Euler’s identity proof for calculus 2 students!

Visit 👉 brilliant.org/blackpenredpen/ to learn more calculus and imaginary numbers (20% off with this link)
-----------------------
The Euler's Identity is considered to be the most beautiful equation in math, which says e^(i*pi)+1=0, t-shirt here: amzn.to/427Seae This video is for calculus 2 students after we cover the Taylor series. We will first do the classic proof of Euler's formula where we use the Taylor series of e^x, sin(x), cos(x), and complex numbers to draw the connection that e^(iz)=cos(z)+isin(z). Then we will plug in pi and get the equation e^(i*pi)=-1, which is the same as e^(i*pi)+1=0. Euler's formula is also the main building block of complex analysis.
Calculus 2 power series pdf 👉 bit.ly/calc2series
See how we can integrate e^(ax)*sin(bx) with Euler's formula and no integration by parts: 👉 • Integral of e^(ax)sin(...
0:00 Proving Euler's Formula e^(iθ)=cos(θ)+i*sin(θ)
4:58 Check out Brilliant
5:52 Proving Euler's Identity e^(i*pi)+1=0
---------------------------------------------
**Thanks to ALL my lovely patrons for supporting my channel and believing in what I do**
AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls
---------------------------------------------
💪 If you would also like to support this channel and have your name in the video description, then you could become my patron here / blackpenredpen
Calculus 2 power series pdf: bit.ly/calc2series

[@blackpenredpen]

Shop an Euler's Identity t-shirt: amzn.to/427Seae

[@WestExplainsBest]

As a teacher, the ability to speed up like that would be amazing.

[@themibo899]

*3 hours later* kid in the back who didn"t listen: yooooo how can you have the sqrt of -1 ??? I just checked on my calculator and it says "undefined" worse even: Everyone in the class: fuck he's right, we think you made a mistake sir

[@anshumanagrawal346]

@@themibo899 lol

[@sudirosudiro4218]

kzbin.info/www/bejne/aYnNlJWIjr6sr7M proof of polynomial (x+a)^n

[@timhaines3877]

I had a physics professor who would just cartoonishly wave his arms and say "as any fool can see!" and then write down the last statement of the derivation. It was his way of not-so-subtly hinting that the physics is more important than the n-th degree of the math- he would show us resources that contained the full derivations later. That might not work in a math class, though.

[@RobinHillyard]

You forgot to mention that Euler’s identity also employs the four principal mathematical operators: addition, multiplication, exponentiation, and equality. ;)

[@blackpenredpen]

Oh yea! Thanks.

[@slavisastojakovic3400]

Whenever he says 'Ladies and gentlemen' I know magic is about to happen.

[@blackpenredpen]

😇

[@plislegalineu3005]

more like maθ

[@eleSDSU]

@@plislegalineu3005 weird way to spell mathgic

[@leziehr]

i have been looking for this equation’s derivation for ages, thank you for this! this really is the most beautiful equation ever.

[@blackpenredpen]

😃

[@okenyodong1420]

Amazing proof, thank you so much. I was struggling to wrap my head around Eulers identity for months.

[@General12th]

If you're ever curious about where those series come from (or you don't feel like remembering them), just derive them yourself! Suppose cos(x) = a + bx + cx^2 + dx^4 + ... . Set x = 0, solve for a, then differentiate both sides with respect to x and solve for b. Differentiate yet again and keep going! Within four or five iterations, you should see a pattern emerge. You can do the same thing for any function, like sin(x) or e^x or ln(1+x). It's much easier than having to remember them.

[@catnip202xch.]

Yeah and it’s the Taylor series and I fucking hate it so bad it’s the bane of my existence lol

[@General12th]

@@catnip202xch. Deriving a Taylor series shouldn't be that hard. Do you know how to differentiate most functions? Can you set x = 0 and solve for coefficients one after another?

[@Goku_is_my_idol]

Taylor series just assumes every function can be expressed as a polynomial function of infinite order. Let f(x)= a0 + a1x + a2x²+.......+anxⁿ+....upto infinity f(0)=a0 f'(0)=a1 f"(0)=2a2 ....... And so on fⁿ(0)=n!an Then just multiply individual an values to xⁿ and thats the mauclaurin series. I should mention this is a specification of Taylor series. The original one's a bit more complicated to derive: f(a+h)=f(a) + hf'(a)+..... Remembering things is hard. Try to comprehend the underlying concept and you wont need memorization.

