17:13 I disagree, it would be true if the exponent was real, but in this case it is imaginary, so the exponential retains a modulus of one. You have to do more work to prove that the integral goes to 0. My guess is that there's something to do with distributions, or Fourier transforms.
I think that the proper way to kill off that integral is going to be related to the Riemann-Lebesgue Lemma, which covers similar integrals with rapidly oscillating phase.
You have to do the complex integration in a pizza sector boundary of angle pi/4 of the function exp(-t²(1+u²))/(1+u²), in the limit t->Inf the angular part and the part in the real axis go to zero so the exponential with the imaginary exp(-it²(1+u²))/(1+u²) goes to zero.
Write the integral as the sum of the integral from 0 to epsilon + integral from epsilon to 1. Then IPP on the epsilon to 1 integral shows that it goes to 0 as t goes to infinity. But the integral from 0 to epsilon is bounded by epsilon. This shows that the original integral is bounded by 2*epsilon for t large enough.
Wow! This was amazing. There's one step that seems to be missing. You have two options for sqrt(-i) that are negatives of each other. The way I justify your choice is by rewriting the sin(x^2) integral as twice an integral from 0 to oo, then noticing it must be positive by viewing it as an alternating series of the areas under each hump. Come to think of it, of course the integrals for cos(x^2) and sin(x^2) are equal since sin(x^2) = cos(x^2 - pi/2) and a translation by pi/2 doesn't matter when the bounds are infinite.
Your English is quite good. I once had a professor with so thick a Polish accent that I couldn't understand half of what he said. Fortunately, it was a math class so I was able to follow most of what he said based on what he wrote on the board. Great videos.
HOOOOOOOOLY FUCK!!! limit of exp(-i*x^2) x-> inf is not zero, it's undefined. cos(x^2) -i*sin(x^2) does not converge at inf! You were just lucky that you got the correct solution. Fix it babe. I honestly did not expect you to make such a mistake.
@@PapaFlammy69 No sir. It does not converge. For the limit to exist, both the real and imaginary parts individually must converge and obviously the limit of cos(x^2) or sin(x^2) do not exist at infinity. I tried Matlab, Mathematica and Maple and none of them give an answer of zero. If you have proof for what you claim, please share it with me.
@@erfanmohagheghian707 See Liam Clink's comment above: essentially, as t increases, the number of osciallations of the exponential inside the u-interval [0,1] increases, and so 0 becomes a better and better approximation. I agree with you, that the OP should have pointed this out.
I love this guy - I suck at filming videos and speaking English. Very funny and humble. Nicely presented. You took the time to clearly step by step explain the approach here.
Not to party poop, but cant you solve the integral of e^(-ax^2) with the polar coordinate transformation for the Gaussian integral, get sqrt(pi/a) and just plug in a=i?
It is usually solved via your method, but I think papa flammy just wanted to use another method that was more interesting and/or fun. IMO I find leibniz rule as a more elegant and fun way to solve the integral even tho polar coordinates may be more efficient. After all one of the cool things about math is that there are almost infinite paths u can take but they all end up at the same place.
I think it can not be done. Because e^(-ax^2) tends to zero when a>0, but you can't say i>0. That is the reason I am looking for a solution that does not involve e^(-i x^2)->0 or e^(-i x)->0.
Your videos are AWESOME. I have a degree in mathematics and I've been on the alternative side of the internet and life for over a decade. Therefore, your usage of profanity, dank usage of memes (no normie shit), and some tough integral bois compel me to watch your videos. Thank you for existing :o The substance, thoroughness, and demeanor make these videos so fun to watch and enjoy!
17:20 Uh exp(-i*inf) is not zero. The complex exponential oscillates and never changes amplitude. Just like cos(x) doesn't tend to zero as x tends to infinity. If your integral was on the imaginary line and not the real line then I think you could say it goes to zero, but why are you saying that an oscillatory integrand tends to zero?
I had to go to class but I was going to say this earlier. I thought about it some and realized that as t increases, the number of cycles in the integration interval increases such that the amplitude varies progressively less with u and it becomes a progressively better approximation to say that each cycle integrates to zero
I mean, I understand what for it is done and how, just not sure how assuming t = 0 would not result in division by zero ( u = x/t)? Do we absolutely not care what values can t take once we move to u at 6:50?
