College Math Teaching

February 19, 2012

Divergent Improper Integrals: change of variables to an unbounded integrand.

Filed under: calculus, integrals, pedagogy, improper integrals, change of variable, integration by substitution — collegemathteaching @ 10:41 pm

This post was motivated by a student question: my student wanted help with the following problem:

\int^{\infty}_{1} \frac{x^2}{\sqrt{x^3 +1}} dx
Of course the idea is to do a substitution: u = x^3 + 1 which transforms the integral into \frac{1}{3} \int^{\infty}_{2} \frac{1}{\sqrt{u}} du 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 x, \frac{x^2}{\sqrt{x^3 +1}} \approx \frac{x^2}{\sqrt{x^3}} = \sqrt{x} and of course \int^{\infty}_1 \sqrt{x} dx diverges because the integrand does not go to zero (in fact, is unbounded!) as x 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 [1, \infty ) , is there a change of variables which will change the integrand to either an unbounded function on [1, \infty ) 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 \int^{\infty}_1 \frac{1}{x^r} dx
If 0 < r < 1 , let k > \frac{1}{1-r} . Then the change of variable u = x^{\frac{1}{k}} transforms \int^{\infty}_{1} \frac{1}{x^r} dx to k \int^{\infty}_1 u^{k(1-r) -1} du and of course u^{k(1-r) -1} is unbounded on [1, \infty) .

If 1 < r let k < \frac{1}{1-r} < 0 . Then \int^{\infty}_1 \frac{1}{x^r} dx is transformed into |k|\int^{1}_{0} u^{-1+k(1-r)} du 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 \int^{1.5}_{1} 3x^{\frac{1}{2}} dx and \int^{(1.5)^3}_1 \frac{1}{\sqrt{u}} du . The transformation is accomplished by using u =x^3 . Note how the transformation stretches the interval of integration to account for the function “shrinkage”.

On the other hand, using u^{-2}=x transforms \int^{\infty}_1 \frac{1}{x^2} dx into 2\int^{1}_{0} u du

February 16, 2012

The “equals” sign: identities, equations to be solved and all that…

Filed under: basic algebra, calculus, differential equations, logic, mathematical ability, mathematics education, pedagogy, student learning, vector spaces — collegemathteaching @ 7:09 pm

Here is the sort of thing that got me thinking about this topic: a colleague had a student complain about how one of her quiz problems was scored. The problem stated: “show that \sqrt{2} + \sqrt{3} \neq \sqrt{5} “. She was offended that her saying “\sqrt{x} + \sqrt{y} \neq \sqrt{x+y} ” wasn’t enough to receive credit and would NOT take his word for it. In fact, she took this to the student ombudsman!!!

But that raised the question: “what do we mean when we tell our students “\sqrt{x} + \sqrt{y} \neq \sqrt{x+y} “?

Of course, there are some central issues here. The first issues is that our “sure of herself” student thought that “\sqrt{x} + \sqrt{y} \neq \sqrt{x+y} ” meant that this relation is NEVER true for any choice of x, y , which of course, is false (e. g. let y = 0 and x \ge 0 .) In fact, \sqrt{x} + \sqrt{y} \neq \sqrt{x+y} is the logical negation of the statement \sqrt{x} + \sqrt{y} = \sqrt{x+y} ; the latter means that “this statement is true for ALL x, y and its negation means “there is at least one choice of x, y 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, \sqrt{x} + \sqrt{y} \neq \sqrt{x+y} means that \sqrt{x} + \sqrt{y} and \sqrt{x+y} 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 x^2 + 3x + 2 = 0 or y^{\prime \prime} + y = 0 ?
These are NOT equalities in the space of functions; the first means “what values in the domain does f^{-1}(0) take given f(x)=x^2 + 3x + 2 and the second asks one to find the inverse image of 0 for the operator D^2+1 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.

February 12, 2012

Mathematical Research at “Teaching Institutions”

Filed under: academia, editorial, mathematics education, pedagogy, research — collegemathteaching @ 1:03 am

I read the blogs of a few academics and I had to wince a bit when I read that the push to make professors teach a higher load was happening even in Canada.

