5 ms·
Fundamental Theorem of Calculus
- bikrampanda 6mo agoWhat is the font used on the site?
- genezeta 6mo agoAlegreya
- crispyambulance 6mo agoThat font, and how it's integrated with the math looks amazing. Katex for the math?
- viscousviolin 6mo agoSeems like Katex from the scripts getting loaded. I love the design too, kinda medieval-chic.
- eru 6mo agoLooks more early modern to me. :)
- viscousviolin 6mo agoToday I learned there's a CSS property for styling the first letter of a paragraph, neat. (https://css-tricks.com/almanac/properties/i/initial-letter/ https://css-tricks.com/almanac/properties/i/initial-letter/) --edit: The font used for those initials is called Goudy Initialen: https://www.dafont.com/goudy-initialen.font https://www.dafont.com/goudy-initialen.font
- quibono 6mo agoI love this -- I'll have to do something like that for my site. I always liked the big initials on the start of a paragraph. Though it feels a bit more prose-applicable than for non-fiction writing.
- emmelaich 6mo agoInspection suggests "https://online-fonts.com/fonts/alegreya https://online-fonts.com/fonts/alegreya"
- thaumasiotes 6mo agoAlso https://online-fonts.com/fonts/goudy-initialen https://online-fonts.com/fonts/goudy-initialen
- dalvrosa 6mo agoAlready replied :) The source code of the website is open if you wanna check it out!
- bikrampanda 6mo agoThanks. Your website looks really nice!
- dalvrosa 6mo agoGlad that you like it :)
- WCSTombs 6mo agoTechnically "it depends on the browser settings," but the body font Alegreya is served directly by the site, so I think it would be the one used in almost all cases. The math fonts used in the formulas are just the ones provided by KaTeX, which I think are just TeX's default math fonts.
- mellosouls 6mo agohttps://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculu...
- emacdona 6mo ago> f is Riemann integrable iff it is bounded and continuous almost everywhere. FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities". I like that characterization b/c it seems more precise than "almost everywhere", but I've heard both. I mention that because when I read the first footnote, I thought this was a mistake: > boundedness alone ensures the subinterval infima and suprema are finite. But it wasn't. It does, in fact, insure that infima and suprema are finite. It just does NOT ensure that it is Riemann integrable (which, of course the last paragraph in the first section mentions). Thanks for posting. This was a fun diversion down memory lane whilst having my morning coffee. If anyone wants a rabbit hole to go down: Think about why the Dirichlet function [1], which is bounded -- and therefore has upper and lower sums -- is not Riemann integrable (hint: its upper and lower sums don't converge. why?) Then, if you want to keep going down the rabbit hole, learn how you _can_ integrate it (ie: how you _can_ assign a number to the area it bounds) [2] [1] One of my favorite functions. It seems its purpose in life is to serve as a counter example. https://en.wikipedia.org/wiki/Dirichlet_function https://en.wikipedia.org/wiki/Dirichlet_function [2] https://en.wikipedia.org/wiki/Lebesgue_integral https://en.wikipedia.org/wiki/Lebesgue_integral
- mjdv 6mo ago> FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities". It is not: for example, the piece-wise constant function f: [0,1] -> [0,1] which starts at f(0) = 0, stays constant until suddenly f(1/2) = 1, until f(3/4) = 0, until f(7/8) = 1, etc. is Riemann integrable. "Continuous almost everywhere" means that the set of its discontinuities has Lebesgue measure 0. Many infinite sets have Lebesgue measure 0, including all countable sets.
- emacdona 6mo agoAh, thanks for the clarification! Would it have been accurate then to have said: "iff it is bounded and has countable discontinuities"? Or, are there some uncountable sets which also have Lebesgue measure 0?
- 6mo ago
- mchinen 6mo agoI've studied the proofs before but there's still something mystical and unintuitive for me about the area under an entire curve being related to the derivative at only two points, especially for wobbly non monotonic functions. I feel similar about the trace of a matrix being equal to the sum of eigenvalues. Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.
- sambapa 6mo agoYou meant antiderivative?
- ironSkillet 6mo agoIt is not determined by the derivative, it's the antiderivative, as someone else mentioned. The derivative is the rate of change of a function. The "area under a curve" of the graph of a function measures how much the function is "accumulating", which is intuitively a sum of rates of change (taken to an infinitesimal limit).