Your English is great, your video quality is great, your mathematics is great. I enjoyed this a lot, especially because I could follow you, for the most part.
I attempted this integral on my own and I was able to do it in less than 5 mins. You just need to change exp[-i(x)^2] to exp[-(√i*x)^2] and let u=√i*x. After that you get a constant multiplied by the gaussian integral. Then just calculate i^(-0.5) and you get the final result.
@@michelkhoury1470 See the previous few comments. That substitution alters the bounds to complex numbers, so you DON'T end up with the Gaussian integral. You need complex analysis to prove that the result is nonetheless the same.
I don't know if this is valid or not, as it seems too easy, but can't you just get the same result in a few lines by substituting into the Gaussian integral (known value of sqrt(pi))? If we let x = r * e^(i * pi / 4); dx = e^(i * pi / 4) dr; then: sqrt(pi) = e^(i * pi / 4) * (integral e^(-ir^2) dr) Which then splits up into the relevant sin and cos integrals, and real and imaginary parts can be compared. Is there something wrong with this approach?
is this complete? I think one discussion is missed. There are two values z such that z^2 = -i, how do you know you've chosen the correct one? Maybe you can argue that the integral of cos(x^2) should be positive, but I don't have a convincing argument tbh.
Great video as always. :D You mention at the end that you were working on this for a long time. How exactly did you go about figuring out a way to solve for these. Like how did you get to your initial I(t)? Specifically how to recognize that putting the t in the bounds is better than, say, multiplying some function in terms of t to the integrand. Like I know a common way for the dirichlet integral (sin(t)/t) is to multiply in an e^(-st) and differentiate that under the integral sign, and that one is kinda justified from the laplace of sin(t)/t, but for this video it just seems so Out of Nowhere. This is really a genius way to approach it.
@@TheUltimateMCguy to be honest it is just prediction with trial and error, it's like deeply understanding what you will be doing, then defining some function that would fit certain conditions. so mostly from practice, prediction and guess work.
Your 10:03 observation makes a derivative with respect to t of (1+u^2) like if the latrer was a constant ! But u depends on t, so why are you just carrying it down from e^(-i*t^2(1+u^2)) ?
At 6:00 you have shown I'(t) =2× I(t)×exp(-it^2), maybe integration factor or separation of variables would have been nicer? (I am sure someone smart will correct me, I am just eating dinner whilst watching papa flammable)
Would it work to substitute u=x(sqrt(i)), then say that this is the Gaussian integral which equals sqrt(pi), which gives sqrt(pi/i), or -sqrt(pi)sqrt(i)i. This is sqrt(2pi)/2-i(sqrt(2pi)/2). The integral can be proved with a well-known trick of multiplying by the same integral, but using a different variable and converting to polar coordinates.
No. Such a substitution shifts the integration bounds from the real to some other axis in the complex plane, so you don't actually end up with the Gaussian integral. You can calculate it like in this video, or using complex analysis. Only then can you know for certain that the the formal substitution happens to give the correct answer. See e.g. th-cam.com/video/XDUZ8bGdvYs/w-d-xo.html
exp(-it²) does not tend to 0 as t-> +∞, but rather precesses around the unit circle indefinitely. I think you need a bit more work to prove the integral at 17:22 vanishes...
It would be nice if you explained in short why one would want to solve this integral especially since you already mentioned it's importance for physics
When I opened the video I thought that since: int[exp(-a x^2)] from - inf to inf = sqrt(pi/a), a = i, the integral should be sqrt( - pi * i), I discarded that option given the length of the video... (imagined dark stuff happening) and it turned our to be correct . I have always used that result for integrations in quantum mechanics, which is quite classical (the result for the gaussian). Anyways, interesting video.
Yes, it turns out to be true, but you must prove it, since you are shifting the bounds from the real to some other axis in complex plane the with that substitution. That proof can be done as in this video, or by complex analysis, e.g. as in th-cam.com/video/XDUZ8bGdvYs/w-d-xo.html
Why not just use the standard Gaussian, where integral of e^-ax^2 is sqrt(pi/a). Then integral of e^-ix^2 is sqrt(pi/i)=sqrt(-ipi)=(1-i)sqrt(pi/2). Then the Fresnel integrals will correspond to the real and imaginary parts respectively.
