At my house, when someone makes a mistake like that we say, “That’s it. You’re fired!” Of course we are always kidding and nobody ever gets fired including you. Keep up the great work!
For anyone wondering, if you were to evaluate the integral at 10:39, it would evaluate to 2sin(pi(n-m))/(n-m). This is defined for all values of n and m except where n = m. For every other value, n-m becomes a whole number, and sin(any whole number*pi) = 0.
You, sir, are amazing. Besides the mastery of complex math and the Fourier transform, I cannot get over your mastery of the black and red pens! Fantastic video!
With this formula, the Fourier series for e^x in the interval (-π, π) is more obvious, since the coefficient sequence is easier to calculate. One has that the integrand is e^[(1 - im)x] which is anti-differentiated to [(1 + im)/(1 + m^2)]e^[(1 - im)x]. Evaluate this at the boundaries and subtract to get [(1 + im)/(1 + m^2)][(e^π - e^(-π))][(-1)^m]. The real and imaginary parts gives you the cosine and sine coefficients of the real Fourier series, respectively. Yet this is neater and easier to obtain. You should make a video on the Fourier transform and its inverse. It would complete this series, and it also relates to the Laplace transform as well, which you have covered in this channel already.
Complex is better; you don't need to remember which coefficients have a 1/π and which have 1/2π. It also just works if f is a complex-valued function, and you just don't get all conjugate pairs.
Also, I realize that if the real Fourier series uses the expression a0/2 + S, then you don't have to remember which coefficients have 1/pi and which have 1/2pi either, because they all have 1/pi in that version.
Your explanation is very clear you make it look easy.. that means you did a great job. I will change my slides according to your video ;) thank you so much, on behalf of me and my students!
I prefer the complex Fourier Series. It helped me solve many engineering problems in noise and vibration control. It also help massively with laser optics design. Everywhere I turned where a complex signal was there, the complex Fourier helped a lot, especially since I was collecting the total vector length at the end and was not interested in the phase at all.
Another rather interesting bit is that, believe it or not, tan(x) has a Fourier series. The trick is, it's not convergent (and someone mentioned some stuff about trying to find a Fourier series for 1/x below; that's similar, but tan(x) is arguably a more natural choice because it is periodic), and as a result one needs to reinterpret it using a suitably appropriate definition of extended summation, similar to what was done for the video summing 1 + 2 + 3 + 4 + ... to -1/12, only one does not need a method as sophisticated as Ramanujan summation. In particular, such divergent Fourier series should be interpreted with what is called Fejer summation, which is basically a specific application of Cesaro summation, the simple "averaging" method that is used to sum Grandi's series, 1 - 1 + 1 - 1 + 1 - 1 + ... , to 1/2, to the Fourier series. (Fejer summation can also be used to improve the convergence of some convergent but otherwise "bad" Fourier series like that for the square wave - it "smooths out" the Gibbs phenomenon [the "spikiness" near the discontinuities] and thus allows it to approximate the square wave much better.) In particular, the series for tan(x) is tan(x) = 2 sum_{n=1...inf} (-1)^(n-1) sin(2nx) (divergent) Would be good material to go over in a video, I think, including the derivation (which requires the Cauchy principal value, I think, to make sense of the "bad" integrals across the poles of tan(x).).
All the other terms in the summation evaluated to 0 because of the integral but only when n=m, the integral is non-zero (i.e, 2π) and so we get only one Cn as all the others in the sum go to 0.
I have a question on Fourier transform ( FT). Why in FT, the sign of the parameter of the exponential function is negative ( exp(-i2\pi nk/N))? Why don't we use the positive sign in FT and negative sing in the inverse one? Is it just by definition, or there is a mathematical reason for that?
@@angelmendez-rivera351 I mean the formula, brief explanation about it and where does it come from, you know like the same he did with the Fourier series but this time with the Fourier transform
@@Jacob-uy8ox Fourier Transform is just the analogue of the Fourier Series, except we let the period of the function go to infinity (so we aren't restricted to decomposing only a certain interval).
Am I correct when I presume that: A: if and only if Cn is the complex conjugate of C-n for all n, is f(x) real on the domain -pi to pi? B: such Fourier series always describe a periodic function with period 2pi that between -pi and pi (but not necessary at pi and -pi itself) is equal to f(x)?
