@@MsJavaWolf I was the student in class with the lowest grade and asked the teacher about it. She was happy asf when I asked. mind you I dropped out later. Here I am back again after 3 years.
Yeah, this is one of my favorite derivations. Talk about how coordinates on the unit circle in the complex plane correspond to sin and i*cos, do the McClaurin series, and turn it into e^iΘ.
This is probably the best explanation for a high school class, since they haven’t learned about Taylor Series yet. The explanation that makes the most sense to me involves using Maclaurin Expansion to form Euler’s theorem: e^(jx) = cosx + jsinx, and then simply plugging in pi for x, cos(pi) =-1, jsin(pi) = 0. Therefore, e^(j*pi) = -1. For those that don’t know, any function can be written as an infinite Maclaurin series, for example, e^x = Σ(x^n)/n!, from n= 1 to n= infinity. This is called Maclaurin expansion (called this because the series is centered at 0). To save time typing, the Maclaurin expansion of e^(jx) = the Maclaurin expansion of cosx + jsinx. This theorem is incredibly important for electrical engineering. AC circuits can be solved much easier thanks to this theorem and complex numbers. We love this theorem because it combines our 2 favorite functions: e and sinusoids (sine and cosine). We love e because it makes our differential equations easy to solve. We love sinusoids because in real life AC voltage is generated in cosine and sine waves (there are other waves, but let’s only focus on sinusoids). Euler’s theorem marries these 2 functions. The math now becomes a lot easier because we have a way to convert e functions to sinusoid functions. Edit: corrected a fuck up
I think you may have missed where I said just that. It's called Maclaurin expansion. I didn't type it out because I was feeling lazy. Basically, you can take any function and represent it as an infinite series. You need to know derivative calculus to understand this explanation. You use this formula: Σ(nth derivative(a)*(x-a)^n)/n!, from n=0 to n= infinity Now, that's probably very hard to read, but its the only way I can type it. I'll explain it. This is called a Taylor series. Taylor series are used to approximate functions. When n= infinity, the function is exactly the same as the sum of the series. I'm going from n=0 to n=3 because that will get the point across. The a value is where the function is centered at. If a = 0, then the Taylor series is called a Maclaurin series. Macalurin expansion (or Taylor expansion if a != 0) is just expanding the series. Let's use e^(jx) as our example. I'm not going to do cosx or jsinx because that would take too long. You can do it on your own as practice if you really want to or look it up. First you take the nth derivative, from n= 0 to n = 3 and plug in the a value. In this case a= 0. n= 0 derivative: just the function at 0: e^(0*j) = e^0 = 1 n= 1 derivative: je^(0*j) = j n= 2 derivative: j^2*e^(0*j) = -1 n= 3 derivative: j^3*e^(0*j) = -j Now you multiply each derivative term by ((x-a)^n)/n!, where n = whatever derivative term you are on. Add up each derivative term. Remember, a = 0 and 0! = 1. approximation of e^(jx) = 1(x-0)^0/0! + j(x-0)^1/1! + -(x-0)^2/2! + -j(x-0)^3/3! more simplified: 1 + jx - (x^2)/2 - j(x^3)/6 Basically, when you do this process for cosx and jsinx, you find that e^(jx) = cosx + jsinx. You can thank Euler for discovering this. Hope you had fun reading this. Edit: corrected a fuck up
There's actually a really nice very short proof of Euler's formula that does not require the series definitions that goes as follows: Define a function f : C -> C such that f(z) = e^(-iz) (cosz + i sinz). [Notice f is entire].Then differentiating by the product rule we get: f'(z) = -ie^(-iz) (cosz+ i sinz) + e^(-iz) (-sinz +icosz) = e^(-iz) [(-icosz + sinz) + (-sinz + i cosz)] = e^(-iz) [0] = 0 Hence f is a constant function. Now just compute f(0) = e^(-i(0)) (cos 0 + i sin0) = (1)(1 + i(0)) = 1 Hence for all z we have f(z) = 1 = e^(-iz) (cosz + i sinz). Then just multiply both sides by e^(iz) and you get the result.
Euler’s identity does NOT come from Taylor Series, although you can derive it from them. You can also prove it with basic integration and differentiation, but it is not how you conceptually understand it. It really comes from more advanced algebra. 3blue1brown (math TH-camr) has a pretty great video on it.
