Hey! I realise I've jumped into using contour integrals but I thought this would be a good time to check - do you enjoy these kinds of videos? I am absolutely up for doing a few videos going over the basics of contour integration if that would be useful. Let me know!
I tend to think that Complex Analysis fixes the broken theory of the Real line. It brings a fundamental and beautiful appreciation of 'reality' to our eyes.
I always enjoyed this algorithm. my favorite class in undergrad was a complex calculus elective. Went to grad school for engineering and never saw the stuff again 😭
"Applied Complex Variables" by John W. Dettman (Dover Publishers) is a great read (The Math Sorcerer has a video on it.): the first part covers the geometry/topology of the complex plane from a Mathematician's perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicist's perspective. I've used Smith Charts (RF/microwave engineering) for years, but learned from Dettman that the "Smith Chart" is an instance of a Möbius Transformation. For practical reasons, a typical "Math Methods for Physics & Engineering" course introduces the Cauchy-Riemann Conditions, Conformal Mapping, Contour Integrals and applications of the Residue Theorem, but has to omit a lot interesting details. The Schaum's Outline on "Complex Variables" is a great companion book for more problems/solutions and content.
Love using complex analysis to solve real problems. It's my favourite way to solve them. Been years since I did it but always love seeing it. Other than the exams I wish I was still studying Maths.
Usually solving this with the Weierstrass substitution would be standard; using complex contours might actually take longer! Still, am enjoying the good video.
@@Tosi31415 That's fair! I love contour integrals so any chance I get to shortcut with them is my favourite but I definitely see the appeal of that method too!!
I am elated that maths video titles are following viral video templates, what a time to be alive. Nice video! As a physicist me no maths good. I am unfamiliar with Cauchy's residue theorem but nice to see an example of it
@@drover7476 Yes! Got to have the catchy titles these days... glad you enjoyed! If you'd like more complex analysis content let me know. Thanks for the comment!!
2:39 I was lost after this point, I'm not familiar with contour integration/Residue theorem. Would be nice to see some videos going over the basics though!
Man, I like this video, and I can't wait for the next one. A nice recap of complex analysis. Just, please, more words with "r" - I love your pronunciation!
@@Deepnil Hey! I use a canon DSLR on a tripod with a mic attached. I'm away at the moment so not sure of the exact model but I can let you know in a few weeks when I'm back!
This was great i am not familiar with complex values yet but they seem to be an integral part math. This video was so interesting if you are studying in school right now what field are you hoping to go into
@@O_of_1 Hoping to learn as much maths as I possibly can and see where it takes me! I'd love to be an academic some day. I'm so glad you enjoyed this! If I do more complex analysis in the future do you think some videos going over the basics would be helpful or is diving straight in best? Thank you!
@@OscgrMaths The Prime Number Theorem in particular might benefit :-) I am not being snotty, You full well know these things and how they work, I just think that most folk don't undestand these things. Sure Phycicists and Engineers go by the book and calculate but do they realy understand? (Bernhard Riemann)
@@OscgrMaths honestly I feel like you would be great at teaching the basics and it would serve nicely as one could reference your own video on complex analysis to help understand the more intricate complex integration problems you may cover . Overall great channel and I'm loving the content
Strange that this approach leads to same result as doing it straight without complex analysis. The 2 pi constant f.e. is just like "I'm a full circle, I'm everywhere" 🤔 Also strange that Integrating a straight line equals the real value of the contour of a circle, while it's clearly a different route, it kindof gives same result.
