Euler's Identity With a Limit!

  Рет қаралды 5,520

Mu Prime Math

Mu Prime Math

2 жыл бұрын

e^x limit for real numbers: • THE FACT
Euler's formula for the complex exponential can be proved in many different ways, such as with Taylor series. Here we find Euler's identity using a limit equation for e^x and the polar form of complex numbers!
Calculus Problems playlist: • Calculus Problems
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Music: OcularNebula - The Lopez

Пікірлер: 35
@nabla_mat
@nabla_mat 2 жыл бұрын
Mind blowing! I find it more elegant than the “infinite series proof”
@alaechoulli6111
@alaechoulli6111 2 жыл бұрын
Loved the explanation! Keep going !
@MyOneFiftiethOfADollar
@MyOneFiftiethOfADollar Жыл бұрын
What you proved is sometimes presented as the definition of e^x where e is the only constant such that d/dx(e^x) = e^x Still nice to see complex variable proof of it.
@paolopalmisano827
@paolopalmisano827 2 жыл бұрын
This is originally after Euleur dimostration with Power series
@MathSolvingChannel
@MathSolvingChannel 2 жыл бұрын
nice video!
@JorgeMartinez-bs6iz
@JorgeMartinez-bs6iz 2 жыл бұрын
We need more complex analysis proofs! Great vídeo btw.
@aashsyed1277
@aashsyed1277 2 жыл бұрын
Cool video!
@mathismind
@mathismind 2 жыл бұрын
Nice one!
@jorgeluisjaramillogaitan7906
@jorgeluisjaramillogaitan7906 2 жыл бұрын
Thanks a lot, it´s a nice video.
@schweinmachtbree1013
@schweinmachtbree1013 2 жыл бұрын
bravo!
@michaelbaum6796
@michaelbaum6796 Жыл бұрын
Great👍
@rodbhar6522
@rodbhar6522 2 жыл бұрын
It's driving me crazy watching you put that pen cap on and off a million times. lol
@MyOneFiftiethOfADollar
@MyOneFiftiethOfADollar Жыл бұрын
So focus more on his excellent proof😂
@guill3978
@guill3978 2 жыл бұрын
E
@MichaelRothwell1
@MichaelRothwell1 2 жыл бұрын
Very cool to spot that this limit formula for e^x can also be applied to complex arguments. My only problem is that you omitted the proof (using your definition of e^x) that the derivative of e^x is e^x. This is of course needed to prove that the derivative of its inverse function ln x is 1/x, which you need for your proof.
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
As shown in the video linked in the description, the limit formula for e^x can be proved equivalent to more standard definitions of e^x for real arguments, from which the derivative formulas follow.
@MichaelRothwell1
@MichaelRothwell1 2 жыл бұрын
@@MuPrimeMath That's fine then, thanks.
@barryzeeberg3672
@barryzeeberg3672 2 жыл бұрын
just wondering: at around 9:55, you are taking the ln of your expression. but I wonder if this is circular reasoning, as this operation assumes the existence and properties of ln, which is equivalent to assuming the existence and properties of e . But the proof you are establishing is the definition of e , so I am not sure whether using that definition and property constitute a circular argument? Other than that, I think it is a great video proof. Also, I noticed that the property of cis(a) x cis (b) = cis (a+b) is highly suggestive of the behavior of e (this is probably not just a coincidence, but is likely to be a core feature of using cis to prove something about e) ?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
This proof is not about defining the number e. We already know what e is, we already know e^x for real values of x, and we already know about ln and properties of ln. This video is about finding the value of e^(iθ). But the ln at 9:55 only involves real numbers, so we don't have to worry about not knowing e^(iθ) yet!
@barryzeeberg3672
@barryzeeberg3672 2 жыл бұрын
@@MuPrimeMath thank you, I appreciate the clear explanation
@ruffledraphael
@ruffledraphael 2 жыл бұрын
Hi, as we know Euler's number is defined as n→inf where n is whole number. You changed the variable from n to x where x is from real number set, how is this still valid? Doesn't it contradict with Eular's number's definition? Would you explain please I am a high school student.
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
If a limit expression has a certain value over the real numbers, then it has the same value over the whole numbers. This is a result from calculus that you can prove using the epsilon-delta definition of a limit! Basically, the reason it's true is that every whole number is also a real number.
@sea34101
@sea34101 2 жыл бұрын
All the functions in this proof are continuous so you can do this change.
@ruffledraphael
@ruffledraphael 2 жыл бұрын
Thanks a lot. I have another question, you used the cis function from the identity but isn't the whole point of the proof is to show that you can prove the identity using limits? Why use cis function on the first place since we are proving the thing that there should be a cis function?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
cis is just defined as cos(θ) + i sin(θ), so we know that it exists. The point of the proof is to show e^(iθ) = cis(θ).
@ccg8803
@ccg8803 2 жыл бұрын
Why Lim(cis (x)) = cis(Lim(x))? I mean, why are we able to switch those things?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
There is a general result from calculus: if f is continuous and lim g(x) exists, then lim f(g(x)) = f(lim g(x)). cis is continuous, so we can swap it with the limit!
@ccg8803
@ccg8803 2 жыл бұрын
@@MuPrimeMath thanks!!
@ameroamigo1
@ameroamigo1 Жыл бұрын
Starting about 6:17 I think he ignores the i in the i(theta)
@MuPrimeMath
@MuPrimeMath Жыл бұрын
In that case I'm computing the angle of the number in the complex plane, so I'm considering theta as a length in the imaginary direction.
@XYZ-yn2hn
@XYZ-yn2hn 2 жыл бұрын
very thorough explanation but why does it feel like a circular reasoning? or maybe something too familiar that makes it sound so true.
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
It might feel circular because we introduce cos(x) + i sin(x) at the beginning! Remember that every complex number can be written as r*cis(θ); that's just polar coordinates. The key for Euler's identity is proving e^(iθ) = cis(θ).
@angeldude101
@angeldude101 2 жыл бұрын
I felt it getting a bit circular when I noticed that various exponent laws were getting proved for cis(x). cis(a)cis(b) = cis(a+b) is just the same as (e^a)(e^b) = e^(a+b), and cis(a)^n = cis(an) is the same as (e^a)^n = e^(an). The kicker is that I've seen exponentiation extended to the reals be defined precisely through these relationships.
@chrisgreen_1729
@chrisgreen_1729 2 жыл бұрын
Another great explanation from Mu Prime Math, and I like the way you don't skip over the technical details. Carl Friedrich Gauss was reported to have commented that if Euler's identity was not immediately apparent to a student upon being told it, that student would never become a first-class mathematician. I wonder how many students these days think an 18 minute video is "immediately apparent"?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
There is no direct citation for this quote besides a book that says "I wouldn't put it past him", so I would take it with a grain of salt!
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