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At first I was amazed that he can do backwards writing so neatly. Then realised he just flipped the video

@HogTieChamp

3 жыл бұрын

I was amazed but then you ruined the magic for me!!

@manasaprakash7125

3 жыл бұрын

What????

@offbeatstuff8473

2 жыл бұрын

I was just going to comment the same thing.

@umershaikh7179

2 жыл бұрын

that is pretty obvious...

@ParagPardhiNITT

2 жыл бұрын

@@manasaprakash7125 sarcasm dude 😅

Mathematicians: Look at my integral of my dreams. Physicists: Cool. But does that serve any purpose? Mathematicians: NO, but look at it. It's so magical. ;p

@123akash121

3 жыл бұрын

truest thing i have heard

@mathieuaurousseau100

3 жыл бұрын

Next century physicist : hey guys, you will never believe what weird function I'm trying to integrate today

@jimschneider799

3 жыл бұрын

@@mathieuaurousseau100 - this century's pure mathematics is next century's applied mathematics, because of those meddling physicists.

@BriTheMathGuy

3 жыл бұрын

😂So True!

@Ascientistsjourney

3 жыл бұрын

@@BriTheMathGuy woah you saw my comment. Thanks bro you made my day 😊

Holy cow that’s the prettiest integral I have ever seen

@BriTheMathGuy

3 жыл бұрын

I think so too!

@mathe.dominio4765

2 жыл бұрын

👌

@turbostar101

2 жыл бұрын

And he’s doing it backwards!

@eduferreyraok

2 жыл бұрын

I would took a little twist over the improper integral, by applying a laplace transform which matches with the definition : F(s) = L { f(t) } = integral from 0 to inf of f(t). e^(-st) dt .

Never have I understood "Sufficiently advanced math is indistinguishable from magic" more than this very moment.

@GreenCaulerpa

3 жыл бұрын

Except the original quote was “Any sufficiently advanced technology is indistinguishable from magic” from Arthur C. Clarke‘s book „Profiles of the Future: An Inquiry into the Limits of the Possible“ (1962). But I agree this integral is pretty much nightmare stuff if you haven‘t seen once how to solve it.

@tnk4me4

3 жыл бұрын

@@GreenCaulerpa Yes thank you for explaining the joke. You get an internet cookie. Congratulations.

@GreenCaulerpa

3 жыл бұрын

@@tnk4me4 yummy, thanks for that cookie!

@rmxevbio5889

2 жыл бұрын

@@GreenCaulerpa nice quote!

2:40 dude nice thank you for being aware that you can’t just interchange infinite sums and integrals willy nilly.

@HeinrichHartmann

3 жыл бұрын

He did not give an argument, though. He just mentioned "uniform convergence". But why would this sum converge uniformly? ln(x) has a singularity at 0, so I am not sure about uniform convergance on [0,1].

@grekiki

3 жыл бұрын

@@HeinrichHartmann Series for e^x converges absolutely

@markusdemedeiros8513

3 жыл бұрын

​@@HeinrichHartmann I can try to fill in the details for anyone interested: x log(x) is bounded on (0,1]: I will not do this here but it is concave up, has a minimum, and the limit at both 0 and 1 is 0. Therefore there's some closed interval containing all values of x log x for x in (0,1]. The power series of e^x converges uniformly on any closed subinterval of it's interval of convergence R, so the series for e^(x log x) converges uniformly for x in (0,1].

@holomurphy22

3 жыл бұрын

@@markusdemedeiros8513 One could just say that x log(x) is continuous on (0,1] and can be extended continuously to [0,1] as it converges to 0 in 0. The extended function is bounded because of 'extreme value theorem' and thus x log(x) is bounded on (0,1] I may be misspelling things a bit

@onradioactivewaves

3 жыл бұрын

@@markusdemedeiros8513 thanks, I appreciate that summary.

That was NOT the result I was expecting form this. Absolutely beautiful

@BriTheMathGuy

3 жыл бұрын

Glad you enjoyed it!

@kaasmeester5903

2 жыл бұрын

It is. But I still hate integrals :) I never had much issues with other mathematics (up to a masters in EE) but integrals always turn into these crappy little puzzles that apparently I'm just to dumb to solve.

