42 ms·

I liked this simple calculus exercise

496 points by kamaraju 4y ago

ryan-duve 4y ago
This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the answer. Like the words the author uses for the integral, it's all valuable technique.
- tonyarkles 4y ago
  Alternatively, x = sin(theta) and arcsin(sin(theta)) = theta, depending on the limits. It’s a cool technique, thanks for the reminder!
- 77pt77 4y ago
  Dude. That's just a direct application of the inverse function theorem. Even geometrically you can see that swapping the axis (x and y) gives you the desired result.
  - CGamesPlay 4y ago
    GP just happened to be one of the lucky 10,000 on that day in math class. https://xkcd.com/1053/ https://xkcd.com/1053/
- saghm 4y ago
  In turn, this reminds me a bit of a calculus problem I saw freshman year of college that I ended up sharing with my old high school calculus teacher because it similarly looks intractable until you make a "simplification". The problem was to find the integral of `x/(x + 1)` (I forget the exact bounds, presumably from 0 to x with respect to x). The trick is that this the same as `(x + 1 - 1)/(x + 1)`, which you can then split into `(x + 1)/(x + 1) - 1/(x + 1)`, which you can then integrate much more easily.
  - markisus 4y ago
    Why not directly u substitute u = x+1
  - Spivak 4y ago
    I liked these problems because they make you feel clever figuring them out but hated them because they were a mountain of tedious algebra.
- mydogcanpurr 4y ago
  The approach is less mysterious if you recognize it as going back to definitions, which is a common problem solving technique.
- ogogmad 4y ago
  One technique for finding the derivative of the sin function is to find the derivative of arcsin first. The arcsin function can be expressed as the area of a certain figure, and therefore admits an expression as an integral. From this, the derivative of arcsin is immediate. Finally, apply your technique to arcsin(sin(x)) = x and obtain the derivative of sin. A similar technique finds the derivative of exp(x) from ln(x), by defining the latter as the integral of 1/x.
  - selimthegrim 4y ago
    ChatGPT?
    - ogogmad 4y ago
      :P See https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/1230225#1230225 https://math.stackexchange.com/questions/75130/how-to-prove-... The OP uses the derivative of sin to find the derivative of arcsin. My point was that the other way round is sort of natural as well. Thought I'd share. The broader point is that you can obtain all the classic definitions and identities of the elementary functions [1] by combining only (i) the four arithmetic operations, (ii) integration, and (iii) taking inverses of functions (iv) the constants 0 and 1. I guess that's a bit cool. This works fine over the real numbers. Over the complex numbers, this might require Riemann surfaces to work. e^z in complex analysis is usually defined by its Taylor series, instead of as the inverse function of the natural log (whose domain is actually a Riemann surface). [1] - https://en.wikipedia.org/wiki/Elementary_function https://en.wikipedia.org/wiki/Elementary_function
- mjd 4y ago
  This is a great example. Thanks very much!
- justeleblanc 4y ago
  That's how you find any reciprocal function's derivative. If g = f^{-1}, then g'(y) = 1/f'(g(y)).
  - Spivak 4y ago
    By your rule g = arcsin f = sin g'(x) = 1 / cos(arcsin(x)) But that's not the usual form. Can you show that the denominator is sqrt(1 - x^2)? Doing that requires you make the same observation as in the parent post.
    - Tainnor 4y ago
      > Can you show that the denominator is sqrt(1 - x^2)? That's a trig identity that is easily derivable: draw a right triangle with hypotenuse 1 and one side being x. Its one angle will be arcsin(x) by definition. Then by Pythagoras the remaining side is sqrt(1-x^2), which is the cosine of that angle. I guess GP's point was that the implicit differentiation trick by using the inverse is generalisable and not only restricted to trigonometric functions. Obviously, which simplifications you can then additionally make will depend on the function in question.
noobcoder 4y ago
It's easy to fall into the trap of relying on rote memorization of integration rules, but problems like (⋆) force students to truly understand the concepts behind the math.
- mkl 4y ago
  I don't think that's true. Floor is a piecewise function, so you follow the rule for integrating piecewise functions and break it into a sum of integrals of each piece, then follow the rules for those (they're all basically the same, so you don't need to do 1000 of them). You don't need to think about periodic functions at all.
  - chii 4y ago
    > they're all basically the same > You don't need to think about periodic functions at all. except you just described that which is called period...so it's actually good for a student to notice these things, and use the correct term, so that they can associate the name with the idea.
    - mkl 4y ago
      I don't think many would notice doing it that way, as it's disguised by the different limits and constant factors. The indefinite integrals are what is basically the same on the surface. There's more to periodicity than different sections being similar on the surface, and there's certainly no need for them to use the correct terms.
nateb2022 4y ago
Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal. Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?
- monoprotic 4y ago
  > Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2. > And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.
  - nateb2022 4y ago
    Thank you!
- echojc 4y ago
