## Thursday, October 12, 2006

### Calculus

Calculus is the second-greatest technological invention by mankind (second only to the printing press). When Isaac Newton and Gottfried Leibniz independently invented the calculus, building on the work of earlier mathematicians like Descartes and Barrow, they showed themselves to be responsible for the modern technological age. In particular, the Fundamental Theorem of the Calculus has extremely far-reaching consequences. It is easily the most important theorem in all of mathematics.

I love calculus; it's my favorite mathematics. Aside from simply having the very best name of any branch of math, it's so powerful! There's so much you can do with it. And it is very beautiful.

So I wanted to put forth a few calculus challenges that were a little off the beaten track. Enjoy!

1. Assume a function f is thrice differentiable. If, at a point a we have f''(a) = 0 and it is not the case that f'''(a) = 0, what can you say about f at a? Supposing the second derivative is zero, and the third derivative is also zero, what changes? This is known as the Third Derivative Test.

2. Are endpoints critical points? This is a neat question, asked by one of my students at Tech. The answer is not in any calculus book that I am aware of, though the question is clearly in the realm of calculus.

3. Without looking it up, do you know how to integrate sec(x)?

4. Integrate sqrt(1 - x^2) by using trig substitution, and check your answer using implicit differentiation. (Ok, this one is a bit more mainstream; I just like trig substitution).

5. Examine the following famous ODE, known as the time-independent Schroedinger equation: f''(x)=2m ( V(x) - E ) f(x) / hbar^2, where hbar is Planck's constant, and m is mass. V(x) is simply a function of x called the potential, and E is the energy, a constant. Assume that f is continuous and differentiable everywhere on the real line, and is also nonzero. Show that for bound states (fancy way of saying that f has a finite integral over the entire real line), it must be that E is greater than the minimum value of V(x) for all x.

6. Also in the above problem with the Schroedinger equation, show that if V(x) is an even function, then f can be taken to be either even or odd.

Numbers 5. and 6. are due to Griffiths' Introduction to Quantum Mechanics, 1995, p. 24.

Have fun!

Visit Math Help Boards for friendly, free and expert math help.

At 10/12/2006 08:10:00 PM ,  Susan said...

I feel like I'm letting you down :(, but I honestly can't do a single one of those problems. I finished multi-variable calculus freshman year, and haven't taken a calculus class since. I've used the basics of calculus quite a bit since, but nothing like what is required for those problems!

Calculus is fascinating, I will grant you, and I was sort of hoping for a calculus tutoring student this year, so my memory would be refreshed, but at the same time, I knew I would have to prep for that, and I don't have to spend out-of-tutoring time prepping for any of my other classes I tutor, which is nice.

Really, though, doesn't Linear Algebra or Abstract beat Calculus? :) And Geometry beats them all. . .

At 10/12/2006 08:24:00 PM ,  Adrian C. Keister said...

Aww, Susan! Can't you take a crack at number 4. at least? You can do it! Granted, there are subtle things in the other ones, but that one is pretty straight-forward.

Here's a hint on 4.: draw a right trangle, like so:

/|
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Let the angle in the lower left be theta. Label the hypotenuse 1, and either of the other two sides gets to be x. Pythagoras tells us that the last side has to be sqrt(1 - x^2). That means you can translate the entire integrand plus the dx differential into the theta domain. Then do the integral in the theta domain, and finally transfer back to the x domain. Comprendo?

I'd like to see if anyone else, like maybe John, would take a crack at those. However, if you'd like your own personal set of solutions, just let me know. ;-)]

In Christ.

At 10/12/2006 08:48:00 PM ,  Adrian C. Keister said...

And Calculus definitely beats everything. Period. No question. Linear algebra is second, though, in usefulness. Abstract and Geometry, while certainly having their applications, tend to apply in extremely advanced situations that I don't even begin to understand. They certainly have not had quite the generally felt (if not understood) effect that calculus and linear have had.

In Christ.

At 10/12/2006 09:25:00 PM ,  Susan said...

My dad took a crack at #1-4, but he's never had Quantum Mechanics, so he ignored the last two. Have I told you that he actually taught my college calculus classes at the local community college? So literally all the calculus I know, I know from Father Dear :).

I'm transcribing his jottings for him. He's not beside me to clarify, so if I explain something wrong, 'tisn't his fault. And his notes were written in haste, so no promises.

#1

Let f(x) = x^3 and g(x) = x^5. Both are thrice differentiable.

For f(x), the second derivative at x = 0 is 6x = 6(0) = 0, but the third derivative is 6.

But then look at g(x). The second derivative there is 20x^3 = 0 when x = 0, but also the third derivative is 60x^2, which is 0 when x = 0.