What's most fascinating is the way he looks to be writing from right to left for us. It's surely inverted but stil.. Thanks for the vid

As an engineering student my first instinct was to use a euler's method of approximation cause "fuck that work" LOL

@adamuhaddadi5332

3 жыл бұрын

stupid approximateurs >:(

@bowenjudd1028

3 жыл бұрын

It’s ancient, but it works

@chungus478

2 жыл бұрын

You know you're an engineer when using π=3 does not seem like an approximation

@bowenjudd1028

2 жыл бұрын

@@chungus478, and a mathematics or physics student if it does.

Damn i got stuck watching this video and the integral of e^-x^2 in loop because at the end of each video the guy says “click on the video on the screen” and its an infinite loop :D

@BriTheMathGuy

3 жыл бұрын

You've fallen into my trap!!

@Muhahahahaz

4 ай бұрын

Oh no… I actually just arrived at this video from a different video, but I could end up in the same loop as well Next step: make sure that every sequence of video links eventually leads to this specific loop. Reminds me of the Collatz Conjecture… 🤔

Nice result, but now you should explain what is the value of the infinite sum 🛡️

@johannes8144

3 жыл бұрын

It's maybe a bit late, but the value is round about 1.2912859970626636

@zebran4

2 жыл бұрын

@@johannes8144 Thank you! Did you compute that analyticaly or numericaly?

@polychromaa

2 жыл бұрын

@@zebran4 It’s not possible to compute the value analytically as of this moment.

@user_2793

2 жыл бұрын

@@zebran4 By analytically you mean in terms of "non trivial" functions/expressions? If so it's very unlikely this can be expressed like that, just as a gut feeling

@zebran4

2 жыл бұрын

@@user_2793 Yes. By trivial expresions too.

we need more integrals like this, this is amazing

Wow, this was much better than i expected! Truly beautiful!

I really admire the way you explain, not in a hurry

Love your content! You can really feel your love for the math

@BriTheMathGuy

3 жыл бұрын

Glad you enjoy it!

Well done! This is really amazzzing !

The actually important explanation for interchanging sum and integral is brushed away like nothing. This took away the beauty of it.

Thank you for sharing this beauty. Keep shining brother

@BriTheMathGuy

3 жыл бұрын

You bet!

Love the way you speak and write.

@BriTheMathGuy

3 жыл бұрын

Thanks very much and thanks for watching!

Wow, that was sum-thing else; thank you so much for sharing!

@BriTheMathGuy

2 жыл бұрын

Glad you enjoyed it!

This was the cutest introduction of solution I have ever seen in addition to the handsomeness of the one who introduced it. 😅🤭 Bravo!

Even if you know all of these properties, there is so much knowledge that goes into applying them in ways that are helpful. Can't imagine figuring this out!

I have my term exams in few days and watching this is satisfying ❤️

@tamazimuqeria6496

3 жыл бұрын

Same here, good luck

@sourabhparadeshi4162

3 жыл бұрын

@@tamazimuqeria6496 good luck

@BriTheMathGuy

3 жыл бұрын

Best of luck all!!

@heh2393

3 жыл бұрын

How was it?

Extraordinary! I didn't see it coming.

I am genuinely getting addicted to your videos !

@BriTheMathGuy

3 жыл бұрын

Glad you like them!

Amazing video!!

Friggin high school maths still giving me headache. Good job

This channel is amazing !!!!!!

I'm in the sophomore year so I understand anything when start caculus, but I still loving your content, Ive always been ahead of the current math subject of my school so I tjink that watchint this will also help a bit more. For now I'm studying analytical geometry, is easy and I like, and calculus I'll some time soon

I find this so pretty. Almost like discrete sum (over all integers) of sinx/x = pi and integral (-inf to +inf) of sinx/x also equals pi. Amazing and yet baffling.

An effective channel. Thank you

@BriTheMathGuy

3 жыл бұрын

Glad you think so!

Wow, this is my kind of rollercoaster I enjoyed during lockdown, thanks math man

@BriTheMathGuy

2 жыл бұрын

Glad to hear it!

Great video, cool result. Thanks for this.

@BriTheMathGuy

2 жыл бұрын

Glad you liked it!

