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.
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.
@@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.
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!
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.
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.
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?
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...).
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 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.
Well, is no one going to draw the roughly pentagonal contour traced by those five complex vectors? Ending on the real line, a little to the left of zero?
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
@@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.
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
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 🤣🤣
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.
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.
@@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.
it _really_ was more _complex_ than we all _imagined_ 😂
😂😂😂
oof.jpg
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!
This result kept me awake for 2 days💀 in the end....I just had to accept it😂😂
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.
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.
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?
You'd have to put Im(e^ix) inside the natural logarithm and you would get nowhere from there
@@daveydd Well, I'd then expand Im(e^ix) into (e^ix - e^-ix)/2i...
Hi,
"terribly sorry about that" : 1:48 , 4:38 , 5:29 , 7:39 ,
"ok, cool" : 2:37 , 6:55 .
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...).
The real and imaginary parts of Li1(e^-2i), Li2(e^-2i) and Li3(e^-2i) all don't have closed forms.
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
So creative, I think you have a lot of imagination
At 7:50, it's not 1/4, it's 1/2 as you have factored i/2 earlier
The i/2 is only factored in the first sum
What in the dilogarithm of e^2i is this?
Quite similar to my reaction 💀
Can you please justify your use of the log expansion at 4:08? Your expansion is valid for abs(z)
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?
Ofcourse its supposed to be real bro I'm talking about the final closed form😂
@@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.
cant we use by parts on this?
Well, is no one going to draw the roughly pentagonal contour traced by those five complex vectors?
Ending on the real line, a little to the left of zero?
In 7:40, the second sumand is + and the third is -
oh shit this was wild especially the part did not expect it to be REAL!
I think the 'i' in the denominator of sin might be a convention. These conventions have really messed me up.
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
@@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.
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
@@raddish4440 absolutely true. But think, if we did this all Physics teachers would stop saying there are no 'i''s in physics.
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?
Great vid! But when is the identity ln(e^x) = x valid in the case of complex numbers?
Well, always. ln(x) and exp(x) are each others inverse, also when the argument is complex.
😂 Thx for real value... 👍
i have a very difficult eq diff : y+1 = ( x^n ) * exp(-x)
i need the solution please 🙂
Try using integration by parts.
@@TH-cam_username_not_found how ?
@@borhenbouchniba wait, is this even an equation? what does y+ 1 mean ?
@@TH-cam_username_not_foundwe have f(x) = ( x^n ) * exp(-x) - 1 i need to find the reciprocal function of f
@@TH-cam_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
it's a little joke
Use a 3rd degree Lagrange polynomial. 99.8% accurate
ok cool
First for the first time. Thank you for your job !
Second😂