Big thanks to blackpenredpen for hosting my video! My channel features a bunch of interesting topics in maths and physics. I'm still finding my feet right now, but I hope to have some video series that cover mathematical topics in detail (group theory, complex analysis, ...) as well as some curiosities (pythagoras's theorem proofs, the hidden link between electricity and magnetism, ...)
I assume at 4:20 you implicitly used small angle approximation formula. Besides, your announcement seems like a full duplicate of 3Blue1Brown video (especially latest analogy with optics and symmetrical spaces).
@@sergpodolnii3962 I think that depends on where you choose the first linear segment to be - I chose it so that it was vertical - so pi/2 is the correct angle. Yeah, my solution was very similar to 3b1b, but was my own solution and came out first. I probably wouldn't have made so much of a mess of the inscribed angle theorem if I had known the answer!
I'm pretty sure you can do the same by turning the complex curvy integral into an integral with real bounds. basically, if a curve is defined by the function f(x) that takes real values and outputs complex values and by the bounds a and b which are real. the integral of g(x)dx over that curve is the integral from a to b of g(f(x))f'(x)dx. so in this case the integral from 0 to 2π of 1/(e^ix)*i*e^ix*dx. | | V the integral of i*dx from 0 to 2π indeed, you get 2πi
My favourite maths book is 'Visual Complex Analysis'! I'm going to be producing a lot more videos that will hopefully share my excitement for the subject!
Thank you very much for explaining the residue theorem in a intuitive manner. This is my first exposure to the subject and it's making me excited to take complex analysis, whenever I get to it. I can't wait to start solving some integrals with this!
I have a list of things I wanna do when I grow up and start earning. There are 2 UA-cam channels that are on that list of which I want to become a patron of : Blackpenredpen and Kurzgesagt.
To understand how the complex numbers C form an inner product vector space, it suffices to see that a + bi = (a, b). Then we can defined the complex product in terms of matrix-vector products. Let M = [(1, 0); (0, -1)] and N = [(0, 1); (1, 0)], the permutation matrix. Then the complex number product between complex numbers x = (a, b) and y = (c,d) is given by the vector xy = (a’Mb, a’Nb), with a’Mb and a’Nb representing nonstandard (non-Cartesian inner products).
Took a little bit of time not to be distracted by the accent. I am used to listening to that accent giving an excellent lecture on ancient Briton or the flora and fauna of the Amazon rain forest. Lol. But, it was an excellent demo. BPRP, do you own version please. I could listen to you all day! You should do a little stand up comedy on the side!
I was just studying residues lately... It is not so simple to fully teach these things in one video because imagine that you have an essential singularity, then you must expand the function in a laurent series to find the residue. Or if you have a pole of order 4 for example... Complex analysis is quite the endeavor
I prefer to look at this in another manner, dz/z the numerator dz rotates at exactly the same rate as 1/z but in the opposite direction so dz/z does not rotate and the integral is linear adding up to 2pi for a circle circumference. For other situations where the product F(z) dz, rotates, the summation is zero. most of the time, It is like a person going out for a walk if he changes direction to go back to the same spot he started then the integral is zero, but if he does not change direction then he covers some distances in his integral.
This is fascinating. My favourite mathematical problem is below. And it relates, I seem to think, to this video explanation of Cauchy's Residue Theorem. But, for the life of me, I cannot understand why my problem has such a 'nice' answer, which it does. You can solve it geometrically (Desmos will get you a quick answer) or algebraically, more satisfying, but you need to know L'Hospital's Rule. Here it is: imagine an n-sided regular polygon with an incircle and a circumcircle. Let's define A as the area between the circumcircle and the polygon, and B as the area between the polygon and the incircle. What is the ratio of A to B as n tends to infinity? I'd love it if someone can enlighten me as to why this problem has the 'nice answer' that it does. Many thanks
Can you find x when cosxsinx=x ? I know that x=0 from the intersection of the graph of both functions but is there any way we can use algebra to solve for x here ?
