"i don't like to be in the bottom, but i don't care" cracked me up

@RamAurelius

2 жыл бұрын

*Bprp wants to know your location*

@dougr.2398

2 жыл бұрын

@@RamAurelius 😳😳😳😱😱😱😱😂🤣😂

The result π/4 is valid on (0, π). On (-π, 0) the value is instead -π/4. At x=kπ the result is 0. The sum is 2π-periodic in x.

@jelmerterburg3588

2 жыл бұрын

Yeah, you'd almost think that this infinite sum is a square wave ;)

At 5:11 we have to be more careful, when splitting up the series into two we can no longer guarantee convergence (as a simple counterexample, neither of them converge when x = 0, which is when the initial series evaluates to 0 and not pi/4).

Here's how I did it. Consider the exponential series sum from n=0 to infty of e^(i(2n+1)x)/(2n+1). The sine series we're interested in is the imaginary part. If we differentiate the exponential series term by term wrt x we get a geometric series which turns out to be real-valued. So the sine series has derivative 0 and hence is constant. Hence we can set x to a convenient value, say pi/2, and get 1-1/3+1/5-... = pi/4

@wolfbirk8295

Жыл бұрын

What about convergence ?....

Ah yes, the Fu-Yei Series. Another member of the family.

@bprpfast

2 жыл бұрын

Omg 😂

Highly satifying.

This is such a beautiful sum!

Applied math hat on killed me🤣

Thank you!

I really liked your videos and now with these emojis and all these jokes i love them even more hahaha. Great job man :D

Thanks for the video very interesting sum!

@silentintegrals9104

2 жыл бұрын

toally agree!!

This result needs a bit more care with the complex arguments. I remember experimenting with similar series as an undergrad and if i recall correctly these kind of functions are only piecewise constant with jumps at certain angles. In particular if you take the derivative you‘d get a sum of cosines which can be zero infinity and neg. infinity with the latter two cases at the positions of the jumps

Infinite series and complex numbers. What a cool pair!! :D

I'm amazed Dr. Peyam how you keep able coming up with such varied examples. Also the gong at the beginning made me chuckle a bit.

@drpeyam

2 жыл бұрын

Thank you!!

@dougr.2398

2 жыл бұрын

Going, GOING…. GONG!!! (= going minus « i »)

Thanks for watching

If you integrate with repect to x repeatedly and plug in nice values of x, you can get 1+1/2^k+1/3^k+... for k even and 1-1/2^k+1/3^k-... for k odd. 😁

@drpeyam

2 жыл бұрын

Beautiful!!!!

Are you trying to convince us that sin(0) + sin(3*0)/3 + sin(5*0)/5 + ... equals pi/4?

@drpeyam

2 жыл бұрын

Yep, on (0,pi)

Great result! But a little worry about those steps when substituting those complex entries, it needs to justify those argument angles 😶

Taking undergrad PDEs rn, recognized this sum from the thumbnail given how many times I’ve had to write down the Fourier series for 1 🤣

Capo total, mon ami! 😄😄😄

This sum is the Fourier series of the function f(x) = {-π/4 for -π < x < 0, +π/4 for 0 < x < +π}. So the result should be 0 for x = nπ, -π/4 for (2n-1)π < x < 2nπ, π/4 for 2nπ < x < (2n+1)π for any n integer.

@drpeyam

2 жыл бұрын

Yep, so it’s pi/4 on (0,pi), just as expected

3:22 HE SAID IT!!

Dear Dr Peyam. Decompose the function f(x)=π/4, given in the interval (0,π), into a Fourier series in terms of sines. You will get, π/4= ∑(from n= 1 to ∞)sin((2n-1)x)/(2n-1) =sinx+sin3x/3x +sin5x/5x...... On this path you will find an infinite number of "discoveries".

@drpeyam

2 жыл бұрын

Correct

Actually, you should use alternately the principal logarithm and the other one. I recognized it as the Fourier transform of a square wave.

awesome video

I first encountered this type of series in a boundary value problem

6:42 finally a moment in this video where I can relate with you😅🥰

Amazing inverse-Fourier deduction.

An oddly satisfying video

Perfect!! What about heavenly even satisfying sum?

simply amazing. this is the man that can find the last digit of π

Ja,ja,ja. Excelente! :D

Thanks

@drpeyam

2 жыл бұрын

Thanks so much for the super thanks!!! 😍😍😍

11:16 there should be e^(ix). I hoped the answer would've been complex :D

What if we considered a different branch cut at the end?

@drpeyam

2 жыл бұрын

Probably cancels out

As soon as you mentioned the power series for arctanx, I though of De Moivre's Theorem, (cosx+isinx)^n = cos(nx)+isin(nx) because then you could input e^(ix) into the arctan expansion and then write e^((2n+1)ix) as cos((2n+1)x) + isin((2n+1)x). The problem is that arctan alternates, so you probably would have to use artanh instead, so that might be more difficult. I haven't finished the video yet because my phone's at 2% but I'll be back to see if my prediction was useful or not.