For forty years I've been wondering whether there is any natural number for which, if it is written down in English, all of its letters occur in alphabetical order.
Jack Sainthill I think that we can use proof by exhaustion to prove there is almost no such natural number with the property you describe. The reason I say we can prove it by exhaustion is that there is only a finite number of cases to consider. There are two types of names a natural number can have. One type is the positional type. This type of name is a name given to a number based on the multiples of powers of ten in its decimal expansion. As such, if we can find that any individual part of the name already is not in alphabetical order, we can discard every number whose name contains this part. This means we only need to consider the names of the powers of ten individually, and the names of the digits, individually. The only exception to this rule is the names of the multiples of 10 less than a hundred. Also, we shall consider a separate case of proof once we get to a million. To start with, we have the digits one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. None of these has a name whose letters are arranged in English alphabetical order. Also, thirteen through nineteen fail to have such a name, because the suffix teen does not have its letters arranged in alphabetical order. Twenty, thirty, Forty, Fifty, Sixty, Seventy, Eighty, Ninety, Hundred. 40 has an English name whose letters are arranged in alphabetical order. However, this may be the only natural number with this property. Now we are at naming powers of 10 individually. Hundred, Thousand, Myriad, Crore, and Lakh fail to have this property. Any number whose name contains these names as parts of it is discarded. Additionally, any number whose name ending contains the suffixes -lion and -liard is discarded. Since every number after million and milliard contains a name containing those suffixes, every number afterwards is discarded, except for those numbers whose primary name is not a positional type name, but the second type name, which I call the special type. These are numbers which are large, or very large, whose name is special given their significance in googology, and do not use positional names as their primary name. Example of this would be the google and the googleplex. Nonetheless, the number of these special names is finite in the literature of googology, and to my understanding, none of these names satisfy the property. As such, 40 is the only natural number which has an entire name in English consisting of letters whose arrangement is in English alphabetical order.
It would be nice if you could come of with a function that models your amount of subscribers as a function of time. I wonder whether it's exponential or linear. You could analyse it a bit to make a prediction about how many subs you might have at the end of the year. Also you could derive it to get a function to describe the rate of subscriptions and do some other cool stuff that you came up with if you pleased. I think it'd be kinda cool
I think the growth is half-exponential, meaning it is between exponential and polynomial. More precisely, if f(t) = a*b^t is an exponential function, then the growth is given by a function g(t) with the property that g[g(t)] = f(t). In other words, g(t) is a half-exponential function, because it is the compositional square root of the exponential function, the half-iterate of it.
He has already made several videos about the factorial function for non-natural values. In summary, the factorial function evaluated at some complex number z is equal to Pi(z), where the function Pi(z) is equal to the integral from t = 0 to t grows infinite of (t^z)*e^(-t)
In fact, for the bonus gravy, if you have a sequence characterized by having m 1s followed by having m -1s followed by having m 1s again, with a repeating cycle, then the formula for such a sequence is (-1)^floor(n/m) = cos(π·floor(n/m))
@@lucasdepetris5896 There is no good way to explain it, because the formula is one of those formulas that cannot be derived by solving simple equations. You either notice the pattern, or you don't, and if you don't, there is pretty much nothing you can do about it. What I noticed is that the sequence 1, -1, 1, -1,... etc., which is very similar to your sequence, has formula (-1)^n. This one is just common knowledge, and you realize it easily by trying to calculate every power of -1. But the one difference is that I need to be able to repeat every n instead of just counting forward. In other words, instead of plugging in n = 0, n = 1, n = 2, etc., I need to be able to plug in n = 0, n = 0, n = 1, n = 1, n = 2, n= 2, etc. to get your sequence. In other words, I need some sort of increasing step function. And there really are only two step functions that satisfy these properties that can do the job: ceiling(n/2), and floor(n/2). And you get floor(n/2) after testing both on the sequence. That is how I came up with an answer. I really just followed a pattern and reasoned through it slowly and borrowed from other knowledge, but you cannot derive the solution systematically. It's just impossible. Especially if you don't know the floor and ceiling functions to begin with.
