Great idea! You use anime girls to teach math. One girl is the teacher and another one is the student. The best part is that the student try to solve the problem (and make small mistake) and the techer only support her (explain mistake using example).
This is called the Socratic method. As you can guess, it originated with Socrates. It was originally used for argumentation / debate, but here it's being applied to education.
I'm 5th semester in CS/Math and I never thought about that way to generalize the binomial formula. I vaguely knew about the Gamma function before, but this is such an approachable way to generalize, very well explained.
This is an unexpected way to present generalized binomial coefficients and expansion. I appreciate that our amine hosts acknowledged that this was an intuitive approach and not a rigorous proof; a proof is probably beyond the scope of an 11-minute video.
I love that little text at the end of every episode saying you gained this or you unlocked that, it feels like I streaming some math skills and actually getting something
wow I didn't expect this much intuition for the part I was struggling during complex analysis in Mathematics for Theoretical Physics course. Thank you.
I just learned that Zundamon isn’t a girl, but a bow (and a dish, and ultimately a fairy) and is referred to as 'it.' Later on, it became a Vocaloid, free for commercial use-and here we are.
This is a refreshing format in which to package math education. I appreciate what you're doing Zundamon. My own desire to eventually produce some interesting math videos has me contemplating what style I'd like to utilize. It's reassuring and encouraging to know that someone else out there has figured out a novel approach. Subscribing to your channel!!
Omg I remember watching a video about how ancient Babylonians calculated square roots and the expanded square root of this video explains why the way they do it works! I love when I manage to connect the dots after watching math vids lol
Thank you for another great video! But I don't feel you have fully answered the question. Your series expansion only converges for |x| < 1, but the original problem statement does not have this restriction. So further, you should write ( x + 1 ) ^pi = x^pi * ( 1 + 1/x ) ^pi, and the series expansion for this would converge for |x| > 1. Finally supplement with 1^pi = 1 and leave the x = - 1 case for another day.
I went to the Taylor expansion of (1+x)^n immediately when I saw this problem but hadn't considered the binomial coefficient for non-natural numbers or where n < k.
My first attempt on this problem would be just to go with the exponential. E.g. (x+1)^π = e^(π ln (x + 1)) = 1 + π ln (x + 1) + (π ln (x + 1)) ^ 2 / 2 + (π ln (x + 1))^3 / 6 + ... In this approach we can use complex numbers, there is no limitation for the argument x. An alternative approach could be in picking an infinite series representation of π and then (x+1)^π becomes an infinite product (x+1)^s1 * (x+1)^s2 * ... where si come from that series representation of π (infinite sum). Now I'm going to watch the video.
I expected that Taylor Expansion would be used by thinking f(x)=(1+x)^pi and f(a)=f(0)=1. But this is another way of using binomial theorem by expanding integers into real numbers. Wow!
This is at the same time a so weird but also so interesting and unique way to teach a math, and I am loving it. Interaction between characters makes this even more enjoyable and helps understand the topic. I believe this style can help somone get invested in math. Keep up the good work!
One approach you can take is to consider Leibniz' formula for pi: 4 - 4/3 + 4/5 - 4/7 +...=8/3 + 8/35 + 8/99 +... (x+1)^pi can then be presented as a product series. If we take t as (x+1)^8, then it equals (t^1/3)*(t^1/35)*(t^1/99)*...
since x + 1 = e^{ln(x+1)}, we could therefore convert (x+1)^\pi into e^{ln(x+1)\pi}, which the series expansion at x = 0 looks quite obvious. but it's still quite interesting if we see it from the perspective of binomial theorem, i think maybe the gamma function could also be required to be introduced here.
I loved the explanation. Also Just checked out your original japanese channel. Please try to recreate those old videos on this channel too. I'd love to see more such videos, it really helps.
pls keep the english version cos i enjoy listening to the videos like audio, with japanese i have to focus on watching even tho i might be doing somethin else like multitasking
I got a question! I came up with this on my own, but it seems the good people at Math Stack Exchange have (partly) figured it out. The "Generating Function" of a series is a function with coefficients equal to the series. For example, take the series 1/0!, 1/1!, 1/2!, 1/3!, 1/4!,.... The generating function for this series is 1/0! * x⁰+ 1/1! * x¹ + 1/2! * x² + 1/3! * x³ +... = e^x. What function is its own generating function? i.e. f(x) = f(0)x⁰+f(1)x¹+f(2)x²+... related question, what function has the property that f(n) = d^n/dx^n f(x) at 0 (i.e. f(0) = f(0), f(1) = f'(0), f(2)=f''(0), f(3)=f'''(0), and so on).
