Of course the idea is to do a substitution: which transforms the integral into which diverges. So far, so good. But then I told him one of my calculus tips: “it is often a good idea to try to guess the answer ahead of time” and then pointed out that for large values of and of course diverges because the integrand does not go to zero (in fact, is unbounded!) as tends to infinity.

Then I realized that a change of variables had taken an unbounded function to a bounded one…though one which did not produce a convergent improper integral.

That lead to the natural question: if one has an integrand which is positive but monotonically decreasing to zero on , is there a change of variables which will change the integrand to either an unbounded function on or at least one that does not decrease to zero?

I admit that I have not answered this question yet, nor have I looked it up. But I can answer this question for a certain class of functions:

Theorem

Given
If , let . Then the change of variable transforms to and of course is unbounded on .

If let . Then is transformed into which is an integral of a bounded function over a bounded region.

In short, one class of functions whose improper integral diverges can be transformed to functions that tend to infinity and the class of functions whose integrals converge can be transformed into functions which are bounded over a bounded interval.

Here is such an example: We show the equivalent integrals and . The transformation is accomplished by using . Note how the transformation stretches the interval of integration to account for the function “shrinkage”.

On the other hand, using transforms into

But that raised the question: “what do we mean when we tell our students ““?

Of course, there are some central issues here. The first issues is that our “sure of herself” student thought that “” meant that this relation is NEVER true for any choice of , which of course, is false (e. g. let and .) In fact, is the logical negation of the statement ; the latter means that “this statement is true for ALL and its negation means “there is at least one choice of for which the statement is not true. “Equal” and “not equal” are not symmetric states when it comes identities, which can be thought of as elements in the vector space of functions.

So, means that and are not equal in function space, though they might evaluate to the same number for certain choices in the domain.

So, what is the big deal?

Well, what about equations such as or ?
These are NOT equalities in the space of functions; the first means “what values in the domain does take given and the second asks one to find the inverse image of 0 for the operator where the domain is the set of all, say, twice differentiable functions.

But, but…would the average undergraduate student understand ANY of this? My experience tells me “no”; hence I intentionally allow for this vagueness and only address it as I need to.

Now as a professor, I don’t have a “dog in this hunt” so to speak because I teach at a 10-12 hour per semester load institution and I am grateful to have the job. But as a citizen and as a mathematician, I think that the research intensive universities have a place and that it would be a colossal mistake to turn them into “teaching institutions”. We need places that generate knowledge, and one isn’t going to be able to generate top-level mathematical knowledge (say, at the level that gets published in the Annals of Mathematics or in Inventiones mathematicae if one isn’t devoted to keeping current and active on a full time basis.

Now my institution does have a research requirement for tenure and promotion and I’ve developed a modest publication record and am still working to add to my publication list. And yes, many of my colleagues have published far more than I have and I salute them for it.

But let’s face facts: mathematical research at institutions like ours consists of
1. tackling spin-off problems from areas opened some time ago
2. working with another medium level scholar in another discipline to solve some of the mathematical problems related to that discipline
3. working on “nice to know” things that one discovers (or rediscovers) when preparing for class.

As a colleague at a similar institution said: “I dabble here and there; there just isn’t time to learn something that takes 5 years to master”.

The fact is that teaching classes, doing service and meeting with perplexed students soaks up the vast majority of one’s time. Add to that the fact that one’s “upper division” class might be a class that one has never taught or one that you last taught a decade ago; in reality one has to relearn much of the material that has faded from memory.

Then, there is the basic brain rot that occurs from mostly dealing with trying to explain to students that . One also has the baby-sitting of getting the poorer performing students to not text in class and to explain to them why their course average of 65 doesn’t entitle them to a B (or even a C) and to do so in a way that doesn’t have their parents complain to the department chair.

Then there is the brain atrophy that comes from not reading anything hard for months at a time; then when you try to read something hard you often only have a few minutes of uninterrupted time to do so.

Hence the research that you can do, while it can require cleverness, really can’t require that you master the new sophisticated techniques.

On a side note: this is part of the purpose of my writing this blog; it encourages me to learn stuff that is “new to me” or “what I should have learned a long time ago.” After this next round of exams, I hope to talk about quartic splines that produce increasing, convex curves.

But often if I stick with it, it doesn’t take 30 “f*cking” minutes.

Here is an example: I was trying to remember how to calculate and was trying to remember instead of think. I looked at an old calculus book…no avail…then I was shamed into thinking. About 2-3 minutes later it struck me:
“you know how to simplify don’t you?”

Problem solved…shame WIN.

of course things like are easily converted to things like , etc.

This leads to another point. Often when we teach we use the “conjugate trick” which only works for square roots. The above method works for the other fractional powers.

It isn’t every day that someone in the mainstream media brings up calculus.

I consider the following to be minimal prerequisites:

Basic calculus

1. limits (epsilon-delta, 2-d limits)

2. limit definition of the derivative

3. basic calculus differentiation and integration formulas:
chain rule, product rule, quotient rule, integration and differentiation of polynomials, log, exponentials, basic trig functions, hyperbolic trig functions, inverse trig functions.