Thank you for your interesting video. May I ask you two additional exercises? 1) The pole is on the contour. I never understand if I have to include of or not the pole with a small semicircle on the neighbourhood of the pole or if it is the same to exclude this or not. 2) An integrale with an hypersingularity, not a simple pole. Thank you
Usually when there is a pole on the contour, I would suggest using a different substitution that results in a different contour. As far as I know, there is no one-size-fits-all solution for poles located directly on the contour itself. However, it definitely makes a difference whether you include it with a small semi-circle around the pole or not. I assume that by hypersingularity, you mean a pole of order greater than 1. The formula for calculating the residue of a pole is in fact more complicated than presented in this video. The residue of a pole z0 of order n is Res(f(z),z0)=(1/(n-1)!)*lim[z->z0](d^{n-1}/dz^{n-1}((z-z0)^n*f(z))). This results in the simpler formula presented in this video when plugging in n=1. The example exercise given at the end includes a pole of order 13, so the more general formula is necessary.
Really awesome and enjoyable video! Actually I was waiting to see a new video, and finally you did. Dealing with this integral with complex methods is much easier than real methods, I believe. To solve this with real methods, I think we can use tangent half-angle substitution, also known as Weirstrass substitution. For enjoyment, I will give it a try. For the challenge question, I could do it with the same method, and got 5pi/4. Thank you so much of making such content, please keep it up.
Thanks so much for the kind comment! Your answer to the challenge is very close which suggests you must have the right method. Let me know if you want the answer or if you share your method I can give you some help!
@@حسينالقطري-ب8ص The way I solved it I had to know that if the function is in the form of a series then the residue at the pole is just the coefficient of the term 1/z . This is because the definition of the residue of a simple pole is the coefficient of that term in the expansion of the function - hope that helps! Let me know if you have more questions.
@OscgrMaths First thought: We know that sin(x) and cos(x) lie between -1 and 1 for all real values of x. Therefore, raising their product to the power 6 will result values that are positive and closer to zero. Hence the value of the integral is about zero. Actual thought: (some mess😄)=5pi/512.
@@MathWithAnE I learnt a lot of stuff on contour from qncubed3 (excellent channel) and he often does his capital gammas like that 😭 it means when I'm using gamma anywhere else (like in gamma function) I do it normally but for contours specifically I do it that way. Thanks for the comment!!
Although Geometric Probability is a very simple concept, I wonder if you could find a problem with integrals that is otherwise unsolvable without geometric probability. You can deal with really complex probabilities using a geometric analogy, and it can be pretty simple to deal with "infinite fractions" (that is, finding area's without needing to integrate). I wonder if you could combine this with trig functions to find curves on a graph without Calculus. It would be quirky and irregular, but maybe it would be fun. 😄👌👍
I think you have made a sign error somewhere, as, intuitively the function is strictly positive. Solving with substitution, the answer is indeed 2pi/3. I however don't really know where this occurred. Great video nonetheless!
@justsomerowingguy-fn7mq Hey, the function is actually always negative! for the expression 1/(4cosx-5) consider the max and min values of cosx (1 and -1). At -1 the function evaluates to -1/11 and at 1, it is equal to -1. This means the integral should be negative. Try plotting it in desmos if you want to verify this! Thanks for the comment and support - let me know if you have any more questions.
If you use the complex expressions of cos and sin and substitute e^itheta as z, you'll have to use the more general formula for finding the residue of a pole z0 of order n. The pole in this video was of order 1, which gives this simple formula of Res(f(z), z0)=lim[z->z0]((z-z0)*f(z)), but in general, for a pole z0 of order n, the formula is Res(f(z), z0)=(1/(n-1)!)*lim[z->z0](d^{n-1}/dz^{n-1}((z-z0)^n*f(z))), which collapses to the previous formula when plugging in n=1. In this case, when using the substitution above, the order of the pole is 13.
I found that primitives of 1/(4cos(x)-5) are -2/3tan^-1(3tan(x/2))+C which are not continuous at x=pi. Is it problematic if my primitives aren’t continuous on [0,2pi]?
Not by any stretch of the imagination are these hieroglyphics Gammas. I had ancient Greek in school for 4 years, I would remember that. 🤣 Lowercase gamma looks like a "y", uppercase Gamma like an "F," but without the short horizontal stroke. Please practice.