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 \int e^x dx \neq \frac{e^{x+1}}{x+1} + C . 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.

February 7, 2012

Forgotten Basic Algebra: or why we shouldn’t rely on the “conjugate trick”

Filed under: calculus, derivatives, how to learn calculus, pedagogy, elementary mathematics, basic algebra — collegemathteaching @ 7:01 pm

I’ll admit that, after 20 years of teaching at the university level, I sometimes get lazy. But…as I age, I must resist that temptation even though at times I find myself muttering “I don’t have 30 extra f*cking minutes to figure out how to do this…”

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 lim_{z \rightarrow w} \frac{z^{1/3} - w^{1/3}}{z - w} 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 \frac{u - v}{u^3 - v^3} don’t you?”

Problem solved…shame WIN.

of course things like lim_{z \rightarrow w} \frac{z^{7/8} - w^{7/8}}{z - w} are easily converted to things like \frac{u^7 - v^7}{u^8 - v^8} , etc.

This leads to another point. Often when we teach lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} we use the “conjugate trick” which only works for square roots. The above method works for the other fractional powers.

January 17, 2012

Applications of calculus in the New York Times: Comparative Statics (economics)

Filed under: applications of calculus, applied mathematics, calculus, economics, mathematical economics, media, news — blueollie @ 5:37 pm

Paul Krugman has an article that talks about the economics concept of comparative statics which involves a bit of calculus. The rough idea is this: suppose we have something that is a function of two economics variables f(x,y) and we are on some level curve: f(x,y) = C_1 at some point (x_0, y_0, f(x_0, y_0) = C) . Now if we, say, hold y constant and vary x by \Delta x what happens to the level curve C_1 ? The answer is, of course, C = C_1 + (\Delta x) \frac{\partial f}{\partial x} (x_0,y_0) + \epsilon where \epsilon is a small error that vanishes as \Delta x goes to zero; this is just multi-variable calculus and the idea of differentials, tangent planes and partial derivatives. The upshot is that the change in C , denoted by \Delta C is approximately (\Delta x) \frac{\partial f}{\partial x} (x_0,y_0) .

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

January 12, 2012

So you want to take a course in complex variables

Filed under: advanced mathematics, analysis, applied mathematics, calculus, complex variables, derivatives, differential equations, integrals — collegemathteaching @ 5:30 pm

Ok, what should you have at your fingertips prior to taking such a course?

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 \delta is any piecewise smooth curve running from z_1 to z_2 then \int_{\delta} e^z dz = e^{z_2} - e^{z_1} . But in many cases, the similarity vanishes and more care must be taken.

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

2. No non-constant everywhere differentiable function is bounded; compare that to f(x) = \frac{1}{1+x^2} in calculus.

3. Integrals can have some strange properties. For example, if \delta is the unit circle taken once around in the standard direction, \int_{\delta} Log(z) dz 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 x = 0 ) for \frac{1}{1+x^2} 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 f(x) = x^{4/3} which has one derivative but NOT two derivatives at x = 0

There is much more, of course.

December 9, 2011

Striking a balance between precision and being intelligible

Filed under: advanced mathematics, applied mathematics, matrix algebra, numerical methods, pedagogy, vector spaces — collegemathteaching @ 2:54 am

Ok, what do we mean by: x + 2 = 1 ? Now, what do we mean by (A+B)x + (B-A) = 1 ? Of course, the answer is “it depends”. The most common use of the first “equation” is “find the real number x such that that number added to 2 equals 1.” In the second case, the most common use is “find real numbers A, B such that this equation is true for all real x .

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 \int \int_{A} f dx dy = \int \int_{A} f dy dx but \int \int_{A} f dx\wedge dy = -\int \int_{A} f dy\wedge dx 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 AX = B where X is a column vector of variables and B 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 A = LDU where D was a diagonal matrix with non-zero entries, L, U 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 a_{1,1}x_1 +a_{1,2}x_2 = 3, a_{2,2}x_2=5 we don’t count the “coefficient times the variable” as a multiplication. But once we solve and obtain x_2 = \frac{5}{a_{2,2}} we then count operations involving x_2 . (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.