[@Muslim_011]

Brilliant 👍

[@eleSDSU]

Bold of you to believe I can do this for any function :3

[@user-yr8lm3sv4s]

new inspiration on the formula,thank u

[@dr.rahulgupta7573]

Excellent presentation of the topics in a beautiful manner by an experienced teacher . Vow !

[@jordimayorgisbert6490]

Amazing !!! I don’t know that development of e^z using Taylor Series. Thank you !!!!

[@JSSTyger]

I agree that this is the most beautiful equation in math. I learned it in my fifth and final year at the university, taking a course i didnt even need. Needless to say, I am quite grateful I took that course.

[@Mark-nm9sm]

I am watching your 100 infinite series little by little and the difference in hair style is so nuts compared to now. Also the mic is now a pokeball amazing

[@idrisShiningTimes]

This equation id really beautiful, like no words to describe this masterpiece

[@fred-ev7zd]

THIS EQUATION IS SO FUN!!!

[@michapodlaszuk9025]

Wow, it's really cool how easily I can show people that I'm done with the proof, thanks

[@sahincpa1230]

Very very nice presention. I highly appreciated you sir.

[@valerium730]

The second you wrote down "1-(1/2!)t^2" i immediately knew what was up, and it's beautiful.

[@valerium730]

like no joke i actually shouted "no way!" out loud the second i saw it

[@meiljohnbueno3149]

I really admire those people who really understand maths and its complexities. Wisdom is really a gift ❤️❤️

[@eleSDSU]

You might be confusing knowledge with wisdom.

[@octs609]

@@eleSDSU No they are definitely wise.

[@idjles]

Not only does it have the 5 most important numbers, it also has the 4 most important operations: equality, addition, multiplication and exponentiation.

[@stevenlin2005]

I’m a 10th grade student in Taiwan, and I chose pre-calculus as my elective course (it’s really PRE-calculus, we won’t even mention Epsilon-Delta definition of a limit this semester!) Though I can’t fully understand most of your calculus videos, I still find them interesting! And I wish to have a teacher like you ：）

[@uchihasasuke7436]

Make sure you know unit circle very well haha 加油同学

[@blackpenredpen]

See how we can integration e^(ax)*sin(bx) with Euler's formula and no integration by parts: 👉 kzbin.info/www/bejne/e5DIlH-Nn8Slma8

[@rosaorwinter910]

I have a genuine question for you, if e^(iθ)=cos(θ)+i*sin(θ) then wouldn't that mean that if you plug in 2π you would get +1, meaning that i×2π=ln(1) so therefore i=0?

[@mazroatussaadah617]

O km momok

[@abdelhadiech-chaibi2921]

Can you calculate integral of exp(x) over t^2 +1

[@velivelmu8530]

another one I liked, in latex: f(x) = e^{-ix}(\cos x + i \sin x) \\ f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x) \\ f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - e^{-i x}(i \cos x + i^2 \sin x) \equiv 0 \\ f^{\prime}(x) = 0 \;\;\; \forall \; x \in \mathbb{R}\Rightarrow f(x) \text{ is a constant} \\ f(0) = e^{0}(\cos 0 + i \sin 0) = 1 \cdot(1+0) = 1 \Rightarrow f(x) = 1 \;\;\; \forall \; x \in \mathbb{R} \\ \\ 1 = e^{-ix}(\cos x + i \sin x) \Rightarrow e^{ix}=\cos x + i \sin x \;\;\; \forall \; x \in \mathbb{R}

[@AirshipToday]

@@rosaorwinter910 No, because if you were to divide both sides by i aka multiply by -i you get 2π=0 and understand that ln(x) has infinite solutions

[@RomanoPRODUCTION]

It's beautiful ❤️

[@user-tt7wu2zh7h]

The beauty of math in one episode

[@carultch]

Are you proud to see mathematics making your alphabet famous?