Comparing the two graphs, there really is no noticable difference at those two points, so he's not sure what you're getting at there.

[Edit by Susan: Are you meaning something simple like a change of degree?? - and I actually understood what I just typed above, just to clarify. I understand derivatives, it's integrals that are long-gone]

#2

Critical points occur when the derivative is 0 or undefined. This may or may not be an endpoint, depending on the function.

#3

The key to this is multiplying sec(x) by [(sec(x)+tan(x))/(sec(x)+tan(x))], which of course doesn't change the value, since it is just equal to one. The beauty of this trick is that you end up with a du/u set-up and it integrates nicely to ln(abs(sec(x)+tan(x)))+C. I'm Father Dear's witness that he didn't use a Calculus book to come up with that. I proctored his work :).

#4

[Edit by Susan: This one is going to be fun to transcribe - NOT!]

The integral of sqr(1-x^2)dx = the integral of sqr(1-sin^2(theta))*cos(theta)*d-theta, since x = sin(theta) and dx therefore = cos(theta)d-theta.

The integral of sqr(1-sin^2(theta))*cos(theta)*d-theta = integral of sqr(cos^2(theta))*cos(theta)*d-theta = integral of cos^2(theta)*d-theta = integral of (1 + cos(2*theta)/2 times d-theta = integral of (1/2)*d-theta + integral of (cos(2*theta))/2 times d-theta = (1/2)*theta + (1/4)*sin(2*theta) + C = (1/2)*theta + (1/4)*2*sin(theta)*cos(theta) + C.

Here he has a nifty right triangle drawn, with legs of length x and sqr(1 - x^2) and angle theta between hypoteneuse and the second leg.

1/2)*theta + (1/4)*2*sin(theta)*cos(theta) + C = (1/2)*sin^-1(x) + (1/4)*2*(x/1)*(sqr(1-x^2))/1 + C = (1/2)*sin^-1(x) + (x*sqr(1-x^2))/2 + c

At 10/12/2006 10:05:00 PM ,  Adrian C. Keister said...

#1. Definitely not getting the idea. Try looking at inflection points, and whether those two given situations guarantee an inflection point or no. Quarter credit for looking at some examples.

#2. Hint: to be more general, suppose you take a function differentiable everywhere on the real line, such as x^2, and artificially restrict the domain to [-1,1]. Now clearly the origin is a critical point. What about -1 and 1? Are they critical points or not? Quarter credit.

#3. Full credit. It's a dirty trick, isn't it? I've always wondered who in tarnation dreamed that one up.

#4. Half credit. You've definitely got to read the directions! Hint: work is correct, so far.

#5. No credit, even for humility. ;-)] Try examining the sign of the function f versus its second derivative.

#6. No credit. Use definitions of even and odd, and try changing the sign of x in the ODE and see what you get.

In Christ.

At 10/13/2006 04:08:00 AM ,  John Dekker said...

No, I'm out. There was definitely a time when I could do 3 and 4. 1 seems vaguely familiar.

I did study some quantum physics in second year, and I do remember Schroedinger's equation. But it's fading fast.

Still, there are enough mathematical blogs around. :) I was going to put up a problem on my blog, but I'd have to make it a w-h-o-l-e lot easier. ;)

Anyway, just as Mathematics is Queen of the Science, Number Theory is Queen of Mathematics. :)

At 10/13/2006 07:36:00 AM ,  Meg Garrison said...

You put calculus and the printing press in the category? The printing press was an invention. Calculus was a discovery of a system that God had already created. Just nitpicking. ;-)

At 10/13/2006 07:37:00 AM ,  Meg Garrison said...

*in the same category

At 10/13/2006 09:07:00 AM ,  John Dekker said...

Well, for that matter, the "invention" of the printing press was really the discovery of something God had invented as well. :)

At 10/13/2006 10:19:00 AM ,  Susan said...

This comment has been removed by a blog administrator.

At 10/13/2006 10:20:00 AM ,  Susan said...

In what way is the printing press a discovery, John? I'm not seeing that :). Calculus is a system that God created, but a printing press was a man-made tool, although of course God ordained the invention and gave man the ability to invent it. But I still think the two things are distinct and that the printing press was not a discovery :).

Father Dear is out earning daily bread at the moment, but I'm sure he'll look at your grading later today, Adrian. I have no comment, since this is above my head anyway. Or rather, no longer in my head :).

At 10/13/2006 12:35:00 PM ,  zan said...

?

At 10/13/2006 01:09:00 PM ,  Lydia said...

No entiendo!
Habla usted ingles?

At 10/13/2006 05:25:00 PM ,  Susan said...

Here we go:

#1

Dad's really not sure what you're getting at when you ask what changes??

