@@imTyp0_well I think you’d have to take it as a fact that we can extend the definition of the Taylor series expansion for e^x to complex numbers. Unless there’s a way to derive that too?
One of the nicest proofs I know of is to first define e as the limit of (1+1/n)^n as n goes to infinity. From that, one can replace n with n/x to get e=lim (1+x/n)^(n/x), so e^x=lim(1+x/n)^n for all positive x. Then extend that to negative x. Then take that to be the definition of e^x for any complex x. That gives e^(ix)=lim (1+ix/n)^n, and now the right side is basically what you get when you take more and more arcs of size x/n. That gives an angle of x on the unit circle, so it's cos x+i sin x. Of course, it takes some work to justify all that, but it's really the only intuitive explanation for e^(ix)=cos x+i sin x that I know.
Sitting in class and feeling stupid vs. Watching Papa and feeling smart. => prof. ○ Papa = e with Papa ○ e =Papa => everybody needs a Papa for math Q. E. D.
The superior proof is to multiply the best prime number by pi and set it to be phi. You can easily calculate that -1 = -1. It's called incomplete induction and it's really poweful. PS: Don't forget to feed your pet. As you surely remember, it eats Happy Meals.
differentiate y=exp[ix] two times, and you get y''=-y So, already have that it's equivalent to y=Acosx + Bsinx then you just need the y[0]=1 and y[pi/2]=i to determine A=1,B=i Over
I have had brilliant for some time now and I'm nothing but impressed at the quality and creativity of the courses. I highly recommend it to anyone who wants to learn math up to college math level or wants to supplement their own studies. At the very least brilliant will be a perfect explainer for the things you may not understand from a textbook.
Unfortunately, I think the best definition of e^z is the power series definition, from which we define cos(z) and sin(z). In that context, there is no proof needed. The complicated part is to then show all the properties of cos and sin that were obvious by definition. Just showing e^z is 2*pi*i periodic is already a somewhat complicated proof, but it gives maybe the best definition of pi out there.
its so cool to get this super simple proof that doesn't rely on series. the taylor series propf is cool too but i love euler formula even more now ive seen this diff eqs proof 😊
It is trivial to show e^x > 0 for all x, however, without Euler it is not clear that we can state e^(iφ) != 0 for all φ. Can you re-work the proof by dividing both sides by (cosφ + isinφ) which can be shown to never equal zero?
well after applying the quotient rule you basically get the same expression in the numerator as he got in the video just with every sign flipped because the part where you differentiate the exponential now doesn‘t have the -1 from the inner derivative of -i*phi (bc it‘s +i*phi) and the other part where you differentiate the denominator has a - from the formula for the quotient rule. It‘s not hard (and I‘m unsure if necessary) to prove that the denominator cos^2 - sin^2 + 2*i*sin*cos ≠0
@@giansieger8687 Since (cosφ + isinφ) is never 0, (cosφ + isinφ) ^2 cannot be zero. Also, cos^2φ - sin^2φ + 2*i*sinφ*cosφ =(cos2φ + isin2φ), as it's the double angle formula. See de Moivre's theorem
Well the exponential is usually introduced in its most accessible form, which as a solution to a differential equation over the reals. The problem with Euler's formula is that it involves things that are easy to introduce (i, pi, the exponential, sin and cos as geometric functions), but those definitions are not really great ways of defining those functions for more advanced usage. The best definition of exp is probably the power series, which gives us the definition of sin and cos. The difficult part is to then prove that sin and cos defined as such are actually the periodic functions we are used to, which is not trivial at all.
While I love ur vids flammy, I can't but smirk at these proofs of the euler foormula, since in most cases e, sin, cos are introduced as power series, where it is completely apparent that this formula holds anyways.
Your delivery doctor lost his license because he could not help but smirk at that ugly kid that popped out. Unless you can prove the infinite power series representations of sine and cosine, then you construe as “apparent” is vacuous.
Hello mister, while I appreciate your effort of putting math videos on UA-cam, you seemed to have missed the point - if I define sine and cosine as power series there is nothing there to be proven.
@@BedrockBlocker sort of like defining God as Love. Then one can declare God loves you. No proof required. My turn: I define you as stupid. Therefore you are stupid.
At least for me, sine and cosine weren't _introduced_ as power series. I'd bet that most people learned sin(x) = opposite/hypotenuse, not that: sin(x) = sum, for n from 1 to infinity, of [(-1)^n * x^(2n+1)]/(2n+1)! And even if they did, they would need to know that the sequence x, (-x^3/3!), (x^5/5!) ... is absolutely convergent. (Yes, they'd be right even if they didn't know about that, but a "proof" based on a false assumption isn't a proof.) I checked a couple precalc textbooks for anything about absolute convergence _or_ definitions of trig functions as power series, but couldn't find either of those topics. And, at least in my state, trig is either covered in precalc or before it.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
What a coincidence, my maths teacher explained Euler's formula today and he told us just to take it as fact then this drops...