@@omarelataly3628 1/2sin(2x) = 0 so sin(2x) = 0. sin t = 0 at t = 0, π, 2π, 3π etc. So t is any integer multiple of π. Since t = 2x, x can be any integer multiple of π/2
Ben Durham No, you solved the wrong equation. He asked to solve sin(x)cos(x) = x, not sin(x)cos(x) = 0. He is asking for the solution of the equation sin(2x) = 2x, which is x = 0, and I am fairly certain this is the only possible solution. If we let x be complex, then we should substitute y = 2x and instead solve sin(y) = y for complex y. This means sin(y) = sin[Re(y)]cos[i·Im(y)] + sin[i·Im(y)]cos[Re(y)] = sin[Re(y)]cosh[Im(y)] + i·sinh[Im(y)]cos[Re(y)]. Then the solution to sin(y) = y is given by the solutions to the system of equations Re(y) = sin[Re(y)]cosh[Im(y)] and Im(y) = sinh[Im(y)]cos[Re(y)]. Now, since sinh[Im(y)] = cosh[Im(y)]tanh[Im(y)], we can say Im(y) = ([Re(y)]/sin[Re(y)])tanh[Im(y)]cos[Re(y)]. We can rewrite this as Im(y)/Re(y) = tanh[Im(y)]cot[Re(y)] or as tan[Re(y)]/Re(y) = tanh[Im(y)]/Im(y). By now, it may simply be convenient to let Re(y) = z and let Im(y) = f[Re(y)] = f(z) for some yet-to-be-found f. Then this simply becomes tan(z)/z = tanh[f(z)]/f(z). Keep in mind that f : R/{0} -> R/{0}, so f(z) = i·z is not a possibility. If there exists some f such that for some value of z, the equation above has solutions, then you will find other solutions to sin(y) = y other than y = 0. Otherwise, y = 0 is the only solution, and I am afraid this latter is true. This can be seen by letting f(z) = c·z and solving z as a function of c or c as a function of z after substitution in the previous equation. What you will need to find is c·tan(z) = tanh(c·z) and c(z) in the solution set, where c is real.
Omar Magdy You cannot use much algebra since finding solutions to sin(y) = y is equivalent to finding an inverse to g(y) = sin(y) - y or to h(y) = y - sin(y), which is non-elementary.
Your videos are awesome I have a doubt which i have been trying for it about 1 week but not getting it. Question is related to trigonometry Q) Tan(x+100°)=Tan(x)+Tan(x+50°)+Tan(x-50°). I will be very thankful if u or anybody could help me out! Thanks in advance
we have 0 1 2 3 4 5 6 . we take 4 of them at the same time . what are all the possibilities to make a round number ? (You can't choose the same number more than 1 time)
@@arequina OK, I just did. Sub z=R*e^(i*theta). Then dz = R*i*e^(i*theta) dTheta. If you plug them in, lots of stuff cancels and you get integral of i * dTheta which is 2*pi*i because you integrate from 0 to 2 pi radians. I get the same result in about 4 lines. Explain why this is wrong or why it is harder?
@@angelmendez-rivera351 I think all you need is the knowledge that z = R * e^(i * theta) which I think is taught fairly early when you start studying complex numbers. It is just some trig. In fact he even mentions that z = R * e^(i * theta) in the video.
Herbert Susmann Yes, but that is not sufficient information to perform the integration. You still would need to use the fact that for the given contour, r = 1, C(z) = e^i·z, and that the integrand is 1/z·C’(z), but the latest fact comes directly and necessarily as a theorem on line integrals, which requires covering some vector calculus, and that would make the video unnecessarily long. For an introductory video on complex analysis and residues, this video presents a better approach.
Hello blackpenredpen just wanted to say I love your videos! I'm currently struggling with a limit, and that is "lim x→∞ ((2x - 3)/(2x + 1))^(x)". If you wish, please solve in an upcoming video of yours. Thanks in advance for you time!