[@hakimalaarossi7632]

Thank you ❤❤❤

[@randomlyselected4550]

आशीर्वाद from India ✋🏻

[@thxmori]

BPRP, I have been following u for o long time, can you describe Catalan’s constant that my teacher ask me to study it.😄

[@iandmetick07]

Hey Steve I love your videos, I would like you do Reimann hypothesis. What's it and what is the problem there and what's so difficult to solve it 😂

[@huizylove]

Do you have a cal2 review? I need it your the best!

[@blackpenredpen]

I got you. 100 calc 2 problems. kzbin.info/www/bejne/gajcnJKjqcqshMU

[@huizylove]

@@blackpenredpen Thank you! 😊

[@GaryFerrao]

OMG the sudden close up of your face for the sponsor message was so creepy lol 😅😂

[@AznJsn82091]

I will get a tattoo of that in the future!

[@sanjeevkr.1497]

Love you sir

[@bogdyee]

There was a problem once that I found but I could not understand the answer and why it works. The problem sounded like this. Given a board of M rows an N columns find the number of ways to tile an MxN rectangle with 1x2 (they can be rotated) dominoes. (something related to aztec treasure problem but on a rectangular board) and the formula that I found on a forum is like (I tested it with a program and it does indeed give the right answer for several M and N test cases). 2^(M*N/2) * ( cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) ) ^ (1/4) over all m,n (integers) in the range 0

[@sr.tarsaimsingh9294]

Everybody knows they are watching KZbin, But I know I am watching a Legend. 😁

[@bhavydugar6665]

What I like best is that he is one of the few Americans to correctly pronounce Euler’s name is

[@rogo7330]

2:07 that pause was funny, sorry)

[@blackpenredpen]

Lol

[@154bharshbhanushali9]

Hello Sir , Facing a lottt of difficulty in solving Definite Integration of Greatest Integer Function questions , Plz If possible then , Make a separate video for it 🔥🔥🔥🔥🔥🔥🔥🔥🔥

[@lekiwi_4145]

Thx

[@mikethegamedev]

Good vid

[@meesbaker]

Would it be a proper derivation to say the space in the circle with r=1 equals pi, so integral from 0-1 of sqrt(1-x^2) = pi/4. Then complex substitution gives pi/4 = i/2*ln(-i) which can be transformed into the euler formula e^ipi = -1

[@JESUSisLORDDd344]

would there be a time when u do more induction

[@rasseboiii3505]

Is there any way possible to solve the equation x^^x=4? The answer is 2 obviously, but can we confirm thats the only one? Talking about tetrarion, not just regular exponentiation

[@dimastus]

Well, it is good and valid proof/derivation, but I think, that using infinite series in that case is like shooting at fly with a bazooka. Why not to use limit definition of e^x and geometric meaning of complex number operations?

[@volodymyrgandzhuk361]

It would be too boring

[@DawidEstishort]

Doesn't the geometric interpretation come from the fact that you can represent e^iz as cosz+isinz? I mean without it we could represent a complex number as a point in coordinate system, but we wouldn't know where e^iPi is.

[@dimastus]

@@DawidEstishort no, geometric interpretation can be derived from trigonometry. Any complex number is defined by real and imaginary parts or x and y coordinates, like vectors. But you can go to polar coordinates r and t. Then real part x = r*cos t , imaginary part y = r*sin t. So we have z = x + iy = r(cos t + i sin t). If you try to multiply two complex numbers with coordinates x1 y1 and x2 y2 (or r1 t1 and r2 t2) in cartesian you will get a mess, but in polar, after some trigonometric exercise, you will get a new number with r = r1*r2 and t= t1 + t2. (Lengths are multiplied and angles are added). With this as a starting point, you can use the limit definition. e^it=lim (1+it/n)^n So you need to multiply the number 1+it/n by itself n times and tend n to inf. Now recall, lengths are multiplied so the resulting length is lim sqrt(1+t^2/n^2)^n = 1, and angles are added so lim (t/n)*n =t. And so e^it=1*(cos t + i sin t). Hope, I've convinced you.