4. Fundamental Theorem of calculus.

5. Sequences (convergence)

6. Series: geometric series test, ratio test, comparison tests

7. Power series: interval of convergence, absolute convergence

8. Power series: term by term differentiation, term by term integrals

9. Taylor/Power series for 1/(1-x), sin(x), cos(x), exp(x)

Multi-variable calculus

1. partial derivatives

2. gradient

3. parametrized curves

4. polar coordinates

5. line and path integrals

6. conservative vector fields

7. Green’s Theorem (for integration of a closed loop in a plane)

The challenge
Some of complex variables will look “just like calculus”. And, some of the calculations WILL be “just like calculus; for example it will turn out if is any piecewise smooth curve running from to then . But in many cases, the similarity vanishes and more care must be taken.

You will learn many things such as:
1. The complex function is unbounded!

2. No non-constant everywhere differentiable function is bounded; compare that to in calculus.

3. Integrals can have some strange properties. For example, if is the unit circle taken once around in the standard direction, depends on where one chooses to start and stop, even if the start and stop points are the same!

4. You’ll come to understand why the Taylor series (expanded about ) for has radius of convergence equal to one…it isn’t just an artifact of the trick used to calculate the series.

5. You’ll come to understand that being differentiable on an open disk is a very strong condition for complex functions; in particular being differentiable on an open disk means being INFINITELY differentiable on that open set (compare to which has one derivative but NOT two derivatives at

There is much more, of course.

In short, we are using the equal sign very differently: in the first case we are using it as the equivalence relation in the field of real numbers. In the second case, we are really talking about vector space equivalence.

We see this multiple use in calculus all the time; for example but Of course, the first is the usual non-oriented integral that we talk about in calculus courses (absolute values of the Jacobians!) and the latter is the oriented integral that we use for 2-forms, which, when you think about it, is the logical extension of the usual calculus I definite integral.

There are certainly more examples.

What got me to thinking about this was an office hour encounter I had with a numerical methods student (a good student who is doing solid work in the course). We were talking about various methods of solving the matrix problem where is a column vector of variables and is the “answer” vector of numbers. We were discussing the number of operations (multiplications/divisions and additions/subtractions) required to obtain a solution if we had that where was a diagonal matrix with non-zero entries, are lower and upper triangular matrices (respectively) with 1′s on the diagonal.

She kept on being off by a peculiar factor on the multiplication count.

Eventually we figured out the problem. When we converted the matrix equations to equations, she was counting the matrix entry multiplied by the unsolved for variables as a multiplication. Why? Well, once we solved for the variable we then counted operations with it AFTER it had been “solved for”. Example: given we don’t count the “coefficient times the variable” as a multiplication. But once we solve and obtain we then count operations involving . (of course, the diagonal elements are non-zero).

It is clear why we do this: prior to being solved for, the variables are really storage locations, and we are interested in counting the numerical operations that can contribute to round off error. But when we think about it, we are actually distinguishing between several types of multiplications: matrix multiplication, scalar multiplication in a vector space between a vector and a scalar, and the scalar (numerical) multiplication.

However, explaining that in class might lead to confusion among the students; it is probably best to bring this up only if someone is confused about it.

The language of mathematics can be so subtle that sometimes it probably good pedagogy to speak a bit informally, at least to beginning students.

But adding the particular solution to the particular solution yields a general solution with arbitrary constants which can be solved for to meet a given initial condition.

So how does one obtain a particular solution?

Students almost always learn the so-called “method of undetermined coefficients”; this is used when the driving function is a sine, cosine, , a polynomial, or some sum and product of such things. Basically, one assumes that the particular solution has a certain form than then substitutes into the differential equation and then determines the coefficients. For example, in our example, one might try and then substitute into the differential equation to solve for and . One could also try a complex form; that is, try and then determines and then uses the real part of the solution.

A second method for finding particular solution is to use variation of parameters. Here is how that goes: one obtains two linearly independent homogeneous solutions and then seeks a particular solution of the form where and where is the determinant of the Wronskian matrix. This method can solve differential equations like and sometimes is easier to use when the driving function is messy.
But sometimes it can lead to messy, non transparent solutions when “undetermined coefficients” is much easier; for example, try solving with variation of parameters. Then try to do it with undetermined coefficients; though the answers are the same, one method yields a far “cleaner” answer.

There is a third way that gives a particular solution that meets a specific initial condition. Though this method can yield a not-so-easy-to-do-by-hand integral and can sometimes lead to what I might call an answer in obscured form, the answer is in the form of a definite integral that can be evaluated by numerical integration techniques (if one wants, say, the graph of a solution).

This method is the Convolution Method. Many texts introduce convolutions in the Laplace transform section but there is no need to wait until then.

What is a convolution?
We can define the convolution of two functions and to be:
. Needless to say, and need to meet appropriate “integrability” conditions; this is usually not a problem in a differential equations course.

Example: if , then . Notice that the dummy variable gets “integrated out” and the variable remains.