Justin Chan 2x - 3 = (2x + 1) - 4, so (2x - 3)/(2x + 1) = [(2x + 1) - 4]/(2x + 1) = 1 - 4/(2x + 1). Now let y = -4/(2x + 1), which implies 2x + 1 = -4/y, which implies x = -(2/y + 1/2). Therefore, [1 - 4/(2x + 1)]^x = (1 + y)^[-(2/y + 1/2)] = [(1 + y)^(-2/y)][(1 + y)^(-1/2)]. Limits are multiplicative, so you can obtain the product of the limits of each factor to get the limit of the product, which is what you want, because they are the same. Therefore, we can find the limits of (1 + y)^(-1/2) and of (1 + y)^(-2/y) separately and multiply them to get the answer you desire. The easier limit to calculate is the limit of (1 + y)^(-1/2) = 1/sqrt(1 + y). As x -> ♾, y -> 0-, and lim 1/sqrt(1 + y) (y -> 0-) = 1/sqrt(1 + 0) = 1. Now, to calculate lim (1 + y)^(-2/y) (y -> 0-), notice that (1 + y)^(-2/y) = [(1 + y)^(1/y)]^(-2). Now let z = -1/y. As y -> 0-, z -> ♾, so we want lim [(1 - 1/z)^z]^2 (z -> ♾). Since f(t) = t^2 is a continuous, we can find the limit of (1 - 1/z)^z and then square it to get the desired limit, which actually would be the answer since the answer is just this multiplied by 1. (1 - 1/z)^z = e^[ln(1 - 1/z)/(1/z) = e^(-[-ln(1 - 1/z)/(1/z)]) = e^(-ln[1/(1 - 1/z)]/[1/z]) = e^(-ln[z/(z - 1)]/[1/z]) = e^(-ln[1 + 1/(z - 1)]/[1/z]) = e^[-ln([1 + 1/(z - 1)]^z)] = ([1 + 1/(z - 1)]^z)^(-1). Let z = α + 1, such that as z -> ♾, α -> ♾, with ([1 + 1/(z - 1)]^z)^(-1) = [(1 + 1/α)^(α + 1)]^(-1). Thus, we want to lim (1 + 1/α)^(α + 1) (α -> ♾) and raise it to -2 to get the solution to your problem. Now notice that (1 + 1/α)^(α + 1) = [(1 + 1/α)^α](1 + 1/α). One last time, we can find the product of the individual limits to get the limit of the product. Αs α -> ♾, 1 + 1/α -> 1. lim (1 + 1/α)^α (α -> ♾) = e by definition, so the product of the limits is e, and as such, the solution to your problem is e^(-2) = 1/e^2.
I don’t understand what the integral is summing up. I want to see a curve under which we are calculating an area. I don’t understand what it means to just put C as the starting point. Where is 1/z? What is happening here?
ophello There is no curve under which you are calculating the area under because the integration is over the complex numbers, not the real numbers. Unlike integration over the real numbers, there is basically no unambiguous physical-geometric interpretation to complex integrals.
jhony angarita El que la circunferencia esté definida por una curva cerrada no indica que la circunferencia no es el perímetro de un círculo, pues estrictamente hablando, nada en la definición del perímetro dice que contorno del cual se toma el perímetro debe tener curvatura cero. Así que no importa que el círculo no sea un polígono.
así los otros polígonos por más que sus lados sea pequeños siguen siendo líneas rectas no hay que confundir medida con forma ala distancia un cuadrado Seve como un punto
jhony angarita Cierto, pero no estamos trabajando con los polígonos mismos, sino con el límite de la secuencia de los polígonos, y esa límite sí incluye curvas. Si no sabes lo que es un límite, entonces ponte a estudiar y deja de criticar.
cuando existe un ángulo Ai una horizontal una vertical una diagonal dicha diagonal es el radio para los polígonos y para la circunferencia los ángulos nunca llegarán al límite para dar la medida exacta de la curva
You said i was the variable of the summation going from 1 to N, then you used the same letter to represent the imaginary unit. Perhaps it might be better not to use i as the summation variable. And what about Delta i, which i was that?