  I think this kind of problem is less about maths and more about how one might approach an unfamiliar problem. Not knowing where to begin is normal. What you're looking to do is to build up the intuition for how you can break down the problem into smaller pieces so that you can investigate its properties. x/2 - floor(x/2) is the natural place to start because it's the smallest independent piece of the equation. Take a couple of minutes to plot this on a graph for a small range of values, like 0 <= x <= 6 (deciding what range to check is also part of your problem solving skillset). With this, you can calculate and sketch out e^(above result) on a graph. Finally, knowing the principle that a definite integral calculates the area under the curve, you should be able to use your sketch to reason out how to calculate the entire original integral. Hopefully you can see how solving this kind of problem isn't about knowing anything about this particular problem, but simply investigating it without any prior expectations, which is why the author thinks this is an interesting exercise for students.
laurieg 4y ago
I did well in high school math. These days, when something involving algebra, trigonometry, geometry etc comes up I feel like I have a good understanding of it but my calculus seems weak to non-existent. I'm not sure if it's how I was taught, how I studied it or something else but calculus always seemed like a huge step change in difficulty. That said, I love how this article gives practical hints on how to replicate the insight and solve the question, rather than just the insight itself.
- xivzgrev 4y ago
  I think it’s a common feeling. Even though they are both “math”, they feel like different skill sets.
- mtlguitarist 4y ago
  I find personally that my math ability is set to approximately 2-3 levels "below" the highest level math I completed, and I've seen hold for others. I have an applied math bachelors so I've taken analysis, dynamical systems, and other high level math classes, but I find that the stuff that I actually remember at a level to pass undergrad exams is up to linear algebra or maybe a little more advanced. Of course if I were to relearn it'd be much faster, but years of being a software engineer have caused me to forget all that stuff.
  - physicles 4y ago
    I was literally about to write this same comment. I did physics so I finished some group theory and complex analysis, but 20 years later all that stuff is gone because I never applied it in other classes. Only the stuff that kept coming up, like calc 1/2 and first order diffeq, really stuck.
    - kaba0 4y ago
      With that said, does anyone know of a good method to relearn math efficiently? I found it to be really hard to self-learn any math topic, most books repeat everything from the basics at the beginning like what a set is, and then suddenly turn into ultra-advanced with “the proof is trivial” all around.
arjie 4y ago
Ha ha, sadly this can be transformed into a symbol manipulation answer as well. I know because this (stated slightly differently) is one of the questions in my 12th standard (senior year high-school equivalent) Mathematics I class. Here's someone writing it out on video on a tutoring site https://www.doubtnut.com/question-answer/int050exdx-where-x-denotes-the-fractional-part-of-x-630440021 https://www.doubtnut.com/question-answer/int050exdx-where-x-... You have to spot the period, but x - floor(x) is called "fractional part of x" where I come from and is a named function which everyone is familiar with. Then, without knowing the area-under-the-curve interpretation, one can blindly apply another symbol-manipulation tool: the summing of integral over a period.
- calf 4y ago
  Floor functions trigger my fear instinct, but at least with this question I could just sit and visualize what the graph looks like, e.g. the basic sawtooth function from first year engineering.
- planede 4y ago
  That's just an algebraic application of the same idea. I don't think it's sad.
navels 4y ago
Reminds me of college when I said to my Real Analysis professor "that's a neat trick". His response: "It's not a trick, it's a method." :-)
- lordnacho 4y ago
  I would classify a trick as something that happens to work but isn't rigorous. Like treating dy/dx as a fraction sometimes works, but only under certain conditions.
  - azalemeth 4y ago
    What's your favourite example of where it doesn't work? Physics is full of quasi-infinistesimal quantities and I always like good counter examples (ideally without invoking something like the blamange function or similar....)
    - lordnacho 4y ago
      Off the top aren't there differential equations that are inseparable? ie you cannot just pretend x and y are x(t) and y(t).
    - deleted 4y ago
      [deleted]
    - contravariant 4y ago
      It basically works without issue in 1d, simply because dx and dy can be considered modular forms and dy is exactly dx times the derivative dy/dx. You can even put an integral sign in front of them and calculate the corresponding integral. Where this doesn't work is if you have more than 1 dimension. Then you need to deal with the added complexity of integrating modular forms and the fact that in 2D you don't have df = (df/dx) dx but df = (df/dx) dx + (df/dy) dx. The chain rule also changes into a matrix product, rather than a simple dz/dx = dy/dx dz/dy.
    - supernewton 4y ago
      Somewhat silly example: given a plane in 3 dimensions defined by x+y+z=1, we have ∂x/∂y * ∂y/∂z * ∂z/∂x = -1.
ajeet_nathawat 4y ago
I loved maths in school and unfortunately didnt pursue maths but solved this problem while sipping coffee and listening to cornfield chase, I realised why I loved maths. the solution is so simple and so intuitive if you solve with graph.
aashutoshrathi 4y ago