@@Appilesh I confirmed that, yes, you need the Cauchy principal value for the real part of the coefficients. The imaginary part is always integrable, but the real part is not.
try to find three numbers a b c such that all three are prime and satisfy the equation a² + b² = c² please, someone help I'm having problems with this, although I'm pretty sure there are numbers like this save my soul ;-;
Black Pen Red Pen long ago made a video on finding a set of equations that generates every Pythagorean triple, each for every input you make. The generator proves that there is no solution to the your problem, meaning there is no Pythagorean triple in which every number is prime. It is impossible.
You are the best teacher........#bprpyay!!!!!!!!!!! So..... Please integrate e^x^x^x^e^i ...... ....... I think it would be great!!!!😅🤗🤗🤗🤗🤗🤗🤗🤗 And it's time for you to have a haircut
What does the ........ stand for? Also, I’m fairly certain there is nothing you can do to integrate that function using elementary functions. In fact, even with known special functions, it may be impossible.
Aaaaaah i struggle when i want to turn the domain of the function [-pi;pi] to [-L;L] to make it match physics and signal analysis =( I have troubles turning the nx into 2pi*frequency*time =( Yay^(-1) :'((((
At 8:15, n should go from negative inf to postive inf. Sorry I missed the negative sign.
At my house, when someone makes a mistake like that we say, “That’s it. You’re fired!” Of course we are always kidding and nobody ever gets fired including you. Keep up the great work!
No you didn’t miss any sign you were testing our attention 🙈🙈
Because of conjugacy, can we represent them in C_n ?
The fast and the fouriers.
He won't approximate the solution...sorry bro
His positivity and enthusiasm is contagious! I find myself smiling while he explains everything clearly and simply. Love him.❤
For anyone wondering, if you were to evaluate the integral at 10:39, it would evaluate to 2sin(pi(n-m))/(n-m). This is defined for all values of n and m except where n = m. For every other value, n-m becomes a whole number, and sin(any whole number*pi) = 0.
The traditional blackboard and chalk style is more engaging than just a screen. Your personality is encouraging us to learn.
You, sir, are amazing. Besides the mastery of complex math and the Fourier transform, I cannot get over your mastery of the black and red pens! Fantastic video!
you deserve a 1M subscriber by the end of the year, not just 400K.
He got it
With this formula, the Fourier series for e^x in the interval (-π, π) is more obvious, since the coefficient sequence is easier to calculate. One has that the integrand is e^[(1 - im)x] which is anti-differentiated to [(1 + im)/(1 + m^2)]e^[(1 - im)x]. Evaluate this at the boundaries and subtract to get [(1 + im)/(1 + m^2)][(e^π - e^(-π))][(-1)^m]. The real and imaginary parts gives you the cosine and sine coefficients of the real Fourier series, respectively. Yet this is neater and easier to obtain.
You should make a video on the Fourier transform and its inverse. It would complete this series, and it also relates to the Laplace transform as well, which you have covered in this channel already.
I was trying to find this derivation for ages after my university lecturers seemed to ignore it. Thanks, you absolutely nailed it!
Complex is better; you don't need to remember which coefficients have a 1/π and which have 1/2π. It also just works if f is a complex-valued function, and you just don't get all conjugate pairs.
iabervon I like it more too.
This is true, although there is pedagogical utility to the real-valued Fourier series.
Also, I realize that if the real Fourier series uses the expression a0/2 + S, then you don't have to remember which coefficients have 1/pi and which have 1/2pi either, because they all have 1/pi in that version.
Wow, here comes heavenly sent lecturer. Thank you very much
It's January 1st, 2020 and, yes, we met his goal of 400k subs (he has 409k today) ☺️ I'm very happy for you bro!
Watching all these maths videos to relax is making me want to return to university :D ... So much better than Netflix
Steve calculated that by the rate that the channel is growing, we will most certainly reach 400 000 subscribers, he is just trying to be humble.
Your explanation is very clear you make it look easy.. that means you did a great job. I will change my slides according to your video ;) thank you so much, on behalf of me and my students!
Glad it was helpful!
I prefer the complex Fourier Series. It helped me solve many engineering problems in noise and vibration control. It also help massively with laser optics design. Everywhere I turned where a complex signal was there, the complex Fourier helped a lot, especially since I was collecting the total vector length at the end and was not interested in the phase at all.
How is this guy not at 1 million subscribers, it's been 3 years since this video!
now 1.03
best teacher i ever had
best explanation on youtube bro!!