Hello, try to destroy this integral with complex analysis, I will be attentive, thank you, greetings from Bogota Colombia,,,integrate ((Sin(e^(x^(4)))) from 2 to infinity)
I think you might be accidentally writing a delta, because you have no idea how anything works: i.pinimg.com/736x/86/fd/30/86fd30a1bfa4d53b31d9ec891e82d721--greek-alphabet-hand-written.jpg
you also demonstrate that you don't understand what suffix-s means in English, when you corrected yourself for saying 'residues'. what you said in speech was perfectly correct, because the meaning of '-s' is to despecify everything it c-commands, which is logically what you wanted in the context of what you were talking about at the time. being stupid, you probably think that '-s' is a plural marker. never realizing that it appears when we're speaking of an entire category, when we're speaking of a lack of something, and it not only can appear on verbs, but it can be the entire verb in the form of 'is'. - Cats have tails. = A cat has a tail. - Cats run fast. - No cats run fast. - That cat runs fast. - That cat is fast. like I said... you clearly have no idea how anything works.
Hey! I realise I've jumped into using contour integrals but I thought this would be a good time to check - do you enjoy these kinds of videos? I am absolutely up for doing a few videos going over the basics of contour integration if that would be useful. Let me know!
please do a vid on the basics that would help 🙏
@@MathCastt Okay great will do!!
I tend to think that Complex Analysis fixes the broken theory of the Real line. It brings a fundamental and beautiful appreciation of 'reality' to our eyes.
Yes!!!!! You should!!!
I like your technique of wiping the board clean and moving on.
I always enjoyed this algorithm. my favorite class in undergrad was a complex calculus elective. Went to grad school for engineering and never saw the stuff again 😭
You should try Quantum Physics. Complex values are essential to it.
"Applied Complex Variables" by John W. Dettman (Dover Publishers) is a great read (The Math Sorcerer has a video on it.): the first part covers the geometry/topology of the complex plane from a Mathematician's perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicist's perspective. I've used Smith Charts (RF/microwave engineering) for years, but learned from Dettman that the "Smith Chart" is an instance of a Möbius Transformation.
For practical reasons, a typical "Math Methods for Physics & Engineering" course introduces the Cauchy-Riemann Conditions, Conformal Mapping, Contour Integrals and applications of the Residue Theorem, but has to omit a lot interesting details.
The Schaum's Outline on "Complex Variables" is a great companion book for more problems/solutions and content.
@@douglasstrother6584 Nice! Thanks for the comment this is great.
My lecturer use "Complex Analysis 9th Edition" by James Brown and Ruel Churchill
@@f.r.y5857 That's a classic.
I'm reading through "Functions of a Complex Variable" by Thomas M. MacRobert (1950) which I found at my public library.
Love using complex analysis to solve real problems. It's my favourite way to solve them. Been years since I did it but always love seeing it. Other than the exams I wish I was still studying Maths.
@@Unchained_Alice Me too! There's something so satisfying about using complex analysis.
Usually solving this with the Weierstrass substitution would be standard; using complex contours might actually take longer! Still, am enjoying the good video.
Good solution, hovever using clever substitutions and algebric manipulation remains the best and most fun method for me
@@Tosi31415 That's fair! I love contour integrals so any chance I get to shortcut with them is my favourite but I definitely see the appeal of that method too!!
That's a pretty good exercise! Thanks for sharing!
@@bernardmarquot996 Thank you!
Been 10 years since I graduated and I dont think I would pass that many exams if I had to do them now. I remember the concepts but not the details
Nicely done! This is transporting me back 1/2 century.
@@DavidMFChapman Glad you enjoyed!
I instantly thought of Weierstrass substitution personally
I am elated that maths video titles are following viral video templates, what a time to be alive. Nice video! As a physicist me no maths good. I am unfamiliar with Cauchy's residue theorem but nice to see an example of it
@@drover7476 Yes! Got to have the catchy titles these days... glad you enjoyed! If you'd like more complex analysis content let me know. Thanks for the comment!!