That was pretty plain. Short, but to explain something plain, you have to have understood or practiced it pretty fine ! I have seen 15min videos out there. So, thanks =) (Also a comment there below is comprehensive)
huh? I think Eddie is trying to explain to high school kids, as in reality you probably would not need this equation unless you do some major in mathematics at a university.
@@MrYoumitube I did a Major in Mathematics at university. :P I still think it would be easier to conceptualize if it was explained that the number line is on the X-axis and the pure imaginary line is on the Y-axis of a coordinate grid. All highschoolers know the Cartesian coordinate system, so explaining the notation shouldn't alienate them and should assist with conceptualizing the idea of rotation.
They might be taking a different syllabus. For example, I at 15 years old learnt radians, but they have e^i(pi), implying they have e, which isn't even taught to us in this school.
Based on the syllabus here in Australia ide say these are graduate students who took a simpler maths and then did this bonus course as introductory to higher level maths which explains why they tackle concepts like imaginary numbers (higher levels) while still straying from ideas like radians (lower level)
It may have been an idea here to accentuate the central importance of the unit circle and associating the symbolic expression e^(i(theta)) with the visual image of the unit circle. This could be linked to the trig based observation that x=cos(theta) + y=sin(theta) traces out the unit circle(we don't need to go through the rigours of taylor series in order to introduce the unit circle association with e^(i(theta)), do it later for the benefit of the rigorous questioning mind). Writing down the inevitably approximate numeric expression for e^1 is both irrelevant and misleading. And so is suggesting that there are a bunch of complicated pieces that need to be assembled to have any hope of understanding what isn't such a complicated idea after all. A lot of intuitive insight can be gained by simply ensuring a strong visual appreciation of the role the unit circle can play as a central mathematical object. For example, not realizing that a fourier transform is nothing more than binding a function to a unit circle, as opposed to a straight line like one is used to due to repetitious cartesian graphing, causes much unnecessary confusion for those continuing in their mathematical studies. Use of the circle in the fourier transform both naturally symbolizes and functionally captures the periodicity aspect we seek to extract. It all makes sense if only the appropriate perspectives are built up and communicated in the first place. The worst thing you can do to a sinusoidal function is butcher it by immediately seeking to represent it in its linear Cartesian aspect. Yet that is exactly what we do, presumably trading the more appropriate and insightful perspective, for one of mere ingrained familiarity.
No, i stress only the importance of setting up an initial association of e^(i(theta)) with the unit circle and explaining, "When you see expressions with e^(i(theta)) you should think circle".This introduces the idea that mathematical expressions are capable of talking about objects and are not just lifeless instruction sets. It is most certainly not beyond the scope of the kids to create an association between the expression e^(i(theta)) and the already well known image of the circle. Where I add explanation to justify the central importance of the circle it is not intended as part of the explanation to the children. The graphical representation is seemingly no more complex then the use of taylor series, yet this is used in high school, presumably with the idea that the overly cumbersome trig functions are somehow simpler despite their inherent transcendental nature.
BeeThat Guy Okay, I can see where you're coming from but don't forget that e and i are new to the pupils, so to say these two meaningless things create a circle is kinda meaningless
It becomes meaningful by virtue of having the meaning provided. Such is the nature of learning. The time to express the mere association is short. The image it provides however, can be long lasting and remains in the mind beyond the condensed time frame of the class room, whereby it can be further pondered at the students leisure.
Normally, you learn Euler's formula when you are in trigonometry since it involves the trigonometric form of a complex number. It is taught right around the time when you graph trigonometric equations, which gives rise to the need for complex numbers. I find it a little strange that you were learning that in a calculus class since calculus address real numbers provided that you were not taking complex analysis, which is beyond the scope of calculus AB.
It is not even about identity. It is about the definition of raising any number albeit e to an imaginary power. If we settle for the idea that an imaginary number is not really that imaginary, but just a real number in two dimensional space that has a rotation component counterclockwise, e to the power of an imaginary number, say x, means just a real number that can be represented as a vector in polar coordinates (r, x) where r has a magnitude that is equal to e and making an angle x with the x-axis.