Eh? It’s super easy to show with Taylor series, which I assume you see before complex numbers.
@@imTyp0_ In Australia, we do complex numbers first, but I do know the Taylor series too
My math textbook also says this, which is super silly
@@imTyp0_well I think you’d have to take it as a fact that we can extend the definition of the Taylor series expansion for e^x to complex numbers. Unless there’s a way to derive that too?
@@Mortgageman145at least you do taylor series in the new syllabus, we didnt even do them in the old maths syllabus in aus
One of the nicest proofs I know of is to first define e as the limit of (1+1/n)^n as n goes to infinity. From that, one can replace n with n/x to get e=lim (1+x/n)^(n/x), so e^x=lim(1+x/n)^n for all positive x. Then extend that to negative x. Then take that to be the definition of e^x for any complex x. That gives e^(ix)=lim (1+ix/n)^n, and now the right side is basically what you get when you take more and more arcs of size x/n. That gives an angle of x on the unit circle, so it's cos x+i sin x.
Of course, it takes some work to justify all that, but it's really the only intuitive explanation for e^(ix)=cos x+i sin x that I know.
What is "nice" about it other than it is the proof you understand best?
@@MyOneFiftiethOfADollar using limits to prove it is pretty cool mate
eee to the aa faa - papa flammy
circa 1943, colorized
Sitting in class and feeling stupid vs. Watching Papa and feeling smart. => prof. ○ Papa = e with Papa ○ e =Papa => everybody needs a Papa for math Q. E. D.
This proof was nicer than the integral of floor(sqrt x) mod (floor(x)) from 1 to 25
1 to 25* cant take mod 0
@@hyperpsych6483 let p mod 0 = q st p is in the set of real numbers B) solved
@@hyperpsych6483 Yeah that makes sense thinking about it now...
Ye, I had a typo on my board and didn't notice it throughout the video and while computing ^^'
I’m taking complex variables right now and it’s one of my favorite math classes. We’re learning contour integrals right now.
The superior proof is to multiply the best prime number by pi and set it to be phi. You can easily calculate that -1 = -1. It's called incomplete induction and it's really poweful.
PS: Don't forget to feed your pet. As you surely remember, it eats Happy Meals.
differentiate y=exp[ix] two times, and you get y''=-y
So, already have that it's equivalent to y=Acosx + Bsinx
then you just need the y[0]=1 and y[pi/2]=i to determine A=1,B=i
Over
I have had brilliant for some time now and I'm nothing but impressed at the quality and creativity of the courses. I highly recommend it to anyone who wants to learn math up to college math level or wants to supplement their own studies. At the very least brilliant will be a perfect explainer for the things you may not understand from a textbook.
False, the shortest proof is to just write "this proof is trivial and left as exercise for reader".
I actually have an even shorter and more marvelous proof, but it is unfortunately too short to fit in the margins of a UA-cam comment.
@@albertrichard3659 good ol' fermat
he must be laughing in his tomb rn looking at us trying to find his former proof
Unfortunately, I think the best definition of e^z is the power series definition, from which we define cos(z) and sin(z). In that context, there is no proof needed. The complicated part is to then show all the properties of cos and sin that were obvious by definition. Just showing e^z is 2*pi*i periodic is already a somewhat complicated proof, but it gives maybe the best definition of pi out there.
Papa Flammy can never disappoint🗣🗣🗣🗣🗣🗣🗣🗣🗣🗣🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Absolutely
I have a faster one!
Theorem: e^iφ = cos(φ) + i sin(φ)
Proof: This is left as an exercise for the reader.
its so cool to get this super simple proof that doesn't rely on series. the taylor series propf is cool too but i love euler formula even more now ive seen this diff eqs proof 😊
3:21 American accent slipping away gradually :)
Thanks
Thanks a lot Charlie!!!!
euler as F1 racer hits harder when you prove it with Macla(u)ren series
04:43: that is really. really. pretty.
It is trivial to show e^x > 0 for all x, however, without Euler it is not clear that we can state e^(iφ) != 0 for all φ. Can you re-work the proof by dividing both sides by (cosφ + isinφ) which can be shown to never equal zero?