6 months ago I had the assignment of solving these integrals in complex analysis. I solved the integral from 0 to infinity of sin(x^2)/x using the Feynman technique, no complex analysis used, and it was valid somehow. Where the fuck were you back then? lol
Mann könnte einfach folgendes argument verwenden das Integral von minus unendlich bis unendlich von e hoch minus a x hoch 2 dx entspricht der Wurzel aus Pi durch a. Wenn a gleich i ist. Dann multiplizier ich Nenner und Zähler mit i durch. Das lässt sich dann einfach umformen
Nein, weil solch eine Transformation die Inegrationslimiten verschiebt von der reellen Achse zu irgendeine andere Achse in der komplexen Zahlebene, d.h. man erhält nicht das Standard Gaussche Integral.
Writing math by hand is a bad thing to do to new generations. Get a decent symbolic math editor/solver and at least make the symbols standard. Do you trust programmers who make canned math manipulation tools, or some old guy named Leibniz or someone else? You cannot let yourself get excited, and start rushing. When you think you are close to a good results, that is the time to be more careful, not less. These are routine methods and issues in time series analysis. The integrals themselves, for different signals can be valid or not, depending on whether the original assumptions when the methods and steps were invented, match the conditions in the real world. Particularly, your approach is more like a one or few independent signal situation, where assumptions about normally distributed errors from a simple model make sense. But throw in multiple independent signal sources and it will not add up properly. Stop wasting your time and the time of anyone you might want to help or teach. Put what you learn into the computer, so your equations can be "compiled", verified, traced step by step, audited, reproduced, compared, shared, and used in real calculations. Bourbaki had a good idea, but they and you both fail to see that humans, with their finite memory, can never completely represent some solutions where the number or steps and the number of pieces is beyond human capability. And, groups of humans get stupider as there are more humans trying to communication, not smarter. For relatively small problems, there is a sweet spot where a group of humans can get more done then one alone. But that number is small. On the Internet most of the problems, issues and opportunities require tens of thousands, tens of millions or billions of humans to cooperate - losslessly, open, auditable - in order to achieve some kind of global optimization for the human species. The language AIs are not good mathematicians. The space of things possible with their representation of knowledge is much too limited. In the same way, hand mathematics on a blackboard that keeps disappearing will only work on tiny problems. You might think you put in a lot of work and were doing wonderful things when you could match up some things you have memorized and combine things in new ways. But relatively simple algorithms can whip through those same explorations, and never make a mistake, in a few seconds of computer time. When I started college I mentioned one day to my room mate that I had made a perfect score in my Chemistry SAT. From where I came from that was a good thing. But he said he had not only made perfect score on the main test, but on 4 of the specialized tests as well. In my second year, I shared a group with 5 other students (individual rooms and shared kitchen and living room). One of my mates had memorized every integral and could combine things he had memorize quickly and efficiently. But I knew the books he had read, and had memorized much of that. I vowed then that was a losing battle. At most I could remember a million such thing, exactly, and their derivations, and the history and context of their development. But, eventually, with life getting more complex, and most large problems facing the human species requiring the usual tens of thousands or tens of millions of humans and their supporting computer software and data - to "solve". I am not saying to give up. Mostly, video is linear. It is one dimensional, and will not efficiently emulate multidimensional problems in efficient notions and visualizations. It is possible to put complex problems into videos, but they only give clues to humans (and a very limited set of AIs). The proper tool for most problems now is the computer, not a blackboard. I have tried to encourage groups who say they want to make tools for symbolic mathematics to help the human species. They all start out saying "open" and then as soon as they get it working, they lock it down and charge what market will bear. They solve a problem or two and that just becomes marketing case studies. They help a few organizations as an "expert" or "consultant" and that just requires more advertising and marketing than real work. It is hard to judge your age. I would say I am