(It's a rather general question. . . )

#2

Father Dear stands by his original response, which really came pretty much straight from his Calculus book. We have a book at home (the one I used in college) and one at school that he teaches from. Both say the same thing. I'll quote from the one I have handy:

The number c in the domain of f is called a critical point of f if either
>> f'(c) = 0, or
>> f'(c) does not exist

Incidentally, the above came from page 144.

Your "hint" of taking y = x^2 and restricting it does not have anything to do with the above definition of a critical point. Do your books speak in more ambiguous terms, perhaps?

I think I know what you're getting at, by the way :). Your idea is that a critical point is only a critical point if it would be so if the graph continues in a like manner? Perhaps? But I'm going to go with p. 144 :-D.

#4 He's not sure the point of implicitly differentiating here. You usually do that when an equation is not solved for y, which this one is, so what is the point of this? I would say there is considerably more of a reason to rationalize cube roots than there is to differentiate this equation :-D, and you never finished your problem either.

At 10/13/2006 06:17:00 PM ,  Susan said...

Okay, I've been brainstorming on 1. I still say your question is too general. What changes? It depends :-D. In the case of x^3 and x^5, the degree changes. But I am brainstorming over a rather interesting way to generate the scenario for any function.

Look at a generic function f(x), s.t. f''(a) = 0 and f'''(a) is not equal to zero. Assuming the triple derivative exists (I can't think of a function whose triple derivative is a vertical line), then there must be some k in the reals so that f'''(a) = k. Now take your function f(x), and alter it to create g(x), s.t. g(x) = f(x) - f'''(a)*e^x. This function will necessarily have to have g'''(a) = 0, unless I am missing something. But I'm not sure how useful that is :). I can't think more right now. I have to go.

Wait, wait, that fails with the double derivative, though. Hmmm.

At 10/13/2006 08:47:00 PM ,  Adrian C. Keister said...

Ok, I'm going to post the complete answer to the first problem, so SPOILER.
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#1. In the first case, we are guaranteed an inflection point at a(point where the concavity changes from positive to negative or vice versa). That is because the second derivative is zero, which is necessary but not sufficient. The third derivative being nonzero clinches it, because then the second derivative, at a, cannot simply graze the x axis, but must cross it confidently. Ergo, the concavity must change.

In the second part, the Third Derivative Test fails. We could have it concave up, as in x^4 at the origin, or concave down, as in -x^4 at the origin, or an inflection point, as in x^5 at the origin. In order to find out what is happening, you have to resort to the full-blown Second Derivative Test, wherein you check the concavity on either side of a, close enough not to have run into another zero of f''.

#2. Your definition of a critical point is certainly correct and is what every calculus book says, but you cannot simply repeat that definition and appreciate the subtle things going on in this situation. Recall that we have the function x^2 with artificially restricted domain [-1,1]. Question: does the two-sided derivative exist at -1, 1? If so, are they critical points? If not, are they critical points? Your statement "if the graph continues in a like manner" is getting warmer.

#4. You have to use implicit differentiation in order to differentiate the arcsin function, at least, the first time you differentiate it. If you know the formula, then of course, in practice, you don't have to re-derive the formula. The way I stated the problem indicates I meant for the person doing the problem to derive that formula from scratch. The point of differentiating is to check your integration to make sure it's correct. That's all. And you can certainly choose not to do so, at the expense of half the credit. Just like I refused to finish your problem, and got my grade knocked down from a 90 to a 67. Hey, it's my problem, right? ;-)]

In Christ.

At 10/13/2006 09:39:00 PM ,  John Dekker said...

The printing press always existed in the mind of God. :) Scientific and technological developments involved thinking God's thoughts after him.

Remember "the invention of the Cross"? The word originally meant "discovery" - they're two sides of the same coin.

At 10/14/2006 03:46:00 PM ,  Susan said...

Dad understands your explanation of #1. Let's just say, it's sort of, kind of clear to me :).

I'm still not sure what you're getting at with #2. Also, does this matter? ;) Or rather, what difference does this distinction make in the realm of calculus? I'm curious.

John, I still think you're really stretching it with your analysis of discovery/invention. Calculus was a fully created, functional, consistent system without any help from man. Man was passive in its creation. Thus it was discovered, not invented by man. But the printing press, while certainly always existing in the mind of God, came about by man's active role in creating it. Thus it was invented, not just discovered.

At 10/14/2006 04:26:00 PM ,  Adrian C. Keister said...

About #2. This distinction is not at all important when it comes to practice. The only situation in which it would come up is in the closed-interval method for maximizing or minimizing a function. And there, you always check the endpoints and the critical points. So whether or not endpoints are critical points doesn't matter, because you check them anyway. However, for theoretical purposes, it is good to have these things down.