I wish I had math teacher like this when I was a kid. (My algebra II prof was an embittered stutterer) I love math now in my mid 40s but have no foundation. :(
Yeah to understand these things first they have to understand the relationship between a circle, sin and cos waves, then factorials and laws of indices and imaginary number i and how it was made. Fun fact: Euler's number wasn't invented by Euler himself, it was invented by Jacob Bernoulli. But it's extremely amazing because it made euler's identity possible you wouldn't believe how beautiful Euler's identity is until you saw a awesome video by Mark Newman he's made the best explanation I've seen on Euler's identity with a perfect explanation. He also explains the Fourier Transform very well too (Fourier transform for those that do not know is what makes the internet and all signal reliant devices possible, without it there'd be no telephones, wifi, internet, text messages etc, not even a gaming console lol)
Could you make a longer explanation on this, I've read the description so I know why it's a short video, but if you can then can you make a longer explanation
I have a question too. How is it possible a positive number raised to any number be equal to a negative number? Something is very wrong with that math we all have been taught and future generations will be laughing.
Why wouldn’t it be possible? He just showed that e^iπ = -1. That’s just what happens if you raise a number to a complex number: you get a rotation. And future generations will just keep benefiting from complex numbers. We already use them all the time to solve tons of real world problems, so they will probably be thankful that they can have a lot of useful technology thanks to the mathematicians that dared to explore complex numbers
@@AndresFirte All the things you think it's done thanks to complex numbers can also be done by vector analysis. Complex numbers are a hoax, invented math just for publishing papers. Tell me, can you graph an equation containing "i"?
@@pelasgeuspelasgeus4634 oh really? Can you multiply a vector (2,1) by another vector (3,4)? Can you divide them? No. Because vectors are neither a ring nor a field. On the other hand, complex numbers are indeed vectors, but they are also a ring and a field. So there’s stuff that can be done with complex numbers that can’t be done with the vectors from R×R And yes, you can graph complex numbers, there’s many ways of graphing them. Welch Labs have many videos showing graphs of them, 3Blue1Brown too
what level are these students? i thought they would be about 17 but they haven’t learned about radians so are they younger than 15?? or maybe in their country they learn about them later? i don’t know if someone does please enlighten me
Bro,I guess a^x=1,does not necccesarily imply that x=0,if we bring complex numbers in the discussion !Not sure,but I mean if a^x =1,THEN x can be 2i(pi)/ln(a) What say ??
First things first: you’re wrong about your graph of e. As n increases, e also increases, but levels out at 2.71828... “e” does not increase exponentially like your graph shows. Second: Is it possible to factor Euler’s Formula like some quadratic equations? Factor e^(i*pi) + 1 = 0 into the form of (px +r)(qx + s) = 0?
Not exactly, that is just an approximation. E is, similarly to Pi, infinitely long. You could approximate it further by adding more terms to this sum: e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ... and keep going. The more terms you add, the closer to e you get. But the only way to get the exact value of e would be to add up all the terms, up to infinity, which is not practically possible, so you can only approximate it closer and closer. Right now we have around 500 billion digits of e known, so I guess you could say that is sufficient enough approximation...
Well, do you know calculus, trigonometry, algebra 2 and probably complex numbers? If not, I'd suggest you take a deep look into these topics before even concidering Euler's Identity. As he said, it's like building a house with no foundation.
It's 2.71828... And it is based on a special case of the exponential function, which is its own derivative. It's also based on compounding a 100% growth rate continuously, and how it compares to a 100% growth rate compounded annually.
Let's me show you guys that all you guys have been copying and trying to prove nonsense to be valid E^(πi) +1 = 0 ---> E^(πi) = - 1 Taking both sides to 2nd power gives: (E^(πi))^2 = ( - 1)^2 E^(2πi) = 1 Taking natural logarithm of both sides gives: Ln (E)^(2πi) = Ln(1) (2πi)*Ln(E) = Ln(1) 2πi (1) =0 Taking both sides to 2nd power gives: 4π^2(I)^2 = 0^2 4π^2(-1) =0 -4π^2 = 0 which is impossible In other words , the Euler's formula E^(πi) +1 = 0 is completely false or a mathematically nonsensical statement
I have no idea what that means, as my maths is very rudimentary, but if it's true, it sounds like something that mathematicians and maybe physicists really ought to know about.