3.14159... es la mitad de la sumatoria de los lados de un polígono regular de radio igual a 1 este perímetro está compuesto de rectas una línea curva no está compuesto de rectas y 2π no es el valor correcto
jhony angarita Lo que dices es incorrecto. En primer lugar, 3.14159... no significa nada porque más allá del elipsis puede seguir cualquier secuencia de dígitos arbitraria, y por tanto, tal expresión no representa un número, sino como menos un conjunto de números. Lo segundo es que tal sumatoria depende del polígono en sí mismo. El círculo no es un polígono, pero sí es el límite de la secuencia de los polígonos regulares, y como la integral por definición también es un límite y no una mera sumatoria, esto implica por necesidad lógica y axiomática que la calculación debe ser dada estrictamente por un límite, y este límite lo que da el resultado de 2πi. El resultado es demostrablemente correcto, y el que seas ignorante en el ámbito del análisis complejo y del cálculo no justifica que digas bobadas como las que dices aquí y que para encima taches a alguien más de incorrecto.
You have the coolest friends. Again, "The Big Bang Theory" is wrapping up, so maybe there could be a show about you guys solving equations and raising hell in biker bars. Yes I mean you'll bring a white board to biker bars and do calculus there.
Sorry about that, I think it would have taken another 10 minutes at least to explain what a residue is. I just wanted to get across the idea that they are associated with points where the function goes to +/- infinity.
Big thanks to blackpenredpen for hosting my video! My channel features a bunch of interesting topics in maths and physics. I'm still finding my feet right now, but I hope to have some video series that cover mathematical topics in detail (group theory, complex analysis, ...) as well as some curiosities (pythagoras's theorem proofs, the hidden link between electricity and magnetism, ...)
I'm assuming your makeup lady is still on strike....
I assume at 4:20 you implicitly used small angle approximation formula. Besides, your announcement seems like a full duplicate of 3Blue1Brown video (especially latest analogy with optics and symmetrical spaces).
@@jibran8410 Haha yes! She'll make me look respectable in the next video!
@@sergpodolnii3962 I think that depends on where you choose the first linear segment to be - I chose it so that it was vertical - so pi/2 is the correct angle. Yeah, my solution was very similar to 3b1b, but was my own solution and came out first. I probably wouldn't have made so much of a mess of the inscribed angle theorem if I had known the answer!
I'm pretty sure you can do the same by turning the complex curvy integral into an integral with real bounds.
basically, if a curve is defined by the function f(x) that takes real values and outputs complex values and by the bounds a and b which are real.
the integral of g(x)dx over that curve is the integral from a to b of g(f(x))f'(x)dx.
so in this case
the integral from 0 to 2π of 1/(e^ix)*i*e^ix*dx.
| |
V
the integral of i*dx from 0 to 2π
indeed, you get 2πi
Cauchy's Residue Theorem, and the techniques that make use of it, are the closest thing to magic that I know of in mathematics.
I love that what he is doing is going from an integral, to essentially the definition of an integral. I love it.
I don't recommend using i as an index of a series when dealing with complex numbers. It can create confusion.
I had the same thought
It was really confusing
Pls more Complex Analysis visual way!!! That thing is so rare on yt...
My favourite maths book is 'Visual Complex Analysis'! I'm going to be producing a lot more videos that will hopefully share my excitement for the subject!
Yea, its really difficult to find beginner friendly material on contour integration.
Woooooow!!! 👏 👏 👏
Thank you very much for explaining the residue theorem in a intuitive manner. This is my first exposure to the subject and it's making me excited to take complex analysis, whenever I get to it. I can't wait to start solving some integrals with this!
Would like a bit more explanation about residues
Yeah sorry it all got a little rushed at the end there. I didn't want to abuse blackpenredpen's generosity with a 20 minute video!
