Integrals of functions with nice inverses

This idea started as a bit of a joke:

https://twitter.com/lightediand/status/1437181306526371845

Of course, for readers of this blog: easy-peasy. so the integral is transformed into and so we’ve entered the realm of rational functions. Ok, ok, there is some work to do.

But for now, notice what is really doing on: we have a function under the radical that has an inverse function (IF we are careful about domains) and said inverse function has a derivative which is a rational function

More shortly: let be such that then:

gets transformed: and then and the integral becomes which is a rational function integral.

Yes, yes, we need to mind domains.

An alternative to trig substitution, sort of..

Ok, just for fun:

The usual is to use which transforms this to the dreaded integral, which is a double integration by parts.
Is there a way out? I think so, though the price one pays is a trickier conversion back to x.

Let’s try so upon substituting we obtain and noting that alaways:

Now this can be integrated by parts: let

So but this easily reduces to:

Division by 2:

That was easy enough.

But we now have the conversion to x:

So far, so good. But what about ?

Write:

Now multiply both sides by to get and use the quadratic formula to get

We need so and that is our integral:

I guess that this isn’t that much easier after all.

Logarithmic differentiation: do we not care about domains anymore?

The introduction is for a student who might not have seen logarithmic differentiation before: (and yes, this technique is extensively used..for example it is used in the “maximum likelihood function” calculation frequently encountered in statistics)

Suppose you are given, say, and you are told to calculate the derivative?

Calculus texts often offer the technique of logarithmic differentiation: write
Now differentiate both sides:

Now multiply both sides by to obtain

\

And this is correct…sort of. Why I say sort of: what happens, at say, ? The derivative certainly exists there but what about that second factor? Yes, the sin(x) gets cancelled out by the first factor, but AS WRITTEN, there is an oh-so-subtle problem with domains.

You can only substitute only after simplifying ..which one might see as a limit process.

But let’s stop and take a closer look at the whole process: we started with and then took the log of both sides. Where is the log defined? And when does ? You got it: this only works when .

So, on the face of it, is justified only when each .

Why can we get away with ignoring all of this, at least in this case?

Well, here is why:

1. If is a differentiable function then
Yes, this is covered in the derivation of material but here goes: write

Now if we get as usual. If then and so in either case:

as required.

THAT is the workaround for calculating where : just calculate . noting that

Yay! We are almost done! But, what about the cases where at least some of the factors are zero at, say ?

Here, we have to bite the bullet and admit that we cannot take the log of the product where any of the factors have a zero, at that point. But this is what we can prove:

Given is a product of differentiable functions and then

This works out to what we want by cancellation of factors.

Here is one way to proceed with the proof:

1. Suppose are differentiable and . Then and
2. Now suppose are differentiable and . Then and
3.Now apply the above to is a product of differentiable functions and
If then by inductive application of 1.

If then let as in 2. Then by 2, we have Now this quantity is zero unless and . But in this case note that

So there it is. Yes, it works ..with appropriate precautions.

Calculus series: derivatives

Reminder: this series is NOT for the student who is attempting to learn calculus for the first time.

Derivatives This is dealing with differentiable functions and no, I will NOT be talking about maps between tangent bundles. Yes, my differential geometry and differential topology courses were on the order of 30 years ago or so. 🙂

In calculus 1, we typically use the following definitions for the derivative of a function at a point: . This is opposed to the derivative function which can be thought of as the one dimensional gradient of .

The first definition is easier to use for some calculations, say, calculating the derivative of at a point. (hint, if you need one: use then it is easier to factor). It can be used for proving a special case of the chain rule as well (the case there we are evaluating at and for at most a finite number of points near .)

When introducing this concept, the binomial expansion theorem is very handy to use for many of the calculations.

Now there is another definition for the derivative that is helpful when proving the chain rule (sans restrictions).

Note that as we have . We can now view as a function of which goes to zero as does.

That is, where and is the best linear approximation for at .

We’ll talk about the chain rule a bit later.

But what about the derivative and examples?

It is common to develop intuition for the derivative as applied to nice, smooth..ok, analytic functions. And this might be a fine thing to do for beginning calculus students. But future math majors might benefit from being exposed to just a bit more so I’ll give some examples.