That's why people call it i You can't really take away 5 apples when you have 3, it doesn't make sense, but you know 5-3=-2 There is no ratio in existance that equals sqrt(2), but yet, you accept it as a real number, just like 1 and 2 Now think about sqrt(-1), you have no way to write it, but, you can still imagine such a thing exists, and to continue with that idea you will find it useful to name it, just like you did in the first 2 cases He made an 8 video series about this identity that starts from a point that I assume you are in, and goes through all the steps to reach the final goal of this beautiful identity th-cam.com/video/8Y2eIPCg7uc/w-d-xo.html
It's not defined in the real numbers, just like negatives aren't defined in the natural numbers, fractions aren't defined in the integers, and pi and square root of 6 aren't defined in the rational numbers. Each one definition of numbers had too many undefined solutions, so bigger definitions were made to accommodate. When you do a square root of a negative number, you can use the number i and continue with your work in the complex numbers.
This is basically my reaction as a prof when a question comes up in class that is WAY down the road! I call it "spoiler alert." Anyway, awesome :-)
It's cool of you to do that, I guess many teachers will not ever bother.
Hahahah that's actually kinda funny and cool. I want my future prof's to do that
@@MsJavaWolf I was the student in class with the lowest grade and asked the teacher about it. She was happy asf when I asked. mind you I dropped out later. Here I am back again after 3 years.
Just bust out the taylor series. They can handle it.
Bitches love Taylor series.
Yeah, this is one of my favorite derivations. Talk about how coordinates on the unit circle in the complex plane correspond to sin and i*cos, do the McClaurin series, and turn it into e^iΘ.
He explains the Euler identity in an 8 part video, one of which talks about the taylor series
That’s like using trig sub to derive the formula for area of a circle
Excellent explanations ! you should do more!
This is probably the best explanation for a high school class, since they haven’t learned about Taylor Series yet. The explanation that makes the most sense to me involves using Maclaurin Expansion to form Euler’s theorem:
e^(jx) = cosx + jsinx,
and then simply plugging in pi for x, cos(pi) =-1, jsin(pi) = 0. Therefore, e^(j*pi) = -1.
For those that don’t know, any function can be written as an infinite Maclaurin series, for example, e^x = Σ(x^n)/n!, from n= 1 to n= infinity. This is called Maclaurin expansion (called this because the series is centered at 0). To save time typing, the Maclaurin expansion of e^(jx) = the Maclaurin expansion of cosx + jsinx.
This theorem is incredibly important for electrical engineering. AC circuits can be solved much easier thanks to this theorem and complex numbers. We love this theorem because it combines our 2 favorite functions: e and sinusoids (sine and cosine). We love e because it makes our differential equations easy to solve. We love sinusoids because in real life AC voltage is generated in cosine and sine waves (there are other waves, but let’s only focus on sinusoids). Euler’s theorem marries these 2 functions. The math now becomes a lot easier because we have a way to convert e functions to sinusoid functions.
Edit: corrected a fuck up
Sure, but then you're anwering it with another question : WHY is e^(jx)=cosx +jsinx? Because you read it somewhere?
I think you may have missed where I said just that. It's called Maclaurin expansion. I didn't type it out because I was feeling lazy. Basically, you can take any function and represent it as an infinite series. You need to know derivative calculus to understand this explanation. You use this formula:
Σ(nth derivative(a)*(x-a)^n)/n!, from n=0 to n= infinity
Now, that's probably very hard to read, but its the only way I can type it. I'll explain it. This is called a Taylor series. Taylor series are used to approximate functions. When n= infinity, the function is exactly the same as the sum of the series. I'm going from n=0 to n=3 because that will get the point across. The a value is where the function is centered at. If a = 0, then the Taylor series is called a Maclaurin series.
Macalurin expansion (or Taylor expansion if a != 0) is just expanding the series. Let's use e^(jx) as our example. I'm not going to do cosx or jsinx because that would take too long. You can do it on your own as practice if you really want to or look it up.