We used to get lot of such tricky stuff during the preparation of IIT-JEE here in India, and I'm telling you if you don't understand Area under curve is integral, you can't touch most of the questions. But I get your point, if you are interested in such questions, you should checkout IIT JEE mathematics question, you'll love them
- never_inline 4y ago
  I am from a rural region where IIT-JEE is not that popular, and I liked working these physics and maths problems initially. Unfortunately the competition has become so intense you practically need coaching (which is expensive) and dedicate lot of time, at that point it becomes grunt work. There are many "tricks" and "shortcuts" taught in these coachings which doesn't exists in normal NCERT syllabus. Needless to say I didn't do very well.
lordnacho 4y ago
This is why I enjoyed doing math contests. You always got these problems that illuminated how things actually work, and the answer is always some set of basics applied to elegantly solve it. I'm actually looking for a set of such problems as I think it's a lot better than grinding out hundreds of quadratics or polynomial derivatives and such. I found the AOSP stuff already, wonder if there's other good sources.
- apocadam 4y ago
  Here's one such place - http://www.math.toronto.edu/oz/turgor/archives.php http://www.math.toronto.edu/oz/turgor/archives.php
de6u99er 4y ago
Nowadays I use Wolfram Alpha for stuff like this.
- keithalewis 4y ago
  That and ChatGPT got yer Rightin' an 'Rithmatic covered. Watcha gonna do 'bout yer Readin'? Then we ain't need no schoolin'
  - nh23423fefe 4y ago
    Living in the past suits you.
cperciva 4y ago
First rule of integrating non-analytic functions: If they're analytic everywhere in the interval in question except a finite number of points, split the integral and compute it one analytic segment at a time! (Second rule: If the function is non-analytic at an infinite number of points, you probably still want to compute it one segment at a time, but adding them back together afterwards may get messy.)
j7ake 4y ago
I am a visual person. A good strategy for me is to plot the integrand to gain insight to the problem.
vrglvrglvrgl 4y ago
[dead]
ggrelet 4y ago
What does the [x/2] notation mean, here?
- elseweather 4y ago
  It's floor(_) - as in, floor(1.999) = 0, but floor(2.001) = 2. If you look carefully the upper flange of the [] square brackets is missing, which makes it a floor. https://en.wikipedia.org/wiki/Floor_and_ceiling_functions https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
  - _dain_ 4y ago
    > floor(1.999) = 0 it's 1 not 0
- le_zonzon 4y ago
  It's the floor function https://en.wikipedia.org/wiki/Floor_and_ceiling_functions https://en.wikipedia.org/wiki/Floor_and_ceiling_functions (Had to check the TeX code to figure that out, MathJax let you do that with a right click on the equation.)
- itengelhardt 4y ago
  Thank you for asking this question. I didn't understand the notation either
- mjd 4y ago
  Sorry about this, I added an explanatory note to the article.
bmacho 4y ago
I think this exercise is dumb. Not interesting, not challenging, not useful, not anything. I'd feel offended, if someone actually approached me with it.
lysozyme 4y ago
My introduction to calculus was “Calculus Made Easy” by Silvanus P. Thompson and I always liked math profs who actively worked to show math for what it is: useful, beautiful but not about the symbols or the jargon. “Any fool can calculate!” I think is what he says in the book. I did some math in college and when I started knowing how to analyze the behavior of functions (and developing the mental math tools to imagine what they look like without having to actually draw them) that’s when I felt like I was kinda getting it
- signa11 4y ago
  i loved that book. unpopular opinion: martin-gardner's intro really ruined the start of the book (for me at least). i just ignored all of it, and was a happy camper. other than that, i.a.maron, piskunov, g.n.berman are all heavy but excellent texts on this beautiful subject.
tachyon5 4y ago
Great post! It really drives home the point that understanding the core concepts in calculus is way more important than just memorising formulas and mechanically applying them. The example problem shows how visualising and breaking down a seemingly complex integral can actually reveal its simpler underlying structure. This reminds me of the need to be adaptable and versatile when tackling math problems, since relying solely on known techniques can limit your ability to solve more complex or unfamiliar problems. Educators should help students focus on developing a deep understanding of math concepts and honing problem-solving skills, rather than just bogging them down in calculations.
einpoklum 4y ago
> If some expression looks complicated, try graphing it and see if you get any insight into how it behaves. This is not always a good idea. Some functions have complicated behavior that makes them either plain hard to draw (e.g. sin(1/x) near 0), or reach very high values but also be near 0, or be otherwise tricky.
- mjd 4y ago
  There is nothing that is always a good idea.
red_admiral 4y ago
Here's another exercise (resp. exam question) that tests understanding: given a sketch of a curve in a graph, roughly sketch the derivative (or integral). The number of otherwise good students who go "but I can't do the derivative without the formula?" suggests we need more questions like this.
- xnorswap 4y ago
  Also a good question to test intuition is being asked to sketch e^sin(x) and sin(e^x).
  - bheadmaster 4y ago
