# In calculus, which questions can the naive ask that the learned cannot answer?

Number theory is known to be a field in which many questions that can be understood by secondary-school pupils have defied the most formidable mathematicians' attempts to answer them.

Calculus is not known to be such a field, as far as I know. (For now, let's just assume this means the basic topics included in the staid and stagnant conventional first-year calculus course.)

What are

1. the most prominent and

questions that can be understood by those who know the concepts taught in first-year calculus and whose solutions are unknown?

I'm not looking for problems that people who know only first-year calculus can solve, but only for questions that they can understand. It would be acceptable to include questions that can be understood only in a somewhat less than logically rigorous way by students at that level.

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Does the Riemann Hypothesis fit? Also, can you offer some kind of loose umbrella definition of what the field of calculus would be? I can see number theory as a field---someone might say "I do research in number theory". But does anyone say "I do research in calculus", rather than, say, in analysis? – alex.jordan Aug 3 '14 at 1:08
I'm a bit hazy on the details, but there was an exercise in Lang's Calculus --- something about the convergence or continuity or differentiability of some power series involving trig functions --- that no one could do, and led to a research publication by the person who eventually solved it. – Gerry Myerson Aug 3 '14 at 3:09
@alex.jordan : I'm referring to those topics included in the conventional first-year calculus course: limits, derivatives, integrals, differential equations, infinite series, geometric questions most neatly dealt with by means of those topics, questions of physics or astronomy or geodesy most neatly so dealt with, areas of science in which differential equations or integrals arise$\ldots\ldots$ ${}\qquad{}$ – Michael Hardy Aug 3 '14 at 3:35
Just a thought.. How about infinite series whose convergence can be proved but whose limits don't have a closed form - or one has not been found yet. Students tend to look at series as just infinite sums and may ask if the sum exists then what is it? I can't remember any good elementary examples though. – Ishfaaq Aug 3 '14 at 4:29
@Ishfaaq $\zeta$ at odd integers is a good example, at least to within our current understanding. – Ian Aug 3 '14 at 5:29

1) Convergence of the Flint Hills series

$$\sum_{n=1}^\infty \frac{1}{n^3 \sin^2 n}$$

is unknown. One can also ask the same question with different exponents - see this paper for more details.

2) Closely related (although $\liminf$ is typically not covered in first year calculus courses, it's too not much of a stretch): whether or not

$$\liminf_{n \to \infty} |n \sin n| = 0$$

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+1 - great exaples. Of course, these arguably turn out to be number theory questions in disguise... :-) – Steven Stadnicki Aug 3 '14 at 15:50
Does this have something to do with hills made of flint, or is it named after two people? – Michael Hardy Aug 3 '14 at 18:22
. . . I see: It's named after a town in Kansas. – Michael Hardy Aug 3 '14 at 18:30
Just for those who do not open the link in this answer, it is clearly said in the paper that the series is named for Flint Hills (a region in Kansas, USA), and it is hinted that the series' name may come from the monograph The Mathematics of Oz: Mental Gymnastics from Beyond the Edge by C. A. Pickover. – Jeppe Stig Nielsen Aug 3 '14 at 23:28
@BenCrowell: I don't think a naive first-year calc student would be at all likely to come up with any of these answers! The intent of the question (as I read it) is that this is an open problem which is easily understood by a first-year student. – zcn Aug 4 '14 at 17:25

While it's certainly got a number-theoretic aspect to it, I'd consider Euler's constant $\gamma = \lim\limits_{n\to\infty}\left(\sum_{k=1}^n\frac1k - \ln n\right)$ to be on-topic for a first-year calculus course (since it features limits, logarithms, and even a fairly simple series), and one of the most fundamental questions one can ask about it — "is this number rational or not?" — is still completely open.

