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.
Yeah, 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.
I'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.
It 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).
> 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]
> 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.
The indicator function of the Cantor set is Riemann integrable. Like you said, though, the Dirichlet function (which is the indicator function of the rationals) is not Riemann integrable.
The reason is because the Dirchlet function is discontinuous everywhere on [0,1], so the set of discontinuities has measure 1. The Cantor function is discontinuous only on the Cantor set.
"Almost everywhere" is precisely defined, and it is broader than that. E.g. the real numbers are almost everywhere normal, but there are uncountably many non-normal numbers between any two normal reals.
Good 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.
> 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.
You could se if it helps with https://betterexplained.com/calculus/lesson-1/ or https://youtu.be/WUvTyaaNkzM
Yeah, 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.
https://calculusmadeeasy.org/1.html
https://en.wikipedia.org/wiki/Calculus_Made_Easy#:~:text=Cal...
1910 book, but actually does the job well
Khan academy
I'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.
[delayed]
It 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).
You meant antiderivative?
> 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
[2] https://en.wikipedia.org/wiki/Lebesgue_integral
> 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.
Ah, 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?
No, it's really sets of measure zero. The Cantor set is an example of an uncountable set of measure 0: https://en.wikipedia.org/wiki/Cantor_set
The indicator function of the Cantor set is Riemann integrable. Like you said, though, the Dirichlet function (which is the indicator function of the rationals) is not Riemann integrable.
The reason is because the Dirchlet function is discontinuous everywhere on [0,1], so the set of discontinuities has measure 1. The Cantor function is discontinuous only on the Cantor set.
Likewise, the indicator function of a "fat Cantor set" (a way of constructing a Cantor-like set w/ positive measure) is not Riemann integrable: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...
No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".
The Cantor set is uncountable and has Lebesgue measure 0.
"Almost everywhere" means "everywhere except on a set of measure 0", in the Lebesgue measure sense.
Here's an example of a Riemann integrable function w/ infinitely many discontinuities: https://en.wikipedia.org/wiki/Thomae%27s_function
Anyone interested in this should check out the Prologue to Lebesgue's 1901 paper: http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.p...
It gives several reasons why we "knew" the Riemann integral wasn't capturing the full notion of integral / antiderivative
"Almost everywhere" is precisely defined, and it is broader than that. E.g. the real numbers are almost everywhere normal, but there are uncountably many non-normal numbers between any two normal reals.
"almost everywhere" can mean the curve has countably infinite number of discontinuities
“Almost everywhere” is a mathematical term and can mean two things (I think):
- except finitely many, or
- except a set of measure zero.
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculu...
What is the font used on the site?
Today 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/)
--edit: The font used for those initials is called Goudy Initialen: https://www.dafont.com/goudy-initialen.font
That font, and how it's integrated with the math looks amazing. Katex for the math?
Seems like Katex from the scripts getting loaded. I love the design too, kinda medieval-chic.
Looks more early modern to me. :)
Inspection suggests "https://online-fonts.com/fonts/alegreya"
Also https://online-fonts.com/fonts/goudy-initialen
Alegreya
Good 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.