How to prove Eulers

Question

How to prove Eulers

nick1337 · Accepted Answer

Step 1Proof:Consider the functionf(t)=e−it(cos⁡t+isin⁡t)for t∈R. By the product rulef′(t)=e−it(icos⁡t−sin⁡t)−ie−it(cos⁡t+isin⁡t)=0identically for all t∈R. Hence, f is constant everywhere.Since f(0)=1, it follows that f(t)=1 identically.Therefore,eit=cos⁡t+isin⁡tfor allt∈R,as claimed.

Vasquez · Accepted Answer

Step 1 Assuming you mean eix=cos⁡x+isin⁡x, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real x in a first-year calculus context, and the MacLaurin series for ez, trusting that it converges for pure-imaginary z since this result requires complex analysis. The MacLaurin series: sin⁡x=∑n=0∞(−1)n(2n+1)!x2n+1=x−x33!+x55!−⋯ cos⁡x=∑n=0∞(−1)n(2n)!x2n=1−x22!+x44!−⋯ ez=∑n=0∞znn!=1+z+z22!+z33!+⋯ Substitute z=ix in the last series: eix=∑n=0∞=1+ix+(ix)22!+(ix)33!+⋯ =1+ix−x22!−ix33!+x44!+ix55!−⋯ =1−x22!+x44!+⋯+i(x−x33!+x55!−⋯) =cos⁡x+isin⁡x

alenahelenash · Accepted Answer

Well this question actually boils down to How is the complex exponential defined? Here is my view of this problem: Let f(x+iy)=ex(cos⁡(y)+isin⁡(y)) Then f has continuous partial derivatives fx and fy, and verifies the Cauchy-Riemann equations, thus it is analytic. Moreover, for any z1, z2∈C we have f(z1+z2)=f(z1)f(z2) Last but not least f(x)=ex for all x∈R In particular we showed that ex can be extended to an analytic complex function, and the theory tells us that such extension is unique. Now, since f(z) is the unique analytic extension of ex to the complex plane, and it also satisfies the exponential relation f(z1+z2)=f(z1)f(z2), we call this function ez

How to prove Euler's formula: e^{i\phi}=\cos(\phi)+i\sin(\phi)?

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