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This is my favorite answer so far, because it can be readily understood at the first-year calculus level and it's a well known hard problem. – Michael Hardy Aug 3 '14 at 18:28
How many years in the electric chair would I be sentenced to if I suggested $\displaystyle\text{writing }\left( \sum_{k=1}^n - \int_1^n dk \right)\frac 1 k\text{ ?}$ ${}\qquad{}$ – Michael Hardy Aug 3 '14 at 23:56
@MichaelHardy Only as many as you'd get for saying "A (less than or equal to) B" when you mean "(A less than B) or (A equal to B)". – algorithmshark Aug 4 '14 at 7:03
@MichaelHardy I would let you off with a warning (in particular, a type mismatch warning). My compiler may not be so nice however. An expression of that type could be well defined. Suppose that you have two operators that take functions and return numbers (the first and second derivative, which we will denote $D_1,D_2$. Then one very often will write operators $D_2-D_1$, and even expression $(D_2-D_1)f$. The main problem with the expression is that $\Sigma$ is an operator on sequences, and $\int$ is an operator on $\mathbb{R}$-valued functions, so you have an implicit type conversion> – Baby Dragon Aug 5 '14 at 2:30
@Baby: that said, if you're willing to have type coercion anywhere in your language, then coercing a function with domain $\mathbb{R}$ to a sequence (i.e. a function with domain $\mathbb{N}$), or in general any subset of the original domain, is one of the less controversial coercions you might perform :-) – Steve Jessop Aug 5 '14 at 17:18

One result that may surprise most calculus students is that there is no algorithm for testing equality of real elementary expressions. This then implies undecidability of other problems, e.g. integration. These are classical results of Daniel Richardson. See below for precise formulations.

$\qquad\qquad$

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Wait, what ? Example: cos(x) is identical to sin(pi/2 - x). And with this small set of compositions (+-* etc.),constants and elementary functions of exp(x), sin(x)... there exist functions f(x) and g(x) where it is impossible to prove that they are identical (f(x) = g(x)) ?! Inconceivable ! – Thorsten S. Aug 5 '14 at 0:03
@ThorstenS. There is no (recursive) algorithm to decide equality of elements in said class of expressions (see the Wikipedia page on "undecidable problems" for the general notions). – Bill Dubuque Aug 5 '14 at 0:06
Wow, I am definitely impressed...+1. – Thorsten S. Aug 5 '14 at 0:07
As a (negative) application, the Risch algorithm, which is supposed to decide whether a function admits a closed-form antiderivative (and computes it if it exsts), cannot decide in general exactly because of this. The proof of its termination relies on one being able to tell whether a given expression is zero. – Alexandre C. Aug 5 '14 at 16:56

A common calculus problem is to prove that the series $1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots$ converges. More difficult is to find the sum:

$$1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}.$$

Much more mysterious is the series:

$$1 + \frac{1}{2^3} + \frac{1}{3^3} + \cdots = 1.2020569\ldots$$

After a few hundred years of searching, it seems unlikely that there is a simple closed form. This number is known as Apéry's constant or just $\zeta(3)$.

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So what's the real problem here? – Martin Brandenburg Aug 5 '14 at 15:26
I suppose the problem I had in mind was finding a closed form, or showing that such a closed form doesn't exist, for some definition of "closed form". For example, does it form an algebraically independent set with $\pi$? It's known to be irrational, but whether it is transcendental or not is open. – Jair Taylor Aug 6 '14 at 3:02
I think the real problem could be "Evaluate $\zeta(3)$." This seems like a great answer: as a question it has the same form as any number of routine problems a freshman calculus student would be assigned. The experts can't answer is....but there is no impossibility proof or proof of non-elementarity in any sense whatsoever. (Unless you think an irrational number is "not elementary".) We simply don't know the answer! – Pete L. Clark May 10 '15 at 1:18

Many indefinite integrals fit this, the poster child being $\int \exp(-x^2)\,dx$ Easy to ask, hard to answer (unless you count the error function as an answer, but in this case that sounds like giving a name to the unknowable).

