Small remark to 4:38. f(x) = 1/((x-1)(x-e)) is actually continuous on its whole domain. Since it is not defined in either 1 or e it cannot be discontinuous at those points. Only if you defined it to be some real number in those points, it would be discontinuous. This is because in the definition of continuity, only points from the domain are considered for the condition |x - x0| < delta => |f(x) - f(x0)| < epsilon
No. Because the zero function is continuous no matter on what subset of say R you define it. In fact any map between topological spaces that maps all values to a single point is automatically continuous. You can not say a functions continuity is violated on a point on which the function isn't even defined
My Calculus Professor (Tony Tromba, UC Santa Cruz 1981) dropped the Dirichlet Function on us at the end of a Friday lecture to give something to discuss at Happy Hour.
2:59 by the way, you forgot to mention 1 thing: e is not only irrational, e is trancendental. That means e², e³, e⁴ etc. cannot be rational numbers. (By the way, this is not true for all irrational numbers. For example: sqrt(2) is irrational, but sqrt(2)²=2, which is rational.)
I would like to mention the “periodic but no smallest period” has the counterexample f(x)=0. Here’s another related counterintuitive fact: All linear functions are the sum of two periodic functions.
wait, what? you said that lim_{k->inf}(k!x) is an integer iff x is rational. But for x=1/2, lim_{k->inf}(k! * 1/2) = infinity, which isn't an integer. Is there some more precise way of phrasing this? (I'm guessing "eventually all terms become integers" or something)
The x is inside the cosine function hence it's limited by -1 and 1 and in the limit even if the cosine is 0.99999999 as j tends to infinity it goes to 0, so only from the rationals can we get 1 for the expression
Great video, but in 5:49 you say that a similar method could be used to create a function that is continuos only on the integers. I remember reading that due to Baire's cathegory theorem cannot be continuos on infinite sets with zero measure like the rationals, is that the case? If so, why does it fail on the rationals but not in the integers even thought the latter also have 0 measure?
I have heard the name Dirichlet before but I've never had a close look at the Dirichlet function. It's truly a crazy function. But what I found most fascinating is that it's not just one of those artificial functions where you define the different parts but that you get this weird beast from what looks like this ordinary double limit! Cool video! PS: Did you make a mistake at 5:36? You say that the same argument applies for the "e case". But is this correct since e is not a rational number? Wouldn't the limit be 0 when x goes to e because when x=e the lower definition (for x not element of Q) applies?
Hi! I was referring to the artificial function defined as you see on the screen 5:36, not the original Dirichlet function. You are right that when x=e then f(e) is indeed 0. And by definition of the limit, the limit of f(x) when x- > e is also 0.
My calculus lecturer gave us the task to proof whether or not the dirchelet function is a regulated function. So its kinda funny to see this video a week later :)
1:17 I get what you're trying to say with lim k to infinity (k factorial * x) is in Z, but that limit (unless x is 0) would actually be positive or negative infinity. I think it would be better to say something along the lines of (for all k bigger than a certain natural number N (k!x) is an integer)
Fun, but funnier when you try to do something like the following: Write the sine function buuuut at x=69 is 69 aaaand can only be written as a combination of elementary functions and limits (no use of the brackets)
Hi bro. Click the `cc` button on the lower right corner, select `auto-translate` and find the language you are using. Most of the time, it is good enough.
But... There are infinitely more irrational numbers than there are rational numbers, I'm sure I learned that some years ago. So shouldn't there be examples for two irrational numbers so close that there's no rational number in between? And if not, why not?
Good question. You can construct a rational number between any two irrational numbers. Because I can't type LaTeX in the comment, here is the link: math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals
The cardinality of the set of irrational numbers is higher than the cardinality of the set of rational numbers, which is a more precise way of saying that there are more irrational numbers than there are rational numbers, so yes to your first question. But to your second question, no, there can't be two irrational numbers so close that there's no rational number in between them. Given any two distinct irrational numbers a and b, the distance d between them is given by d = |a - b|. Since a and b are distinct, a - b is not zero, so d > 0. However small d is, you can always find a fraction 2^(-n) that is smaller than d for a large enough positive integer n. Given that consecutive integer multiples of 2^(-n) are always a distance 2^(-n) apart, which is smaller than d, it must be true that at least one such integer multiple of 2^(-n) falls within the interval between a and b. And any integer multiple of 2^(-n) is a rational number.