That is a very remarkable and beautiful result.

That was beautiful - and scary!

The screen inversion to get his writing right totally blown my mind to the point that I’m unable to focus on what he says.

That answer is beautiful.

Could never work that out myself but it fun to look at.

Math is so beautiful!!

Good explanation!

@BriTheMathGuy

3 жыл бұрын

Glad you think so!

what a beauty!

I've just finished with my Advanced Higher Mathematic course... just re-watching some of these videos for some good memories..

@BriTheMathGuy

2 жыл бұрын

Great job!

You made it so simple :)

@BriTheMathGuy

3 жыл бұрын

Glad you think so!

Got a similar problem in a calc 2 exam, I was very confused and thought it was unsolvable, still processing how to get a numerical value for the solution, very nice video!

@marshian__mallow2624

2 жыл бұрын

For an integral like that. You don’t get a numerical value

Omg the twist at the end is quite a shocker

I am a university student in Korea. I was always interested in math, and I happened to see your KZread while I was looking for a related KZread while preparing for a math test. I think there are a lot of fun and informative contents. I hope your KZread will be better and I will continue to look for it often. Thank you!

@limagabriel7

2 жыл бұрын

do u guys learn calculus in high school in korea?

@user-yz4fo2rp9z

2 жыл бұрын

@@limagabriel7 Yes, I do learn, but for example, in the case of calculus that utilizes two or more variables, I learn properly in college.

@uggupuggu

Жыл бұрын

Why are you named Apple Boss

I'm here to comment just to make your video more popular

@BriTheMathGuy

3 жыл бұрын

Thanks so much!

Glass pane works really well. If you can dim the lights over your hand it will be much better.

Awesome!!! 😊

Awesome vid! Good job!

@BriTheMathGuy

2 жыл бұрын

Thanks for the visit!

blew my mind. Never seen summation and integrals after each other.

@BriTheMathGuy

3 жыл бұрын

Pretty cool right?

@joshuaisemperor

3 жыл бұрын

@@BriTheMathGuy yeah but it also feels intimidating for someone who still has to pass his Calc 2.

@BriTheMathGuy

3 жыл бұрын

You can do it though!

I honestly think I'm more impressed by how good you are at writing backwards. LOL! Good video

@destructiveodst1199

3 жыл бұрын

He’s not writing backwards it’s just mirrored lol

@Unifrog_

3 жыл бұрын

I'm impressed by how well he can write mirrored then /jk

7:08 moment of satisfaction

I like your funny words, magic man.

Totally astounding

I think what is amazing is that the integral of x^x within the same limits gives the same summation but with a (-1)^n, hence having alternating plus and minsu. So the integral of this video outputs a greater value than integral of x^x within the same limits, which makes sense. Because x^-x is bigger than x^x in this interval of 0 to 1.

This might help me with a problem I’m working on

I love the fact that a video about calculus was interrupted by an ad that talks about partials (dentiures).

@BriTheMathGuy

3 жыл бұрын

😂

@ejb7969

3 жыл бұрын

That's because calculus is a subject you can really sink your teeth into! And if anyone is thinking "That joke really bites", I beat you to it. Chew on that one!

I would be unable to do it by myself without guidance But the whole video was a beautiful journey where I was smiling at each new trick Just disappointed it didn't arrived to some usual function development

Something that surprised me more than the continuous sum being equal to the discrete sum is the bounds of those sums. The continuous sum of x^(-x) from 0 to 1 equals the discrete sum of n^(-n) from 1 to infinity... *SAY WHAT?!?!?*

I tried thinking about this in a different way. I began by viewing the original (improper) integral as something I will call L (i.e., limiting sum for the improper integral). I take log(L) and then move the log operation on the inside of the integration. I doubt this obeys all the rules for logarithmic operations on (improper?) integrals. So now I am integrating the function -x log(x) dx on the same upper and lower bounds and still calling this L. The indefinite integral of this is computed to be (x^2)/4 - (1/2)(x^2) log x. Evaluating this at the limits gives 1/4 (the limit for the second term can be evaluated at the lower bound using rules for indeterminate forms and evaluates to a limiting value of 0, there from the right. Anyway, the upshot is that L = 1/4 which makes the original integral e^(1/4) or approximately 1.28, which is close to the result from the derivation in the video, but not identical. Why is this even close? I know something I've done must be wrong, probably because the integration must invoke the complex log function in some way, at least at the lower bound of integration.

