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I was looking into the proof of Euler's formula based with Maclaurin Series and am confused on how one gets from ($\ast$) to ($\ast\ast$):

$$\begin{align*}\cos x & =\sum\limits_{n=0}^{+\infty}\frac {(-1)^n x^{2n}}{(2n)!}\\ \sin x & =\sum\limits_{n=0}^{+\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)!}\\ \cos x+i\sin x & =\sum\limits_{n=0}^{+\infty}\frac {(-1)^n x^{2n}}{(2n)!}+\sum\limits_{n=0}^{+\infty}\frac {i(-1)^n x^{2n+1}}{(2n+1)!}\\ & =\sum\limits_{n=0}^{+\infty}\frac {i^{2n} x^{2n}}{(2n)!}+\sum\limits_{n=0}^{+\infty}\frac {i^{2n+1}x^{2n+1}}{(2n+1)!} \tag{$\ast$} \\ & =\sum\limits_{n=0}^{+\infty}\frac {(ix)^n}{n!} \tag{$\ast\ast$}\\ & =e^{ix}\end{align*}$$

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  • $\begingroup$ Please, avoid the use of images. Use Math Jax and type your equations $\endgroup$
    – jjagmath
    Jul 7 at 0:01

1 Answer 1

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Observe that

  • $S_1:=\sum_{n=0}^\infty a_{2n}=a_0+a_2+a_4+\ldots$.
  • $S_2:=\sum_{n=0}^\infty a_{2n+1}=a_1+a_3+a_5+\ldots$.
  • $\sum_{n=0}^\infty a_{n}=a_0+a_1+a_2+a_3+\ldots = S_1+S_2$.

Here: \begin{align*} \cos(x) + i\sin(x)&=\sum_{n=0}^{\infty} \dfrac{i^{2n}x^{2n}}{(2n)!} + \sum_{n=0}^{\infty}\dfrac{i^{2n+1}x^{2n+1}}{(2n+1)!}\\ &= \sum_{n=0}^{\infty} \dfrac{(ix)^{2n}}{(2n)!} + \sum_{n=0}^{\infty}\dfrac{(ix)^{2n+1}}{(2n+1)!} \\ &= \sum_{n\text{ is even}} \dfrac{(ix)^{n}}{n!} + \sum_{n\text{ is odd}}^{\infty}\dfrac{(ix)^{n}}{n!} \\ &= \sum_{n=0}^{\infty} \dfrac{(ix)^n}{n!} \end{align*}

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