Shortly after being introduced to this function, my calc 2 teacher posed to the class to give an example of a function continuous at only 1 point. His solution was the Dirichlet-adjacent function f(x)={x if rational, -x iff irrational}. It isnt hard to see that f(0) has a straightforward delta epsilon proof.
@@ron-math your accent is on the thicker side so words like graph become gruff between 1:45 and 1:55 I feel are the most noticeable, like the word atoms sounds like items, those are just some examples
Now I am become counter example, destroyer of proofs.
Hidden destroyer of the freshmen who got pranked by their seniors.
It is not a proof, if it is rebuttable
throughout discrete math and real analysis, i alone am the counter example
Moreover, Dirichlet's function is integrable by Lebesgue but not by Riemann.
Yep. I wanted to include this point in the video as well.
Funny how a huge troublemaker to the riemann integral follows immediately from definition of the Lebesgue integral actually
@@fakezpredI mean the proof for it not being Riemann-integrable is pretty trivial too so I wouldn't call it a huge trouble maker
@@lperezherrera1608 by troublemaker I mean being nonintegrable with the riemann definition, not the proof.
Because the set of rationals has a measure equal to zero. So Dirichlet's function is the zero function almost everywhere.
Small remark to 4:38. f(x) = 1/((x-1)(x-e)) is actually continuous on its whole domain. Since it is not defined in either 1 or e it cannot be discontinuous at those points. Only if you defined it to be some real number in those points, it would be discontinuous. This is because in the definition of continuity, only points from the domain are considered for the condition |x - x0| < delta => |f(x) - f(x0)| < epsilon
Yes. Great catch!
It was mind-blowing to discover in my calculus 1 class that f(x) = 1/x is continuous, along with all elementary functions.
I had a smile on my face this whole video :)
I hope more people come across this channel! Can't wait for the next upload
Yes this is a really cool video, and I’m also a fan of Micaiah!
Another example of a periodic function with no smallest period is f(x) = 0
Also another example for two discontinuous functions with a continous sum is any discontinous function f and -f
@@comedyfriendsenglish what if the function is discontinuous because it's not defined at some points? Then the sum will still be discontinuous.
No. Because the zero function is continuous no matter on what subset of say R you define it. In fact any map between topological spaces that maps all values to a single point is automatically continuous. You can not say a functions continuity is violated on a point on which the function isn't even defined
My Calculus Professor (Tony Tromba, UC Santa Cruz 1981) dropped the Dirichlet Function on us at the end of a Friday lecture to give something to discuss at Happy Hour.
Is that the Tromba who coauthored a vector calc textbook with Marsden?
@@SedgeHermit Yes.
"Mathematics is difficult, even for mathematicians." ~ Reinhold Böhme, quoted in Appendix A of "Vector Calculus" (2nd Edition).
Next, the Cantor set indicator function. Discontinuous at an uncountable set of points but still riemann integrable.
This function is a smart bomb on "Let epsilon > 0"
2:59 by the way, you forgot to mention 1 thing: e is not only irrational, e is trancendental. That means e², e³, e⁴ etc. cannot be rational numbers. (By the way, this is not true for all irrational numbers. For example: sqrt(2) is irrational, but sqrt(2)²=2, which is rational.)
You did pass a lot of joy, thanks!
Amazing! Both the original construction and your explaination.
Cheers!
Excellent video! Thank you.
I would like to mention the “periodic but no smallest period” has the counterexample f(x)=0.
Here’s another related counterintuitive fact: All linear functions are the sum of two periodic functions.
I don't think you consider 0 as a period. Otherwise all functions are periodic.
@@farfa2937no, but 0 has no smallest period
f(0) is not period because there's no p for which f(0)≠f(p+0).
1. That's an example, not a counterexample.
2. How?
f(x) = 0
then f(x) = f(x + k) = 0
where k is the interval, and therefore f(x) = 0 is periodic
Conway's base 13 function is discontinous everywhere.
On my to-do list :)
This funxtion absolutely carried me through undergrad
Amazing ! Well done
Thank you! Cheers!