    Intuition tells me e^sin(x) would look similar to an ordinary sinusiod, except its range would be between e^-1 and e, and its shape would not be smooth as a sinusiod. I have no idea what the shape would look like, and I'm a visual learner when it comes to mathematics. I think most of these questions are not measuring intuition per se, but rather has the tested person previously seen such functions plotted on a graph. Either that, or my mathematical intuition has got rusty from years of code monkeying.
    - soVeryTired 4y ago
      It wouldn’t look too dissimilar to a standard sinusoid, just shifted upwards a bit. In the range [-1, 1], e^x is reasonably similar to 1 + x.
    - madcaptenor 4y ago
      You're basically right. Here's the graph: https://www.desmos.com/calculator/lf3xerswdy https://www.desmos.com/calculator/lf3xerswdy
      - bheadmaster 4y ago
        Ah, now it makes sense. The slope of the sine is "squashed" down by the x->e^x mapping. Perhaps (regularly) seeing functions plotted on graphs is a necessary precondition to maintain intuition :)
physicles 4y ago
When I saw the equation referred to as (*), I had a flashback to those problem sets with *hard and **harder problems. ** problems often required some real out-of-the-box thinking. I wasn't always able to solve those, but it was so satisfying when I did (usually after an hour or two of struggle).
- mjd 4y ago
  I just love that little five-pointed star symbol. I also like labeling formulas with playing card suits. I think they are easier for the reader to distinguish when they are looking back for the labeled formula. A while back I had the idea of marking erroneous formulas with a red spade. I'm going to try doing that again because it's hilarious.
kolbe 4y ago
This would make for a good mental math interview problem. Thanks.
postsantum 4y ago
Slightly off-topic: didn't know what ⌊x/2⌋ is Google: x squared (???) GPT: The expression ⌊x/2⌋ represents the greatest integer that is less than or equal to x/2. It is called the floor function of x/2. For example, if x=5, then ⌊x/2⌋ = ⌊5/2⌋ = 2. If x is an even integer, then ⌊x/2⌋ = x/2. If x is an odd integer, then ⌊x/2⌋ = (x-1)/2.
- bluepod4 4y ago
  I was aware of the floor function (and the corresponding ceiling function) since I’m a software engineer. But I wasn’t aware that you could graph it. It never came up in high school or college math. And I never thought about it. Of course, it makes sense now that I’ve seen it.
  - wholinator2 4y ago
    It's also sometimes referred to as a step-function though i believe that name encompasses many more functions than just the floor
    - hgsgm 4y ago
      It's piecewise-linear and periodic. Also called sawtooth. Step functions are piecewise constant. It would not be fun to climb steps of this shape.
    - bluepod4 4y ago
      Yeah, in college I’ve worked with step/unit functions in both my “differential equations” and “signal and systems” classes for sure. But it never occurred to me and it was never presented by professors or TAs that they were essentially the same as floor/ceiling functions. Or maybe I just forgot since it’s been more than a decade?
  - hgsgm 4y ago
    Can you name a single-variable function that software engineers use that can't be graphed? How would such a function be useful?
privacyking 4y ago
That technique won't scale if you're trying to integrate some crazy complex function.
- eesmith 4y ago
  A "calculus tyro", as a general rule, is not trying to integrate some crazy complex function as part of a homework assignment.
- xigoi 4y ago
  Integrating a piecewise continuous function by splitting it is used all the time.
- justeleblanc 4y ago
  Exam questions have two wonderful property: they have an answer, and the person who wrote it knows that you have the knowledge to find the answer.
  - mrguyorama 4y ago
    Nope, good teachers understand when the question they wrote thinking to test certain concepts falls flat, because of some unforeseen oversight or quirk. Something being clear and intuitive to a math professor is not the same as something being clear and intuitive to the students. The best professors strike and ignore questions that "failed", ie didn't accurately test what they thought they would.
    - justeleblanc 4y ago
      I don't quite see how that's relevant to what I said.
      - mrguyorama 4y ago
        >the person who wrote it knows that you have the knowledge to find the answer. This isn't always true. The person who wrote it only knows what they taught, not what you have learned. The point of the test is to find what you have learned. This is just a nitpick I think.
aqme28 4y ago
As with so many integration exercises, this seems more like an exercise not in integrating, but in your understanding of the relevant functions.
ziroshima 4y ago
This is surely a stupid question: In the article, the graph sure looks like a right triangle, with a base of 2 and a height of 1. Wouldn't the area under this curve (from 0-2) be ~1?
- mabbo 4y ago
  That's a graph of `x/2 - floor(x/2)`. The question is about `e^(x/2 - floor(x/2))`.
- dhosek 4y ago
  Ah, but that function is the input to another function and it’s that other function that’s getting integrated.
ketzu 4y ago
The thing I dislike about many maths problems (including many proposed in this thread) is taking the wrong initial approach can make it take forever. Finding the right trick to solve something can feel enlightening, but in my experience it feels mostly frustrating if you waste 10x the time by taking one wrong step in the beginning. After watching Michael Penns youtube channel [1] for some time now, and he loves the floor function, I recognized what was going on - and wondered how I could prove this is 1000 times the simple function beyond just stating it. [1] https://www.youtube.com/@MichaelPennMath https://www.youtube.com/@MichaelPennMath