Great point! I didn't consider that when watching the video.
well after applying the quotient rule you basically get the same expression in the numerator as he got in the video just with every sign flipped because the part where you differentiate the exponential now doesn‘t have the -1 from the inner derivative of -i*phi (bc it‘s +i*phi) and the other part where you differentiate the denominator has a - from the formula for the quotient rule. It‘s not hard (and I‘m unsure if necessary) to prove that the denominator cos^2 - sin^2 + 2*i*sin*cos ≠0
@@giansieger8687 Since (cosφ + isinφ) is never 0, (cosφ + isinφ) ^2 cannot be zero. Also, cos^2φ - sin^2φ + 2*i*sinφ*cosφ =(cos2φ + isin2φ), as it's the double angle formula. See de Moivre's theorem
Good one papa flammy
Could you please make a video about incomplete gamma function?
Hey papa what's your intro song?
Want to know it
well since pi is defined to be the smallest real number greater than zero such that Re(e^i pi) = 0
we are done in one line
Pls more vids like this!
Based. No taylor series needed.
I need an Algebruh shirt 😩
available at my Spring shop :v
Nice proof for this holomorphic function.
This man is a walking green flag - I had no clue he and my fav cooking channel is the same person!
:DDD
Yo that was pretty good
I thought that euler formula was the definition of the complexe exponential? At least it's how it has been introduced to me.
Well the exponential is usually introduced in its most accessible form, which as a solution to a differential equation over the reals. The problem with Euler's formula is that it involves things that are easy to introduce (i, pi, the exponential, sin and cos as geometric functions), but those definitions are not really great ways of defining those functions for more advanced usage.
The best definition of exp is probably the power series, which gives us the definition of sin and cos. The difficult part is to then prove that sin and cos defined as such are actually the periodic functions we are used to, which is not trivial at all.
Here's a nice proof I like:
Let θ(z) = cos(z) + i•sin(z) 'cause why not 🙃
Differentiate both sides
d/dz•θ(z) = d/dz•(cos(z) + i•sin(z))
Blah blah blah, too lazy to write
d/dz•θ(z) = -sin(z) + i•cos(z) = i^2•sin(z) + i•cos(z) = i•(cos(z) + i•sin(z)) = i•θ(z)
Shuffle them around
θ^-1(z)•dθ(z) = i•dz
Integrate both sides
∫ θ^-1(z)•dθ(z) = ∫ i•dz
ln(θ(z)) + x_0 = i•z + x_1 'cause why not 🙃
Blah blah blah, same as before
ln(θ(z)) = i•z + x_2
e^(ln(θ(z))) = e^(i•z + x_2)
θ(z) = x_3•e^(i•z)
We know θ(0) = cos(0) + i•sin(0) = 1
θ(0) = x_3•e^(i•0)
1 = x_3•1
x_3 = 1
We get θ(z) = 1•e^(i•z) = e^(i•z)
Done
I hate my life typing on a phone
While I love ur vids flammy,
I can't but smirk at these proofs of the euler foormula, since in most cases e, sin, cos are introduced as power series, where it is completely apparent that this formula holds anyways.
Your delivery doctor lost his license because he could not help but smirk at that ugly kid that popped out.
Unless you can prove the infinite power series representations of sine and cosine, then you construe as “apparent” is vacuous.
Hello mister, while I appreciate your effort of putting math videos on UA-cam, you seemed to have missed the point - if I define sine and cosine as power series there is nothing there to be proven.
@@BedrockBlocker sort of like defining God as Love.
Then one can declare God loves you. No proof required.
My turn: I define you as stupid. Therefore you are stupid.
At least for me, sine and cosine weren't _introduced_ as power series. I'd bet that most people learned sin(x) = opposite/hypotenuse, not that:
sin(x) = sum, for n from 1 to infinity, of [(-1)^n * x^(2n+1)]/(2n+1)!
And even if they did, they would need to know that the sequence x, (-x^3/3!), (x^5/5!) ... is absolutely convergent. (Yes, they'd be right even if they didn't know about that, but a "proof" based on a false assumption isn't a proof.)
I checked a couple precalc textbooks for anything about absolute convergence _or_ definitions of trig functions as power series, but couldn't find either of those topics. And, at least in my state, trig is either covered in precalc or before it.
@@MyOneFiftiethOfADollar You really are hilarious xD
what's more flammable than Chlorine Trifluoride?
joe
This only proves that "(e^{-it})^{-1}=cos(t)+i sin(t)" hehe
nice
"e to the A PHA"
❤❤❤
Ee to da a fa?
VG Jens
Isn’t this circular logic?
no, why?
Its another way to demonstrate using ODE.
That’s cool at all, but it involves knowing the derivative of sine X
Not first this time.... I love you
e to the ah phah?