about 50 years older than you. If you look at your own skills and potential and look ahead at your own future, knowing what you know and learning as fast as you can, think what you will be in 10 years or 5 decades. But you cannot take it with you, and all that will evaporate when you die one day. Putting things on paper is a losing battle. If you fill hundreds of paper books (or PDF which is effectively paper), you have a horrible way to try to share what you have memorized. If you write the perfect book on some topic, and a million people read it, they all have to take that information into their heads, match it to anything in their memory that might fit, and then recreate all that you have written down on paper, by hand. They have to re-invent what you wrote, because the form = paper does not itself do anything. On the Internet one of the saddest things is Wikipedia and all those ARXIV preprint servers. Both have LOTS of equations, data, graphs, charts, tables visualizations, units constants dimensions properties -- all only in form where humans are required to use it at all, and where only humans working on paper can verify, combine, compare and apply. That can change a little, but the real answer, at least a small step for the human species and its survival - is to put symbolic equations and relations into shareable and immediately computer usable form that humans can query and guide and interact with. Teach the computers who teach generations of students, not humans who will eventually forget and who always die eventually, Richard Collins, The Internet Foundation
Hi Flammy, I've typeset your proof into some lecture notes I plan to give to my students. take a look drive.google.com/file/d/1pzCIUBSU29YzPPdXKxsfhCfZwNsT7nu6/view?usp=sharing and please tell me how you would like me to cite you. The proof is on page 198, in section A.5. There is a small accreditation for you on page 199, but I'm very happy to expand it as you like. Great work, BTW.
![Don't be fooled! Integrating a resistant Integral! [ e^cos(x)cos(sin(x)) from 0 to 2pi ]](/img/n.gif)
17:13 I disagree, it would be true if the exponent was real, but in this case it is imaginary, so the exponential retains a modulus of one. You have to do more work to prove that the integral goes to 0.
My guess is that there's something to do with distributions, or Fourier transforms.
I think that the proper way to kill off that integral is going to be related to the Riemann-Lebesgue Lemma, which covers similar integrals with rapidly oscillating phase.
Yes I agree with you. There is no attenuating part but argument.
You have to do the complex integration in a pizza sector boundary of angle pi/4 of the function exp(-t²(1+u²))/(1+u²), in the limit t->Inf the angular part and the part in the real axis go to zero so the exponential with the imaginary exp(-it²(1+u²))/(1+u²) goes to zero.
Write the integral as the sum of the integral from 0 to epsilon + integral from epsilon to 1.
Then IPP on the epsilon to 1 integral shows that it goes to 0 as t goes to infinity. But the integral from 0 to epsilon is bounded by epsilon.
This shows that the original integral is bounded by 2*epsilon for t large enough.
Wow! This was amazing. There's one step that seems to be missing. You have two options for sqrt(-i) that are negatives of each other. The way I justify your choice is by rewriting the sin(x^2) integral as twice an integral from 0 to oo, then noticing it must be positive by viewing it as an alternating series of the areas under each hump.
Come to think of it, of course the integrals for cos(x^2) and sin(x^2) are equal since sin(x^2) = cos(x^2 - pi/2) and a translation by pi/2 doesn't matter when the bounds are infinite.
Your English is quite good. I once had a professor with so thick a Polish accent that I couldn't understand half of what he said. Fortunately, it was a math class so I was able to follow most of what he said based on what he wrote on the board. Great videos.
germanenglish, best english change my mind
before this "c" I really thought that the constant wasn't important
HOOOOOOOOLY FUCK!!! limit of exp(-i*x^2) x-> inf is not zero, it's undefined. cos(x^2) -i*sin(x^2) does not converge at inf!
You were just lucky that you got the correct solution. Fix it babe. I honestly did not expect you to make such a mistake.
@@PapaFlammy69 No sir. It does not converge. For the limit to exist, both the real and imaginary parts individually must converge and obviously the limit of cos(x^2) or sin(x^2) do not exist at infinity. I tried Matlab, Mathematica and Maple and none of them give an answer of zero. If you have proof for what you claim, please share it with me.
@@erfanmohagheghian707 See Liam Clink's comment above: essentially, as t increases, the number of osciallations of the exponential inside the u-interval [0,1] increases, and so 0 becomes a better and better approximation. I agree with you, that the OP should have pointed this out.