Really nice results - I assume there is no closed form for the sum, but it made me a bit surprised at the end that you never touched on that topic.

@assasin1992m

3 жыл бұрын

There is, it equals sin(pi) / gamma(pi/2)

@captainhd9741

3 жыл бұрын

@@assasin1992m What is sine doing here? 🤔

@captainhd9741

3 жыл бұрын

@@assasin1992m makes me wonder if there is a complex extension for z^(-z) integral

@ha14mu

3 жыл бұрын

Isn't sin(pi) 0?

@assasin1992m

3 жыл бұрын

@@ha14mu yes, but the limit toward pi in this expression converges to a non zero result

What I don’t understand is how mathematicians make such amazingly leaps such as the various substitutions to get to the answer.

@braedenbertz1063

2 жыл бұрын

Its a lot of trial and error, looking at past results and seeing if there are parallels, and a lot of luck :)

Would be interesting if the series which resulted from this integral converges to some value :) next challenge ? ;)

uniform convergence is not sufficient to invert limit and integral, because the integration interval is not a segment (ln is not defined as 0)

*We can keep going on exploring & doing maths .. cuz it only demands three qualities of our mind* 1. *Curiosity to know* 2. *Using only knowledge i.e. No belief system* 3. (most important) *Focused mind to dig deep into the question*

Does the final infinite sum converge? Awesome integral btw!

@BriTheMathGuy

3 жыл бұрын

Thanks! and yes it most certainly does! (around 1.29 or so)

@sophiophile

3 жыл бұрын

@@BriTheMathGuy is there an exact identity for what it converges to, or did you just get this by approximation?

@leofisher1280

3 жыл бұрын

@@sophiophile there is no closed form for it sadly so all you can do is solve it numerically.

@davidgillies620

3 жыл бұрын

The good news is the convergence is extremely rapid. The first ten terms of the sum give you the value of the integral to about 3 parts in a trillion.

@olbluelips

2 жыл бұрын

@@tBagley43 almost all this kind of stuff has no closed form

Nice solution. 👌👏

@BriTheMathGuy

3 жыл бұрын

Thank you! Cheers!

thank you

@BriTheMathGuy

3 жыл бұрын

Welcome!

Please upload videos on IMO problems too they are also very deep

It's even crazier how fast it converges. For the first 7 values of n you literally have n digits of precision, after that it the rate of precision keeps getting higher.

@captainhd9741

3 жыл бұрын

Care to share an example? I am admittedly too lazy to figure out the value of the sum and how fast it gets to these values.

@jackweslycamacho8982

3 жыл бұрын

@@captainhd9741 use desmos and input sum for sum and int for integral

@captainhd9741

3 жыл бұрын

@@jackweslycamacho8982 I prefer Wolfram but good idea!

@MarioRossi-sh4uk

3 жыл бұрын

@@captainhd9741 1 1 2 1.25 3 1.28703703703704 4 1.29094328703704 5 1.29126328703704 6 1.29128472050754 7 1.29128593477322 8 1.29128599437787 9 1.29128599695904 10 1.29128599705904 11 1.29128599706255 12 1.29128599706266 13 1.29128599706266

A continuous sum becomes a discrete sum. Totally wish you extended the video by 1 minute to really nail that in for the younger audience that may be casually watching this fantastic puzzle

0:17 Well, we don't _have_ to. The power rule gives xx^(x-1) = x^x, the exponential rule gives ln(x)x^x, so the total derivative is the sum: x^x + ln(x)x^x.

@qq3088

3 жыл бұрын

That works for x^x and x^(-x). But does this work for any derivative of f(x)^f(x)? Or only those cases?

@EpicMathTime

3 жыл бұрын

@@qq3088 It generally works. It doesn't have to be exponentiation, and the functions don't need to be the same. It's a general property of differentiation that is used extensively. In other words, every derivative of a function with multiple instances of x can be realized as the sum of all "partial derivatives" with respect to each instance of x.