@@stemcell7200 I read this " golem.ph.utexas.edu/category/2019/11/random_permutations_part_2.html " and still trying to understand residue
I have a list of things I wanna do when I grow up and start earning. There are 2 UA-cam channels that are on that list of which I want to become a patron of : Blackpenredpen and Kurzgesagt.
Anwesha Guha aww thank you!!
To understand how the complex numbers C form an inner product vector space, it suffices to see that a + bi = (a, b). Then we can defined the complex product in terms of matrix-vector products. Let M = [(1, 0); (0, -1)] and N = [(0, 1); (1, 0)], the permutation matrix. Then the complex number product between complex numbers x = (a, b) and y = (c,d) is given by the vector xy = (a’Mb, a’Nb), with a’Mb and a’Nb representing nonstandard (non-Cartesian inner products).
Took a little bit of time not to be distracted by the accent. I am used to listening to that accent giving an excellent lecture on ancient Briton or the flora and fauna of the Amazon rain forest. Lol. But, it was an excellent demo. BPRP, do you own version please. I could listen to you all day! You should do a little stand up comedy on the side!
Lovely explanation. Perfect
Just as I am learning about this in class. Thanks for sharing it was very clear!
Higgsino physics : ))))
How’d your class go?
Cauchy residue theorem? More like “Cool, I’m glad to hear about ‘em”…your explanations that is. Thanks so much for sharing!
why in the minute 3:35 you multiply the value of the function with delta i? why can't you just add only the mean values of the function?
I was just studying residues lately... It is not so simple to fully teach these things in one video because imagine that you have an essential singularity, then you must expand the function in a laurent series to find the residue. Or if you have a pole of order 4 for example... Complex analysis is quite the endeavor
The summation index i is easily confused for the imaginary number i, I’m sure I’m not the only one who was caught off guard by that several times
I prefer to look at this in another manner,
dz/z the numerator dz rotates at exactly the same rate as 1/z but in the opposite direction so dz/z does not rotate and the integral is linear adding up to 2pi for a circle circumference. For other situations where the product F(z) dz, rotates, the summation is zero. most of the time,
It is like a person going out for a walk if he changes direction to go back to the same spot he started then the integral is zero, but if he does not change direction then he covers some distances in his integral.
Very clear and very nice
Could you elaborate on what even is a 'contour integral' ?
I just subscribed. Brings back memories.
arequina yay thank you!!
It is really awesome!
Excellent explanation.
One doubt
How 1/A2 is becomes 1/A e^I psi (Apsi e^ipi/2 psi
This is fascinating. My favourite mathematical problem is below. And it relates, I seem to think, to this video explanation of Cauchy's Residue Theorem. But, for the life of me, I cannot understand why my problem has such a 'nice' answer, which it does. You can solve it geometrically (Desmos will get you a quick answer) or algebraically, more satisfying, but you need to know L'Hospital's Rule. Here it is: imagine an n-sided regular polygon with an incircle and a circumcircle. Let's define A as the area between the circumcircle and the polygon, and B as the area between the polygon and the incircle. What is the ratio of A to B as n tends to infinity? I'd love it if someone can enlighten me as to why this problem has the 'nice answer' that it does. Many thanks
Thankyou.very helpful video
7:35 isn't the singularity sepossed to move?
Can you find x when cosxsinx=x ?
I know that x=0 from the intersection of the graph of both functions but is there any way we can use algebra to solve for x here ?