[@DawidEstishort]

That makes sense. I didn't notice limits work nicely with lengths of vectors here. Although I think in this case series approach is more elagant. Using this approach he just had to use the definitions of functions as a series and everything else was shown in the video. I think if he wanted to use lim approach in the video he would need to spend too much time with trigonometric calculations to show where everything comes from (like showing why multiplicating complex numbers in polar coordinates works how it works). Cheers

[@dimastus]

@@DawidEstishort indeed, moreover, this video is targeted to calc 2 students. So, yes, I guess series approach is better when you study them. I just really like proofs with geometric origin, even if they are longer or not intuitive. It's like some kind of art or sport for me.

[@neowise9264]

6:28 Yes senpai

[@naeemalqise8978]

A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 I need the solution please

[@crochou8173]

hi ive been wondering at what value A will A^x=logA(x) have a single solution maybe make a video out of it? thanks

[@andrewkarsten5268]

Is this log base A of x or log of Ax? Log base 10 or e if that’s the case?

[@muhammadmahmoodalam7221]

What is this man? @blackpenredpen Should we consider pie 3.14159.... or 180 degree angle? Then e to the power i0 will be equal to 1. And put something else if you want a different result of your choice.

[@juniorisrrael3184]

Esto prueba que e^x es una función sin ceros en todo C.

[@skltfz4997]

is e^z just taylor series proof for a=0 case? or if we proof a = 0 is true then it’s true for every a (in the case a= pi)?

[@ewthmatth]

e^i*tau=i^4 * *runs away* *

[@bhavydugar6665]

Is this the only way to prove as we can use the analytical way of splitting the graph of e^ix into the real and imaginary part and obtain cos(x) as the real part and sin(x) as the imaginary part and then obtain this identity

[@3manthing]

No. I know of a proof via calculus. If you consider a function f(t)=e^(it)/(cos(t)+i sin(t)). f is a nice function, since cos(t) and sin(t) can not be both zero, for any t, so our function is devoid of poles and discontinouties. Taking a derivative f'(t)=(i e^(it)(cos(t)+i sin(t))-e^(it)(-sin(t)+i cos(t)))/(cos(t)+i sin(t))^2 ; f'(t)=(i e^(it)cos(t)-e^(it)sin(t))+e^(it)sin(t)-i e^(it)cos(t))/(cos(t)+i sin(t))^2 f'(t)=(i ( e^(it)cos(t)-e^(it)cos(t)) - (e^(it)sin(t)-e^(it)sin(t)))/(cos(t)+i sin(t))^2 f'(t)=(i ( 0 ) - ( 0 ))/(cos(t)+i sin(t))^2 f'(t)=(0)/(cos(t)+i sin(t))^2=0 This implies f is a constant for all t (f(t)≡A, A is some constant). But if f is constant for every t, we can evaluate it at any t, to see what A is. Since f(0)= e^(0 i)/(cos(0)+i sin(0)) and i 0=0 i=0,e^(0)=1, cos(0)=1, sin(0)=0 so f'(0)=1/(1+i 0)=1/1=1. So f is constantly 1, so e^(it)/(cos(t)+i sin(t))=1, or e^(it)=cos(t)+i sin(t). ▧

[@herbie_the_hillbillie_goat]

The rabbit hole goes deeper. Euler's identity contains the five fundamental mathematical constants, three fundamental operations (addition, multiplication, exponentiation) and equality. Of course, this invariably brings up the tau debate and the fact that e^iτ=1.

[@terzonxmrphy4119]

this man planning to become an assailant sage or something?DAM!!!

[@onyxa3144]

I'm going for a math minor in college and I can say I'm definetly using this video to impress my family and friends

[@rajatchaudhary6246]

Sir make a group in which you solve our query

[@jagatiello6900]

7:22 Ladies and gentlemen: Euler's Identity!

[@wigfranz]

We know that e^(iø) = cos ø + i. sin ø and when we substitue ø = 2π, we get e^(i.2π) = 1 Say, take "ln" to both side, we get ln e^(i. 2π) = ln 1 = 0 i. 2π = 0 How could it be? Can you explain this case, why i. 2π can be 0?