For your second question, I found 4 fonctions wich satisfies the conditions : It's the fonction f(x) = e^(zx) where z is approximatively 0.32 +- 1.34i or 2.06 +- 7.59i
@@catmacopter8545 i thought again about your question. If you want a function to be its own generating function, the fact that that it is a generating dunction that means it is a power series function, or so-called analytic function. That means that its coefficients ( f(0), f(1), f(2), ...) must limit to 0 (otherwise the series would diverge and wouldn't be a well-defined function in the first place!)
@@quantumgaming9180 Well, firstly with the Taylor series, we would have f(x) = sum f(n)/n! x^n. This ressemble a lot the exponential series. So I tried f to be the form of e^(zx). If you plug this into the formula, you get e^(zx) = e^(xe^z). So I need to find the complex number z such that z = e^z. By writing z = a+ib and identifying the real and imaginary part, you get two equations : e^a cos(b) = a e^a sin(b) = b Thus tan(b) = b/a. Plug that into the second equation : e^(b/tanb) sinb = b. Then with wolfram alpha, I could find the 4 solutions b of this equation and thus get the four pair (a,b) that satisfies this problem
Hey bro, in case you havent gotten this comment yet, I can feel that metans voice has been altered a little maybe zundamons too. Let me first tell you that you did a brilliant work they are musch better then what we had previously so thank you!. But stil I think its good with zundamon but I think metan should be speaking a little fast, it looks slow so it makes it feel robotic. Maybe just me but I wanted to let you know!
I enjoyed the faster pace as I could still follow along at this speed but the pacing of how they talked is sometimes unnatural. I think this speed is good for learning but the previous speed was very nice to relax to
You can use the Gamma function. (n k) = n! / (n-k)! / k!. Factorials can be replaced by Gamma functions. So (n k) = Gamma(n+1) / Gamma(n-k+1) / Gamma(k+1) This works for complex numbers too, so funny things can be done with it.
Here in the UK, we are taught the formula expansion of (ax+bx)^n, where n is a negative or a non-integer, as part of a our A-level syllabus (A-levels are the equivalent of high schools in regions outside of UK). This meant it was recognizable that for any value of n in (ax+bx)^n, (ax+bx)^n = 1 + nx + ... (n(n-r+1)/r!) * x^r and for n=π, substitute π into the equation. We have recently covered this material, so if there was anyone in my class who did not grasp the concepts before, they should surely be able to comprehend it now once I send them this video.
The actual proof involves finding a power series for (1 + x)^k. If we write the power series, we are have this formula: f(x) = Sigma(Cn. x^k, n=0, infinity). Next, we have to use the power series for 1/(1+x) = Sigma(x^n, n=0, infinity), and by manipulating, we find the formula above.
Just put it with exponential, like (x+1)^pi = e^(ln(x+1)^pi) = e^(pi*ln(x+1)), would be way more easier. I know it would only work if x > 1, but it will probably give a more important idea of what it represent at first for people that doesn't have that much knowledge of math. Think of this more of an approach than a result.
![[Eng Sub] Can You Expand x+1 Raised to an Irrational Power?](/img/1.gif)
I never believed that anime girls would teach me math
yeah well i watched the same character play factorio
Green is actually a femboy...
there is a large overlap between mathematicians and weebs
@@azurev2258 flammable maths has several uwu video so.......
They teach it pretty well too, my teacher sure didn't start with such an intuitive explanation of nCk
I expected that gamma function was introduced, but oh well... after all the Γ(x+k)/Γ(x) = x*(x-1)*...*(x-k+1)
same
Fr
Didn't everyone expect that?
It's like trying to kill a snake with a cannon, or whatever the phrase is.
However with Gamma function one would have some issue to expand with _exactly_ pi terms... This problem does not occur with the binomial serie
i Wonder who's target audience here?
me
I'm the target audience
We are the target audience
@@mateusvmv based weeb math enjoyer
I knew about the Maclaurin expansion but never thought it can be interpreted as binomial coefficient like this.
Me
Great idea!
You use anime girls to teach math.
One girl is the teacher and another one is the student.
The best part is that the student try to solve the problem (and make small mistake) and the techer only support her (explain mistake using example).
This is called the Socratic method. As you can guess, it originated with Socrates. It was originally used for argumentation / debate, but here it's being applied to education.
Nope, not a great idea. It is annoying AF.