First you take the nth derivative, from n= 0 to n = 3 and plug in the a value. In this case a= 0.
n= 0 derivative: just the function at 0: e^(0*j) = e^0 = 1
n= 1 derivative: je^(0*j) = j
n= 2 derivative: j^2*e^(0*j) = -1
n= 3 derivative: j^3*e^(0*j) = -j
Now you multiply each derivative term by ((x-a)^n)/n!, where n = whatever derivative term you are on. Add up each derivative term. Remember, a = 0 and 0! = 1.
approximation of e^(jx) = 1(x-0)^0/0! + j(x-0)^1/1! + -(x-0)^2/2! + -j(x-0)^3/3!
more simplified: 1 + jx - (x^2)/2 - j(x^3)/6
Basically, when you do this process for cosx and jsinx, you find that e^(jx) = cosx + jsinx. You can thank Euler for discovering this. Hope you had fun reading this.
Edit: corrected a fuck up
Julian Bell you can prove it pretty easily with no expansions.
Lyuboslav Angelushev no, it's a simple proof.
Hmm. I'm pretty sure that not all functions have maclaurin series. That is why we also have things like Fourier' series.
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such a brilliant and hilarious way to explain exponential growth!
I probably would have used a snowball rolling down a hill as an analogy.
There's actually a really nice very short proof of Euler's formula that does not require the series definitions that goes as follows:
Define a function f : C -> C such that f(z) = e^(-iz) (cosz + i sinz). [Notice f is entire].Then differentiating by the product rule we get:
f'(z) = -ie^(-iz) (cosz+ i sinz) + e^(-iz) (-sinz +icosz) = e^(-iz) [(-icosz + sinz) + (-sinz + i cosz)] = e^(-iz) [0] = 0
Hence f is a constant function. Now just compute f(0) = e^(-i(0)) (cos 0 + i sin0) = (1)(1 + i(0)) = 1
Hence for all z we have f(z) = 1 = e^(-iz) (cosz + i sinz). Then just multiply both sides by e^(iz) and you get the result.
The identity comes from complex variables field.
e^(i*pi) = cos(pi) + isin(pi) = -1
More generally:
e^x = cos(x) + sin(x)
Your formula has a tiny typo:
e^(i*x) = cos(x) + i * sin(x)
Mr Huy, the latter formula that you've written is not true
it't actually i(iota) that enables the euler's identity
Euler’s identity does NOT come from Taylor Series, although you can derive it from them. You can also prove it with basic integration and differentiation, but it is not how you conceptually understand it. It really comes from more advanced algebra. 3blue1brown (math TH-camr) has a pretty great video on it.
That was pretty plain. Short, but to explain something plain, you have to have understood or practiced it pretty fine ! I have seen 15min videos out there.
So, thanks =)
(Also a comment there below is comprehensive)
I've watched many explanations for this, and this video is by far the shortest and most simple.
The class wasn't ready for this
Nice explanation.....as my head did not explode.
Thank you for this! I just subscribed 🙂
Should first explain the complex plane briefly as an extension of the number line, so that rotating in it makes more sense.
huh? I think Eddie is trying to explain to high school kids, as in reality you probably would not need this equation unless you do some major in mathematics at a university.
@@MrYoumitube I did a Major in Mathematics at university. :P
I still think it would be easier to conceptualize if it was explained that the number line is on the X-axis and the pure imaginary line is on the Y-axis of a coordinate grid. All highschoolers know the Cartesian coordinate system, so explaining the notation shouldn't alienate them and should assist with conceptualizing the idea of rotation.
That's every mathematician's dream innit, I wish some random person would ask me math questions too
e^(i*pi) = cos (pi) + i*sin (pi) = - 1+0 = -1
How old are the students since they havent learnt about radians, imaginary numbers etc..?
They might be taking a different syllabus. For example, I at 15 years old learnt radians, but they have e^i(pi), implying they have e, which isn't even taught to us in this school.