[@gamerff5454]

Taking the natural logarithm (\(\ln\)) of both sides in \(e^{i2\pi} = 1\) yields \(i2\pi = 0\), which is not true. The issue here is that the natural logarithm is a multivalued function for complex numbers, and in this case, it leads to an incorrect result. Thats what chatgpt says

[@berdly3367]

E^(i*pi) + 1 = 0 literally reminded me of quadratic

[@_Domo_]

Well this is epic

[@avunz125]

That's a nice validation. but how did we discovered that? Did someone just randomly tried to taylor expand e^ix? Probably not. How to invent math, that is the question.

[@ImTheRisingPhoenix]

@blackpenredpen Hello, after watching the video I made a little google search about the topic and people seem to say that the identity is "Euler's proof of God" and they are saying it because the equation is made from all the basic constants from every field of mathematics, could you elborate on that or is it just because the beauty in that math Euler said that in order for this to be true God must be real?

[@NonTwinBrothers]

This one has captions let's goooo

[@28sarthakdwivedi44]

Sir can u visit India once ? There are so many here to appreciate your work especially me.....

[@miradota347]

Ladies and gentlemen, we got em.

[@Mono_Autophobic]

I have a music video of this equation too

[@ernestgrigoras7658]

I know this might be offtopic...but is there any solution for x^(x+1)=(x+1)^x ?

[@ronaldjensen2948]

Notice 2^3 < 3^2 but 3^4 > 4^3. Since both functions are continuous between x=2 and x=3 they must intersect in that range. Therefore, there is (at least) one solution.

[@dhrubajyotiray1214]

isn't this maclern's series

[@vaughanwilliamson173]

...and when you sketch e^(i pi) on the complex number plane with the unit circle drawn, e^(i pi) + 1 = 0 is just so obvious!

[@mrocto329]

buuut you have to prove that r(cos x + isin x) = e^ix which comes from taylor series.

[@johndoyle2347]

Yes, the recursive form of Euler's Identity should be considered the most beautiful and important of lone equations, but the solution to the Basel problem and its reciprocal to define the moments in time where black holes get crushed into dark matter as part of a Big Bounce event, well...

[@brouquier7172]

If you use tau (equal to 2*pi) instead of pi then you get e^(i*tau) = 1, which to be honest looks even better!

[@PwerGuido]

It does not

[@brouquier7172]

@@PwerGuido I think you'll find that it does.

[@debajyotidas1]

“Everybody knows that i^2=1”. That moment when nobody in my grade knows what I is...

[@sonalinegi9651]

It also can be solved as following 🙃😅 e^i@ =cos@+isin@ (euler's identity) So, @=π (here) e^iπ= cosπ+isinπ , (sinπ=0 & cosπ=-1) Hence, e^iπ= -1 Now, e^iπ+1= -1+1=0 hence proved.

[@user-kw2xs5rs2f]

thats only for one value and thats in fact not a proof, by that logic x^2 = 2x if we only consider x=0 and x=2

[@varun3282]

Then, e^ipie must be =-1 right.

[@GUTY1729]

HERMOSO

[@RukkuChinnavera]

2:12 he paused I laughed

[@janmamu8721]

E^(i•tau)

[@xevira]

not just irrational... _transcendental._

[@parinvivek4289]

1:10 is it just me who got jump scare?

[@maxv7323]

I don't think most calc 2 students can show that the derivative of e^z is e^z given that the understanding of the derivative that we learn kinda relies on us using the reals (I think)

[@General12th]

Doesn't the limit definition of the derivative apply in the complex world too? I don't see why not.

[@andrewkarsten5268]

@@General12th you have to be very careful how you do that. In many cases yes, but you need complex analysis to actually show why it’s reasonable and the answers we get. You cannot just assume that things play nice like that, you first need to prove it before you use it.

[@andrewkarsten5268]

@Max V you’re right, the definitions taught up to calc 2 and 3 are dependent on using real number inputs. You need complex analysis to extend the definitions into the complex world before you can start taking derivatives of functions with complex inputs.

[@BilalAhmed-on4kd]

Why radians and not degrees

[@dingle37]

I’m an algebra I student trying to understand as best I can... 😅

[@alexmoga8654]

Why is this vid unlisted?

[@officiallyaninja]

how did you find this video 2 days early?