None of the anime girls look like proper students. The hair are way too smooth, they respond too quickly and there are no bags under their eyes.
@@baselinesweb that's a great idea, but not for the ancient
@@Kirillissimus that's how they're look preety
My brain is nourished
I'm 5th semester in CS/Math and I never thought about that way to generalize the binomial formula. I vaguely knew about the Gamma function before, but this is such an approachable way to generalize, very well explained.
Love the 4th wall break. Nice video, refreshing some math I haven't thought about in a while...
math if it was good:
it is good
@@mzg147math if it was peak
People only see Peak as Peak when it involves Anime or Women lol
This is an unexpected way to present generalized binomial coefficients and expansion. I appreciate that our amine hosts acknowledged that this was an intuitive approach and not a rigorous proof; a proof is probably beyond the scope of an 11-minute video.
amene*
MY CALCULUS GRADE MIGHT BE SAVED 🗣🔥
We can finally claim watching more anime would make us smarter! :D
genuinely one of the best math channels
I love that little text at the end of every episode saying you gained this or you unlocked that, it feels like I streaming some math skills and actually getting something
wow I didn't expect this much intuition for the part I was struggling during complex analysis in Mathematics for Theoretical Physics course. Thank you.
Incredible video! I really love the step by step without tedious repetitions of intermediate steps
Wait. Then we can make continuous pascal triangles?
Nah bc anything other than natural numbers just makes an infinite Pascal plane
Yes, I think Veritasium has a video about that
@@blueslime5855what's the name of the video please?
@@blueslime5855 oh really? I might have to check that out
@henrynagel2658 it's called: The discovery that transformed pi
amazingly clear explanation of extending exponents to real numbers
Zundamon lore...
@@Saikiisdone profile picture checks out
@@obz1357but that's just a theory...
Well, you are matpat, so, maybe make a video about it?
@@gmdFrame retired
I just learned that Zundamon isn’t a girl, but a bow (and a dish, and ultimately a fairy) and is referred to as 'it.'
Later on, it became a Vocaloid, free for commercial use-and here we are.
We got anime girls teaching math before GTA 6
This is a refreshing format in which to package math education. I appreciate what you're doing Zundamon. My own desire to eventually produce some interesting math videos has me contemplating what style I'd like to utilize. It's reassuring and encouraging to know that someone else out there has figured out a novel approach. Subscribing to your channel!!
The type of content I never knew I needed, thank you so much for this ❤
The creators invested a lot of work into this video. Congratulations! Thank you for making this video ❤
This channel teaches math better than literally any other channel I’ve seen
Omg I remember watching a video about how ancient Babylonians calculated square roots and the expanded square root of this video explains why the way they do it works! I love when I manage to connect the dots after watching math vids lol
Thank you for another great video! But I don't feel you have fully answered the question. Your series expansion only converges for |x| < 1, but the original problem statement does not have this restriction. So further, you should write ( x + 1 ) ^pi = x^pi * ( 1 + 1/x ) ^pi, and the series expansion for this would converge for |x| > 1. Finally supplement with 1^pi = 1 and leave the x = - 1 case for another day.
The case missing is not 1^pi but 2^pi
@@mutenfuyael3461 ohhh sorry I didn't notice! 2^pi is real anyway and 0^pi we know to be zero, so there are no missing cases!
actually on second thoughts, zero to power pi is zero or undefined? I think irrational powers of zero are undefined...
@@TheDannyAwesome
I believe that it is mostly accepted that for any real x, 0^x is 0 unless x=0 (0^0 = 1).
@@KioKah strictly positive x
0:55 those are the numbers in pascals triangle!
Blew my mind! Incredibly well done 😊
this is amazing and you are brilliant! I never thought I need this!
how international these comments are lmao
Очень)
@@koishi0 why do Russians add a paranthesis at the end of the sentence?
@@Prabhu108. It's like a smile ':)' but we only write ')'
de todo el mundo
Wsup from morocco 🗣
This is just what I needed! I was struggling to expand a sine function raised to a power
i love this series so much, my brain is braining now
my first instinct is to use taylor expansion, it's amazing how the results are the same XD
I went to the Taylor expansion of (1+x)^n immediately when I saw this problem but hadn't considered the binomial coefficient for non-natural numbers or where n < k.
New voices, I like it! The pronunciation is super clean now, it's less jarring. (Although Zundamon sounds older now)
My first attempt on this problem would be just to go with the exponential.