November 3, 2011

Finding a Particular solution: the Convolution Method

Filed under: advanced mathematics, applied mathematics, calculus, derivatives, differential equations, integrals, Laplace transform — collegemathteaching @ 4:08 pm

Background for students
Remember that when one is trying to solve a non-homogeneous differential equation, say:
y^{\prime \prime} +3y^{\prime} +2y = cos(t) one finds the general solution to y^{\prime \prime} +3y^{\prime} +2y = 0 (which is called the homogeneous solution; in this case it is c_1 e^{-2t} + c_2 e^{-t} and then finds some solution to y^{\prime \prime} +3y^{\prime} +2y = cos(t) . This solution, called a particular solution, will not have an arbitrary constant. Hence that solution cannot meet an arbitrary initial condition.

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, e^{at} , 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 y_p = Acos(t) + Bsin(t) and then substitute into the differential equation to solve for A and B . One could also try a complex form; that is, try y_p = Ae^{it} and then determines A 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 y_1, y_2 and then seeks a particular solution of the form y_p = v_1y_1 + v_2y_2 where v_1 = -\int \frac{f(t)y_2}{W} dt and v_2 = \int \frac{f(t)y_1}{W} dt where W is the determinant of the Wronskian matrix. This method can solve differential equations like y^{\prime \prime} + y = tan(t) 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 y^{\prime \prime} + 4y = cos(5t) 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 f and g to be:
f*g = \int_0^t g(u)f(t-u)du . Needless to say, f and g need to meet appropriate “integrability” conditions; this is usually not a problem in a differential equations course.

Example: if f = e^t, g=cos(t) , then f*g = \frac{1}{2}(e^t - cos(t) + sin(t)) . Notice that the dummy variable gets “integrated out” and the variable t remains.

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

The Convolution Method
If y(t) is a homogenous solution to a second order linear differential equation that meets initial conditions: y(0)=0, y^{\prime}(0) =1 and f is the forcing function, then y_p = f*y is the particular solution that meets y_p(0)=0, y_p^{\prime}(0) =0

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

Suppose we want to solve y^{\prime \prime} + y = tan(t) . The homogeneous solution is y_h = c_1 cos(t) + c_2 sin(t) and it is easy to see that we need c_1 = 0, c_2 = 1 to meet the y_h(0)=0, y^{\prime}_h(0) =1 condition. So a particular solution is sin(t)*tan(t) = tan(t)*sin(t)= \int_0^t tan(u)sin(t-u)du = \int_0^t tan(u)(sin(t)cos(u)-cos(t)sin(u))du = sin(t)\int_0^t sin(u)du - cos(t)\int_0^t \frac{sin^2(u)}{cos(u)}du = sin(t)(1-cos(t)) -cos(t)ln|sec(t) + tan(t)| + sin(t)cos(t) = sin(t) -cos(t)ln|sec(t)+tan(t)|

This particular solution meets y_p(0)=0, y_p^{\prime}(0) = 0 .

Why does this work?
This is where “differentiation under the integral sign” comes into play. So we write f*y = \int_0^t f(u)y(t-u)du .
Then (f*y)^{\prime} = ?

Look at the convolution integral as g(x,z) = \int_0^x f(u)y(z-u)du . Now think of x(t) = t, z(t) = t . Then from calculus III: \frac{d}{dt} g(x,z) = g_x \frac{dx}{dt} + g_z \frac{dz}{dt} . Of course, \frac{dx}{dt}=\frac{dz}{dt}=1 .
g_x= f(x)y(z-x) by the Fundamental Theorem of calculus and g_z = \int_0^x f(u) y^{\prime}(z-u) du by differentiation under the integral sign.