How about Weirstrass Function which is continuous but non differentiable everywhere
on the to-do list.
wait, what? you said that lim_{k->inf}(k!x) is an integer iff x is rational. But for x=1/2, lim_{k->inf}(k! * 1/2) = infinity, which isn't an integer. Is there some more precise way of phrasing this? (I'm guessing "eventually all terms become integers" or something)
The x is inside the cosine function hence it's limited by -1 and 1 and in the limit even if the cosine is 0.99999999 as j tends to infinity it goes to 0, so only from the rationals can we get 1 for the expression
Do the Devil's Staircase function now! Also, I recommend the book 'Counterexamples in Analysis'
Great video, but in 5:49 you say that a similar method could be used to create a function that is continuos only on the integers. I remember reading that due to Baire's cathegory theorem cannot be continuos on infinite sets with zero measure like the rationals, is that the case? If so, why does it fail on the rationals but not in the integers even thought the latter also have 0 measure?
Good point! Rationals are dense while integers are not.
I have heard the name Dirichlet before but I've never had a close look at the Dirichlet function.
It's truly a crazy function. But what I found most fascinating is that it's not just one of those artificial functions where you define the different parts but that you get this weird beast from what looks like this ordinary double limit!
Cool video!
PS: Did you make a mistake at 5:36? You say that the same argument applies for the "e case". But is this correct since e is not a rational number? Wouldn't the limit be 0 when x goes to e because when x=e the lower definition (for x not element of Q) applies?
Hi!
I was referring to the artificial function defined as you see on the screen 5:36, not the original Dirichlet function. You are right that when x=e then f(e) is indeed 0. And by definition of the limit, the limit of f(x) when x- > e is also 0.
My calculus lecturer gave us the task to proof whether or not the dirchelet function is a regulated function. So its kinda funny to see this video a week later :)
1:17 I get what you're trying to say with lim k to infinity (k factorial * x) is in Z, but that limit (unless x is 0) would actually be positive or negative infinity. I think it would be better to say something along the lines of (for all k bigger than a certain natural number N (k!x) is an integer)
Right. Good point.
Fun, but funnier when you try to do something like the following:
Write the sine function buuuut at x=69 is 69 aaaand can only be written as a combination of elementary functions and limits (no use of the brackets)
This man like godel and hypasus 😂
0:23, for number 4 can't you just have the constant function c?
As mathematicians will say, it is trivial 🤣
A Weierstrass function is also an answer to most of those questions
On the todo list.
Now find a function that is discontinuous at any rational number but continuous otherwise
Can the greatest integer function be the answer to 2nd question ❓⁉️
Mind be more specific? You mean the ceil() function?
He be like Gojo fr
Herkes için altyazı lütfen
Hi bro.
Click the `cc` button on the lower right corner, select `auto-translate` and find the language you are using. Most of the time, it is good enough.
But...
There are infinitely more irrational numbers than there are rational numbers, I'm sure I learned that some years ago.
So shouldn't there be examples for two irrational numbers so close that there's no rational number in between?
And if not, why not?
Good question. You can construct a rational number between any two irrational numbers. Because I can't type LaTeX in the comment, here is the link: math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals
The cardinality of the set of irrational numbers is higher than the cardinality of the set of rational numbers, which is a more precise way of saying that there are more irrational numbers than there are rational numbers, so yes to your first question.
But to your second question, no, there can't be two irrational numbers so close that there's no rational number in between them. Given any two distinct irrational numbers a and b, the distance d between them is given by d = |a - b|. Since a and b are distinct, a - b is not zero, so d > 0. However small d is, you can always find a fraction 2^(-n) that is smaller than d for a large enough positive integer n. Given that consecutive integer multiples of 2^(-n) are always a distance 2^(-n) apart, which is smaller than d, it must be true that at least one such integer multiple of 2^(-n) falls within the interval between a and b. And any integer multiple of 2^(-n) is a rational number.
Thank you! @@omp199
@@ron-math You're welcome.
Shortly after being introduced to this function, my calc 2 teacher posed to the class to give an example of a function continuous at only 1 point. His solution was the Dirichlet-adjacent function f(x)={x if rational, -x iff irrational}. It isnt hard to see that f(0) has a straightforward delta epsilon proof.
This function is cursed, don't use it
Okay the way you pronounced continuous and other things are just wrong. Except that the video seems great
Can you help me identify the list of mispronunciations? Thanks!
The pronunciation is fine. Your ears function poorly.
@@ron-math your accent is on the thicker side so words like graph become gruff between 1:45 and 1:55 I feel are the most noticeable, like the word atoms sounds like items, those are just some examples
Thank you so much for pointing it out! I will definitely work on it in the future videos.@@finthechat2717