I saw this derivation eons ago; let's see if I can recall. Taylor series expansion: f(x)=f(0)+f'(0)x+f''(0)x^2/2!+f'''(0)x^3/3!+f''''(0)x^4/4!+... e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...; sin(x)=x-x^3/3!+x^5/5!-x^7/7!+...; cos(x)=1-x^2/2!+x^4/4!-x^6/6!+... e^{ix}=1+ix+(ix)^2/2!+(ix)^3/3!+(ix)^4/4!+(ix)^5/5!+(ix)^6/6!+(ix)^7/7!+...=(1-x^2/2!+x^4/4!-x^6/6!+...)+i(x-x^3/3!+x^5/5!-x^7/7!+...)=cosx+i*sinx And there we have it, which is so much easier after someone else has figured it out ahead of time.
On differentiating you treated i=sqrt(-1) as a real number. Why? How was it validated that differentiation of i is the same as a real number?
Differentiation is done here with respect to theta. So, i=sqrt(-1) is treated as a constant, not as a real number.
@GLabsPlus i=sqrt(-1) is a constant but differentiation of it hasn't been validated. You won't find it in any math book.
I saw this derivation eons ago; let's see if I can recall. Taylor series expansion: f(x)=f(0)+f'(0)x+f''(0)x^2/2!+f'''(0)x^3/3!+f''''(0)x^4/4!+...
e^x=1+x+x^2/2!+x^3/3!+x^4/4!+...; sin(x)=x-x^3/3!+x^5/5!-x^7/7!+...; cos(x)=1-x^2/2!+x^4/4!-x^6/6!+...
e^{ix}=1+ix+(ix)^2/2!+(ix)^3/3!+(ix)^4/4!+(ix)^5/5!+(ix)^6/6!+(ix)^7/7!+...=(1-x^2/2!+x^4/4!-x^6/6!+...)+i(x-x^3/3!+x^5/5!-x^7/7!+...)=cosx+i*sinx
And there we have it, which is so much easier after someone else has figured it out ahead of time.
That's right!
The next video goes through the proof of Euler's Formula using the Taylor Series.
ua-cam.com/video/vLgDGCChOoU/v-deo.html
I really like your writing font, looks neat
Thank you!