Probably Grade 10. Credibility: I'm an Australian student
Here, we learn about radians in 10th grade, and we learn about e and i in 12th grade
Based on the syllabus here in Australia ide say these are graduate students who took a simpler maths and then did this bonus course as introductory to higher level maths which explains why they tackle concepts like imaginary numbers (higher levels) while still straying from ideas like radians (lower level)
@@stellarose611 He is a high school teacher
Superb sir ur amazing and your unique sir
Wish I had a teacher like this
You sir are extremely underrated
It may have been an idea here to accentuate the central importance of the unit circle and associating the symbolic expression e^(i(theta)) with the visual image of the unit circle. This could be linked to the trig based observation that x=cos(theta) + y=sin(theta) traces out the unit circle(we don't need to go through the rigours of taylor series in order to introduce the unit circle association with e^(i(theta)), do it later for the benefit of the rigorous questioning mind). Writing down the inevitably approximate numeric expression for e^1 is both irrelevant and misleading. And so is suggesting that there are a bunch of complicated pieces that need to be assembled to have any hope of understanding what isn't such a complicated idea after all. A lot of intuitive insight can be gained by simply ensuring a strong visual appreciation of the role the unit circle can play as a central mathematical object. For example, not realizing that a fourier transform is nothing more than binding a function to a unit circle, as opposed to a straight line like one is used to due to repetitious cartesian graphing, causes much unnecessary confusion for those continuing in their mathematical studies. Use of the circle in the fourier transform both naturally symbolizes and functionally captures the periodicity aspect we seek to extract. It all makes sense if only the appropriate perspectives are built up and communicated in the first place. The worst thing you can do to a sinusoidal function is butcher it by immediately seeking to represent it in its linear Cartesian aspect. Yet that is exactly what we do, presumably trading the more appropriate and insightful perspective, for one of mere ingrained familiarity.
BeeThat Guy I think this is beyond the scope of the kids he's teaching. Learning e^theta graphical relationships comes at early university
No, i stress only the importance of setting up an initial association of e^(i(theta)) with the unit circle and explaining, "When you see expressions with e^(i(theta)) you should think circle".This introduces the idea that mathematical expressions are capable of talking about objects and are not just lifeless instruction sets. It is most certainly not beyond the scope of the kids to create an association between the expression e^(i(theta)) and the already well known image of the circle. Where I add explanation to justify the central importance of the circle it is not intended as part of the explanation to the children. The graphical representation is seemingly no more complex then the use of taylor series, yet this is used in high school, presumably with the idea that the overly cumbersome trig functions are somehow simpler despite their inherent transcendental nature.
BeeThat Guy Okay, I can see where you're coming from but don't forget that e and i are new to the pupils, so to say these two meaningless things create a circle is kinda meaningless
It becomes meaningful by virtue of having the meaning provided. Such is the nature of learning. The time to express the mere association is short. The image it provides however, can be long lasting and remains in the mind beyond the condensed time frame of the class room, whereby it can be further pondered at the students leisure.
BeeThat Guy I agree, but it would be to little avail I would think, given the condensed time frame
I’m curious what math class this is, because I just finished calculus AB and we just learned about that formula.
Normally, you learn Euler's formula when you are in trigonometry since it involves the trigonometric form of a complex number. It is taught right around the time when you graph trigonometric equations, which gives rise to the need for complex numbers.
I find it a little strange that you were learning that in a calculus class since calculus address real numbers provided that you were not taking complex analysis, which is beyond the scope of calculus AB.
maths extension 2, offered to students in year 12 in high school (or earlier if they accelerate)
Could you share a video on FOURIER SERIES?
Thank you!
I guess that’s one approach, not how I learned it when I was 14.
It is not even about identity. It is about the definition of raising any number albeit e to an imaginary power. If we settle for the idea that an imaginary number is not really that imaginary, but just a real number in two dimensional space that has a rotation component counterclockwise, e to the power of an imaginary number, say x, means just a real number that can be represented as a vector in polar coordinates (r, x) where r has a magnitude that is equal to e and making an angle x with the x-axis.
Someone here watched too much Monogatari...
Ayy that’s me as well
Being Sodachi is suffering
I wish I had math teacher like this when I was a kid. (My algebra II prof was an embittered stutterer) I love math now in my mid 40s but have no foundation. :(
Yeah to understand these things first they have to understand the relationship between a circle, sin and cos waves, then factorials and laws of indices and imaginary number i and how it was made. Fun fact: Euler's number wasn't invented by Euler himself, it was invented by Jacob Bernoulli. But it's extremely amazing because it made euler's identity possible you wouldn't believe how beautiful Euler's identity is until you saw a awesome video by Mark Newman he's made the best explanation I've seen on Euler's identity with a perfect explanation. He also explains the Fourier Transform very well too (Fourier transform for those that do not know is what makes the internet and all signal reliant devices possible, without it there'd be no telephones, wifi, internet, text messages etc, not even a gaming console lol)
I wish i had a teacher like you in school
I learned radians when i was like 15 and wasn't in an advanced math how old are your students?
trev janke radians are typically taught at age 16/17 in the UK, this could be true of Australia, too
I live in Australia and learnt them when I was 14/15.
