Neat!

This is probably not the point of the video, but if you have a function f : [0,1] U {2} -> R, then take f(x) = 0 if x is irrational, 1 if x is rational(both inside [0,1]) and let f(2) = 2 say, this function is continuous at 2 only.

Okay, I did this function before watching the video. It's domain are rational numbers. First we have to create the function LD(q) which returns the last digit of q. This is ofc discontinuous everywhere. F(q) = LD(q) • q This is discontinuous everywhere except in q=0 😎

could you also have a function only be continuous on an open or closed interval? I think a piecewise function could give the answer yes to one of them at least. what about only continuous in countably infinitely many points between 0 and 1?

1/x is continuous, but you need to wrap around infinity. that's not really possible for us to do, but the line does continue like any other continuous function.

2:46 8 is less than 9

Very nice problem and solution! I used logs base 2 to solve the problem (I see that @derwolf7810 used logs base 3, quite a similar approach). To solve the challenge given at the end of the video, to compare ⁵(3⁴) with 3^(⁵4), I had to cheat a bit and use a calculator: ln(81^81)=81 ln(81)≈355.950 ln(4^256)=256 ln(4)≈354.891 So 81 ln(81)>256 ln(4) (but by a small amount!) 81^81>4^256 81^81>4^4^4 81^81^81>4^4^4^4 81^81^81^81>4^4^4^4^4 4×81^81^81^81>4^4^4^4^4 3^(4×81^81^81^81)>3^4^4^4^4^4 81^81^81^81^81>3^4^4^4^4^4 ⁵(3⁴)>3^(⁵4) I tried to compare 81^81 and 4^4^4=4^256 without a calculator. The nearest I got was to show that 81^80<4^4^4 as follows: 243<256 so 3⁵<4⁴ so 81=3⁴=(3⁵)^(4/5)<(4⁴)^(4/5)=4^(16/5)=4^3.2 So 81^80<(4^3.2)^80=4^(3.2×80)=4^256, and 81^80<4^256. Even so, as we saw using a calculator, 81^81>4^256, so the two numbers are pretty close (on a logarithmic scale).

I first thought about sqrt(-abs(x)), what about that?

very helpful this will make me move up another set (hilsdon)

xdirichlet(x)?

the next question would be could a function be differentiable at only 1 point?

My favorite example of a function with degenerate continuity properties is this: If x is irrational, let f(x) = 0. If x is rational and nonzero, write x = p/q, where p and q are coprime and q is positive. Then set f(x) = 1/q. Finally, set f(0)=1. This function ends up being continuous at every irrational point, and discontinuous at every rational point.

Great video, I learned a lot through this! Another very interesting thing I see in this expansion is the fact that it simultaneously also displays the binomial expansion for the term (1 - x)^n, so for example n=2: 1 - 2x + x² n=3: 1 - 3x + 3x² + x³ n=4: 1 - 4x + 6x² - 4x³ + x⁴ and so on.

I entered in the video already knowing the answer, but the explanation was really good.

This is insane! I was bored through a couple math classes, so my paper was filled with this exact idea a couple months ago!

I had a hunch that the right expression might be larger, since there is an entire extra exponentiation basically from the getgo. I know this isn't rigirous at all and there are definitely counterexamples, but still can give you a guess to disprove/prove.

I always like when you backwards and disappear out of the screen. It reminds me of the Homer Simpson Meme where he disappears in the bushes 🙂

Hello sir, please how to solve this équation floor (x)= sin(x), thanks

Next pentation 😂

tbh im predicting its just gonna be the same thing but negative

A pretty similar alternative, though a rougher estimation: (2^3)^^4 = 2^3^log_3(log_2(8^8^8^8)) = 2^3^log_3(3 8^8^8) = 2^3^(1 + log_3(8) 8^8) = 2^3^(1 + log_3(8) 2^24) < 2^3^(log_3(9) 2^24 + log_3(9) 2^24) = 2^3^(2^26) < 2^3^(3^27) = 2^3^(3^3^3) = 2^(3^^4)

How I hand wave inequalities involving iterated exponentiation of similar looking expressions: Imagine if multiplication was addition, and exponentiation was multiplication. Tetration would then be something similar to exponentiation. Basically, I take the logarithm. Just as there is iterated exponentiation, there's also iterated logarithms. It's not a perfect fit, but it often does the job.