So we let x = t, z = t and we see \frac{d}{dt} (f*y) = f(t)y(0) + \int_0^t f(u) y^{\prime}(t-u) du which equals \int_0^t f(u) y^{\prime}(t-u) du because y(0) = 0 . Now by the same reasoning \frac{d^2}{dt^2} (f*y) = f(t)y^{\prime}(0) + \int_0^t f(u) y^{\prime \prime}(t-u) du = f(t)+ \int_0^t f(u) y^{\prime \prime}(t-u) du because y^{\prime}(0) = 1 .
Now substitute into the differential equation y^{\prime \prime} + ay^{\prime} + by = f(t) and use the linear property of integrals to obtain f(t) + \int_0^t f(u) (y^{\prime \prime}(t-u) + ay^{\prime}(t-u) + by(t-u))du = f(t) + \int_0^t f(u) (0)du = f(t)

It is easy to see that (f*y)(0) = 0. Now check \frac{d}{dt} f*y(0) = f(t)y(0) + \int_0^0 f(u) y^{\prime}(t-u) du = 0 .

October 31, 2011

Differentiation Under the Integral Sign

Filed under: advanced mathematics, applied mathematics, calculus, derivatives, differential equations, integrals, Laplace transform — collegemathteaching @ 10:05 pm

Suppose we have F(s) = \int_a^b f(s,t)dt and we’d like to know what \frac{d}{ds} F is.
The answer is \frac{d}{ds}F(s) = \int_a^b \frac{\partial}{\partial s} f(s,t)dt .

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 f(s,t) to make this result true: continuity of f(s,t) and \frac{\partial}{\partial s} f(s,t) on some rectangle in (s,t) 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 M = max{|f|} on some interval (a,b) , then |\int_a^b f(t)dt| \leq M |b-a| and the Mean Value Theorem.

Now recall from calculus: \frac{d}{ds} F =lim_{s_0 \rightarrow s} \frac{F(s_0)-F(s)}{s_0 - s} = lim_{s_0 \rightarrow s} \frac{1}{s_0 -s} \int_a^b f(s_0,t)-f(s,t) dt =lim_{s_0 \rightarrow s} \int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s_0} dt

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 \frac{f(s_0,t)-f(s,t)}{s_0 - s} remind you of anything? Right; this is the fraction from the Mean Value Theorem; that is, there is some s* between s and s_0 such that \frac{f(s_0,t)-f(s,t)}{s_0 - s} = \frac{\partial}{\partial s} f(s*,t)

Because we are assuming the continuity of the partial derivative, we can say that for s sufficiently close to s_0 , |\frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t)| < \epsilon

This means that | \int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t) dt | < \int_a^b |\frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t)| dt < \epsilon (b-a)

Now realize that \epsilon can be made as small as desired by letting s_0 get sufficiently close to s so it follows by the \epsilon-\delta definition of limit that:
lim_{s_0 \rightarrow s}\int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t) dt=0 which implies that
lim_{s_0 \rightarrow s}\int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s}dt -\int_a^b \frac{\partial}{\partial s} f(s,t) dt=0
Therefore lim_{s_0 \rightarrow s} \frac{F(s_0)-F(s)}{s_0 - s} - \int_a^b \frac{\partial}{\partial s} f(s,t) dt=0
So the result follows.

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

October 10, 2011

The Picard Iterates: how they can yield an interval of existence.

Filed under: advanced mathematics, differential equations, numerical methods, numerical solution of differential equations, Taylor Series, uniqueness of solution — collegemathteaching @ 9:33 pm

One of the many good things about my teaching career is that as I teach across the curriculum, I fill in the gaps of my own education.
I got my Ph. D. in topology (low dimensional manifolds; in particular, knot theory) and hadn’t seen much of differential equations beyond my “engineering oriented” undergraduate course.

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 y'= t + y^2 with y(0)=1 and to produce the graph of y(t) from t = 0 to t = 3 .

There is a unique solution to this and the solution is valid so long as the t and y value of the solution curve stays finite; note that \frac{\partial }{\partial y}f(t,y)=2y.

So, is it possible that the y 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.

y_{0}=1

y_{1}=1+\int_{0}^{t}x+1dx=\frac{1}{2}t^{2}+t+1

y_{2}=1+\int_{0}^{t}x+(\frac{1}{2}x^{2}+x+1)^{2}dx=

y_{2}=\frac{1}{20}t^{5}+\frac{1}{4}t^{4}+\frac{2}{3}t^{3}+\frac{3}{2}t^{2}+t+1

The integrals get pretty ugly around here; I used MATLAB to calculate the
higher order iterates. I’ll show you y_{3}

y_{3}=\frac{49}{60}t^{5}+\frac{13}{12}t^{4}+\frac{4}{3}t^{3}+\frac{3}{2}t^{2}+t+1+O(t^{11})

where O(t^{11}) means assorted polynomial terms from order 6 to 11.