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I don't think this counts, because we have proven that no elementary antiderivative exists, which is the extent to which this could be "solved". – Ian Aug 3 '14 at 4:18
@Ian: I think the proof is outside first year calculus, while the question is not. So the pro can solve it, but not explain the solution to the first year student. A good point, though. – Ross Millikan Aug 3 '14 at 4:54
@Ian: the question can be (and often is) asked by an undergrad student, but the answer is way beyond the analysis course. There's a paper called 'Basic Calculus' by Ryan Reich that proves the statement, and it requires knowledge of monomials, transcendental functions and differential fields. Many applied mathematicians don't know this stuff. – Alex Aug 3 '14 at 11:25

Is $e+\pi$ an irrational number?

This is a variation of Steven Stadnicki's answer, but without the need to introduce definitions of new constants, which $\gamma$ presumably could be to a first-year calculus student.

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Students might respond to this by saying it has nothing to do with calculus. Many might contemptuously say only an imbecile would think it has anything to do with calculus. 20-year-olds are like that. They say the question of the volume of a sphere has nothing to do with calculus and is far more basic than that, since they learned the formula in 8th grade and it's insulting to them to suggest that they don't have a perfect understanding of everything there is to know about the volume of a sphere. Then if you ask if they know how to prove that it's $(4/3)\pi r^3$ they say$\ldots\ldots$ – Michael Hardy Aug 4 '14 at 21:49
$\ldots$that it never occurred to them that there was such a thing as _proving_ something like that. In other words, to them, something has something to do with calculus only if it's about something they first heard of in a calculus course. They heard of $\pi$ and $e$ earlier. – Michael Hardy Aug 4 '14 at 21:50
Right, I see your point @Michael. – Daniel R Aug 5 '14 at 6:16
@user21820 : Expecting the kinds of students who are typically encouraged to take calculus today to appreciate real analysis is imbecilic. So is expecting that sort of students to understand or appreciate axiomatic foundations of anything. The present system encourages vast numbers of students who are and who ought to remain outsiders to calculus to show up in the course. The sorts of efforts you recommend are wasted and anyone who didn't figure that ought by 1985 has paid no attention. Ask students who've passed the course that you propose what they learn in that course. The will $\ldots$ – Michael Hardy Jul 24 '15 at 13:10
$\ldots$ say they learned a bunch of incomprehensible stuff. That's not something we should try to change. If we insist on teaching calculus to students who are and who ought to remain outsiders to the field of mathematics, we should not try to change the fact that they are outsiders to mathematics. Long experience --- far too long --- tells us that that is futile and immoral. Rather we should teach a different kind of course, suitable to that kind of student. ${}\qquad{}$ – Michael Hardy Jul 24 '15 at 13:10

Some discussion of the three-body problem, especially if your students have followed a physics course where Newtonian mechanics have been taught, can be relevant:

• It is very easy to state (it is a simple ODE)
• It is useful (think space travel)
• Very few things are known about it, if we compare against the two-body problem: no closed form of trajectories (series exist though), new periodic solutions have been discovered in 2012, boundedness of solutions is difficult to assess, etc.

As a bonus, this can be used as a motivation for the need of asymptotic expansions.

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This is one of the two best answers to this question so far, IMO. – Michael Hardy Aug 10 '14 at 17:53

Whether the real numbers and the theory of limits correctly model the physical universe.

In other words, is spacetime infinitely divisible and mathematically complete, ie a continuum? Or is it discrete, in which case calculus is only a continuous approximation but not the literal truth.

This is an ancient problem that nobody knows the answer to. It relates to Zeno's paradoxes.

To be fair, this is a problem of physics, not mathematics. But it's within the spirit of the question.