I love this guy - I suck at filming videos and speaking English. Very funny and humble. Nicely presented. You took the time to clearly step by step explain the approach here.
=)
thx
Not to party poop, but cant you solve the integral of e^(-ax^2) with the polar coordinate transformation for the Gaussian integral, get sqrt(pi/a) and just plug in a=i?
thats what i did lol
@@PapaFlammy69 Viele Wege fuehren nach Rom
yes you definitely can do this for sure!
It is usually solved via your method, but I think papa flammy just wanted to use another method that was more interesting and/or fun. IMO I find leibniz rule as a more elegant and fun way to solve the integral even tho polar coordinates may be more efficient. After all one of the cool things about math is that there are almost infinite paths u can take but they all end up at the same place.
I think it can not be done. Because e^(-ax^2) tends to zero when a>0, but you can't say i>0. That is the reason I am looking for a solution that does not involve e^(-i x^2)->0 or e^(-i x)->0.
I simply melt when I hear your voice
At 17:35 isn't √ab =√a√b only valid for positive a and b?
Well yes, but actually no
*only valid for a or b, meaning if they are not both negative, you can split the square roots
Your videos are AWESOME. I have a degree in mathematics and I've been on the alternative side of the internet and life for over a decade. Therefore, your usage of profanity, dank usage of memes (no normie shit), and some tough integral bois compel me to watch your videos. Thank you for existing :o The substance, thoroughness, and demeanor make these videos so fun to watch and enjoy!
This are the kind of math channels we were looking for!! Keep up the great work! :D
17:20 Uh exp(-i*inf) is not zero. The complex exponential oscillates and never changes amplitude. Just like cos(x) doesn't tend to zero as x tends to infinity. If your integral was on the imaginary line and not the real line then I think you could say it goes to zero, but why are you saying that an oscillatory integrand tends to zero?
I had to go to class but I was going to say this earlier. I thought about it some and realized that as t increases, the number of cycles in the integration interval increases such that the amplitude varies progressively less with u and it becomes a progressively better approximation to say that each cycle integrates to zero
14:30 Sorry for being dumb and stuff but what's up with making t=0 AND having u = x/t at the same time? Why is this legit for finding C?
I mean, I understand what for it is done and how, just not sure how assuming t = 0 would not result in division by zero ( u = x/t)? Do we absolutely not care what values can t take once we move to u at 6:50?
This is an excellent channel prepared with really high amount of effort clearly. Thank you so much for providing this amazing platform for us!
Your English is great, your video quality is great, your mathematics is great.
I enjoyed this a lot, especially because I could follow you, for the most part.
I attempted this integral on my own and I was able to do it in less than 5 mins. You just need to change exp[-i(x)^2] to exp[-(√i*x)^2] and let u=√i*x. After that you get a constant multiplied by the gaussian integral. Then just calculate i^(-0.5) and you get the final result.
Thomas Q I did the same thing :p
@@michelkhoury1470 See the previous few comments. That substitution alters the bounds to complex numbers, so you DON'T end up with the Gaussian integral. You need complex analysis to prove that the result is nonetheless the same.
@@jimmyb998 yes anyway you can use Laplace transform to prove this
I don't know if this is valid or not, as it seems too easy, but can't you just get the same result in a few lines by substituting into the Gaussian integral (known value of sqrt(pi))?
If we let x = r * e^(i * pi / 4); dx = e^(i * pi / 4) dr; then:
sqrt(pi) = e^(i * pi / 4) * (integral e^(-ir^2) dr)
Which then splits up into the relevant sin and cos integrals, and real and imaginary parts can be compared. Is there something wrong with this approach?
See all the comments above: such a substitution alters the integration bounds, and you don't actually end up with the Gaussian integral.
At 4:00 you can just use the fundamental theorem of calculus , which tells you that what's inside the square is an antiderivative of t->exp(-it^2)
is this complete? I think one discussion is missed. There are two values z such that z^2 = -i, how do you know you've chosen the correct one? Maybe you can argue that the integral of cos(x^2) should be positive, but I don't have a convincing argument tbh.
6:50 you say u = x\t but then later we use I( t = 0 ) to evaluate c. Why is this valid? Would taking the limit as t goes to 0 give the same result?
