Even math nerds look at some equations and say… all hells no.

The algorithm suggested your videos few days ago and now I found my new addiction 🤣🤣

it _really_ was more _complex_ than we all _imagined_ 😂

@maths_505

14 күн бұрын

😂😂😂

@davidblauyoutube

10 күн бұрын

oof.jpg

The first time I read the name maths 505, it sounded in my head like a mathematics course for students in masters 2. I was quite intimidated that this channel will be treating advanced maths and that I was too premature to be here. Even though the course is advanced, I continue watching because my love for mathness was above all. In the beginning it was hard to follow but I continue to follow you. It's more than a year I'm here and in school I'm going ahead(I will obtain my bachelor's degree this year in physics(mechanics) and next year I will be in masters 1) but watching you frequently transformed me into a beast compared to my other classmates. I thank you because my trajectory is more awesome and I think things will have been boring if I didn't know you. Thanks very much. I think through this text, I've formulated almost all my gratefulness towards you.

@maths_505

14 күн бұрын

Dude honestly it's comments like these that make me feel like I've actually done something useful with my life❤️ thank you so much it seriously means alot to me.

@leroyzack265

13 күн бұрын

@@maths_505 thank you too my hero. When other guys propose integrals and differential equation, I don't find attractive but yours are the most attractive and instructive.

Feels good to be one of the "math people of YT". OK cool! 😂 I like the complex trig id sin(x) = i/2 exp(-ix)(1 - exp(2ix)) = exp(-ix + i pi/2)( ... ) = exp(i(pi/2 - x))(1 - exp(2ix)). It's really quite useful sometimes.

bro's like one of those science wiz archetypes in film: lemme just pull up this doohickey for this extraordinary situation! Oh, now we have to deal with this Cthulu monstrosity? I happen to have just the mcguffin to exorcise it! And now all this needs is some basic integration by parts and then slap it onto Wolfram Alpha. I leave figuring out how TF it works as an exercise.

At 7:50, it's not 1/4, it's 1/2 as you have factored i/2 earlier

@madsheller5242

8 күн бұрын

The i/2 is only factored in the first sum

I think this is my first comment on your channel, even though I've watched so many of your videos already. I am just starting my first calculus course at university this year, but I find your videos very interesting! Because of my rather limited knowledge in this area, I was really expecting a nice cleanup at the end - especially from you. But simply realising that this is the final result that cannot be simplified further, I found very interesting...! Thank you for your videos, I'll continue to love them!

@maths_505

13 күн бұрын

This result kept me awake for 2 days💀 in the end....I just had to accept it😂😂

First for the first time. Thank you for your job !

@justmarijn9122

14 күн бұрын

Second😂

somehow and why is it that always the simple looking cute upper esh becomes a murder scene when solving and rarely for the answers. Love ur vids sir

That's quite an interesting result. If you take out the stuff that's purely real or purely imaginary, and write log(1-e^-2i) as Li1(e^-2i), this can be rearranged, ultimately showing that 2*Li1(e^-2i)-2i*Li2(e^-2i)-Li3(e^-2i) has an imaginary part of exactly 2/3, which is just wild. I am not familiar with any linear combination of Li1, Li2 and Li3 of something that has a known closed form, and I'd not be surprised if none exists. Not even WolframAlpha recognizes that the imaginary part is exactly 2/3, insisting on an approximation (0.66666...).

@renerpho

13 күн бұрын

The real and imaginary parts of Li1(e^-2i), Li2(e^-2i) and Li3(e^-2i) all don't have closed forms.

Very interesting integral and fine result. Thank you.

Sometimes you would see a sin(x) and say its the imaginary part of e^ix, but sometimes you instead expand it into (e^ix - e^-ix)/2i. When do we use which version? Is it something like when the trig function is an argument of another function? In that case, when can we bring the real and imaginary part operators outside of an integral?

@davidleanokalocsa

14 күн бұрын

You'd have to put Im(e^ix) inside the natural logarithm and you would get nowhere from there

@renerpho

13 күн бұрын

@@davidleanokalocsa Well, I'd then expand Im(e^ix) into (e^ix - e^-ix)/2i...

What in the dilogarithm of e^2i is this?

@maths_505

14 күн бұрын

Quite similar to my reaction 💀

So creative, I think you have a lot of imagination

oh shit this was wild especially the part did not expect it to be REAL!

Your title card suggests that it's surprising the integral is real. the function is well defined on that interval, so why wouldn't the area under the curve be real?

@maths_505

14 күн бұрын

Ofcourse its supposed to be real bro I'm talking about the final closed form😂

@renerpho

13 күн бұрын

@@maths_505 An example of where a closed form tells you less about the thing you're trying to calculate than the integral you started with. 🙂 Kind of like the cubic formula, which gives you the solution to any cubic equation, but in a form that makes it hard to tell if it's a real number or not.

Can you please justify your use of the log expansion at 4:08? Your expansion is valid for abs(z)

idk how dilogarithm works but i was expecting it to cancel with the exponential somehow

Can we get a separate vid explaining greek alphabet and Lis?

I think the 'i' in the denominator of sin might be a convention. These conventions have really messed me up.

@raddish4440

13 күн бұрын

It gives a hint of a way you can derive it by just looking at the identity though, which can be useful if you're seeing it for the first time or if you are trying to memorize it. I'd argue that rewriting it as -i/2 (exp(... would be less clear despite being mathematically equivalent

@thomasolson7447

13 күн бұрын

@@raddish4440 I mean that I don't think it exists. We put that 'i' there to make math nicer. Since it is there, there are a bunch of trig functions that need to be revised. Like tangent addition. Hyperbolic stuff also gets revised.

@raddish4440

13 күн бұрын

Well if we want to destroy our original definition for the sin function than sure but then we get i's all over the place in our normal trig mathematics! I think it just moves your problem elsewhere. That i being in the complex definition of the sin has some nice geometric meaning as well so I don't think it's an eyesore at all nor does it make this math nicer to work with. But depending on your application maybe it does, I just don't see any reason to make a new definition that is incompatible with existing mathematics

@thomasolson7447

13 күн бұрын

@@raddish4440 absolutely true. But think, if we did this all Physics teachers would stop saying there are no 'i''s in physics.

😂 Thx for real value... 👍

In 7:40, the second sumand is + and the third is -

Great vid! But when is the identity ln(e^x) = x valid in the case of complex numbers?

@madsheller5242

8 күн бұрын

Well, always. ln(x) and exp(x) are each others inverse, also when the argument is complex.

Hi, "terribly sorry about that" : 1:48 , 4:38 , 5:29 , 7:39 , "ok, cool" : 2:37 , 6:55 .

cant we use by parts on this?

ok cool

i have a very difficult eq diff : y+1 = ( x^n ) * exp(-x) i need the solution please 🙂

@YouTube_username_not_found

13 күн бұрын

Try using integration by parts.

@user-di4iq7ec1d

13 күн бұрын

@@KZread_username_not_found how ?

@YouTube_username_not_found

13 күн бұрын

@@user-di4iq7ec1d wait, is this even an equation? what does y+ 1 mean ?

@user-di4iq7ec1d

13 күн бұрын

@@KZread_username_not_foundwe have f(x) = ( x^n ) * exp(-x) - 1 i need to find the reciprocal function of f

@user-di4iq7ec1d

13 күн бұрын

@@KZread_username_not_found i need to find the reciprocal function of f such as f(x) = ( ( x^n ) * exp(-x) ) - 1

please dont say : terribly sorry about that...we understand those small mistakes....jut continu.just. you re great...am math professor from morroco

@waarschijn

14 күн бұрын

it's a little joke