Now, of course, being differentiable at a point means being continuous there (the limit of the numerator of the difference quotient must go to zero for the derivative to exist). And we all know examples of a function being continuous at a point but not being differentiable there. Examples: are all continuous at zero but none are differentiable there; these give examples of a corner, vertical tangent and a cusp respectively.

But for many of the piecewise defined examples, say, for and for the derivative fails to exist because the respective derivative functions fail to be continuous at ; the same is true of the other stated examples.

And of course, we can show that has continuous derivatives at the origin but not derivatives.

But what about a function with a discontinuous derivative? Try for and zero at . It is easy to see that the derivative exists for all but the first derivative fails to be continuous at the origin.

The derivative is at and for which is not continuous at the origin.

Ok, what about a function that is differentiable at a single point only? There are different constructions, but if for rational, for irrational is both continuous and, yes, differentiable at (nice application of the Squeeze Theorem on the difference quotient).

Yes, there are everywhere continuous, nowhere differentiable functions.

When is it ok to lie to students? part I

We’ve arrived at logarithms in our calculus class, and, of course, I explained that only holds for . That is all well and good.
And yes, I explained that expressions like only makes sense when

But then I went ahead and did a problem of the following type: given by using logarithmic differentiation,

And you KNOW exactly what I did. Right?

Note that is differentiable for all and, well, the derivative *should* be continuous for all but..is it? Well, up to inessential singularities, it is. You see: the second factor is not defined for , etc.

Well, let’s multiply it out and obtain:

So, there is that. We might induce inessential singularities.

And there is the following: in the process of finding the derivative to begin with we did:

and that expansion is valid only for
because we need and .

But the derivative formula works anyway. So what is the formula?

It is: if where is differentiable, then and verifying this is an easy exercise in induction.

But the logarithmic differentiation is really just a motivating idea that works for positive functions.

To make this complete: we’ll now tackle where it is essential that .

Rewrite

Then

This formula is a bit of a universal one. Let’s examine two special cases.

Suppose some constant. Then and the formula becomes which is just the usual constant power rule with the chain rule.

Now suppose for some positive constant. Then and the formula becomes which is the usual exponential function differentiation formula combined with the chain rule.

Proving a differentiation formula for f(x) = x ^(p/q) with algebra

Yes, I know that the proper way to do this is to prove the derivative formula for and then use, say, the implicit function theorem or perhaps the chain rule.

But an early question asked students to use the difference quotient method to find the derivative function (ok, the “gradient”) for And yes, one way to do this is to simplify the difference quotient by factoring from both the numerator and the denominator of the difference quotient. But this is rather ad-hoc, I think.

So what would one do with, say, where are positive integers?

One way: look at the difference quotient: and do the following (before attempting a limit, of course): let at which our difference quotient becomes:

Now it is clear that is a common factor..but HOW it factors is essential.

So let’s look at a little bit of elementary algebra: one can show:

(hint: very much like the geometric sum proof).

Using this:

Now as

we have (for the purposes of substitution) so we end up with:

(the number of terms is easy to count).

Now back substitute to obtain which, of course, is the familiar formula.

Note that this algebraic identity could have been used for the old case to begin with.

What is going on here: sum of cos(nx)…

This started innocently enough; I was attempting to explain why we have to be so careful when we attempt to differentiate a power series term by term; that when one talks about infinite sums, the “sum of the derivatives” might fail to exist if the sum is infinite.

Anyone who is familiar with Fourier Series and the square wave understands this well:

yields the “square wave” function (plus zero at the jump discontinuities)

Here I graphed to

Now the resulting function fails to even be continuous. But the resulting function is differentiable except for the points at the jump discontinuities and the derivative is zero for all but a discrete set of points.

(recall: here we have pointwise convergence; to get a differentiable limit, we need other conditions such as uniform convergence together with uniform convergence of the derivatives).

But, just for the heck of it, let’s differentiate term by term and see what we get:

It is easy to see that this result doesn’t even converge to a function of any sort.

Example: let’s see what happens at

And this repeats over and over again; no limit is possible.

Something similar happens for where are relatively prime positive integers.

But something weird is going on with this sum. I plotted the terms with

(and yes, I am using as a type of “envelope function”)

BUT…if one, say, looks at

we really aren’t getting a convergence (even at irrational multiples of ). But SOMETHING is going on!