Let u= xsqrti, du= sqrti dx. Integral becomes: 1/sqrti integral from -inf to +inf of e^-(u^2). That's all.
David amp You have to change integration path to -i*inf to i*inf
I thought the same thing at first but it's not the same integral as a Gaussian just because of that substitution
David amp However if you do parametric differentiation and set the parameter to i, you get the right answer too
@@1495978707 Yes, all of which is part of complex analysis. I believe the intent was to avoid complex analysis :)
20:00 - > For sure I enjoyed a lot. Great work!
de Moivre and Euler are proud of you!!
Just a question, you could have used the other square root of -i, so I imagine there is reason for this integrals to be always positive, isn´t it?
Great video as always. :D
You mention at the end that you were working on this for a long time. How exactly did you go about figuring out a way to solve for these. Like how did you get to your initial I(t)? Specifically how to recognize that putting the t in the bounds is better than, say, multiplying some function in terms of t to the integrand.
Like I know a common way for the dirichlet integral (sin(t)/t) is to multiply in an e^(-st) and differentiate that under the integral sign, and that one is kinda justified from the laplace of sin(t)/t, but for this video it just seems so Out of Nowhere.
This is really a genius way to approach it.
I wonder if anyone could answer this
@@TheUltimateMCguy to be honest it is just prediction with trial and error, it's like deeply understanding what you will be doing, then defining some function that would fit certain conditions.
so mostly from practice, prediction and guess work.
Your 10:03 observation makes a derivative with respect to t of (1+u^2) like if the latrer was a constant ! But u depends on t, so why are you just carrying it down from e^(-i*t^2(1+u^2)) ?
I thought I would never say this about an Analysis proof but this has been very beautiful.
At 6:00 you have shown I'(t) =2× I(t)×exp(-it^2), maybe integration factor or separation of variables would have been nicer? (I am sure someone smart will correct me, I am just eating dinner whilst watching papa flammable)
Do you have this written out to follow along?
Would it work to substitute u=x(sqrt(i)), then say that this is the Gaussian integral which equals sqrt(pi), which gives sqrt(pi/i), or -sqrt(pi)sqrt(i)i. This is sqrt(2pi)/2-i(sqrt(2pi)/2). The integral can be proved with a well-known trick of multiplying by the same integral, but using a different variable and converting to polar coordinates.
No. Such a substitution shifts the integration bounds from the real to some other axis in the complex plane, so you don't actually end up with the Gaussian integral. You can calculate it like in this video, or using complex analysis. Only then can you know for certain that the the formal substitution happens to give the correct answer. See e.g. th-cam.com/video/XDUZ8bGdvYs/w-d-xo.html
exp(-it²) does not tend to 0 as t-> +∞, but rather precesses around the unit circle indefinitely. I think you need a bit more work to prove the integral at 17:22 vanishes...
Is there any other method to find this integral?
I tried to do it using the Gaussian integral.cos(x^2) came out to be right but sin(x^2) came out to be negative sqrt(2 pie).Can anyone help?
If the square root has two possible values how can you know for certain that you have to choose the positive one?
A true whirlwind, leaving my mind a dizzy loop.
I miss these
:'(
It would be nice if you explained in short why one would want to solve this integral especially since you already mentioned it's importance for physics
I loved it even if a triangle sufices i loved it great work. Even for a 4 year video the mathematical quality is top
very elegant :) I wonder what happens when you integrate the series expansion of sin and cos... im going to try that :P
I found the same result (not rigourously I guess) just by making the substitution u^2=ix^2 starting from the original integral
Could you not use the gaussian integral's definition and substitute it as u?
When I opened the video I thought that since: int[exp(-a x^2)] from - inf to inf = sqrt(pi/a), a = i, the integral should be sqrt( - pi * i), I discarded that option given the length of the video... (imagined dark stuff happening) and it turned our to be correct . I have always used that result for integrations in quantum mechanics, which is quite classical (the result for the gaussian). Anyways, interesting video.