lol cos(x)sin(x)=1/2 sin(2x)
@@DarthSagit yes I do know that ... but also when it's in the shape of the double angle formula , how are you gonna solve it ? :)
@@omarelataly3628 1/2sin(2x) = 0 so sin(2x) = 0. sin t = 0 at t = 0, π, 2π, 3π etc. So t is any integer multiple of π. Since t = 2x, x can be any integer multiple of π/2
Ben Durham No, you solved the wrong equation. He asked to solve sin(x)cos(x) = x, not sin(x)cos(x) = 0. He is asking for the solution of the equation sin(2x) = 2x, which is x = 0, and I am fairly certain this is the only possible solution. If we let x be complex, then we should substitute y = 2x and instead solve sin(y) = y for complex y. This means sin(y) = sin[Re(y)]cos[i·Im(y)] + sin[i·Im(y)]cos[Re(y)] = sin[Re(y)]cosh[Im(y)] + i·sinh[Im(y)]cos[Re(y)]. Then the solution to sin(y) = y is given by the solutions to the system of equations Re(y) = sin[Re(y)]cosh[Im(y)] and Im(y) = sinh[Im(y)]cos[Re(y)]. Now, since sinh[Im(y)] = cosh[Im(y)]tanh[Im(y)], we can say Im(y) = ([Re(y)]/sin[Re(y)])tanh[Im(y)]cos[Re(y)]. We can rewrite this as Im(y)/Re(y) = tanh[Im(y)]cot[Re(y)] or as tan[Re(y)]/Re(y) = tanh[Im(y)]/Im(y). By now, it may simply be convenient to let Re(y) = z and let Im(y) = f[Re(y)] = f(z) for some yet-to-be-found f. Then this simply becomes tan(z)/z = tanh[f(z)]/f(z). Keep in mind that f : R/{0} -> R/{0}, so f(z) = i·z is not a possibility. If there exists some f such that for some value of z, the equation above has solutions, then you will find other solutions to sin(y) = y other than y = 0. Otherwise, y = 0 is the only solution, and I am afraid this latter is true. This can be seen by letting f(z) = c·z and solving z as a function of c or c as a function of z after substitution in the previous equation. What you will need to find is c·tan(z) = tanh(c·z) and c(z) in the solution set, where c is real.
Omar Magdy You cannot use much algebra since finding solutions to sin(y) = y is equivalent to finding an inverse to g(y) = sin(y) - y or to h(y) = y - sin(y), which is non-elementary.
I don't know people dislike such vids
EVEN THE THEORY OF EVERYTHING CANT FIND A SOLUTION!!!!!!
Didn't got that 4:16 can anyone explain
Ahhh good memories of complex analysis. Do some partial differential equations!
Your videos are awesome
I have a doubt which i have been trying for it about 1 week but not getting it. Question is related to trigonometry
Q) Tan(x+100°)=Tan(x)+Tan(x+50°)+Tan(x-50°). I will be very thankful if u or anybody could help me out!
Thanks in advance
Rajendra singh rathore
Hmmmm I will think about it!
we have 0 1 2 3 4 5 6 . we take 4 of them at the same time . what are all the possibilities to make a round number ?
(You can't choose the same number more than 1 time)
Great Vid !!!
Why is his channels name stem cell I mean he is making math vids right
Do you know what 'STEM' means?
@@hello_2632 stem cells are a type of cell in your body.
STEM stands for Science, Technology, Engineering, and Maths.
It's very nice
Why not just keep z in polar form and integrate theta from 0 to 2 pi? Seems a lot easier than limit method.
Try it and you'll see why this is the preferred method.
@@arequina OK, I just did. Sub z=R*e^(i*theta). Then dz = R*i*e^(i*theta) dTheta. If you plug them in, lots of stuff cancels and you get integral of i * dTheta which is 2*pi*i because you integrate from 0 to 2 pi radians. I get the same result in about 4 lines. Explain why this is wrong or why it is harder?
He would need to prove that such an integral is equivalent, which is more of a hassle without sufficient knowledge on complex analysis.
@@angelmendez-rivera351 I think all you need is the knowledge that z = R * e^(i * theta) which I think is taught fairly early when you start studying complex numbers. It is just some trig. In fact he even mentions that z = R * e^(i * theta) in the video.