E.g. (x+1)^π = e^(π ln (x + 1)) = 1 + π ln (x + 1) + (π ln (x + 1)) ^ 2 / 2 + (π ln (x + 1))^3 / 6 + ...
In this approach we can use complex numbers, there is no limitation for the argument x.
An alternative approach could be in picking an infinite series representation of π and then (x+1)^π becomes an infinite product (x+1)^s1 * (x+1)^s2 * ... where si come from that series representation of π (infinite sum). Now I'm going to watch the video.
Yes!
I literally learned the binomial theorem a week ago so it's quincidence for such a video to be recommended to me now
2:44 sis taught me the binomial theorem proof that I needed and deserved too 😭 I hate my teachers even more now .
I clicked on this for the equation. I did not expected to be learning from Anime girls. Excellent.
I expected that Taylor Expansion would be used by thinking f(x)=(1+x)^pi and f(a)=f(0)=1. But this is another way of using binomial theorem by expanding integers into real numbers. Wow!
What a great and creative way to explain hard math concepts. You just earned yourself a sub! Keep on ❤
This is at the same time a so weird but also so interesting and unique way to teach a math, and I am loving it. Interaction between characters makes this even more enjoyable and helps understand the topic. I believe this style can help somone get invested in math. Keep up the good work!
this is the single greatest learning experience of my life
Wow, this was awesome! I love the approach this channel is taking, keep it up!
Congrats for π subs
Wow, you're already have an english version. Congratulations😊
this is some quality stuff. keep up the good work!
Strange! I feel like I've just witnessed my year-one granddaughters discussing advanced mathematics.
I love your teaching❤
OH MY GOD BRO THANK YOU FOR MAKING SENSE OUT OF THAT STUPID CONBINQTION THEOREM. THAT WAS A REAL PIECE OF MEAT
It's not that difficult, but I expected it to be more complex, great way to explain.
Certainly the channel of all time ‼🗣✖➖➗🟰
the binomial theorem is what we use to proof the power rule
this is amazing. please do more of these
I didn't sure about mathematical rigor of it 😅. However it's still creative way to calculate it. Those 2d is like new Euler and Newton, I'm serious ✨
Ah yes math lore
?
Dude how the heil do you have that few suscribers? Your content is awesome
i love this channel
英語バージョンがあったとは！英語の勉強にもなって素晴らしい
This is unexpectedly really good! thanks for the vids 😊
Im subbing before this guy/girl becomes famous!
And here I thought gamma functions would be invoked. But it was simpler than all that. Nice video!
Yo maan, this is awesome, continue making ur mathematic videos
Very nice video dealing with a non trivial problem. The presentation does not use sophisticated tools. Very good video.
One approach you can take is to consider Leibniz' formula for pi: 4 - 4/3 + 4/5 - 4/7 +...=8/3 + 8/35 + 8/99 +...
(x+1)^pi can then be presented as a product series. If we take t as (x+1)^8, then it equals (t^1/3)*(t^1/35)*(t^1/99)*...
I am mind blown and I have understood only the first 4 minutes 😮
I didn't even know that Zundamon has an English voicebank lmao
A great math lesson i wish i could study math deep but i am a medical student
really intersting(awa)!That"s an unexpected unfolded
since x + 1 = e^{ln(x+1)}, we could therefore convert (x+1)^\pi into e^{ln(x+1)\pi}, which the series expansion at x = 0 looks quite obvious. but it's still quite interesting if we see it from the perspective of binomial theorem, i think maybe the gamma function could also be required to be introduced here.
I loved the explanation. Also Just checked out your original japanese channel. Please try to recreate those old videos on this channel too. I'd love to see more such videos, it really helps.
wow that is brilliant!
pls keep the english version cos i enjoy listening to the videos like audio, with japanese i have to focus on watching even tho i might be doing somethin else like multitasking
This is cool man keep making these videos
this is fantastic. i learned a lot
This is great, but I wish we get a version with japanese zundamon with English subtitles
The japanese version already has english subtitles (good quality ones, too)
I got a question! I came up with this on my own, but it seems the good people at Math Stack Exchange have (partly) figured it out.
The "Generating Function" of a series is a function with coefficients equal to the series. For example, take the series 1/0!, 1/1!, 1/2!, 1/3!, 1/4!,.... The generating function for this series is 1/0! * x⁰+ 1/1! * x¹ + 1/2! * x² + 1/3! * x³ +... = e^x.