Yes, it turns out to be true, but you must prove it, since you are shifting the bounds from the real to some other axis in complex plane the with that substitution. That proof can be done as in this video, or by complex analysis, e.g. as in th-cam.com/video/XDUZ8bGdvYs/w-d-xo.html
0:10 Could You tell from which subject in physics it comes? Best regards
Feynman path integrals, among other things
Video muito bom, gostei.
Meus professores desconhece esse método de integral.
você que criou sozinho?
Can anyone explain why his profile pic changed and his comments are disabling?
i've read remarks. Exercize super technic and interesting. Bye raw. Thank you very much.
Absolut beeindruckend!
Isn't the reason those two integrals are equal the fact that sinx is just a shift of cosx so the shift is nothing compared to -∞ to ∞
yes, it is.
@@jimmyb998 no it is not. Write f(x) = sin(x+pi/4). You have that the integral from -inf to inf of f(x^2) is sqrt(pi), which is not the same.
@@holomurphy22 you are right! my bad
U just nailed it.
Can we just substitute t = x*sqrt(i) and get the same result because of Gausse integral? Yes, we can. You are really cool, guy, but WTF?
The fact that you can do that substitution, and still get the right answer hinges precicely on the calculation in this video. It isn't true a priori.
VERY NICE! Chapeu!!
wonderful, congratulations
Why not just use the standard Gaussian, where integral of e^-ax^2 is sqrt(pi/a). Then integral of e^-ix^2 is sqrt(pi/i)=sqrt(-ipi)=(1-i)sqrt(pi/2). Then the Fresnel integrals will correspond to the real and imaginary parts respectively.
I think it's only defined like that over the reals.
6 months ago I had the assignment of solving these integrals in complex analysis. I solved the integral from 0 to infinity of sin(x^2)/x using the Feynman technique, no complex analysis used, and it was valid somehow.
Where the fuck were you back then? lol
Great job,,man!!!
You've got a nice German accent, I like it!
You did great job
You could have used multivariable calculus
Do you mind if I show a picture of you to my hairdresser? I would like to have the same hair style as you.
Gute Arbeit!
Why don’t you just use Euler’s formula bruh
Mann könnte einfach folgendes argument verwenden das Integral von minus unendlich bis unendlich von e hoch minus a x hoch 2 dx entspricht der Wurzel aus Pi durch a. Wenn a gleich i ist. Dann multiplizier ich Nenner und Zähler mit i durch. Das lässt sich dann einfach umformen
Nein, weil solch eine Transformation die Inegrationslimiten verschiebt von der reellen Achse zu irgendeine andere Achse in der komplexen Zahlebene, d.h. man erhält nicht das Standard Gaussche Integral.
Writing math by hand is a bad thing to do to new generations. Get a decent symbolic math editor/solver and at least make the symbols standard. Do you trust programmers who make canned math manipulation tools, or some old guy named Leibniz or someone else?
You cannot let yourself get excited, and start rushing. When you think you are close to a good results, that is the time to be more careful, not less.
These are routine methods and issues in time series analysis. The integrals themselves, for different signals can be valid or not, depending on whether the original assumptions when the methods and steps were invented, match the conditions in the real world. Particularly, your approach is more like a one or few independent signal situation, where assumptions about normally distributed errors from a simple model make sense. But throw in multiple independent signal sources and it will not add up properly.
Stop wasting your time and the time of anyone you might want to help or teach. Put what you learn into the computer, so your equations can be "compiled", verified, traced step by step, audited, reproduced, compared, shared, and used in real calculations.
Bourbaki had a good idea, but they and you both fail to see that humans, with their finite memory, can never completely represent some solutions where the number or steps and the number of pieces is beyond human capability. And, groups of humans get stupider as there are more humans trying to communication, not smarter. For relatively small problems, there is a sweet spot where a group of humans can get more done then one alone. But that number is small. On the Internet most of the problems, issues and opportunities require tens of thousands, tens of millions or billions of humans to cooperate - losslessly, open, auditable - in order to achieve some kind of global optimization for the human species.
The language AIs are not good mathematicians. The space of things possible with their representation of knowledge is much too limited. In the same way, hand mathematics on a blackboard that keeps disappearing will only work on tiny problems. You might think you put in a lot of work and were doing wonderful things when you could match up some things you have memorized and combine things in new ways. But relatively simple algorithms can whip through those same explorations, and never make a mistake, in a few seconds of computer time.