Another rather interesting bit is that, believe it or not, tan(x) has a Fourier series. The trick is, it's not convergent (and someone mentioned some stuff about trying to find a Fourier series for 1/x below; that's similar, but tan(x) is arguably a more natural choice because it is periodic), and as a result one needs to reinterpret it using a suitably appropriate definition of extended summation, similar to what was done for the video summing 1 + 2 + 3 + 4 + ... to -1/12, only one does not need a method as sophisticated as Ramanujan summation. In particular, such divergent Fourier series should be interpreted with what is called Fejer summation, which is basically a specific application of Cesaro summation, the simple "averaging" method that is used to sum Grandi's series, 1 - 1 + 1 - 1 + 1 - 1 + ... , to 1/2, to the Fourier series. (Fejer summation can also be used to improve the convergence of some convergent but otherwise "bad" Fourier series like that for the square wave - it "smooths out" the Gibbs phenomenon [the "spikiness" near the discontinuities] and thus allows it to approximate the square wave much better.)
In particular, the series for tan(x) is
tan(x) = 2 sum_{n=1...inf} (-1)^(n-1) sin(2nx) (divergent)
Would be good material to go over in a video, I think, including the derivation (which requires the Cauchy principal value, I think, to make sense of the "bad" integrals across the poles of tan(x).).
what a guy. You hit 400k by the end of the year just as you hoped!
Thank you! I am very happy and grateful for this! : ))
At first, I like the complex one instead of real one.
Second, I'm happy for you because you achieved your gool.
I really like the new blue marker. 😎
your videos and your energy is fantastic!
I'm sorry if this is a dumb question but at 12:29 where did the summation of Cn go?
All the other terms in the summation evaluated to 0 because of the integral but only when n=m, the integral is non-zero (i.e, 2π) and so we get only one Cn as all the others in the sum go to 0.
I’m Japanese, university’s 2nd grade, majoring in physics.
Who cares
Tom Diem
6 people
Yo Ming Hi
l cannot understand the purpose what you want to express.
@@令狐皂白 Um, maybe he's looking for friends?
So cool mr.steve, thanks.
Just show that they’re orthogonal, and then they form a basis under the L^2 inner product for functions with compact support
Great vid! Where did your C0 go at the end tho?
I like both😘 .....today is BLUEPENPINKPEN !! 😨......Yay!!!! 🤘
Extremely well done. Good.
I have a question on Fourier transform ( FT). Why in FT, the sign of the parameter of the exponential function is negative ( exp(-i2\pi nk/N))? Why don't we use the positive sign in FT and negative sing in the inverse one? Is it just by definition, or there is a mathematical reason for that?
AMAZING! thanks for the video! very helpful!
Excellent!
Very good one, thank you so much
0:27 I do like convenience
Complex version rules!
I will like for the amount of work put in this... Even thou i have no clue what was going on here, or what four yay series are
Watch his previous videos for a better understanding of it.
8:17 but i will complain that u did not put - before the infinity
great video - thanks!
thank you so much, keep going
Who came to the unlisted vid from the link in the Lil Integral's description? : )
: )
What is lil means?
why is 'i' poisitive at the final formula f(x)? It should be negative?
Plz can you do a video about system of differential équation second ordre but two unknown ( we use it in fluid mecanics )
What about Fourier transform?
Jacobo Zapata This is related to the Fourier transform.
@@angelmendez-rivera351 I mean the formula, brief explanation about it and where does it come from, you know like the same he did with the Fourier series but this time with the Fourier transform
@@Jacob-uy8ox Fourier Transform is just the analogue of the Fourier Series, except we let the period of the function go to infinity (so we aren't restricted to decomposing only a certain interval).
3blue1brown has a great video on the fourier transform
I’m convinced he’s building up to FT, particularly after watching this video.
Yes, indeed , well done.
Thank you sir
Thanks for this nice explanation😍🥲
why is the conjugate a+ib ignored and simply merged into the equation as if it is a-ib????
idk;(
Thank's bro u are the best please stay on it 😎😍
Amazing! (I prefer the complex version :D )
Am I correct when I presume that:
A: if and only if Cn is the complex conjugate of C-n for all n, is f(x) real on the domain -pi to pi?