@drpeyam

2 жыл бұрын

Nah, arctanh sucks lol

@violintegral

2 жыл бұрын

@@drpeyam well you really did use artanhx when you used the identity 1/iarctan(ix)=artanhx=1/2ln((1+x)/(1-x))

@wolfbirk8295

Жыл бұрын

@@violintegral how is arctan(z) defined ?...

I am from India and preparing for competitive exam. Your videos are so helpful to me ❤️❤️

Here's a challenge!! For alpha>1, study the convergence of the series sum(from n=3 to inf) [ int(from n to n+1) 1/(x^alpha * ln x) dx ] A series of an integral!!

@drpeyam

Жыл бұрын

Also a homework problem 😂

@ianmi4i727

Жыл бұрын

​@@drpeyam A more or less difficult one!!

Lovely Math!

@drpeyam

2 жыл бұрын

Agreeeeed

@silentintegrals9104

2 жыл бұрын

Totally agree!!!

You look great Peyam 😊

@drpeyam

2 жыл бұрын

Thanks so much 🥰

Es muy genial

sorry but can I ask how many answers does ln(-1) has?

@drpeyam

2 жыл бұрын

Infinitely many, but one principal one

What’s the application to Physics ? (My perpetual question)

@drpeyam

2 жыл бұрын

Quantum mechanics lol

@drpeyam

2 жыл бұрын

Also this is the Fourier series of a step function

@dougr.2398

2 жыл бұрын

@@drpeyam please say a few words more? Thank you!

@dougr.2398

2 жыл бұрын

@@drpeyam what kind of a step function? A periodic one or thé Heaviside step function? (Or is that immaterial, somehow?)…..one that goes from zero to one over a finite interval or infinite (infinite interval might require Fourier intégrâmes, no?). Or one That goes from minus one-half to one-half, or minus one to plus one? (Some results then would/should differ by a constant then….)

eugh... even when i know its in complex numbers, writing "ln(-1)" makes me feel filthy

“i” don’t care if it’s top or bottom 😎

I is in denominator but I do not care. Hilarious joke, just love it. I think with knowing of inverse hyperbolic functions all of it would be faster ;) But this explanation just contains them implicitly so it's okay ;)

the results is incorrect for x=0, where did we use that x should be nonzero?

@drpeyam

2 жыл бұрын

x is in (0,pi)

@Czeckie

2 жыл бұрын

@@drpeyam sure, but why? I think the devil is in the detail at 8:15 as the logarithm has branch points at x=+-1.

@goblin5003

2 жыл бұрын

@@Czeckie 4:32 imagine what happens if x=0

No my love! if your sum started at n=0, you would get pi/4. Since it starts at n=1 you get pi/4 -sin(x)

How come? This sum depends on x. The result also has to depend on x

@drpeyam

2 жыл бұрын

Isn’t that amazing? It doesn’t depend on x, the function converges to a constant

@Unchained_Alice

2 жыл бұрын

@@drpeyam It seems to converge to -π/4 for some x values and to 0 for certain ones. The 0 one is clear but the negative one I just tested on my computer so I can't be sure. Is the negative one due to the principal logarithm not being continuous for negative reals?

@drpeyam

2 жыл бұрын

Right, it’s still pi/4 on (0,pi) so it’s ok

you must have had fun editing this video

@drpeyam

2 жыл бұрын

I really did hahaha 😝

@drpeyam

2 жыл бұрын

I loved the cow sound

What was that :0

Yes!, our little friends i's :)

Instead of step 3 I would use arctan a - arctan b = arctan ((a-b)/(1+ab))

@drpeyam

2 жыл бұрын

I heard that might make the denominator 0 though

@AlexAlex-kb4js

2 жыл бұрын

@@drpeyam sure, but arctan (something/0) = arctan (infinity) = pi/2 And we have some 1/2 in front so it should lead as well to pi/4

Yoooooooooooooooo!

Not right demonstration. ln(-1)=ln(exp(i*Pi))=ln(exp(3i*Pi)=ln(exp((2k+1)i*Pi)). But that doesnt mean that i*Pi=3i*Pi..

@drpeyam

2 жыл бұрын

The demonstration is correct, principal logs

Proffesor,,,you will not be able to use the arctangent formula because....here xy =-1 ,,,but tan-1(x) + tan-1(y) = tan-1((x-y)/(1+xy))

5:14 .... .. . . . ... . . . . . . .. . .. . . Wtf

@drpeyam

2 жыл бұрын

I was hoping someone would notice 😂😂😂😂😂

Hola

Those edits 😂

Should a 7th grader be here?

@blableu4519

2 жыл бұрын

why not