding solutions to the system easy. We put our variables into a column vector and write

and from this we find

Thus we can use the variables as parameters on the space of solutions, and define each of the as a function of the .

But in general we don’t have a linear system. Still, we want to know some circumstances under which we can do something similar and write each of the as a function of the other variables , at least near some known point .

The key observation is that we can perform the Gauss-Jordan elimination above and get a matrix with rank if and only if the leading matrix is invertible. And this is generalized to asking that some Jacobian determinant of our system of functions is nonzero.

Specifically, let’s assume that all of the are continuously differentiable on some region in -dimensional space, and that is some point in where , and at which the determinant

where both indices and run from to to make a square matrix. Then I assert that there is some -dimensional neighborhood of and a uniquely defined, continuously differentiable, vector-valued function so that and .

That is, near we can use the variables as parameters on the space of solutions to our system of equations. Near this point, the solution set looks like the graph of the function , which is implicitly defined by the need to stay on the solution set as we vary . This is the implicit function theorem, and we will prove it next time.

as i sit home on a wednesday night pondering the fact that i may never get into college (although this statement seems unreasonable) i decide to create a new blog.

since my other one somehow got deleted. what a shame.

i feel that my life is too busy for my own good.

i’m barely passing AP calculus, and it’s upsetting.

i decided i’m too lazy to use capital letters. my journalism teacher would be so disappointed.

the newspaper, dance, ap classes, drama, everything is just so JUMBLED.

i can’t control it. it sucks. and i constantly keep adding more shit onto my pile. why?

no idea.

it’s a problem.

I started to cover the Fundamental Theorem of Calculus today. I think I displayed a Fundamental Lack of Enthusiasm as I delivered the lecture. I’m going to blame the fact that I taught an extra Calculus lecture on Monday for my officemate, who was out of town, so that by today my teaching stamina was at the point where it would normally be on a Friday.

The Fundamental Theorem of Calculus is such a great theorem that I should be jumping up and down as I talk about it. How can I expect my students to be excited and in awe of it if I myself am not acting the part? My old officemate Nick found the following passage in Howard Eves’ Great Moments in Mathematics after 1650:

Surely no subject in early college mathematics is more exciting or more fun to teach than the Calculus. It is like being the ringmaster of a great three-ring circus. It has been said that one can recognize the students on a college campus who have studied the Calculus – they are the students with no eyebrows. In utter astonishment at the incredible applicability of the subject, the eyebrows of the calculus students have receded higher and higher and finally vanished over the backs of their heads.

Anyway, I’ll be talking about the FTC again on Friday. I’ve come home early this afternoon, so I’ll be sure to have enough time to rest up and be enthusiastic by then.

At last we come to the theorem that I promised. Let be continuously differentiable on an open region , and . If the Jacobian determinant at some point , then there is a uniquely determined function and two open sets and so that

• , and
• 
• is injective on
• is defined on , , and for all
• is continuously differentiable on

The Jacobian determinant is continuous as a function of , so there is some neighborhood of so that the Jacobian is nonzero within . Our second lemma tells us that there is a smaller neighborhood on which is injective. We pick some closed ball centered at , and use our first lemma to find that must contain an open neighborhood of . Then we define , which is open since both and are (the latter by the continuity of ). Since is injective on the compact set , it has a uniquely-defined continuous inverse on . This establishes the first four of the conditions of the theorem.

Now the hard part is showing that is continuously differentiable on . To this end, like we did in our second lemma, we define the function

along with a neighborhood of so that as long as all the are within this function is nonzero. Without loss of generality we can go back and choose our earlier neighborhood so that , and thus that .

To show that the partial derivative exists at a point , we consider the difference quotient

with also in for sufficiently small . Then writing and we find . The mean value theorem then tells us that

for some (no summation on ). As usual, is the Kronecker delta.

This is a linear system of equations, which has a unique solution since the determinant of its matrix is . We use Cramer’s rule to solve it, and get an expression for our difference quotient as a quotient of two determinants. This is why we want the form of the solution given by Cramer’s rule, and not by a more computationally-efficient method like Gaussian elimination.