B: such Fourier series always describe a periodic function with period 2pi that between -pi and pi (but not necessary at pi and -pi itself) is equal to f(x)?
5:30 why this is allowed? and is it remain true?
How to calculate the 0th term from the complex fourier series? We can't just put 0 there right? Or can we?
Thank you!!!!
complex things reflect reality
Complex Fourier is superior 👊
For forty years I've been wondering whether there is any natural number for which, if it is written down in English, all of its letters occur in alphabetical order.
Jack Sainthill I think that we can use proof by exhaustion to prove there is almost no such natural number with the property you describe. The reason I say we can prove it by exhaustion is that there is only a finite number of cases to consider. There are two types of names a natural number can have. One type is the positional type. This type of name is a name given to a number based on the multiples of powers of ten in its decimal expansion. As such, if we can find that any individual part of the name already is not in alphabetical order, we can discard every number whose name contains this part. This means we only need to consider the names of the powers of ten individually, and the names of the digits, individually. The only exception to this rule is the names of the multiples of 10 less than a hundred. Also, we shall consider a separate case of proof once we get to a million. To start with, we have the digits one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. None of these has a name whose letters are arranged in English alphabetical order. Also, thirteen through nineteen fail to have such a name, because the suffix teen does not have its letters arranged in alphabetical order. Twenty, thirty, Forty, Fifty, Sixty, Seventy, Eighty, Ninety, Hundred. 40 has an English name whose letters are arranged in alphabetical order. However, this may be the only natural number with this property.
Now we are at naming powers of 10 individually. Hundred, Thousand, Myriad, Crore, and Lakh fail to have this property. Any number whose name contains these names as parts of it is discarded. Additionally, any number whose name ending contains the suffixes -lion and -liard is discarded. Since every number after million and milliard contains a name containing those suffixes, every number afterwards is discarded, except for those numbers whose primary name is not a positional type name, but the second type name, which I call the special type. These are numbers which are large, or very large, whose name is special given their significance in googology, and do not use positional names as their primary name. Example of this would be the google and the googleplex. Nonetheless, the number of these special names is finite in the literature of googology, and to my understanding, none of these names satisfy the property.
As such, 40 is the only natural number which has an entire name in English consisting of letters whose arrangement is in English alphabetical order.
@@angelmendez-rivera351
Indeed it is, thank you. My forty-year quest is ended. ;)
Jack Sainthill That’s very meta wtf
Now Fourier transform derived from Fourier series pls
Blackpenredpen yay
Are you going to do the complex Fourier series for e^x next?
Rob Maybe, although I basically already covered in the comments of how to use the complex Fourier series with e^x.
Are you going to do some RUclips’s on finite input response filters?
I am to late to see this video 😂
love it❤
The Kronecker Delta function at the end?
Awesome
trés élégant mon cher ami
ur my savior
It would be nice if you could come of with a function that models your amount of subscribers as a function of time. I wonder whether it's exponential or linear. You could analyse it a bit to make a prediction about how many subs you might have at the end of the year. Also you could derive it to get a function to describe the rate of subscriptions and do some other cool stuff that you came up with if you pleased. I think it'd be kinda cool
I think the growth is half-exponential, meaning it is between exponential and polynomial. More precisely, if f(t) = a*b^t is an exponential function, then the growth is given by a function g(t) with the property that g[g(t)] = f(t). In other words, g(t) is a half-exponential function, because it is the compositional square root of the exponential function, the half-iterate of it.
hey !! could you please make a video of the meaning of " i factorial " like...does it have an approach?
He has already made several videos about the factorial function for non-natural values. In summary, the factorial function evaluated at some complex number z is equal to Pi(z), where the function Pi(z) is equal to the integral from t = 0 to t grows infinite of (t^z)*e^(-t)
Thank you so much ! ... This video is so very amazing. (My comment is clean version😊😊)
The complex version is much easier to remember and use for me.
8:18 so fun!
great!
Why did you multiply both sides by e^-imx
Thx!
Why don’t you make a video on fourier transformation?
Could you please upload a video working out a formula for the series 1,1,-1,-1,1,1,-1,-1,... ?
