110

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
  2. the most readily comprehensible

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.

Michael Hardy
  • 1
  • 30
  • 276
  • 565
  • 3
    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? – 2'5 9'2 Aug 03 '14 at 01:08
  • 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 03 '14 at 03:09
  • @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 03 '14 at 03:35
  • 3
    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 03 '14 at 04:29
  • Does [this](http://math.stackexchange.com/questions/188643/calculus-conjecture) count? – Tunococ Aug 03 '14 at 04:41
  • 1
    @Ishfaaq $\zeta$ at odd integers is a good example, at least to within our current understanding. – Ian Aug 03 '14 at 05:29
  • http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics#Analysis – Martin Brandenburg Aug 05 '14 at 15:30
  • Maybe this question is its own answer? – Hagen von Eitzen Aug 05 '14 at 16:34
  • Despite this question's now having +55 votes, probably at least a half-dozen people have down-voted it. Would they please explain their objections? – Michael Hardy Aug 10 '14 at 17:52
  • Are you sure that the new tag fits? The foundations tag is for foundational questions in mathematics, and this does not seem to be entirely the case here. – Asaf Karagila Feb 20 '16 at 16:48
  • @AsafKaragila : It does not fit and I've deleted it. I actually thought I was editing a different question. $\qquad$ – Michael Hardy Feb 20 '16 at 16:49
  • I wonder what the agenda here is, seems sinister. – marshal craft Dec 26 '17 at 20:54

13 Answers13

87

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

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

zcn
  • 14,794
  • 24
  • 53
  • 13
    +1 - great exaples. Of course, these arguably turn out to be number theory questions in disguise... :-) – Steven Stadnicki Aug 03 '14 at 15:50
  • 5
    Does this have something to do with hills made of flint, or is it named after two people? – Michael Hardy Aug 03 '14 at 18:22
  • 3
    . . . I see: It's named after a town in Kansas. – Michael Hardy Aug 03 '14 at 18:30
  • @MichaelHardy: I don't know where the name originates from - [MathWorld](http://mathworld.wolfram.com/CooksonHillsSeries.html) also calls $\displaystyle \sum_{n=1}^\infty \frac{1}{n^3 \cos^2 n}$ the Cookson Hills series (whose convergence is also unknown). It may make for an interesting source hunt – zcn Aug 03 '14 at 19:23
  • 1
    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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Clifford_A._Pickover). – Jeppe Stig Nielsen Aug 03 '14 at 23:28
  • +1, but I don't think a naive first-year calc student would be at all likely to come up with this series. It takes a *lot* of insight to understand why its convergence is difficult to establish, and why the exponents 3 and 2 work. –  Aug 04 '14 at 14:02
  • 4
    @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 04 '14 at 17:25
  • 1
    Actually, asymptotic limits, being the foundation of both derivative and integral calculus as well as the proof behind a lot of trigonometry, were among the first things my teacher covered in high school precalculus. For instance, the definite integral of $f(x)$ over the domain $\{a,b\}$ is basically the limit of the sum of the areas of rectangles of height $f(x)$ and width $(b-a)/n$ for a number $n$ of $x$-values evenly spaced between $a$ and $b$, as $n$ approaches infinity. – KeithS Aug 05 '14 at 16:15
  • 3
    Do you have any resources or pages you can suggest regarding this particular problem? $$\liminf_{n \to \infty} |n \sin n| = 0$$ – Victor Sep 23 '15 at 21:29
  • @Victor: I proved this limit diverges long ago while trying and failing to tackle the version that doesn't have the extra n in there. – Joshua Feb 23 '17 at 16:26
54

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.

Michael Hardy
  • 1
  • 30
  • 276
  • 565
Steven Stadnicki
  • 50,003
  • 9
  • 78
  • 143
  • 3
    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 03 '14 at 18:28
  • Plus it's easy to motivate consideration of $\gamma$ -- and even prove its existence -- by looking at $\ln n$ as an integral. (+1) – Micah Aug 03 '14 at 23:44
  • 46
    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 03 '14 at 23:56
  • 6
    @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 04 '14 at 07:03
  • @algorithmshark how are the two expressions different? – user1299784 Aug 05 '14 at 02:11
  • 5
    @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 05 '14 at 02:30
  • 1
    @MichaelHardy: Your operator is nothing more than the Euler-Maclaurin series :) – Alexandre C. Aug 05 '14 at 16:52
  • 5
    @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 05 '14 at 17:18
  • @MichaelHardy: Your identity looked a bit familiar but made me curious whether this could be generalized, and I toyed around a bit... with again curious heuristics. See my new question http://math.stackexchange.com/questions/891918/limit-of-differences-of-truncated-series-and-integrals-give-euler-gamma-zeta-an – Gottfried Helms Aug 09 '14 at 10:50
  • @algorithmshark they are the same thing. – Miles Rout Aug 12 '14 at 19:57
45

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.