- dalvrosa 6mo agoThanks for bringing some intuition!
- magicalhippo 6mo agoThe antiderivative at x is defined as the area under the curve from 0 to x, which the Riemann sum gives a nice intuition for how you can get from the derivative. So to get the area under the curve between a and b, you calculate the area under the curve from 0 to b (antiderivative at b) and subtract the area under the curve from 0 to a (antiderivative at a). At least that's my sleep deprived take.
- 1980phipsi 6mo agoI took calculus in high school and college, and I don't think any of my instructors explained the intuition as well. So sleep-deprived or not, it's a great one!
- deleted 6mo ago[deleted]
- shmoil 6mo agoGood job, David. Have a lollipop. Now learn & write up the proof that the Henstock-Kurzweil integral integrates _every_ derivative. This is what we had in my calculus class on top of the outdated Riemann integral.
- random3 6mo agoThat how you got the taste for lollipops?
- deleted 6mo ago[deleted]
- EdwardDiego 6mo ago> This post introduces the Riemann integral Sweet! I'm keen to learn about the basic fundamentals of calculus! > For each subinterval ...(bunch of cool maths rendering I can't copy and paste because it's all comes out newline delimited on my clipboard) ... and let m<sub>k</sub> and M<sub>k</sub> denote the infimum and supremum of f on that subinterval... Okay, guess it wasn't the kind of introduction I had assumed/hoped. Very cool maths rendering though. As someone who never passed high school or got a degree thanks to untreated ADHD, if anyone knows of an introduction to the basic fundamentals of calculus that a motivated but under educated maths gronk can grok, I would gratefully appreciate a link or ten.
- moi2388 6mo agoKhan academy
- homeonthemtn 6mo agohttps://en.wikipedia.org/wiki/Calculus_Made_Easy#:~:text=Calculus%20Made%20Easy%20%2D%20Wikipedia https://en.wikipedia.org/wiki/Calculus_Made_Easy#:~:text=Cal... 1910 book, but actually does the job well
- dnemmers 6mo agohttps://calculusmadeeasy.org/1.html https://calculusmadeeasy.org/1.html
- mr_mitm 6mo agoYeah, judging by the terseness, this is clearly aimed at undergrads. Then again, this is covered in literally every calculus class, so I'm not sure who this is supposed to be for.
- chillax 6mo agoYou could se if it helps with https://betterexplained.com/calculus/lesson-1/ https://betterexplained.com/calculus/lesson-1/ or https://youtu.be/WUvTyaaNkzM https://youtu.be/WUvTyaaNkzM
- Delphiza 6mo agohttps://minireference.com/ https://minireference.com/ "The No Bullshit Guide to Math and Physics"
- BlackFly 6mo ago> Let f ... be Riemann integrable and F ... differentiable. What many people don't notice the first time they read this in the fundamental theorem of Calculus is that this is a double criteria. That f needs to be integrable seems like an extraneous point when F is differentiable. This holds also for the Lebesgue integral. The understanding is usually that if F is differentiable then its derivative is integrable, that is, people understand the integral as an anti-derivative but the Riemann/Lebesgue integral version of the fundamental theorem of calculus only proves that if the function you want the anti-derivative of is integrable, so you have this separate requirement to prove that f is integrable having already proven F to be differentiable (to f). However, this theoretical (because if you aren't a mathematician you won't be bothered by this sticking point, you'll just insist that the integral is the anti-derivative when an anti-derivative exists) defect is ameliorated by the Henstock–Kurzweil integral which is (I feel) a lot easier to define and understand than the Lebesgue integral. It is practically the Riemann integral with just a minor tweak: the delta in the delta-epsilon proof is allowed to vary by location (essentially, as you approach non-integrable singularities, you tend the delta towards zero). For the Henstock-Kurzweil integral, if F is differentiable then f is (Henstock-Kurzweil) integrable. This happens because not every derivative is Riemann or Lebesgue integrable, you need a stronger integral.
- ghighi7878 6mo agoHenstock-Kurzweil is a neat teaching trick. Often also because it shows that definition of riemann integration is not the only possible one. It leads a good motivation for lebesque later but also to of importance of spaces.
- viscousviolin 6mo agoDoes it usually get taught in the undergrad maths curriculum?
- asplake 6mo agoI wasn’t taught it, but that was forty years ago