Euler's Formula & Euler's Identity - Proof via Taylor Series

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[bendhamwatt]

My math teacher literally just told me to accept this identity today as it wont be covered in the course. But you flammy boi just saved my curiosity from suffering. Thank you

[AndrewDotsonvideos]

Okay so basically this is e^pic = cospc + isinpc

[patricksalhany]

It is a very beautiful formula by Wheeler, because it combines the 4 main branches in Mathematics: Arithmetics represented by 1, Algebra represented by i, Analysis represented by e and Geometry represented by π.

[dectorey]

Papa flammy lovely derivations 😫😫 a perfect way to wake up for the day

[blownn]

I'm taking calc 2 right now and we are learning about taylor series, and I just heard about this e^ipi thing today and decided to look it up. This proof you just showed me has blown my mind greater than any other proof I've ever seen. I'm literally in awe of the implications of this formula, everything has come together in such a simple and beautiful way 😭 thanks for the video sir

[gskartwii]

Unbelievable! I was just going to attempt this proof method today as an exercise for myself.

[yukhyx]

Thank you. This is the starting point of the Schrödinger wave equation.

[rbradhill]

from flamspastic to flamtastic. short and sweet, nicely done ma boi

[Polaris_Babylon]

I did it a few months ago with taylor series too, i've had to learn how to use taylor polynomials before. Also it's possible to demonstrate the identity with differential equations :O

[Israel..]

You can also derive the polar form of Euler's identity using the Taylor series for log and arctan, it's not quite as sexy but it's still cool.
log(a+bi)
= (1/2)log(a^2 + b^2) + log((1 + i(b/a))/sqrt(1 + (b/a)^2))
= (1/2)log(a^2 + b^2) + log(1 + i(b/a)) - (1/2)log(1+ (b/a)^2)
but
log(1 + i(b/a)) = i(b/a) - (1/2)[i(b/a)]^2 + (1/3)[i(b\a)]^3 - (1/4)[i(b/a)]^4 + etc.
= [(1/2)(b/a)^2 - (1/4)(b/a)^4 + (1/6)(b/a)^6 - etc.] + i[(b/a) - (1/3)(b/a)^3 + (1/5)(b/a)^5 - etc.]
= (1/2)log(1 + (b/a)^2) + iarctan(b/a)
thus
log(a+bi) = (1/2)log(a^2 + b^2) + iarctan(b/a)
now take the exponential
a + bi = sqrt(a^2 + b^2)e^(iarctan(b/a))

[Shaunss]

As someone who only knows calc 2 and a bit of calc 3, this is the best explanation that makes sense for me

[eyuin]

Thanks man, I finally know why this identity is true instead of blindly believing it like gospel.

[prodbytukoo]

entered here just for over-delivering in my first diffeq exam and left with my mind fkn blown, that was beautiful damn

[matthewstevens]

Sorry to say this Euler lovers (myself included...)
*This is Cotes' identity.*

[aknighton]

clear explanation of why this is true and why it is beautiful! short enough so I can understand the whole thing at once! Thank you!

[tylerfusco]

This has got to be my favorite proof of euler’s series, because of its elegance and simplicity

[alanr]

I'm already struggling with Integrals in the first half of Calc 2. But now seeing series has definitely sealed my fate.

[saschatrumper]

I'd like to see some rocket Science: Rocket-equation vs Ziolkovski-Equation. Showing how to put down a DE thoughtful showing the implications of not reaching orbital velocity by using only one stage. : )

[tehflyyar]

Random suggestion lol, but do a video on how you got passionate for maths and how it all began etc. thanks papa :)

[outofthebots]

Those 4 videos together are both such beautiful math and such important math. My 15yr son is autistic and incredibly gifted in math I will get him to watch these videos :)


I love watching such beauty in number, thanks for the videos