As approaches zero, continuity of tells us that approaches , and thus so do all of the . Therefore the determinant in the denominator of Cramer’s rule is in the limit , and thus limits of the solutions given by Cramer’s rule actually do exist.

This establishes that the partial derivative exists at each . Further, since we found the limit of the difference quotient by Cramer’s rule, we have an expression given by the quotient of two determinants, each of which only involves the partial derivatives of , which are themselves all continuous. Therefore the partial derivatives of not only exist but are in fact continuous.

Chapter 4 assignments #4 and #5

CalcCh4Assignments

Voices of San Diego reports on one wild-and-crazy AP Calculus teacher, who’s transformed the often-staid class into one with open enrollment and large lectures, much like a college class. The school has plenty of students who in other settings might be counted out (minority and English-learners), but the students seem to be up for the challenge (test results won’t be in until next summer). Teacher Jonathan Winn combines calculus with lots of drama and a dollop of self-help talk. Here he is explaining the chain rule for differentiation (you have to identify the inner function, called “u” in math-speak):

It’s you! It’s u! We found u! You found u!” Winn shouts. The teens giggle. “You can’t solve a problem until you find yourself.”

Liban Dini is one of the students who speaks up a lot, a Somali refugee with a confident manner who came here without his parents as a child. He wants to pursue a career in business, and says Winn sold him on calculus because he'd save money on college by earning credits now. Yet something more than dollars swayed him.

“It’s hard not to get excited if he’s that excited,” Dini said. “Other people, they don’t think you can handle it. He says,’I know you can pass the test.’ He paused. “I feel like he’s just talking to me sometimes. Sometimes you feel like he’s just looking at you. The inner you.”

Click on the link to the story if for no other reason than to see Winn’s picture.  I had the incalculable good fortune to learn calculus from the legendary Mary Sunseri–the class was scheduled at 8 am to keep the numbers down, but if you wanted a seat in one of the largest lecture rooms on campus, you had to get there on time. I remember her tremendous energy the most – but unlike Winn, she never wore a hat.

Consider the indefinite integral where is a positive real number.

This can be evaluated in a number of ways. Here are two of them along with a nice consequence.

First we’ll treat this as a straight-forward partial fraction decomposition question. We have

where are real numbers to be determined. We have then

.

Letting we find ; letting we find . Thus

Done and done.

Now we’ll treat this using hyperbolic substitutions. Remember the fundamental hyperbolic identity: . From this we can derive the identities and by dividing our fundamental identity by and respectively.

Remember the graph of . We see that the domain of is all real numbers and the range is . Further is one-to-one and so we have a with domain and range all real numbers. Here’s a graph if your memory of hyperbolic tangent is a little fuzzy.

So in the case where we can make the substitution . Then and .

Combining this with our first solution (and setting ) we see that

and since and we see that . This formula for can be derived in other ways of course.

When we make a substitution and the reasoning is similar.

Thus we have

on

and

on

Sorry for the late post. I didn’t get a chance to get it up this morning before my flight.

Brace yourself. Just like last time we’ve got a messy technical lemma about what happens when the Jacobian determinant of a function is nonzero.

This time we’ll assume that is not only continuous, but continuously differentiable on a region . We also assume that the Jacobian at some point . Then I say that there is some neighborhood of so that is injective on .

First, we take points in and make a function of them

That is, we take the th partial derivative of the th component function and evaluate it at the th sample point to make a matrix , and then we take the determinant of this matrix. As a particular value, we have

Since each partial derivative is continuous, and the determinant is a polynomial in its entries, this function is continuous where it’s defined. And so there’s some ball of so that if all the are in we have . We want to show that is injective on .

So, let’s take two points and in so that . Since the ball is convex, the line segment is completely contained within , and so we can bring the mean value theorem to bear. For each component function we can write

for some in (no summation here on ). But like last time we now have a linear system of equations described by an invertible matrix. Here the matrix has determinant

which is nonzero because all the are inside the ball . Thus the only possible solution to the system of equations is . And so if for points within the ball , we must have , and thus is injective.

Chapter 4 Assignment # 3

CalcCh4Assignments