15/
16 In my country
16
e^i.x=cos x + i.sin x (Euler's Complex Number Theorem)
Put x = π and there you go
Should be e^i•x=cos x+ i sin x
@@ozaman-buzaman9300 oops thanks for correcting
Siddhartha Roy wouldnt sin(pi)*i be the same as just sin(pi) since it is 0?
@@thepotato3902 no sin(π) is 0 so sin(π)*i will be 0
Siddhartha Roy so multiplying by i makes no difference right? since 0*i is still zero?
A guy from heaven
Also group theory disproved here's the link
th-cam.com/video/WZHSqoiti-4/w-d-xo.html
Someone asked me a question about Euler's identity = I got bored and taught my 1st year further maths
Beautiful
Could you make a longer explanation on this, I've read the description so I know why it's a short video, but if you can then can you make a longer explanation
He has, check out the 8 part series on the most beautiful identity in math. There he explains complex numbers and this identity
Nice job! I like it!
I have a question too. How is it possible a positive number raised to any number be equal to a negative number? Something is very wrong with that math we all have been taught and future generations will be laughing.
Why wouldn’t it be possible? He just showed that e^iπ = -1. That’s just what happens if you raise a number to a complex number: you get a rotation.
And future generations will just keep benefiting from complex numbers. We already use them all the time to solve tons of real world problems, so they will probably be thankful that they can have a lot of useful technology thanks to the mathematicians that dared to explore complex numbers
@@AndresFirte All the things you think it's done thanks to complex numbers can also be done by vector analysis. Complex numbers are a hoax, invented math just for publishing papers. Tell me, can you graph an equation containing "i"?
@@pelasgeuspelasgeus4634 oh really? Can you multiply a vector (2,1) by another vector (3,4)? Can you divide them? No. Because vectors are neither a ring nor a field.
On the other hand, complex numbers are indeed vectors, but they are also a ring and a field.
So there’s stuff that can be done with complex numbers that can’t be done with the vectors from R×R
And yes, you can graph complex numbers, there’s many ways of graphing them. Welch Labs have many videos showing graphs of them, 3Blue1Brown too
@@AndresFirte I won't discuss anything else with you if you don't answer my question on why the sqrt(-1)*sqrt(-1) is not equal to sqrt((-1)*(-1)).
@@pelasgeuspelasgeus4634 I answered to you 9 hours ago. Perhaps you didn’t get the notification. Go check the replies to that thread
Are they learning complex number & exponential before using radian?
my maths teacher mentioned this but kinda went "meh" like it's not worth wasting time on.
i understood everything and yet i still don’t get it
Holy teacher and students 😂
what level are these students? i thought they would be about 17 but they haven’t learned about radians so are they younger than 15?? or maybe in their country they learn about them later? i don’t know if someone does please enlighten me
The Euler's identity is a beautiful equation but useless
"Nothing useless can be truly beautiful." -- William Morris
But I don’t have a calue
Excellent!! !!
e^iπ = i^2
Super smart
Just wanna know what age students are you teaching?
Cuz I'm 16 and learning the same thing right now...
Who came from 3Blue1Brown after you were brainwashed?
Keşke Türkçe altyazısı olsa!
영어귀가 어두워요 해석 부탁드려요 한국인 수학매니아분들!!ㅋㅋㅋㅋㅋ
But sir if we square both sides it gives us 2πi=0, why sir
Squaring gives e^2iπ = 1
Indian defender e^pie
Bro,I guess a^x=1,does not necccesarily imply that x=0,if we bring complex numbers in the discussion !Not sure,but I mean if a^x =1,THEN x can be 2i(pi)/ln(a)
What say ??
First things first: you’re wrong about your graph of e. As n increases, e also increases, but levels out at 2.71828... “e” does not increase exponentially like your graph shows.
Second: Is it possible to factor Euler’s Formula like some quadratic equations? Factor e^(i*pi) + 1 = 0 into the form of (px +r)(qx + s) = 0?
youre wrong lol e is on the x axis and levels off at 2.71... whilst y increases exponentially and that's exactly what his graph shows.