Herbert Susmann Yes, but that is not sufficient information to perform the integration. You still would need to use the fact that for the given contour, r = 1, C(z) = e^i·z, and that the integrand is 1/z·C’(z), but the latest fact comes directly and necessarily as a theorem on line integrals, which requires covering some vector calculus, and that would make the video unnecessarily long. For an introductory video on complex analysis and residues, this video presents a better approach.
Hello blackpenredpen just wanted to say I love your videos! I'm currently struggling with a limit, and that is "lim x→∞ ((2x - 3)/(2x + 1))^(x)". If you wish, please solve in an upcoming video of yours. Thanks in advance for you time!
Justin Chan 2x - 3 = (2x + 1) - 4, so (2x - 3)/(2x + 1) = [(2x + 1) - 4]/(2x + 1) = 1 - 4/(2x + 1). Now let y = -4/(2x + 1), which implies 2x + 1 = -4/y, which implies x = -(2/y + 1/2). Therefore, [1 - 4/(2x + 1)]^x = (1 + y)^[-(2/y + 1/2)] = [(1 + y)^(-2/y)][(1 + y)^(-1/2)]. Limits are multiplicative, so you can obtain the product of the limits of each factor to get the limit of the product, which is what you want, because they are the same. Therefore, we can find the limits of (1 + y)^(-1/2) and of (1 + y)^(-2/y) separately and multiply them to get the answer you desire.
The easier limit to calculate is the limit of (1 + y)^(-1/2) = 1/sqrt(1 + y). As x -> ♾, y -> 0-, and lim 1/sqrt(1 + y) (y -> 0-) = 1/sqrt(1 + 0) = 1.
Now, to calculate lim (1 + y)^(-2/y) (y -> 0-), notice that (1 + y)^(-2/y) = [(1 + y)^(1/y)]^(-2). Now let z = -1/y. As y -> 0-, z -> ♾, so we want lim [(1 - 1/z)^z]^2 (z -> ♾). Since f(t) = t^2 is a continuous, we can find the limit of (1 - 1/z)^z and then square it to get the desired limit, which actually would be the answer since the answer is just this multiplied by 1.
(1 - 1/z)^z = e^[ln(1 - 1/z)/(1/z) = e^(-[-ln(1 - 1/z)/(1/z)]) = e^(-ln[1/(1 - 1/z)]/[1/z]) = e^(-ln[z/(z - 1)]/[1/z]) = e^(-ln[1 + 1/(z - 1)]/[1/z]) = e^[-ln([1 + 1/(z - 1)]^z)] = ([1 + 1/(z - 1)]^z)^(-1). Let z = α + 1, such that as z -> ♾, α -> ♾, with ([1 + 1/(z - 1)]^z)^(-1) = [(1 + 1/α)^(α + 1)]^(-1). Thus, we want to lim (1 + 1/α)^(α + 1) (α -> ♾) and raise it to -2 to get the solution to your problem.
Now notice that (1 + 1/α)^(α + 1) = [(1 + 1/α)^α](1 + 1/α). One last time, we can find the product of the individual limits to get the limit of the product. Αs α -> ♾, 1 + 1/α -> 1. lim (1 + 1/α)^α (α -> ♾) = e by definition, so the product of the limits is e, and as such, the solution to your problem is e^(-2) = 1/e^2.
@@angelmendez-rivera351 Thanks, had a hard time with this one.
BPRP sir gave a spoiler at the beginning of the video..
Lets rock math
Math is beatiful
you please tell me how to Integrate log(cos x)... Indefinite integration
Wow amazing
The simplest way to get this result is to integrate directly and u get ln(z) you arrive the same
Can u find gamma of z+1/2
la vertical tiende a la medida del radio o límite es como buscar límites a un fratal
I don’t understand what the integral is summing up. I want to see a curve under which we are calculating an area. I don’t understand what it means to just put C as the starting point. Where is 1/z? What is happening here?
ophello There is no curve under which you are calculating the area under because the integration is over the complex numbers, not the real numbers. Unlike integration over the real numbers, there is basically no unambiguous physical-geometric interpretation to complex integrals.