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It's in the spirit of a different question really - one where "calculus" in this one is replaced by "physics" in another. – blue Aug 3 '14 at 7:40
But this isn't really a mathematical question. It's a question of physics. – Michael Hardy Aug 3 '14 at 18:26
@MichaelHardy I already pled guilty to that charge and threw myself on the mercy of the court :-) But if we're talking about a question that would be asked by a beginning calculus student, I felt I was within scope. – user4894 Aug 3 '14 at 18:42
Doesn't the existence of Planck time mean that time is not a continuum? From the wikipedia page "for times less than one Planck time apart, we can neither measure nor detect any change". Or does it just mean that we aren't sophisticated enough to measure? – Chance Hudson Aug 4 '14 at 1:36
@ChanceHudson I take no position. It is a question that could be asked by a calculus student (perhaps one concurrently taking physics) that indeed the most learned among us cannot answer. It's been asked since antiquity and the verdict's not in. The verdict isn't even remotely close to being in. Where's the physical experiment that could decide the question? We're not going to solve it here. – user4894 Aug 4 '14 at 2:12

Maybe explain why an elementary function might not have an elementary antiderivative, for example.

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The original question was easily stated and understood calculus questions whose answers are unknown. Which elementary functions have elementary antiderivatives is known (Risch's decision procedure) -- although the answer is quite complicated. – murray Aug 3 '14 at 15:02

With all the recent advances in understanding infinitesimals, we still don't fully understand why Leibniz's definition of $\frac{dy}{dx}$ as literally a ratio works the way it does and seems to explain numerous facts including chain rule.

To respond to the comments below, note that Robinson modified Leibniz's approach as follows. Suppose we have a function $y=f(x)$. Let $\Delta x$ be an infinitesimal hyperreal $x$-increment. Consider the corresponding $y$-increment $\Delta y=f(x+\Delta x)-f(x)$. The ratio of hyperreals $\frac{\Delta y}{\Delta x}$ is not quite the derivative. Rather, we must round off the ratio to the nearest real number (its standard part) and so we set $f'(x)=\text{st}\big(\frac{\Delta y}{\Delta x}\big)$. To be consistent with the traditional Leibnizian notation one then defines new variables $dx=\Delta x$ and $dy=f'(x)dx$ so as to get $f'(x)=\frac{dy}{dx}$ but of course here $dy$ is not the $y$-increment corresponding to the $x$-increment. Thus the Leibnizian notation is not made fully operational.

Leibniz himself handled the problem (of which he was certainly aware, contrary to Bishop Berkeley's allegations) by explaining that he was working with a more general relation of equality "up to" negligible terms, in a suitable sense to be determined. Thus if $y=x^2$ then the equality sign in $\frac{dy}{dx}=2x$ does not mean, to Leibniz, exactly what we think it means.

Another approach to $dy=f'(x)dx$ is smooth infinitesimal analysis where infinitesimals are nilsquare so you get equality on the nose though you can't form the ratio. On the other hand, Leibniz worked with arbitrary nonzero orders of infinitesimals $dx^n$ so this doesn't fully capture the Leibnizian framework either.

In an algebraic context one may be able to assign a precise sense to the Leibnizian generalized equality using global considerations, but I personally don't know how to do that precisely.

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I know some people who consider $\frac{dy}{dx}$ to be a literal ratio in the hyperreals, so this isn't as surprising. – Ryan Aug 3 '14 at 15:45
To make that work $dy$ has to be different from the $y$-increment, though. – Mikhail Katz Aug 3 '14 at 15:52
In what sense do we not fully understand why Leibniz's ratio forms work? en.wikipedia.org/wiki/…. We can consider $\frac{dy}{dx}$ as a ratio of linear functionals. – nomen Aug 5 '14 at 0:28
@nomen, this is one way of sweeping the problem under the rug, but it was definitely not what Leibniz meant. To Leibniz $dx$ was an infinitesimal increment. – Mikhail Katz Aug 5 '14 at 11:39
@user72694: In what sense is it a "problem"? – nomen Aug 5 '14 at 14:38

For what number $k$ is $\int\left(\sin(x^2) + k \cos(x^2)\right)\ dx$ an elementary function?