496 points by kamaraju 4y ago

This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the answer. Like the words the author uses for the integral, it's all valuable technique.

tonyarkles 4y ago
Alternatively, x = sin(theta) and arcsin(sin(theta)) = theta, depending on the limits. It’s a cool technique, thanks for the reminder!
77pt77 4y ago
Dude. That's just a direct application of the inverse function theorem. Even geometrically you can see that swapping the axis (x and y) gives you the desired result.
- CGamesPlay 4y ago
  GP just happened to be one of the lucky 10,000 on that day in math class. https://xkcd.com/1053/ https://xkcd.com/1053/
saghm 4y ago
In turn, this reminds me a bit of a calculus problem I saw freshman year of college that I ended up sharing with my old high school calculus teacher because it similarly looks intractable until you make a "simplification". The problem was to find the integral of `x/(x + 1)` (I forget the exact bounds, presumably from 0 to x with respect to x). The trick is that this the same as `(x + 1 - 1)/(x + 1)`, which you can then split into `(x + 1)/(x + 1) - 1/(x + 1)`, which you can then integrate much more easily.
- markisus 4y ago
  Why not directly u substitute u = x+1
- Spivak 4y ago
  I liked these problems because they make you feel clever figuring them out but hated them because they were a mountain of tedious algebra.
mydogcanpurr 4y ago
The approach is less mysterious if you recognize it as going back to definitions, which is a common problem solving technique.
ogogmad 4y ago
One technique for finding the derivative of the sin function is to find the derivative of arcsin first. The arcsin function can be expressed as the area of a certain figure, and therefore admits an expression as an integral. From this, the derivative of arcsin is immediate. Finally, apply your technique to arcsin(sin(x)) = x and obtain the derivative of sin. A similar technique finds the derivative of exp(x) from ln(x), by defining the latter as the integral of 1/x.
- selimthegrim 4y ago
  ChatGPT?
  - ogogmad 4y ago
    :P See https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/1230225#1230225 https://math.stackexchange.com/questions/75130/how-to-prove-... The OP uses the derivative of sin to find the derivative of arcsin. My point was that the other way round is sort of natural as well. Thought I'd share. The broader point is that you can obtain all the classic definitions and identities of the elementary functions [1] by combining only (i) the four arithmetic operations, (ii) integration, and (iii) taking inverses of functions (iv) the constants 0 and 1. I guess that's a bit cool. This works fine over the real numbers. Over the complex numbers, this might require Riemann surfaces to work. e^z in complex analysis is usually defined by its Taylor series, instead of as the inverse function of the natural log (whose domain is actually a Riemann surface). [1] - https://en.wikipedia.org/wiki/Elementary_function https://en.wikipedia.org/wiki/Elementary_function
mjd 4y ago
This is a great example. Thanks very much!
justeleblanc 4y ago
That's how you find any reciprocal function's derivative. If g = f^{-1}, then g'(y) = 1/f'(g(y)).
- Spivak 4y ago
  By your rule g = arcsin f = sin g'(x) = 1 / cos(arcsin(x)) But that's not the usual form. Can you show that the denominator is sqrt(1 - x^2)? Doing that requires you make the same observation as in the parent post.
  - Tainnor 4y ago
    > Can you show that the denominator is sqrt(1 - x^2)? That's a trig identity that is easily derivable: draw a right triangle with hypotenuse 1 and one side being x. Its one angle will be arcsin(x) by definition. Then by Pythagoras the remaining side is sqrt(1-x^2), which is the cosine of that angle. I guess GP's point was that the implicit differentiation trick by using the inverse is generalisable and not only restricted to trigonometric functions. Obviously, which simplifications you can then additionally make will depend on the function in question.

noobcoder 4y ago

It's easy to fall into the trap of relying on rote memorization of integration rules, but problems like (⋆) force students to truly understand the concepts behind the math.

mkl 4y ago
I don't think that's true. Floor is a piecewise function, so you follow the rule for integrating piecewise functions and break it into a sum of integrals of each piece, then follow the rules for those (they're all basically the same, so you don't need to do 1000 of them). You don't need to think about periodic functions at all.
- chii 4y ago
  > they're all basically the same > You don't need to think about periodic functions at all. except you just described that which is called period...so it's actually good for a student to notice these things, and use the correct term, so that they can associate the name with the idea.
  - mkl 4y ago
    I don't think many would notice doing it that way, as it's disguised by the different limits and constant factors. The indefinite integrals are what is basically the same on the surface. There's more to periodicity than different sections being similar on the surface, and there's certainly no need for them to use the correct terms.

nateb2022 4y ago

Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal. Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?