pi no es la relación perímetro diámetro pues una circunferencia no es un polígono regular es una curva cerrada
jhony angarita El que la circunferencia esté definida por una curva cerrada no indica que la circunferencia no es el perímetro de un círculo, pues estrictamente hablando, nada en la definición del perímetro dice que contorno del cual se toma el perímetro debe tener curvatura cero. Así que no importa que el círculo no sea un polígono.
así los otros polígonos por más que sus lados sea pequeños siguen siendo líneas rectas no hay que confundir medida con forma ala distancia un cuadrado Seve como un punto
jhony angarita Cierto, pero no estamos trabajando con los polígonos mismos, sino con el límite de la secuencia de los polígonos, y esa límite sí incluye curvas. Si no sabes lo que es un límite, entonces ponte a estudiar y deja de criticar.
bro i saw this video after seeing a lobotomy kaisen edit on maths
7:00 Before this moment I liked the video.....
Jk I still liked it
cool
Very clever...
🅱️esidue theorem 😎😎😎👍
: ))))))
@@blackpenredpen hey, what happened to him? his channel hasn't been updated in the last 2 years
cuando existe un ángulo Ai una horizontal una vertical una diagonal dicha diagonal es el radio para los polígonos y para la circunferencia los ángulos nunca llegarán al límite para dar la medida exacta de la curva
pipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipiipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipipi
pyee~~
You said i was the variable of the summation going from 1 to N, then you used the same letter to represent the imaginary unit. Perhaps it might be better not to use i as the summation variable. And what about Delta i, which i was that?
You're right, It's strange that given the whole alphabet, we seem to use only a very confusing subset of letters - n, N, i, etc. !
cuando se hace los cálculos con polígonos uno de los primeros es el cuadrado. el pentágono exagono
jhony angarita Se escribe hexágono. Además, el inicio de la secuencia misma es irrelevante puesto a que solamente importa el límite de la secuencia.
3.14159... es la mitad de la sumatoria de los lados de un polígono regular de radio igual a 1 este perímetro está compuesto de rectas una línea curva no está compuesto de rectas y 2π no es el valor correcto
jhony angarita Lo que dices es incorrecto. En primer lugar, 3.14159... no significa nada porque más allá del elipsis puede seguir cualquier secuencia de dígitos arbitraria, y por tanto, tal expresión no representa un número, sino como menos un conjunto de números. Lo segundo es que tal sumatoria depende del polígono en sí mismo. El círculo no es un polígono, pero sí es el límite de la secuencia de los polígonos regulares, y como la integral por definición también es un límite y no una mera sumatoria, esto implica por necesidad lógica y axiomática que la calculación debe ser dada estrictamente por un límite, y este límite lo que da el resultado de 2πi. El resultado es demostrablemente correcto, y el que seas ignorante en el ámbito del análisis complejo y del cálculo no justifica que digas bobadas como las que dices aquí y que para encima taches a alguien más de incorrecto.
i root of i
ATMON ATMON There already exists a video of this in Black Pen Red Pen’s channel, and the answer is sqrt(e^π).
More complex analysis!!!!
Terrible Will's vídeo, amaizing explanation
Blackredpen which country???🤔🤔🤔
You have the coolest friends. Again, "The Big Bang Theory" is wrapping up, so maybe there could be a show about you guys solving equations and raising hell in biker bars. Yes I mean you'll bring a white board to biker bars and do calculus there.
Beatifull
Wohooooo
Here before 1K likes (it’s at 999 likes right now lol)
Holomorph
This feels like cheating somehow, but w/e I'm good with it :'D
Second
I found this hard to follow .... plus the references to "resides" seem like so much hand waving ...
Sorry about that, I think it would have taken another 10 minutes at least to explain what a residue is. I just wanted to get across the idea that they are associated with points where the function goes to +/- infinity.
It's not that complex!!!.
I couldn't resist the joke :p
@@stemcell7200 :)
:D
i am the 3141st viewer