Hint: $k$ is rational.

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You're right I miswrote it. [Original had square of sin not sin of x^2 for which k = 1 works.] – Joshua Aug 5 '14 at 15:31
(Also, this is closer to a puzzle - at least your phrasing implies that there is an answer, but just a tricky one, whereas the OP's intent is along the lines of 'unsolved easy-to-state problems in calculus' akin to the numerous problems along the lines of the Goldbach, twin prime, etc. conjectures that litter number theory) – Steven Stadnicki Aug 5 '14 at 15:40
It's trivially true if $k=1$. If there are non-trivial solutions, I'd have said "numbers" rather than "number". ${}\qquad{}$ – Michael Hardy Aug 5 '14 at 15:45
I was playing with forms like this as a potential crypto-trapdoor function. I lost too many of my notes but there were variants of this where attempting to find any k defeated all book techniques. Much later I learned that series integration (which is beyond me) was applicable. – Joshua Aug 5 '14 at 16:02
So, BTW... what is the answer? :-) At least at first glance I don't see one; any expression of the form $\sin(x^2)+k\cos(x^2)$ is going to be effectively equivalent to (a multiple of) an expression of the form $\sin(x^2+t)$ for some $t$ (as a function of $k$) and I don't see any reason to expect an integral for any function of that form... unless it's related to the conjugacy of $x^2-2$ to the cosine-doubling formula? – Steven Stadnicki Aug 5 '14 at 17:57

What is $\int x^x$ ? Although not an open problem, it is hard to think of the correct response right off the top of your head unless you have seen it before.

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Some interesting integrals do not have closed-form solutions.

For example, elliptic integrals can only be approximated using table look-ups. This makes it impossible to exactly calculate the arc-length of part of an ellipse, or the surface area of part of a planet (as represented by an ellipsoid).

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First, you don't need table look-ups to compute elliptic integrals. There are a variety of methods for computing them. Second, in what sense can you not exactly calculate the arc-length of an ellipse but can exactly calculate the arc-length of a circle? – Dan Piponi Aug 5 '14 at 19:15
Given a single (granted, transcendental) number (pi), I can exactly compute the arc-length of a unit circle from, say, latitude = 0 degrees to latitude = 45 degrees. (This is part of the beauty of circles -- a given length of perimeter corresponds to the same change in the angle perpendicular to the surface, no matter which chunk of perimeter is chosen.) I cannot obtain a closed-form solution to the corresponding arc-length of an ellipse. – Jasper Aug 5 '14 at 20:20
@Jasper No closed-form solution != cannot be numerically approximated to arbitrary accuracy by an algorithm (not necessarily involving table lookups) – Thomas Aug 6 '14 at 0:04

The series $1-1/2 +1/3 -1/4+\cdots=\ln2$

1) what on earth makes this go to $\ln2$ only god knows would a first year rookie say 2) The moment his teacher says that the series can be made to go to any number of your choice wouldn't amuse him either because by now he might be thinking his teacher is a philosopher and even the poor Riemann in all his might ,might not make him believe so, at that point in time.. 3) The fact that a series of cute little rationals goes on to converge to an ugly irrational would be one of the wildest ideas that could come up in his brains..Let alone the fact that group theorists have made a clever assumption in the definition of group in the finiteness part of the closure property..because if t were not so even the RATIONALS would raise their hands and say heyyy there I AM NOT A GROUP...

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Indeed, this is a surprising proposition and any perceptive student could wonder about this. But it is not something that the learned cannot answer. $\qquad$ – Michael Hardy Feb 20 at 17:08
You can prove this without calculus, just starting from the fact that $e^x\ge x+1$ for all $x$. (You need the squeeze theorem at the end for the very last step, though.) – Akiva Weinberger May 9 at 23:17