HOLY IVE NEVER SEEN COMPLEX ANALYSIS BUT THAT WAS SO COOL
@@NotGleSki I'M GLAD YOU FOUND IT COOL - I AGREE!!
Bro can mog Isaac Newton
high school math is hard and this harder. but still, love the video. love the hard wor, keep going, and keep confusing me lol
Thanks so much!
This was a great video and I love contour integration but I really struggle to understand it, could you do a video on the basics
@@gjproductions9337 Yeah absolutely! I was thinking that might be my next step with this. Thanks for the comment!!
great segue to complex number integration, for people who haven't had much exposure to that or taken a complex analysis course.
@@xoppa09 Definitely! Thanks for the comment.
Yes boss 👍 nothing like maths in the holiday
Could you try some partial differentials soon ? Would be very much enjoyed
Yes! Thanks for the comment.
This is really complex way but
it can be calculated in elementary way
2:39 I was lost after this point, I'm not familiar with contour integration/Residue theorem. Would be nice to see some videos going over the basics though!
@@alian714 Okay! Thanks for the comment this is great to know.
cosx -> tan^ x/2
Gotta love some complex integration!
@@StarGazer-c3k Definitely ! Thanks for the comment.
I am in high school, and I dont understand anything, but ur energy and passion in ur explanations has earned u a subscriber🎉
@@unamngxale8286 Thanks so much!! Feel free to ask any questions you have.
@@OscgrMaths Will sure do🙏🏾
Bro mogged that integral harder than Newton😂
@@FLASH24x Thanks bro 😂
great exercise, well explained
@@dominiquecolin4716 Thanks so much!
Great channel, subbing and looking forward for more!! 🦕
@@災厄-b9o Thanks so much! So glad you've enjoyed.
Man, I like this video, and I can't wait for the next one.
A nice recap of complex analysis.
Just, please, more words with "r" - I love your pronunciation!
Really good, thanks.
Thanks!
Heya mate! Aweosme stuff! Can I ask what recording equipment you use?
@@Deepnil Hey! I use a canon DSLR on a tripod with a mic attached. I'm away at the moment so not sure of the exact model but I can let you know in a few weeks when I'm back!
@@OscgrMathsfascinating !
This was great i am not familiar with complex values yet but they seem to be an integral part math. This video was so interesting
if you are studying in school right now what field are you hoping to go into
@@O_of_1 Hoping to learn as much maths as I possibly can and see where it takes me! I'd love to be an academic some day. I'm so glad you
enjoyed this! If I do more complex analysis in the future do you think some videos going over the basics would be helpful or is diving straight in best? Thank you!
@@OscgrMaths Poles and Zeroes would be sufficient methinks. They are so glossed over. Understand them and you understand numbers.
@@alphalunamare Good to know, thanks!!
@@OscgrMaths The Prime Number Theorem in particular might benefit :-) I am not being snotty, You full well know these things and how they work, I just think that most folk don't undestand these things. Sure Phycicists and Engineers go by the book and calculate but do they realy understand? (Bernhard Riemann)
@@OscgrMaths honestly I feel like you would be great at teaching the basics and it would serve nicely as one could reference your own video on complex analysis to help understand the more intricate complex integration problems you may cover . Overall great channel and I'm loving the content
Strange that this approach leads to same result as doing it straight without complex analysis. The 2 pi constant f.e. is just like "I'm a full circle, I'm everywhere" 🤔
Also strange that Integrating a straight line equals the real value of the contour of a circle, while it's clearly a different route, it kindof gives same result.
Great video, fellow maths educator! I think you mean (capital) "omega", not "gamma"? At least from the way you've drawn it..
@@inverse_of_zero Thank you!
You are so cool! :)
at 4:09 won't the z be in numerator as well?
Thank you for your interesting video.
May I ask you two additional exercises?
1) The pole is on the contour. I never understand if I have to include of or not the pole with a small semicircle on the neighbourhood of the pole or if it is the same to exclude this or not.