Lucas Depetris Easy: (-1)^floor(n/2) = cos[π·floor(n/2)]
In fact, for the bonus gravy, if you have a sequence characterized by having m 1s followed by having m -1s followed by having m 1s again, with a repeating cycle, then the formula for such a sequence is (-1)^floor(n/m) = cos(π·floor(n/m))
Oh, in case you are not aware of what the floor function is, floor(x) is defined as the largest integer n such that n < x or n = x.
@@angelmendez-rivera351but how did you worked out the formula you gave me, I want to see the full explanation
@@lucasdepetris5896 There is no good way to explain it, because the formula is one of those formulas that cannot be derived by solving simple equations. You either notice the pattern, or you don't, and if you don't, there is pretty much nothing you can do about it.
What I noticed is that the sequence 1, -1, 1, -1,... etc., which is very similar to your sequence, has formula (-1)^n. This one is just common knowledge, and you realize it easily by trying to calculate every power of -1. But the one difference is that I need to be able to repeat every n instead of just counting forward. In other words, instead of plugging in n = 0, n = 1, n = 2, etc., I need to be able to plug in n = 0, n = 0, n = 1, n = 1, n = 2, n= 2, etc. to get your sequence. In other words, I need some sort of increasing step function. And there really are only two step functions that satisfy these properties that can do the job: ceiling(n/2), and floor(n/2). And you get floor(n/2) after testing both on the sequence. That is how I came up with an answer. I really just followed a pattern and reasoned through it slowly and borrowed from other knowledge, but you cannot derive the solution systematically. It's just impossible. Especially if you don't know the floor and ceiling functions to begin with.
7:44 “c”-quence...hehehe
Hoho
I cant understand Cm substitution at the end.
is... is that a thermal detonator?
I think it might be a big problem if f(x)=1/x
2005 VincentChui I think so too since 1/x is bad when x is 0
Would you use Cauchy principal value in this case?
@@runerobin45 - principal, not principle
Appilesh It seems that would be the case, but I am afraid this would make every coefficient equal to 0.
@@Appilesh I confirmed that, yes, you need the Cauchy principal value for the real part of the coefficients. The imaginary part is always integrable, but the real part is not.
What did you study ? (i know math but be more specific please) thanks
any non EE students are watching this video?
: )
I’m a dropout, I think I qualify
astrophysics 🤓
Bro I do mathematics
complex one😍
What is the practical application of fourier series? It is boring if we can not relate it to its
practical application in real life.
Let n = 400K subs till de end of 2019 prove that n = 800k subs at the end of 2021
why is a0 equal to c0 ? cn is not the same thing for n equals 0, 1->+inf or -inf->-1 :\
try to find three numbers a b c such that all three are prime and satisfy the equation a² + b² = c²
please, someone help
I'm having problems with this, although I'm pretty sure there are numbers like this
save my soul ;-;
Black Pen Red Pen long ago made a video on finding a set of equations that generates every Pythagorean triple, each for every input you make. The generator proves that there is no solution to the your problem, meaning there is no Pythagorean triple in which every number is prime. It is impossible.
Complex one
You are the best teacher........#bprpyay!!!!!!!!!!!
So.....
Please integrate e^x^x^x^e^i ......
.......
I think it would be great!!!!😅🤗🤗🤗🤗🤗🤗🤗🤗
And it's time for you to have a haircut
What does the ........ stand for? Also, I’m fairly certain there is nothing you can do to integrate that function using elementary functions. In fact, even with known special functions, it may be impossible.
Is there a way to prove that 1 + 1/2² + 1/3² + .... = pi²/6 with fourier series ????
BELAID Hocine Anis
Yes, you can search "Gucci integral" or check my recent community post.
Parseval’s Theorem
four year series
SAVED MY A$$
Why is it that when n not equal to m the integral is 0?
If you do the calculation, then you will understand why. However, the implication is that the functions are orthogonal as part of a basis
He also made a video on this today or yesterday.
There is a video from 3blue1brown explaining it (or something related to).
shivi mish you can check my previous video in the description
Aaaaaah i struggle when i want to turn the domain of the function [-pi;pi] to [-L;L] to make it match physics and signal analysis =(
I have troubles turning the nx into 2pi*frequency*time =(
Yay^(-1) :'((((
The procedure is the same though. Just integrate by parts.
What happened to "Lil complex"? :(((
Just 29 views?
goat
🤗🤗
Cute😊