monoprotic 4y ago
> Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2. > And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.
- nateb2022 4y ago
  Thank you!
echojc 4y ago
I think this kind of problem is less about maths and more about how one might approach an unfamiliar problem. Not knowing where to begin is normal. What you're looking to do is to build up the intuition for how you can break down the problem into smaller pieces so that you can investigate its properties. x/2 - floor(x/2) is the natural place to start because it's the smallest independent piece of the equation. Take a couple of minutes to plot this on a graph for a small range of values, like 0 <= x <= 6 (deciding what range to check is also part of your problem solving skillset). With this, you can calculate and sketch out e^(above result) on a graph. Finally, knowing the principle that a definite integral calculates the area under the curve, you should be able to use your sketch to reason out how to calculate the entire original integral. Hopefully you can see how solving this kind of problem isn't about knowing anything about this particular problem, but simply investigating it without any prior expectations, which is why the author thinks this is an interesting exercise for students.

laurieg 4y ago

I did well in high school math. These days, when something involving algebra, trigonometry, geometry etc comes up I feel like I have a good understanding of it but my calculus seems weak to non-existent. I'm not sure if it's how I was taught, how I studied it or something else but calculus always seemed like a huge step change in difficulty. That said, I love how this article gives practical hints on how to replicate the insight and solve the question, rather than just the insight itself.

xivzgrev 4y ago
I think it’s a common feeling. Even though they are both “math”, they feel like different skill sets.
mtlguitarist 4y ago
I find personally that my math ability is set to approximately 2-3 levels "below" the highest level math I completed, and I've seen hold for others. I have an applied math bachelors so I've taken analysis, dynamical systems, and other high level math classes, but I find that the stuff that I actually remember at a level to pass undergrad exams is up to linear algebra or maybe a little more advanced. Of course if I were to relearn it'd be much faster, but years of being a software engineer have caused me to forget all that stuff.
- physicles 4y ago
  I was literally about to write this same comment. I did physics so I finished some group theory and complex analysis, but 20 years later all that stuff is gone because I never applied it in other classes. Only the stuff that kept coming up, like calc 1/2 and first order diffeq, really stuck.
  - kaba0 4y ago
    With that said, does anyone know of a good method to relearn math efficiently? I found it to be really hard to self-learn any math topic, most books repeat everything from the basics at the beginning like what a set is, and then suddenly turn into ultra-advanced with “the proof is trivial” all around.

arjie 4y ago

Ha ha, sadly this can be transformed into a symbol manipulation answer as well. I know because this (stated slightly differently) is one of the questions in my 12th standard (senior year high-school equivalent) Mathematics I class. Here's someone writing it out on video on a tutoring site https://www.doubtnut.com/question-answer/int050exdx-where-x-denotes-the-fractional-part-of-x-630440021 https://www.doubtnut.com/question-answer/int050exdx-where-x-... You have to spot the period, but x - floor(x) is called "fractional part of x" where I come from and is a named function which everyone is familiar with. Then, without knowing the area-under-the-curve interpretation, one can blindly apply another symbol-manipulation tool: the summing of integral over a period.

calf 4y ago
Floor functions trigger my fear instinct, but at least with this question I could just sit and visualize what the graph looks like, e.g. the basic sawtooth function from first year engineering.
planede 4y ago
That's just an algebraic application of the same idea. I don't think it's sad.

navels 4y ago

Reminds me of college when I said to my Real Analysis professor "that's a neat trick". His response: "It's not a trick, it's a method." :-)

lordnacho 4y ago
I would classify a trick as something that happens to work but isn't rigorous. Like treating dy/dx as a fraction sometimes works, but only under certain conditions.
- azalemeth 4y ago
  What's your favourite example of where it doesn't work? Physics is full of quasi-infinistesimal quantities and I always like good counter examples (ideally without invoking something like the blamange function or similar....)
  - lordnacho 4y ago
    Off the top aren't there differential equations that are inseparable? ie you cannot just pretend x and y are x(t) and y(t).
  - deleted 4y ago
    [deleted]
  - contravariant 4y ago
    It basically works without issue in 1d, simply because dx and dy can be considered modular forms and dy is exactly dx times the derivative dy/dx. You can even put an integral sign in front of them and calculate the corresponding integral. Where this doesn't work is if you have more than 1 dimension. Then you need to deal with the added complexity of integrating modular forms and the fact that in 2D you don't have df = (df/dx) dx but df = (df/dx) dx + (df/dy) dx. The chain rule also changes into a matrix product, rather than a simple dz/dx = dy/dx dz/dy.
  - supernewton 4y ago
    Somewhat silly example: given a plane in 3 dimensions defined by x+y+z=1, we have ∂x/∂y * ∂y/∂z * ∂z/∂x = -1.

ajeet_nathawat 4y ago

I loved maths in school and unfortunately didnt pursue maths but solved this problem while sipping coffee and listening to cornfield chase, I realised why I loved maths. the solution is so simple and so intuitive if you solve with graph.