There are many properties of convolutions that I won’t get into here; one interesting one is that ; proving this is an interesting exercise in change of variable techniques in integration.

The Convolution Method
If is a homogenous solution to a second order linear differential equation that meets initial conditions: and is the forcing function, then is the particular solution that meets

How might we use this method and why is it true? We’ll answer the “how” question first.

Suppose we want to solve . The homogeneous solution is and it is easy to see that we need to meet the condition. So a particular solution is

This particular solution meets .

Why does this work?
This is where “differentiation under the integral sign” comes into play. So we write .
Then ?

Look at the convolution integral as . Now think of . Then from calculus III: . Of course, .
by the Fundamental Theorem of calculus and by differentiation under the integral sign.

So we let and we see which equals because . Now by the same reasoning because .
Now substitute into the differential equation and use the linear property of integrals to obtain

It is easy to see that Now check .

This is an important result in applied mathematics; I’ll give some applications (there are many!) in our next post. Both examples are from a first course in differential equations.

First, I should give the conditions on to make this result true: continuity of and on some rectangle in space which contains all of the points in question (including the interval of integration) is sufficient.

Why is the formula true? The proof isn’t hard at all and it makes use of the Mean Value Theorem and of some basic theorems concerning limits and integrals.

Some facts that we’ll use: if on some interval , then and the Mean Value Theorem.

Now recall from calculus:

We now employ one of the most common tricks of mathematics; we guess at the “right answer” and then show that the right answer is what we guessed.

We will examine the integrand (the function being integrated). Does remind you of anything? Right; this is the fraction from the Mean Value Theorem; that is, there is some between and such that

Because we are assuming the continuity of the partial derivative, we can say that for sufficiently close to ,

This means that

Now realize that can be made as small as desired by letting get sufficiently close to so it follows by the definition of limit that:
which implies that

Therefore
So the result follows.

Next post: we’ll give a couple of applications of this

Therefore, I learned more about existence and uniqueness theorems when I taught differential equations; though I never taught the existence and uniqueness theorems in a class, I learned the proofs just for my own background. In doing so I learned about the Picard iterated integral technique for the first time; how this is used to establish “uniqueness of solution” can be found here.

However I recently discovered (for myself) what thousands of mathematicians already know: the Picard process can be used to yield an interval of existence for a solution for a differential equation, even if we cannot obtain the solution in closed form.

The situation
I assigned my numerical methods class to solve with and to produce the graph of from to .

There is a unique solution to this and the solution is valid so long as the and value of the solution curve stays finite; note that

So, is it possible that the values for this solution become unbounded?

Answer: yes.
What follows are the notes I gave to my class.

Numeric output seems to indicate this, but numeric output is NOT proof.

To find a proof of this, let’s turn to the Picard iteration technique. We
know that the Picard iterates will converge to the unique solution.

The integrals get pretty ugly around here; I used MATLAB to calculate the
higher order iterates. I’ll show you

where means assorted polynomial terms from order 6 to 11.

Here is one more:

We notice some patterns developing here. First of all, the coefficient of
the term is staying the same for all where

That is tedious to prove. But what is easier to show (and sufficient) is
that the coefficients for the terms for all appear to be
bigger than 1. This is important!

Why? If we can show that this is the case, then our ”limit” solution will have an interval of convergence less than 1. Why? Substitute and see that the sum
diverges because the not only fail to converge to zero, but they
stay greater than 1.

So, can we prove this general pattern?

YES!

Here is the idea: where is a polynomial of order
and is a polynomial whose terms all have order or greater.

Now put into the Picard process:

Note: all of the terms of of degree or higher must come from
the second integral.

Now by induction we can assume that all of the coefficients of the
polynomial are greater than or equal to one.

When we ”square out” the polynomial, the coefficients of the new
polynomial will consist of the sum of positive numbers, each of which is
greater than 1. For the coefficients of the polynomial of
degree or higher: if one is interested in the
coefficient, one has to add at least numbers together, each of which
is bigger than one.

Now when one does the integration on these particular terms, one, of course,
divides by (power rule for integration). But that means that the
coefficient (after integration) is then greater than 1.

Here is a specific example:

Say

Now

Remember that are all greater than or equal to one.

Now

Now when we integrate term by term, we get:

But note that and

Since all of the factors are greater than or equal to 1.

Hence in our new polynomial approximation, the order 4 terms or less all
have coefficients which are greater than or equal to one.

We can make this into a Proposition:

Proposition
Suppose where each

If

Then for all

Proof. Of course, and

Let

Then we can calculate: (since all of the are
defined):

If is odd, then

If is even then

The Proposition is proved.

Of course, this possibly fails for where as we would fail to
have a sufficient number of terms in our sum.

Now if one wants a challenge, one can modify the above arguments to show that the coefficients of the approximating polynomial never get ”too big”; that is, the coefficient of the order term is less than, say, .

It isn’t hard to show that where

Then one can compare to the derivative of the geometric series to show that
one gets convergence on an interval up to but not including 1.