I wish you had used different letters for the formula. Theta for anger and r for magnitude. The -1 on the imaginary line should be - i . Let's use conventional letters. Less confusing that way.
Very good explanation, though I have a problem: e^ℼi = -1, square both sides, e^2ℼi = 1, take logarithms, 2ℼi = ln(1) = 0 ??? What is wrong here anyone?
Hey Graham. There's nothing "wrong" but keep in mind that things get a bit weird when you get into the complex domain. For example, log(e^5)=5 but log(e^5i)!=5. So, because e^{k2ℼi}==0 for any integer k, then the equation works out to be 0==0. When working with complex numbers and exp/log, it's often more insightful to think about numbers as points on the complex plane and apply Euler's formula.
The euler formula can't be right. The left side, exp equation, gives values from 1 to inf by definition. The right side, addition of 2 trig functions which give values from -1 to 1 also by definition. So, how can these sides ever be equal?
Your comment about e^x is true for real-valued x, but for complex-valued numbers, e^{ix} does not grow arbitrarily (where i is sqrt(-1)). In fact, |e^{ix}|=1 for any value of x. So Euler's formula does hold. I don't show the proof of that equality in this video, but it can be demonstrated using Taylor expansion. The basic definition (e^{ik}) is limited to points on the unit circle. A more complete expression is Me^{ik} = M(cos(k) + isin(k)), where M is the "magnitude" or distance from the origin.
@@mikexcohen1 Nope. You got it wrong. How is it proven that Taylor series are valid for imaginary numbers? Also, are you implying that exp function for imaginary numbers doesn't have the same shape as with real numbers?
Nobody yet shown me the intuitive derivative of Euler identity….not even this video…Taylor series is not the derivation…it just corroborates it….I need a geometrical intuitive derivation
You tied so many things together in this wonderful video.
THANK YOU !!
EXCEPTIONAL !!
:)
the most elegant video for the most elegant equation
thank you so much for this
Glad you liked it!
wow sir.....your such an amazing teaher
this is so impacting . thank you Sir
Wonderful presentation!
Fantastic explanations! Thank you.
I just wanna say that you are my new hero!
Hey, now you made me blush ;)
You are a legend! Great explanation and funny too :)
Glad you enjoyed it!
Thanks alot
I finally understand Euler's formula
Awesome.
Great video, thank you for your efforts
My pleasure!
Use tau please
Can this formula be expanded to three dimensions?
hmm.. yes, should be possible when using spherical coordinates.
k must be in rad ? What about degrees
Then you'd need to convert degrees into radians.
God level explanation
Thank you kindly, Paras.
M is definitely for MIke, the magnificent.
:)
The best explanation out there!
thank you soooo much!!
You're welcome soooooo much!
I wish you had used different letters for the formula. Theta for anger and r for magnitude. The -1 on the imaginary line should be - i . Let's use conventional letters. Less confusing that way.
Different conventions in different fields...
Very good explanation, though I have a problem: e^ℼi = -1, square both sides, e^2ℼi = 1, take logarithms, 2ℼi = ln(1) = 0 ??? What is wrong here anyone?
Hey Graham. There's nothing "wrong" but keep in mind that things get a bit weird when you get into the complex domain. For example, log(e^5)=5 but log(e^5i)!=5. So, because e^{k2ℼi}==0 for any integer k, then the equation works out to be 0==0. When working with complex numbers and exp/log, it's often more insightful to think about numbers as points on the complex plane and apply Euler's formula.
@@mikexcohen1 Thanks! - I only learned about Euler's identity yesterday so it's still sinking in.
When this appears on Secondary 3 school syllabus...
(-1)^n=cos(n*pi) + i*sin(n*pi)=e^(i*n*pi)
n*pi=k in this video
very good
The euler formula can't be right. The left side, exp equation, gives values from 1 to inf by definition. The right side, addition of 2 trig functions which give values from -1 to 1 also by definition. So, how can these sides ever be equal?
Your comment about e^x is true for real-valued x, but for complex-valued numbers, e^{ix} does not grow arbitrarily (where i is sqrt(-1)). In fact, |e^{ix}|=1 for any value of x.
So Euler's formula does hold. I don't show the proof of that equality in this video, but it can be demonstrated using Taylor expansion.
The basic definition (e^{ik}) is limited to points on the unit circle. A more complete expression is Me^{ik} = M(cos(k) + isin(k)), where M is the "magnitude" or distance from the origin.
@@mikexcohen1 Nope. You got it wrong. How is it proven that Taylor series are valid for imaginary numbers? Also, are you implying that exp function for imaginary numbers doesn't have the same shape as with real numbers?
OMG THANK YOU!!!!
YOU'RE WELCOME!
Nobody yet shown me the intuitive derivative of Euler identity….not even this video…Taylor series is not the derivation…it just corroborates it….I need a geometrical intuitive derivation
@@Dmitry_Bonch_TaxFree_Economy gross
The thumbnail for this video is terrible, sorry