The answer to #2 is the following: the two-sided derivative does not exist at endpoints. However, because most of the time, if you increased the interval, the derivative would pop into existence, we say that endpoints are not critical points. But you still check them for the procedure I mentioned above.

I do not think that man was passive in the discovery of calculus. It is usually said that Newton invented it, not discovered it. After all, you must have terms defined (limit, derivative, continuity, integral, etc.), that could have been defined in a different way and still produced a consistent system. Also, we can compare the idea of a printing press with calculus. Because the printing press itself is a physical object, and calculus isn't, we might confuse the topic saying that inventions are always physical objects, and discoveries aren't. However, the idea of a printing press is surely in the same category as calculus, don't you think? After all, a software program is not a physical object, but an idea. Do we say that people discovered that algorithm, or invented it? Or maybe a little of both? Similarly, Niagara Falls is a physical object, but that was surely discovered, not invented! I don't think the lines here are black and white, as if invention is totally its own thing, and discovery another, and never the twain shall meet.

In the end, all I was trying to say was that both the printing press and calculus instituted major changes in what people could do. Now certainly you two, John and Susan, can continue arguing about this distinction all you want, and have fun. I'm merely clarifying what I was trying to say.

Toodles.

In Christ.

At 10/16/2006 07:49:00 AM ,  Susan said...

Okay, I'll give you points for defining terms. That certainly made Newton active to an extent. But then, anything that was discovered must be named, so I'm not sure how significant that is. The printing press could have been designed in any number of ways, but calculus was a previous system that merely had to be discovered bit-by-and named. Technique was necessary, yes, but not from-scratch work. Perhaps I'm overanalyzing, though :).

And John and I are not arguing, Adrian! We're debating. *snooty look*

At 10/16/2006 04:45:00 PM ,  Adrian C. Keister said...

Oh, so now I get graded on the post in which I posted some questions for other people? Isn't that a bit backward? I mean, teachers are inerrant, aren't they?

I think you're over-analyzing. I'm not sure your characteristics of the printing press versus calculus are mutually exclusive. How do we know the printing press wasn't invented bit by bit? I'll bet gradual improvements helped it along a bit. Clearly, a Hewlett Packard printer is better than the Gutenberg press by a long shot.

If you're debating, you're making an argument. If you're making an argument, you're arguing. The word "argument" can be a pretty exact synonym for "debate." Ergo, my use of the word "arguing" was not incorrect. So there! *returns snooty look*

In Christ.

At 10/17/2006 11:05:00 AM ,  Susan said...

Calm down, Adrian! *pats on back* I'm not grading you for your post. I'm giving you a "thumbs-up" or kudos for your comment mentioning Newton's semi-active role in Calculus (defining terms). Sheesh.

Okay, I had a mental lapse (blond moment) with argue v. debate. My apologies. My brain told me you said fight, but of course you didn't. You are right that argue and debate are synonymous; my brain just didn't register the correct term. *sigh* I think this is a rather clever breakdown of different types of verbal engagements.

Okay, I think at this point we're just going around in circles with invention and discovery :). Unless either of you two gentlemen care to continue that discussion, of course. . .

At 10/18/2006 10:46:00 AM ,  Adrian C. Keister said...

Very well. *smooths self-ruffled feathers* ;-)]

Yes, yes, another blond moment. You do tend to have quite a few of those, don't you? I wonder if it's something in your water... Or maybe it just has something to do with the color of your hair. Interestingly, young blond women in DC are assumed (by Democrats, at least) to be conservative power people (people like Ann Coulter). Just thought you'd be interested. :-)]

I think you have mentioned that breakdown of different types of verbal engagments before. I found it helpful.

I can TIOC w.r.t. (that's "with respect to", commonly used math abbr.) the Newton thingy.

Would you be interested in the complete solutions? I can email them to you if you like.

In Christ.

At 10/18/2006 06:10:00 PM ,  Susan said...

Oh, it's definitely the water. Metro Atlanta has a high lead and chlorine content, so I'll blame that.

Oh. Except that I filter my drinking water first usually. Hmmm. The filter! It must be poisonous!

Yes, I would be interested in the complete solutions of the calculus problems :) - an e-mail would be great. Maybe it will revive that part of my brain! However, think "She hasn't taken Calculus in 4 1/2 years as you type them" :). Thanks.

At 10/20/2006 05:06:00 PM ,  Susan said...

Attachment received, though I haven't looked at it yet, except to print it off. Thank you much. Is it all right to TIOC? :)

At 10/23/2006 08:09:00 PM ,  Adrian C. Keister said...

Sure, we can TIOC since we're continuing via email.

In Christ.