[@reiterschwein3828]

lol

[@alexmoga8654]

@@officiallyaninja it was linked on his home youtube page before it was posted lmao

[@MadaraUchihaSecondRikudo]

I think what's totally mind boggling about this is that e and pi are supposedly two completely unrelated numbers *found in nature*. e occurs naturally in anything that has to do with exponential growth, and pi occurs naturally in all circles. We did not invent or define those numbers. We observed them. And yet, we've proven that those two numbers are intrinsically linked through the imaginary world.

[@krischan67]

You may move -1 to the left side, I do not :p

[@francescodeleonardis7621]

I remember that shirt, buy white version, from a Simpsons episode.

[@irrafer8663]

After 8 time of calculus 1 this is drug for me

[@michaelmapple8201]

Am I wrong but wasn't there something that needed to be considered when infinate sum's are rearanged like that? Just curious why it is possible in this case while you can also get pure nonsense out of it othertimes.

[@ateium2409]

Did anybody else grab a pen and prove Euler's identity before watching the video ?

[@sakibhasanredwan3468]

I want to purchase brilliant premium but i don't have even any card to pay and also i don't have money.You might be surprised but it is normal in my country like Bangladesh.But i hope ,when i would earn, i will purchase brilliant premium version

[@manangandhi3259]

Here is a queston for you cos2x/1×3 + cos4x/3×5 + cos6x/5×7......infinity

[@lesarXD]

kayra kibar chill dude edit: he deleted most of the comments

[@ruffifuffler8711]

The faceted dimensional transition operator annihilates itself during a transcendental time stretch to remain within the singularity, and create the count, and 1st ordinal..

[@ruffifuffler8711]

The count exists before the 1st ordinal, but has no presence until being recognized by the 1st ordinal, which then: a. bootstraps the count, b. then defines the 5 aspects of the presence of the singularity, c. conducts the euler matriculation of presence to exhaust the count.

[@aymanad3437]

What are you talking about???

[@ruffifuffler8711]

@@aymanad3437 the equation has depth antecedents b4 face value appreciation, and is not an identity.

[@yami_the_witch]

Not gonna lie, relating it to geometry is much easier. Just imagining it in the context of the complex plane makes it hilariously simple.

[@johndoyle2347]

i = the square root of e to the i times pi power. This describes stable particles. Teach correctly, sir.

[@Peter_1986]

"Everybody knows..." is highly context-dependent. You pretty much have to actively choose to study a bunch of optional math courses in high school to even so much as get introduced to imaginary numbers.

[@kehana2908]

hold on what i learned that like 9th grade math

[@niom9446]

6:28 senpai

[@emrahkaraca1982]

Beauty...

[@Rocky4719]

My friends, next time your significant other asks how beautiful you think they are, just tell them they’re as beautiful as Euler’s identity. And then link this video to explain to them what it is

[@user-rz7ek1wx1i]

So 1 = 1! more beautiful

[@carultch]

0! is defined as 1. 1! is therefore 0! * 1, which also equals 1. It is pretty much the base cases of factorials that 0 and 1 factorial, are both equal to 1. Factorial is defined recursively, so it has to reduce to a base case to have any meaning.

[@robertmcknightmusic]

I don't get why Euler's identity is so significant. Like sure, it has five cool features, but there are other irrational numbers. Wouldn't it be more impressive if phi was also part of the equation?

[@ronaldjensen2948]

The Euler identity is fundamental to the Fourier transformation that makes the Internet as we know it actually work.

[@robertmcknightmusic]

@@ronaldjensen2948 ok interesting. Does it have to be E or could we have also made the internet using another irrational number?

[@ronaldjensen2948]

@@robertmcknightmusic If you think about d/dx b^(x) where b is real and b > 1 using the difference equation: lim h->0 [ b^(x + h) - b^(x) ]/h you find this simplifies to: b^(x) * lim h->0 [ b^(h) - 1 ]/h and that: lim h->0 [ b^(h) - 1 ]/h == 1 if and only if b == e This means using the Taylor (Maclaurin) Series to evaluate b^(x) is only tractable when b == e. It can easily be shown that lim n->∞ ( 1 + x/n)^(n) yields the Taylor series for e^(x).

[@robertmcknightmusic]

@@ronaldjensen2948 fair enough. that's over my head, but I believe you. Thanks for your response.

[@JungKookOfficiaI]

W