2) An integrale with an hypersingularity, not a simple pole.
Thank you
Usually when there is a pole on the contour, I would suggest using a different substitution that results in a different contour. As far as I know, there is no one-size-fits-all solution for poles located directly on the contour itself. However, it definitely makes a difference whether you include it with a small semi-circle around the pole or not.
I assume that by hypersingularity, you mean a pole of order greater than 1. The formula for calculating the residue of a pole is in fact more complicated than presented in this video. The residue of a pole z0 of order n is Res(f(z),z0)=(1/(n-1)!)*lim[z->z0](d^{n-1}/dz^{n-1}((z-z0)^n*f(z))). This results in the simpler formula presented in this video when plugging in n=1. The example exercise given at the end includes a pole of order 13, so the more general formula is necessary.
Can you make a video about the challenge you give at the end. Please❤❤❤❤
Or anyone who has solution with details. Can post it
So cool!
@@sebastians7346 Thanks so much! Really glad you enjoyed.
that's insane, great video! I need to make my videos more like this lmfao
Really awesome and enjoyable video!
Actually I was waiting to see a new video, and finally you did.
Dealing with this integral with complex methods is much easier than real methods, I believe.
To solve this with real methods, I think we can use tangent half-angle substitution, also known as Weirstrass substitution. For enjoyment, I will give it a try.
For the challenge question, I could do it with the same method, and got 5pi/4.
Thank you so much of making such content, please keep it up.
Thanks so much for the kind comment! Your answer to the challenge is very close which suggests you must have the right method. Let me know if you want the answer or if you share your method I can give you some help!
@OscgrMaths omg 😂
I did with + between the two functions!
Now will do the multiplication one.
@@حسينالقطري-ب8ص The way I solved it I had to know that if the function is in the form of a series then the residue at the pole is just the coefficient of the term 1/z . This is because the definition of the residue of a simple pole is the coefficient of that term in the expansion of the function - hope that helps! Let me know if you have more questions.
@OscgrMaths First thought:
We know that sin(x) and cos(x) lie between -1 and 1 for all real values of x. Therefore, raising their product to the power 6 will result values that are positive and closer to zero. Hence the value of the integral is about zero.
Actual thought: (some mess😄)=5pi/512.
@@حسينالقطري-ب8ص Wow nice work!
Bro that Γ be wildin💀
NOOOO I was trying out a new style for my capital gamma and it has definitely backfired 😭😭😭
@@OscgrMaths fr bro 😭
Super nice video! Why do you write your Gamma like that tho😭
@@MathWithAnE I learnt a lot of stuff on contour from qncubed3 (excellent channel) and he often does his capital gammas like that 😭 it means when I'm using gamma anywhere else (like in gamma function) I do it normally but for contours specifically I do it that way. Thanks for the comment!!
@@OscgrMaths stick to your own style .. who knows? it might take off 🙂
Although Geometric Probability is a very simple concept, I wonder if you could find a problem with integrals that is otherwise unsolvable without geometric probability. You can deal with really complex probabilities using a geometric analogy, and it can be pretty simple to deal with "infinite fractions" (that is, finding area's without needing to integrate). I wonder if you could combine this with trig functions to find curves on a graph without Calculus. It would be quirky and irregular, but maybe it would be fun. 😄👌👍
Thanks for the comment!! Excellent suggestion.
I think you have made a sign error somewhere, as, intuitively the function is strictly positive. Solving with substitution, the answer is indeed 2pi/3. I however don't really know where this occurred. Great video nonetheless!
@justsomerowingguy-fn7mq Hey, the function is actually always negative! for the expression 1/(4cosx-5) consider the max and min values of cosx (1 and -1). At -1 the function evaluates to -1/11 and at 1, it is equal to -1. This means the integral should be negative. Try plotting it in desmos if you want to verify this! Thanks for the comment and support - let me know if you have any more questions.
wouldnt intergration by parts and algebraic elimination work here?
