Euler's formula e^iθ=cosθ+i*sinθ

Enterprise 2023-07-01 06:15:24 views: null

I believe that most people know the most beautiful formula of the famous mathematics:{\color{Red} e^{i\pi }+1=0}

 Why do you say it is the most beautiful? Because it contains the most basic e in exponents, the most basic i in complex numbers, the most basic π in circular frequency, and the most basic 0 and 1 in natural numbers.

In essence, this formula is {\color{Red} e^{i\theta }=\cos\theta + i\sin \theta } derived from this formula, just replace θ with π. 

So how is this formula obtained? It can be expanded by using the power series in advanced mathematics, and then can be derived.

Considering e^{ix}the ix in it as a whole, according to the McLaughlin expansion e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}+..., replace x with ix to get:

 e^{ix}=1+ix+\frac{(ix)^{2}}{2!}+\frac{(ix)^{3}}{3!}+\frac{(ix)^{4}}{4!}+\frac{(ix)^{5}}{5!}+\frac{(ix)^{6}}{6!}+...

=1+ix-\frac{x^{2}}{2!}-\frac{ix^{3}}{3!}+\frac{x^{4}}{4!}+\frac{ix^{5}}{5!}-\frac{x^{6}}{6!}+...  

We put the one without i on one side and the one with i on the other side, then we can get:

e^{ix}=(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+...)+ i (x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-...)

=\cos x+i\sin x  

So get certified.

(Supplement, why can Taylor expand, this needs to be proved, but it is ignored here)
 

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Origin blog.csdn.net/Sbs5218/article/details/130705399
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