I decided to plot to

Something is going on, though it isn’t convergence. Note: by accident, I found that the pattern falls apart when I skipped one of the terms.

This is something to think about.

I wonder: for all and we can somehow get close to for given values of by allowing enough terms…but the value of is determined by how many terms we are using (not always the same value of ).

The error term and approximation of derivatives

I’ll go ahead and work with the common 3 point derivative formulas:

This is the three-point endpoint formula: (assuming that has 3 continuous derivatives on the appropriate interval)

where is some point in the interval.

The three point midpoint formula is:

.

The derivation of these formulas: can be obtained from either using the Taylor series centered at or using the Lagrange polynomial through the given points and differentiating.

That isn’t the point of this note though.

The point: how can one demonstrate, by an example, the role the error term plays.

I suggest trying the following: let vary from, say, 0 to 3 and let . Now use the three point derivative estimates on the following functions:

1. .

2. .

Note one: the three point estimates for the derivatives will be exactly the same for both and . It is easy to see why.

Note two: the “errors” will be very, very different. It is easy to see why: look at the third derivative term: for it is

The graphs shows the story.

Clearly, the 3 point derivative estimates cannot distinguish these two functions for these “sample values” of , but one can see how in the case of , the degree that wanders away from is directly related to the higher order derivative of .

Approximating the derivative and round off error: class demonstration

In numerical analysis we are covering “approximate differentiation”. One of the formulas we are using: where is some number in ; of course we assume that the third derivative is continuous in this interval.

The derivation can be done in a couple of ways: one can either use the degree 2 Lagrange polynomial through and differentiate or one can use the degree 2 Taylor polynomial expanded about and use and solve for ; of course one runs into some issues with the remainder term if one uses the Taylor method.

But that isn’t the issue that I want to talk about here.

The issue: “what should we use for ?” In theory, we should get a better approximation if we make as small as possible. But if we are using a computer to make a numerical evaluation, we have to concern ourselves with round off error. So what we actually calculate will NOT be but rather where where is the round off error used in calculating the function at (respectively).

So, it is an easy algebraic exercise to show that:

and the magnitude of the actual error is bounded by where on some small neighborhood of and is a bound on the round-off error of representing .

It is an easy calculus exercise (“take the derivative and set equal to zero and check concavity” easy) to see that this error bound is a minimum when .

Now, of course, it is helpful to get a “ball park” estimate for what is. Here is one way to demonstrate this to the students: solve for and obtain and then do some experimentation to determine .

That is: obtain an estimate of by using this “3 point midpoint” estimate for a known derivative near a value of for which (a bound for the 3’rd derivative) is easy to obtain, and then obtain an educated guess for .

Here are a couple of examples: one uses Excel and one uses MATLAB. I used at ; of course and is reasonable here (just a tiny bit off). I did the 3-point estimation calculation for various values of and saw where the error started to increase again.

Here is the Excel output for at and at respectively. In the first case, use and in the second

In the case, we see that the error starts to increase again at about ; the same sort of thing appears to happen for .

So, in the first case, is about ; it is roughly at .

Note: one can also approach by using powers of instead; something interesting happens in the case; the case gives results similar to what we’ve shown. Reason (I think): 1 is easy to represent in base 2 and the powers of can be represented exactly.

Now we turn to MATLAB and here we do something slightly different: we graph the error for different values of . Since the values of are very small, we use a scale by doing the following (approximating for )

. By design, . The graph looks like:

Now, the small error scale makes things hard to read, so we turn to using the log scale, this time on the axis: let and run plot(N, LE):

and sure enough, you can see where the peak is: about , which is the same as EXCEL.

A real valued function that is differentiable at an isolated point

A friend of mine is covering the Cauchy-Riemann equations in his complex variables class and wondered if there is a real variable function that is differentiable at precisely one point.

The answer is “yes”, of course, but the example I could whip up on the spot is rather pathological.

Here is one example:

Let be defined as follows:

That is, if is irrational or zero, and is if is rational and where .

Now calculate

Let be given and choose a positive integer so that . Let . Now if and is irrational, then .

Now the fun starts: if is rational, then and .

We looked at the right hand limit; the left hand limit works in the same manner.

Hence the derivative of exists at and is equal to zero. But zero is the only place where this function is even continuous because for any open interval , .