it's questions like this that push math to the edge, the very essence of calculus and limits is based off of a very fuzzy definition of a point

For the bonus question, i found with laws of logarithms, we start with 81^81^81^81 ( i ignored the 4 times as it is neglibable and wont affect the final awnser) vs 4^4^4^4^4. Using 4 lns, i got 4* ln(3) * (ln81)^3 vs 4 * (ln(4)^4). Then using 81 = 3^4 i got (4^4) * (ln(3))^4 vs (4^4) (ln(2))^4. Then just by looking at it, I found that if everything i did was right, (3^4)^^5 is bigger than 3^(4^^5). Is this right and can you do a proof if possible of generally which is bigger between (a^b)^^c vs a^(b^^c)?

Love the channel. Here's a problem suggestion... 4^5^9 and 5^6^8 both have just over 1 million digits. Which is bigger?

The hardest bonus problem in my 1st analysis class was about proving that not all sets are possible as the set of discontinuity points of a function. It turns out the Baire category theorem limits what is possible! For example you can't make a function that is only continuous on the irrational numbers.

Blimey, a Dr Barker video where I understood all of it.

When you replaced 8 with 9 in the base and powers of the first term, you had prior knowledge or a "sense" that its value was orders of magnitude smaller enough than the second term for the "increase" to be unable to overcome. How does one figure that out for this kind of comparison? Is there an algorithm or some kind of informal "heuristic"?

Now I know how my daughter feels when I try to explain exponents to her

For the first example, I'm confused. If 4 numbers are drawn at random for two slots, wouldn't there be 16 possible combos, so the probability of any given sequence is 1/16? Wait as I'm typing this I realized, is each event not a sequence of two numbers but instead like a "1 or 2" ? Because then drawing a 1 or 2 out of 1,2,3,4 would have a 1/2 probability.. is that right ?

Brilliant!

Would an example of a function continuous at only one point be:- f(x)={x, xEQ {-x, xE/Q Where it's only continuous at 0?

Let the function be x when it is rational x, and -x when irrational numbers f(x) = x, x is rational; -x, x is irrrational It will be continuous only in 0 Yay! I wrote the example when I haven't watched the video and now he introduced almost the same idea!

The definition of continuous i learned in my clac class that it is continuous at f(x) iff f(x) exists, lim as c approaches x of f(c) exists, and that those two equal each other, is this the same or a different definition?

I think your definition of continuous needs to be less than or equal to epsilon for the delta = epsilon example to work because f(delta) = delta = epsilon for rational delta.

This suggests to me that the formal definition of continuity is in need of improvement.

So in between 2 rational numbers, chosen arbitrarily close to each other (whatever you can define this), there is no continuous set of rational numbers. So rational and irrational numbers keep switching to each other for arbitrarily small steps? Sounds very counterintuitive for me... I wonder how we can prove that

You can draw 1/x without lifting your pan. Because you can't draw 1/x :-)

You can also define the value in such a way that the lower limits of integration can be different from each other and it still might converge.

♡

isn't 1/x overall a continuous function as it connects in infinity? I see a lot of people are referring as the value at 0 being undefined, but I always imagined at that point function just goes to infinity and "wraps" around itself

short answer: yes. F.ex. f: R -> R x -> x if x is rational and otherwise x -> -x is only continuous at x = 0.

Nice!

kinda sketchy... Little less sketchy would be using f(x) = {x²,xeQ ; 0,xēQ}...

bro 12 already got his phd. what am i doing with my life?

I haven’t seen the solution yet, my idea is a function defined as follow: If rational, y=x If irrational, y=-x Limit doesn’t not exist everywhere except at 0, and just so happen y(0)=0 Edit: just saw the solution, have to say this is very fun maths snack, thanks for the video

Classic example that every undergraduate encounters on a second course in analysis...

i am not a mathematician, so maybe i have somethinf missing, but this relies in the fact that irrational numbers and rational numbers intercalate each other. But i am not sure that is true. Say you have some irrational number Q and you add an infinitesimally small irrational number q, then wouldn't the sum be irrational and have no rational numbers in between this making this false. Hope i made myself clear :)

I love a good paradox, and the concept of a function being continuous only at a single point is a *great* paradox. I'll be turning this one over in my head for quite a while. Thanks for this video!

You look like a very specific german.