$\qquad\qquad$ enter image description here

Bill Dubuque
  • 257,588
  • 37
  • 262
  • 861
  • 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 05 '14 at 00:03
  • 7
    @ThorstenS. There is no (recursive) algorithm to decide equality of elements in said class of expressions (see the [Wikipedia page](http://en.wikipedia.org/wiki/Undecidable_problem) on "undecidable problems" for the general notions). – Bill Dubuque Aug 05 '14 at 00:06
  • 2
    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 05 '14 at 16:56
34

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 $\zeta(3)$.

Jair Taylor
  • 14,290
  • 5
  • 32
  • 54
  • 6
    So what's the real problem here? – Martin Brandenburg Aug 05 '14 at 15:26
  • 7
    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 06 '14 at 03:02
  • 2
    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 01:18
24

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.

Alexandre C.
  • 351
  • 2
  • 6
22

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).

Michael Hardy
  • 1
  • 30
  • 276
  • 565
Ross Millikan
  • 362,355
  • 27
  • 241
  • 432
  • 28
    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 03 '14 at 04:18
  • 2
    @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 03 '14 at 04:54
  • 1
    @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 03 '14 at 11:25
17

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.

Daniel R
  • 3,133
  • 3
  • 24
  • 39
  • 12
    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 04 '14 at 21:49
  • 5
    $\ldots$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 04 '14 at 21:50
  • Right, I see your point @Michael. – Daniel R Aug 05 '14 at 06:16
  • @MichaelHardy: That's why calculus should begin with axiomatization of the real numbers and requiring students to prove all the basic properties of real arithmetic using only the axioms and logical inference rules. Then only after that can real real analysis be appreciated. – user21820 Jul 24 '15 at 07:19
  • 1
    @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
  • 1
    $\ldots$ 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
  • @MichaelHardy: It's not futile if done right. Just because many axiomatic presentations are terrible didactic approaches does not mean we cannot find good ones. For a concrete example, using ZFC to do real analysis is a stupid idea, but using the standard real field axiomatization, gently introduced via real world analogies such as movement along a line or pouring water, works. The problem is not that students are incapable of doing mathematics but that we do not teach them what they need to start! [continued] – user21820 Jul 24 '15 at 13:25
  • @user21820 : But they are not willing to learn those things, except sometimes for a period of time too short to reach the punch line. – Michael Hardy Jul 24 '15 at 13:28
  • 1
    @MichaelHardy: [continued] It is different in a CS course, where they are forced to learn to write correct programs because the compiler just will not accept anything syntactically incorrect. The language itself is finite and so students very quickly have a good working grasp of what is syntactically meaningful and what is not. These students also have a much better appreciation of a systematic logical presentation of a proof than the students in the Math courses who have always relied on the teacher to assess their work, who is so inconsistent compared to the compiler. [continued] – user21820 Jul 24 '15 at 13:28
  • 2
    @MichaelHardy: [continued] But it can be done. Our goal in education is **not** to reach any so-called punch line. That is fundamentally flawed, which is why any so-called formal approach to mathematics has failed. Our goal is to teach the students what is meaningful to the real-world, starting with logical reasoning and proceeding on to a language that is powerful enough to express mathematical ideas. It is foolish to attempt to rush through real analysis without first slowly building this foundation in logic and mathematical exploration. After that, all subjects will be an enjoyable breeze. – user21820 Jul 24 '15 at 13:32
12

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.