What function is its own generating function? i.e. f(x) = f(0)x⁰+f(1)x¹+f(2)x²+...
related question, what function has the property that f(n) = d^n/dx^n f(x) at 0 (i.e. f(0) = f(0), f(1) = f'(0), f(2)=f''(0), f(3)=f'''(0), and so on).
The cnstant 0 function is an example. Other than this trivial solution I don't know yet
For your second question, I found 4 fonctions wich satisfies the conditions :
It's the fonction f(x) = e^(zx) where z is approximatively 0.32 +- 1.34i or 2.06 +- 7.59i
@@gowipe-grandcross How did you find these functions?
@@catmacopter8545 i thought again about your question. If you want a function to be its own generating function, the fact that that it is a generating dunction that means it is a power series function, or so-called analytic function. That means that its coefficients ( f(0), f(1), f(2), ...) must limit to 0 (otherwise the series would diverge and wouldn't be a well-defined function in the first place!)
@@quantumgaming9180
Well, firstly with the Taylor series, we would have f(x) = sum f(n)/n! x^n. This ressemble a lot the exponential series. So I tried f to be the form of e^(zx).
If you plug this into the formula, you get e^(zx) = e^(xe^z).
So I need to find the complex number z such that z = e^z.
By writing z = a+ib and identifying the real and imaginary part, you get two equations :
e^a cos(b) = a
e^a sin(b) = b
Thus tan(b) = b/a.
Plug that into the second equation : e^(b/tanb) sinb = b.
Then with wolfram alpha, I could find the 4 solutions b of this equation and thus get the four pair (a,b) that satisfies this problem
Hey bro, in case you havent gotten this comment yet, I can feel that metans voice has been altered a little maybe zundamons too. Let me first tell you that you did a brilliant work they are musch better then what we had previously so thank you!. But stil I think its good with zundamon but I think metan should be speaking a little fast, it looks slow so it makes it feel robotic. Maybe just me but I wanted to let you know!
I enjoyed the faster pace as I could still follow along at this speed but the pacing of how they talked is sometimes unnatural. I think this speed is good for learning but the previous speed was very nice to relax to
Zundamon in English is so uncanny...
Great as always tho
onlock learning found a competitor 🥶🔥
Terrific video!
wait my brain cant comprehend i WILL REWATCH IT
Edit : OK I GOT IT NOW second time is the charm rightt
Easy explanation! Taylor expansion is more general.
You can use the Gamma function. (n k) = n! / (n-k)! / k!. Factorials can be replaced by Gamma functions. So (n k) = Gamma(n+1) / Gamma(n-k+1) / Gamma(k+1)
This works for complex numbers too, so funny things can be done with it.
Here in the UK, we are taught the formula expansion of (ax+bx)^n, where n is a negative or a non-integer, as part of a our A-level syllabus (A-levels are the equivalent of high schools in regions outside of UK). This meant it was recognizable that for any value of n in (ax+bx)^n, (ax+bx)^n = 1 + nx + ... (n(n-r+1)/r!) * x^r and for n=π, substitute π into the equation. We have recently covered this material, so if there was anyone in my class who did not grasp the concepts before, they should surely be able to comprehend it now once I send them this video.
I really love this
Yeah, if you take Tailor's series for (a + b)^n you'll get exactly the same
The actual proof involves finding a power series for (1 + x)^k. If we write the power series, we are have this formula: f(x) = Sigma(Cn. x^k, n=0, infinity). Next, we have to use the power series for 1/(1+x) = Sigma(x^n, n=0, infinity), and by manipulating, we find the formula above.
Wish I was high when watched this video. Awesome!
Just put it with exponential, like (x+1)^pi = e^(ln(x+1)^pi) = e^(pi*ln(x+1)), would be way more easier. I know it would only work if x > 1, but it will probably give a more important idea of what it represent at first for people that doesn't have that much knowledge of math. Think of this more of an approach than a result.
When x>1, (1+x)^pi can be written as Σ(nCr(pi,n)x^(pi-n),n,0,inf). This can be proven using the substitution x=1/u
the only things I could say are
1) the constant term is 1
2) the entirety of negative numbers is an asymptote
Aww, this is cute and informative
These videos are cute, but i wish they wouldn’t use synthesized voices. The feeling of the old TTS was classic
I turn from understanding it to not understanding it from straight brainrot💀💀💀
I would love maths more if we had our math lec taken by anime people
Yo i love this type of videos! Can you do a video about integrals and gamma function? I have a test soon 😅
Zundamon and the other girl are so smart
this is so good
imagine a math professor showing this to college students.
Wow I actually learned something damn