Perez Sandra Rodriguez Elizabeth Garcia Maria
Or we know that e^ix= cosx so u can conclude .....
No... e^ix = cosx + isinx.
Ya but sign nπ=0
yes but you were writing the formula with x not π
pls zoom ur camera it not clear enough to see
Pls shut up
I didn't got it 🙄
Anubhav Chandrakar you need to know trigonometry/ unit circle/ radians
La Tortue PGM well thanks, now I understood it!
Sebastian Davidson glad someone actually understood this mess lol
The definition of _i_ is actually *i² = -1* not _i = √(-1)_
BIG DIFFERENCE
T0y Lit there is because if you define i = √(-1) then you can 'prove' stupid things
It is a huge difference as √(-1) is plus or minus i.
UnderdoneElm77 No... Square root of -1 is always i. -1 could equal i^2 or -i^2, not the other way around.
Dawn Ripper
No! You are wrong. Please look up the definition before posting crap like that!
e= 2.718 no?
e is irrational
Baptiste Broyer roundabout that
Not exactly, that is just an approximation. E is, similarly to Pi, infinitely long. You could approximate it further by adding more terms to this sum:
e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ... and keep going. The more terms you add, the closer to e you get. But the only way to get the exact value of e would be to add up all the terms, up to infinity, which is not practically possible, so you can only approximate it closer and closer. Right now we have around 500 billion digits of e known, so I guess you could say that is sufficient enough approximation...
It's close to that, but the actual e continues forever, just like pi and sqrt(2)
bemdav 1/0!?
Tik
Learned nothing..
Well, do you know calculus, trigonometry, algebra 2 and probably complex numbers? If not, I'd suggest you take a deep look into these topics before even concidering Euler's Identity. As he said, it's like building a house with no foundation.
Avana this is to much bro wtf I don’t even like math it’s a load of bs
Avana so no I would not study this shit
@@ghostlygod9845 you watched the video... why?
U
What a rude class, all talking over him
how the heck is e = 1.7182....
It's 2.71828...
And it is based on a special case of the exponential function, which is its own derivative. It's also based on compounding a 100% growth rate continuously, and how it compares to a 100% growth rate compounded annually.
carultch yeah
Let's me show you guys that all you guys have been copying and trying to prove nonsense to be valid
E^(πi) +1 = 0 --->
E^(πi) = - 1
Taking both sides to 2nd power gives:
(E^(πi))^2 = ( - 1)^2
E^(2πi) = 1
Taking natural logarithm of both sides gives:
Ln (E)^(2πi) = Ln(1)
(2πi)*Ln(E) = Ln(1)
2πi (1) =0
Taking both sides to 2nd power gives:
4π^2(I)^2 = 0^2
4π^2(-1) =0
-4π^2 = 0
which is impossible
In other words , the Euler's formula E^(πi) +1 = 0 is completely false or a mathematically nonsensical statement
The logarithm understander has arrived
I have no idea what that means, as my maths is very rudimentary, but if it's true, it sounds like something that mathematicians and maybe physicists really ought to know about.
You're actually wrong. Here's why:
ln(x²) = 2 lnx if and only if x>0. When x0.
You didn't think it through and misled yourself.
The square root of -1 is not defined.
Jeboy it is
Jeboy There are these things called complex numbers...
That's why people call it i
You can't really take away 5 apples when you have 3, it doesn't make sense, but you know 5-3=-2
There is no ratio in existance that equals sqrt(2), but yet, you accept it as a real number, just like 1 and 2
Now think about sqrt(-1), you have no way to write it, but, you can still imagine such a thing exists, and to continue with that idea you will find it useful to name it, just like you did in the first 2 cases
He made an 8 video series about this identity that starts from a point that I assume you are in, and goes through all the steps to reach the final goal of this beautiful identity
th-cam.com/video/8Y2eIPCg7uc/w-d-xo.html
you're like 1000 years late to the party.
It's not defined in the real numbers, just like negatives aren't defined in the natural numbers, fractions aren't defined in the integers, and pi and square root of 6 aren't defined in the rational numbers.
Each one definition of numbers had too many undefined solutions, so bigger definitions were made to accommodate. When you do a square root of a negative number, you can use the number i and continue with your work in the complex numbers.