aashutoshrathi 4y ago

We used to get lot of such tricky stuff during the preparation of IIT-JEE here in India, and I'm telling you if you don't understand Area under curve is integral, you can't touch most of the questions. But I get your point, if you are interested in such questions, you should checkout IIT JEE mathematics question, you'll love them

never_inline 4y ago
I am from a rural region where IIT-JEE is not that popular, and I liked working these physics and maths problems initially. Unfortunately the competition has become so intense you practically need coaching (which is expensive) and dedicate lot of time, at that point it becomes grunt work. There are many "tricks" and "shortcuts" taught in these coachings which doesn't exists in normal NCERT syllabus. Needless to say I didn't do very well.

lordnacho 4y ago

This is why I enjoyed doing math contests. You always got these problems that illuminated how things actually work, and the answer is always some set of basics applied to elegantly solve it. I'm actually looking for a set of such problems as I think it's a lot better than grinding out hundreds of quadratics or polynomial derivatives and such. I found the AOSP stuff already, wonder if there's other good sources.

apocadam 4y ago
Here's one such place - http://www.math.toronto.edu/oz/turgor/archives.php http://www.math.toronto.edu/oz/turgor/archives.php

de6u99er 4y ago

Nowadays I use Wolfram Alpha for stuff like this.

keithalewis 4y ago
That and ChatGPT got yer Rightin' an 'Rithmatic covered. Watcha gonna do 'bout yer Readin'? Then we ain't need no schoolin'
- nh23423fefe 4y ago
  Living in the past suits you.

cperciva 4y ago

First rule of integrating non-analytic functions: If they're analytic everywhere in the interval in question except a finite number of points, split the integral and compute it one analytic segment at a time! (Second rule: If the function is non-analytic at an infinite number of points, you probably still want to compute it one segment at a time, but adding them back together afterwards may get messy.)

j7ake 4y ago

I am a visual person. A good strategy for me is to plot the integrand to gain insight to the problem.

vrglvrglvrgl 4y ago

[dead]

ggrelet 4y ago

What does the [x/2] notation mean, here?

elseweather 4y ago
It's floor(_) - as in, floor(1.999) = 0, but floor(2.001) = 2. If you look carefully the upper flange of the [] square brackets is missing, which makes it a floor. https://en.wikipedia.org/wiki/Floor_and_ceiling_functions https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
- _dain_ 4y ago
  > floor(1.999) = 0 it's 1 not 0
le_zonzon 4y ago
It's the floor function https://en.wikipedia.org/wiki/Floor_and_ceiling_functions https://en.wikipedia.org/wiki/Floor_and_ceiling_functions (Had to check the TeX code to figure that out, MathJax let you do that with a right click on the equation.)
itengelhardt 4y ago
Thank you for asking this question. I didn't understand the notation either
mjd 4y ago
Sorry about this, I added an explanatory note to the article.

bmacho 4y ago

I think this exercise is dumb. Not interesting, not challenging, not useful, not anything. I'd feel offended, if someone actually approached me with it.

lysozyme 4y ago

My introduction to calculus was “Calculus Made Easy” by Silvanus P. Thompson and I always liked math profs who actively worked to show math for what it is: useful, beautiful but not about the symbols or the jargon. “Any fool can calculate!” I think is what he says in the book. I did some math in college and when I started knowing how to analyze the behavior of functions (and developing the mental math tools to imagine what they look like without having to actually draw them) that’s when I felt like I was kinda getting it

signa11 4y ago
i loved that book. unpopular opinion: martin-gardner's intro really ruined the start of the book (for me at least). i just ignored all of it, and was a happy camper. other than that, i.a.maron, piskunov, g.n.berman are all heavy but excellent texts on this beautiful subject.

tachyon5 4y ago

Great post! It really drives home the point that understanding the core concepts in calculus is way more important than just memorising formulas and mechanically applying them. The example problem shows how visualising and breaking down a seemingly complex integral can actually reveal its simpler underlying structure. This reminds me of the need to be adaptable and versatile when tackling math problems, since relying solely on known techniques can limit your ability to solve more complex or unfamiliar problems. Educators should help students focus on developing a deep understanding of math concepts and honing problem-solving skills, rather than just bogging them down in calculations.

einpoklum 4y ago

> If some expression looks complicated, try graphing it and see if you get any insight into how it behaves. This is not always a good idea. Some functions have complicated behavior that makes them either plain hard to draw (e.g. sin(1/x) near 0), or reach very high values but also be near 0, or be otherwise tricky.

mjd 4y ago
There is nothing that is always a good idea.

red_admiral 4y ago

Here's another exercise (resp. exam question) that tests understanding: given a sketch of a curve in a graph, roughly sketch the derivative (or integral). The number of otherwise good students who go "but I can't do the derivative without the formula?" suggests we need more questions like this.