blue
  • 15,578
  • 35
  • 53
user4894
  • 2,759
  • 1
  • 13
  • 22
  • 9
    It's in the spirit of a different question really - one where "calculus" in this one is replaced by "physics" in another. – blue Aug 03 '14 at 07:40
  • 1
    But this isn't really a mathematical question. It's a question of physics. – Michael Hardy Aug 03 '14 at 18:26
  • 2
    @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 03 '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 04 '14 at 01:36
  • 1
    @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 04 '14 at 02:12
  • @ChanceHudson: The planck time does **not** mean some smallest unit of time. Just as one standard deviation does not at all mean the smallest distance between two data points. A particle can essentially be thought of as literally being some kind of distribution over the entire spacetime, rather than being a point. To measure a distance or time interval we need some measurement device, but the device particles themselves are not points and hence do not give arbitrary precision, even though the wave-functions are continuous and theoretically can take any real values. – user21820 Jul 24 '15 at 07:32
  • @ChanceHudson: Even if this basic quantum mechanical explanation of particles (namely the relativistic Schrodinger equation) turns out to be inaccurate, it is indistinguishable from reality in all experimentally testable situations so far and hence in all those situations it will hold that we cannot make simultaneous measurement of two independent observable quantities more accurately than Planck-scale amounts, regardless of scientific advancement. – user21820 Jul 24 '15 at 07:39
  • Actually, we do know: neither space nor time is a continuum. See the work of Dewey B. Larson for elucidation on this: http://enciklopedia-vortaro-de-la-merk-angla.weebly.com/vorto-dewey-b-larson.html – Mike Jones Mar 04 '16 at 23:59
  • @MikeJones Looks cranky to me. – user4894 Mar 05 '16 at 02:08
  • @user4894: Action-at-a-distance doesn’t seem cranky? – Mike Jones Mar 07 '16 at 03:21
9

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

dwarandae
  • 1,248
  • 1
  • 15
  • 30
  • 7
    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 03 '14 at 15:02
6

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.

Mikhail Katz
  • 35,814
  • 3
  • 58
  • 116
  • 1
    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 03 '14 at 15:45
  • To make that work $dy$ has to be different from the $y$-increment, though. – Mikhail Katz Aug 03 '14 at 15:52
  • 1
    In what sense do we not fully understand why Leibniz's ratio forms work? http://en.wikipedia.org/wiki/Differential_(infinitesimal)#Differentials_as_linear_maps. We can consider $\frac{dy}{dx}$ as a ratio of linear functionals. – nomen Aug 05 '14 at 00:28
  • @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 05 '14 at 11:39
  • 1
    @user72694: In what sense is it a "problem"? – nomen Aug 05 '14 at 14:38
  • @nomen, I see that my post wasn't detailed enough so I will try to elaborate. – Mikhail Katz Aug 05 '14 at 14:43
3

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

Hint: $k$ is rational.

Steven Stadnicki
  • 50,003
  • 9
  • 78
  • 143
Joshua
  • 326
  • 3
  • 11
  • You're right I miswrote it. [Original had square of sin not sin of x^2 for which k = 1 works.] – Joshua Aug 05 '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 05 '14 at 15:40
  • 1
    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 05 '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 05 '14 at 16:02
  • @MichaelHardy See the revised version (with the trigonometric functions being functions of $x^2$, rather than squared themselves) - it's by no means trivial any more ($k=1$ certainly doesn't work, since $\sin(\alpha)+\cos(\alpha)$ is just a scaled and shifted version of $\sin(\alpha)$ and $\int\sin(x^2)\ dx$ is well known to be non-elementary). That said, if it has a known answer - and Joshua suggests it does - then it's not really an answer to the Q as asked. – Steven Stadnicki Aug 05 '14 at 16:20
  • 1
    @StevenStadnicki : OK, that's better. – Michael Hardy Aug 05 '14 at 16:26
  • If you asked the question of the learned and series integration doesn't pan out (I recall something about limited domain of series integration) they won't be able to answer it. It meets the literal definition of the question but possibly not what is really being asked. – Joshua Aug 05 '14 at 16:42
  • 1
    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 05 '14 at 17:57
  • It's probably related to a conjugacy of a similar form. Something having to do with causing the extra terms from a multiply rule cancelling each other leaving only two terms each impossible on their own. – Joshua Aug 05 '14 at 18:55
2

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.

Sandeep Silwal
  • 7,853
  • 1
  • 21
  • 44
0

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).

Jasper
  • 384
  • 1
  • 9
  • 1
    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 05 '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 05 '14 at 20:20
  • 2
    @Jasper No closed-form solution != cannot be numerically approximated to arbitrary accuracy by an algorithm (not necessarily involving table lookups) – Thomas Aug 06 '14 at 00:04
  • "can only be approximated using table look-ups." - I'm sure a typical calculus student, altho unable to evaluate an elliptic integral, can be told to perform the arithmetic-geometric mean iteration by hand. – J. M. ain't a mathematician Aug 13 '19 at 13:43