Here is one more:

y_{4}=\frac{17}{12}t^{5}+\frac{17}{12}t^{4}+\frac{4}{3}t^{3}+\frac{3}{2}t^{2}+t+1+O(t^{23})

We notice some patterns developing here. First of all, the coefficient of
the t^{n} term is staying the same for all y_{m} where m\geq n.

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

Why? If we can show that this is the case, then our ”limit” solution \sum_{k=0}^{\infty }a_{k}t^{k} will have an interval of convergence less than 1. Why? Substitute t=1 and see that the sum
diverges because the a_{k} 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: y_{m}=q(t)+p(t) where p(t) is a polynomial of order m
and q(t) is a polynomial whose terms all have order m+1 or greater.

Now put into the Picard process:

y_{m+1}=1+\int_{0}^{t}((q(x)+p(x))^{2}+xdx=

1+\int_{0}^{t}((q(x)^{2}+2p(x)q(x))dx+\int_{0}^{t}(p(x))^{2}+xdx

Note: all of the terms of y_{m+1} of degree m+1 or higher must come from
the second integral.

Now by induction we can assume that all of the coefficients of the
polynomial p(x) 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 (p(x))^{2} of
degree m or higher: if one is interested in the k^{\prime }th
coefficient, one has to add at least k+1 numbers together, each of which
is bigger than one.

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

Here is a specific example:

Say p(x)=a+bx+cx^{2}+dx^{3}

Now p(x)^{2}=a^{2}+(ab+ab)x+(ac+ca+b^{2})x^{2}+(ad+da+bc+cb)x^{3}+\{O(x^{6})\}

Remember that a,b,c,d are all greater than or equal to one.

Now p(x)^{2}+x=a^{2}+(ab+ab+1)x+(ac+ca+b^{2})x^{2}+(ad+da+bc+cb)x^{3}+\{O(x^{6})\}

Now when we integrate term by term, we get:

\int_{0}^{t}(p(x))^{2}+xdx=a^{2}x+\frac{1}{2}(ab+ab+1)x^{2}+\frac{1}{3}(ac+ca+b^{2})x^{3}+\frac{1}{4}(ad+da+bc+cb)x^{4}+\{O(x^{7})\}

But note that ab+ab+1>2,ac+ca+b^{2}\geq 3, and ad+da+bc+cb\geq 4

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 p(x)=\sum_{j=0}^{k}a_{j}x^{j} where each a_{j}\geq 1.

If q(t)=\sum_{j=0}^{2k+1}b_{j}x^{j}=1+\int_{0}^{x}((p(t))^{2}+t)dt

Then for all j\leq k+1,b_{j}\geq 1.

Proof. Of course, b_{0}=1,b_{1}=a_{0}^{2}, and b_{2}=\frac{2a_{0}a_{1}+1}{2}

Let n\leq k+1.

Then we can calculate: (since all of the a_{n-1},a_{n-2},....a_{1} are
defined):

If n is odd, then b_{n}=\frac{1}{n}(2a_{0}a_{n-1}+2a_{1}a_{n-2}+...2a_{\frac{n-3}{2}}a_{\frac{n+1}{2}}+(a_{\frac{n-1}{2}})^{2})\geq \frac{1}{n}(2\ast \frac{n-1}{2}+1)=1

If n is even then b_{n}=\frac{1}{n}(2a_{0}a_{n-1}+2a_{1}a_{n-2}+....2a_{\frac{n-1}{2}}a_{\frac{n+1}{2}})\geq \frac{1}{n}(2\ast \frac{n}{2})=1

The Proposition is proved.

Of course, this possibly fails for b_{n} where n>k+1 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 k^{\prime }th order term is less than, say, k.

It isn’t hard to show that b_{n}\leq \max a_{i}^{2} where i\in\{0,1,2,...n-1\}

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.

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