xnorswap 4y ago
Also a good question to test intuition is being asked to sketch e^sin(x) and sin(e^x).
- bheadmaster 4y ago
  Intuition tells me e^sin(x) would look similar to an ordinary sinusiod, except its range would be between e^-1 and e, and its shape would not be smooth as a sinusiod. I have no idea what the shape would look like, and I'm a visual learner when it comes to mathematics. I think most of these questions are not measuring intuition per se, but rather has the tested person previously seen such functions plotted on a graph. Either that, or my mathematical intuition has got rusty from years of code monkeying.
  - soVeryTired 4y ago
    It wouldn’t look too dissimilar to a standard sinusoid, just shifted upwards a bit. In the range [-1, 1], e^x is reasonably similar to 1 + x.
  - madcaptenor 4y ago
    You're basically right. Here's the graph: https://www.desmos.com/calculator/lf3xerswdy https://www.desmos.com/calculator/lf3xerswdy
    - bheadmaster 4y ago
      Ah, now it makes sense. The slope of the sine is "squashed" down by the x->e^x mapping. Perhaps (regularly) seeing functions plotted on graphs is a necessary precondition to maintain intuition :)

physicles 4y ago

When I saw the equation referred to as (*), I had a flashback to those problem sets with *hard and **harder problems. ** problems often required some real out-of-the-box thinking. I wasn't always able to solve those, but it was so satisfying when I did (usually after an hour or two of struggle).

mjd 4y ago
I just love that little five-pointed star symbol. I also like labeling formulas with playing card suits. I think they are easier for the reader to distinguish when they are looking back for the labeled formula. A while back I had the idea of marking erroneous formulas with a red spade. I'm going to try doing that again because it's hilarious.

kolbe 4y ago

This would make for a good mental math interview problem. Thanks.

postsantum 4y ago

Slightly off-topic: didn't know what ⌊x/2⌋ is Google: x squared (???) GPT: The expression ⌊x/2⌋ represents the greatest integer that is less than or equal to x/2. It is called the floor function of x/2. For example, if x=5, then ⌊x/2⌋ = ⌊5/2⌋ = 2. If x is an even integer, then ⌊x/2⌋ = x/2. If x is an odd integer, then ⌊x/2⌋ = (x-1)/2.

bluepod4 4y ago
I was aware of the floor function (and the corresponding ceiling function) since I’m a software engineer. But I wasn’t aware that you could graph it. It never came up in high school or college math. And I never thought about it. Of course, it makes sense now that I’ve seen it.
- wholinator2 4y ago
  It's also sometimes referred to as a step-function though i believe that name encompasses many more functions than just the floor
  - hgsgm 4y ago
    It's piecewise-linear and periodic. Also called sawtooth. Step functions are piecewise constant. It would not be fun to climb steps of this shape.
  - bluepod4 4y ago
    Yeah, in college I’ve worked with step/unit functions in both my “differential equations” and “signal and systems” classes for sure. But it never occurred to me and it was never presented by professors or TAs that they were essentially the same as floor/ceiling functions. Or maybe I just forgot since it’s been more than a decade?
- hgsgm 4y ago
  Can you name a single-variable function that software engineers use that can't be graphed? How would such a function be useful?

privacyking 4y ago

That technique won't scale if you're trying to integrate some crazy complex function.

eesmith 4y ago
A "calculus tyro", as a general rule, is not trying to integrate some crazy complex function as part of a homework assignment.
xigoi 4y ago
Integrating a piecewise continuous function by splitting it is used all the time.
justeleblanc 4y ago
Exam questions have two wonderful property: they have an answer, and the person who wrote it knows that you have the knowledge to find the answer.
- mrguyorama 4y ago
  Nope, good teachers understand when the question they wrote thinking to test certain concepts falls flat, because of some unforeseen oversight or quirk. Something being clear and intuitive to a math professor is not the same as something being clear and intuitive to the students. The best professors strike and ignore questions that "failed", ie didn't accurately test what they thought they would.
  - justeleblanc 4y ago
    I don't quite see how that's relevant to what I said.
    - mrguyorama 4y ago
      >the person who wrote it knows that you have the knowledge to find the answer. This isn't always true. The person who wrote it only knows what they taught, not what you have learned. The point of the test is to find what you have learned. This is just a nitpick I think.

aqme28 4y ago

As with so many integration exercises, this seems more like an exercise not in integrating, but in your understanding of the relevant functions.

ziroshima 4y ago

This is surely a stupid question: In the article, the graph sure looks like a right triangle, with a base of 2 and a height of 1. Wouldn't the area under this curve (from 0-2) be ~1?

mabbo 4y ago
That's a graph of `x/2 - floor(x/2)`. The question is about `e^(x/2 - floor(x/2))`.
dhosek 4y ago
Ah, but that function is the input to another function and it’s that other function that’s getting integrated.

ketzu 4y ago

The thing I dislike about many maths problems (including many proposed in this thread) is taking the wrong initial approach can make it take forever. Finding the right trick to solve something can feel enlightening, but in my experience it feels mostly frustrating if you waste 10x the time by taking one wrong step in the beginning. After watching Michael Penns youtube channel [1] for some time now, and he loves the floor function, I recognized what was going on - and wondered how I could prove this is 1000 times the simple function beyond just stating it. [1] https://www.youtube.com/@MichaelPennMath https://www.youtube.com/@MichaelPennMath