When I started college I mentioned one day to my room mate that I had made a perfect score in my Chemistry SAT. From where I came from that was a good thing. But he said he had not only made perfect score on the main test, but on 4 of the specialized tests as well. In my second year, I shared a group with 5 other students (individual rooms and shared kitchen and living room). One of my mates had memorized every integral and could combine things he had memorize quickly and efficiently. But I knew the books he had read, and had memorized much of that. I vowed then that was a losing battle. At most I could remember a million such thing, exactly, and their derivations, and the history and context of their development. But, eventually, with life getting more complex, and most large problems facing the human species requiring the usual tens of thousands or tens of millions of humans and their supporting computer software and data - to "solve".
I am not saying to give up. Mostly, video is linear. It is one dimensional, and will not efficiently emulate multidimensional problems in efficient notions and visualizations. It is possible to put complex problems into videos, but they only give clues to humans (and a very limited set of AIs). The proper tool for most problems now is the computer, not a blackboard. I have tried to encourage groups who say they want to make tools for symbolic mathematics to help the human species. They all start out saying "open" and then as soon as they get it working, they lock it down and charge what market will bear. They solve a problem or two and that just becomes marketing case studies. They help a few organizations as an "expert" or "consultant" and that just requires more advertising and marketing than real work. It is hard to judge your age. I would say I am about 50 years older than you. If you look at your own skills and potential and look ahead at your own future, knowing what you know and learning as fast as you can, think what you will be in 10 years or 5 decades. But you cannot take it with you, and all that will evaporate when you die one day.
Putting things on paper is a losing battle. If you fill hundreds of paper books (or PDF which is effectively paper), you have a horrible way to try to share what you have memorized. If you write the perfect book on some topic, and a million people read it, they all have to take that information into their heads, match it to anything in their memory that might fit, and then recreate all that you have written down on paper, by hand. They have to re-invent what you wrote, because the form = paper does not itself do anything. On the Internet one of the saddest things is Wikipedia and all those ARXIV preprint servers. Both have LOTS of equations, data, graphs, charts, tables visualizations, units constants dimensions properties -- all only in form where humans are required to use it at all, and where only humans working on paper can verify, combine, compare and apply. That can change a little, but the real answer, at least a small step for the human species and its survival - is to put symbolic equations and relations into shareable and immediately computer usable form that humans can query and guide and interact with. Teach the computers who teach generations of students, not humans who will eventually forget and who always die eventually,
Richard Collins, The Internet Foundation
😮
es incorrecto da √(2pi)/4
√(2pi)/4 o lo que es lo mismo 1/2(√pi/2)
le recomiendo que lea el libro de Advanced calcculus SPIEGEL
es.wikipedia.org/wiki/Integral_de_Fresnel
confirme usted mismo, un consejo como matemático le sugiero que no confié en la información de cualquier pagina web .
concluyo mal el ejercicio
O that's how you pronounce Fresnel haha
Very-very nice!)
I come from blackpenredpen, good video!
Beautiful
Amazing 😃
No need to do a video about "the" square root of -i, it's very elementary 😉
Keep up the good work 😊
Really beatiful
How are your dogs always doing interesting stuff?
Nice shirt man ;)
Awesome!
Chen lu
par excellence
I hope you can see where it's coming from!!
is that an AFROJACK T-SHIRT???
OMG 👍👍👍
Cool!
like 300!
th-cam.com/video/JaMcsuY9wLU/w-d-xo.html
Bu yöntem daha kısa ve daha güzel
Holy f*ck
Who is jee aspirant here?? 😂
Error, error, . . . , Error
Many errors . . .
It's so stupid.
so inefficient with awful handwriting
Hi Flammy, I've typeset your proof into some lecture notes I plan to give to my students. take a look drive.google.com/file/d/1pzCIUBSU29YzPPdXKxsfhCfZwNsT7nu6/view?usp=sharing
and please tell me how you would like me to cite you. The proof is on page 198, in section A.5. There is a small accreditation for you on page 199, but I'm very happy to expand it as you like.
Great work, BTW.