@qq3088

3 жыл бұрын

@@EpicMathTime l never knew this!

@dawnstudios7813

3 жыл бұрын

@@EpicMathTime "every derivative of a function with multiple instances of x can be realized as the sum of all partial derivatives with respect to each instance of x", damn that looks like a powerful statement. Do you know a proof for this?

@EpicMathTime

3 жыл бұрын

@@dawnstudios7813 The simplest way to see this is to replace each instance of x with a separate variable (say x, y, etc), and take the total derivative with respect to t. Then, set x = y = ... = t. This collapses the total derivative to the special case of the single variable derivative. This idea underpins differentiation very intimately. You're already doing it when you take any derivative, we just don't phrase it that way. For example, let's take the derivative of sin(x)cos(x) using the statement you just quoted. I'll treat the first instance of x as a constant, making sin(x) a "coefficient", so that 'partial derivative' is -sin(x)². Now I'll treat the second instance of x as constant, and likewise, that 'partial derivative' is cos(x)². Hence, the derivative is the sum of the "partials": cos(x)² - sin(x)². Although I phrased it in this different way, what we did there is precisely the product rule. In other words, the product rule itself is a specific instance of doing the quoted statement.

I was gonna discretize the domain and calculate the area by numerical methods.

Uniforme convergence isn't the reason you can do the important early swap sum integral, the hypotesis are : if we note u_n to be the function inside the sum (here x^n/n!) Then we can use the theorem under the conditions that sum(u_n) converges (i believe not even necessarly uniformly), integral(u_n) converges and sum(integral(absolute value(u_n))) converges. Not a lot of these has to do with uniforme convergence

Just wow 🔥🔥🔥🔥

Its intresting how he uses just SMALL PART of BOARD to explain such complex problems whereas for our teacher need two full boards

I freakin love calculus. I thought this was gonna be really scary at first.

damn that's fuckin sick dude!

Newcomer here. How do you do that mirror writing thing?

Extremely interesting indeed.!!!!!

@BriTheMathGuy

3 жыл бұрын

Glad you think so!

@sonarbangla8711

3 жыл бұрын

@@BriTheMathGuy I wonder if these functions need to be analytic or converge?

Man! You love Gamma function so much 🤣🤣🤣🤣🤣

@BriTheMathGuy

3 жыл бұрын

Yes I do!

Amazzing caculuses, guy!!!

Oh, how cool... I like this Prof...

Can you plot this graph with an x domain from negative infinity to positive infinity. The positive values of x are a decreasing function that heads off to zero as x goes large. The negative domain of x will give a range oscillating between positive and negative values

we can easily solve it by taking natural log and apply integratiom by parts

Is there a similar solution for the integral of the same function but fro 1 (or 0) to infinity. Should be convergent quite obviously, but how to calculate it?

I don't know anything about calculus(only took precal my senior year of highschool) and this is scaring me and I have no idea what's going om

How does the result... make sense? I may be lost when ganma function, but the summatory of n^(-n) isn't just number + number + ...? So the derivative will be... 0?

It's ≈ 1.291291≈ 430/333

dude i graduated with my engineering degree why am I still watching Math videos? beautiful vid btw

I have a question Isn't the final summation converged to something like π/ ( something) ?

@alexanderkartun-giles5961

3 жыл бұрын

The sum equals exactly 1.29129

@casual0815

2 жыл бұрын

I think you might me referring to a similar sum: n goes from 0 to infinity, 1/n^2 The sum is equal to pi^2/6.

What does the infinite sum evaluate to though? I looked it up on desmos and the integral evaluates to around 1.29. If we integrate from 0 to infinity, we get something eerily close to 2.

Are you writing in mirror-image, or reflecting the entire video afterwards?

maybe more like a sophomore's nightmare to some i'd imagine

@BriTheMathGuy

3 жыл бұрын

😂

Is the original function continuous from 0 to 1? How does it look like? it's doubtful.

Wow, just wow.

It was a cool solution. The final answer can write as a zeta function.👌

now please do the integral of x^x in -1 to 0 :D

6:15 why did you replace n plus 1 by v? its a constant so does it really matter? i mean sure youve got the summation but with respect to the integration its just a constant? maybe im not smart enough yet to understand this