Very cool video :D
@@beautyofmath6821 Thank you!
Could you post a solution for the problem in the end pls? I’ve tried for 3 days and couldn’t solve it with complex análisis
If you use the complex expressions of cos and sin and substitute e^itheta as z, you'll have to use the more general formula for finding the residue of a pole z0 of order n. The pole in this video was of order 1, which gives this simple formula of Res(f(z), z0)=lim[z->z0]((z-z0)*f(z)), but in general, for a pole z0 of order n, the formula is Res(f(z), z0)=(1/(n-1)!)*lim[z->z0](d^{n-1}/dz^{n-1}((z-z0)^n*f(z))), which collapses to the previous formula when plugging in n=1. In this case, when using the substitution above, the order of the pole is 13.
is ans 5pi/512
@@zeninfx7053 Yes!! Well done.
@@OscgrMaths i used wallis integral though 😅
Why do you only use the center half of the board? You could keep more useful information visible if the rest of the spaced is used.
Yo congrats on 3k... 10k soon?!?!
@@gregoriousmaths266 Hmm.. not sure about soon... but thanks!!
@@OscgrMaths idk man i feel like you were literally just on 2k lol
❤ awesome
Thanks so much, really appreciate it.
Try to make this less confusing for high school students.
This subject is not for high school level.
What if the denominator term was a double root (pole), e.g. (x-1/2)^2, the pole factor would not then cancel nicely in the residue?
I found that primitives of 1/(4cos(x)-5) are -2/3tan^-1(3tan(x/2))+C which are not continuous at x=pi.
Is it problematic if my primitives aren’t continuous on [0,2pi]?
So what is dz/dθ exactly?
Wow
is the answer of the integral from 0 to 2pi of cos^6x*sin^6x = 5pi/1024?
@@fakecreeper9645 Very close!!! you have one two many factors of 2 on your denominator...
Nice vid
Thanks so much!
This is better than Poirot! :-)
@@alphalunamare That's a big compliment!! Thanks a lot for the comment.
Not by any stretch of the imagination are these hieroglyphics Gammas. I had ancient Greek in school for 4 years, I would remember that. 🤣
Lowercase gamma looks like a "y", uppercase Gamma like an "F," but without the short horizontal stroke. Please practice.
Hello, try to destroy this integral with complex analysis, I will be attentive, thank you, greetings from Bogota Colombia,,,integrate ((Sin(e^(x^(4)))) from 2 to infinity)
Very interesting even though you completely lost me at about 10 seconds.
i solve it i get 5pi/1024 not 5pi/512 can any one figure out why?
Can you show your solution?
hang on... can we just have a whole video about why you can't write the letter gamma?
I think you might be accidentally writing a delta, because you have no idea how anything works:
i.pinimg.com/736x/86/fd/30/86fd30a1bfa4d53b31d9ec891e82d721--greek-alphabet-hand-written.jpg
you also demonstrate that you don't understand what suffix-s means in English, when you corrected yourself for saying 'residues'.
what you said in speech was perfectly correct, because the meaning of '-s' is to despecify everything it c-commands, which is logically what you wanted in the context of what you were talking about at the time.
being stupid, you probably think that '-s' is a plural marker. never realizing that it appears when we're speaking of an entire category, when we're speaking of a lack of something, and it not only can appear on verbs, but it can be the entire verb in the form of 'is'.
- Cats have tails. = A cat has a tail.
- Cats run fast.
- No cats run fast.
- That cat runs fast.
- That cat is fast.
like I said... you clearly have no idea how anything works.
Please don’t sell out to using clickbait terms like DESTROYED to get page views. NO SUBSCRIPTION
@@MyOneFiftiethOfADollar But it did destroy though...
I agree. It is childish.
Great job but your capital gammas are horrible!!(imo)😒
NEVER write "×" for multiplication man .... this is not elementary school! please!
Teta, Eta, nemam pojma, Pi. Scenne searching stupid MyFace.