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I just read a proof on ProofWiki that proves Euler's formula, but I can't seem to understand what is done in this following step:

$$\sum\limits_{n=0}^\infty\left(\frac{(i\theta)^{2n}}{(2n)!}+\frac{(i\theta)^{2n+1}}{(2n+1)!}\right) =\sum\limits_{n=0}^\infty{\frac{(i\theta)^n}{n!}}$$

Could anyone help me understand this equality step by step?

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  • $\begingroup$ Note that 1st and 2nd term on LHS are even and odd terms on the RHS respectively. Then the equality is follows. $\endgroup$ – Lion May 28 '14 at 1:42
  • $\begingroup$ As always I've tried to keep my answer as simple as possible. I am the local worshipper of simplicity. $\endgroup$ – Michael Hardy May 28 '14 at 1:53
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Sure. You understand, I assume, that the first term is the $\cos$ function, and the second term is the $\sin$ function. Try plugging in values of $n$, and see what emerges. $$n=0: \frac{(i\theta)^{2\cdot0}}{(2\cdot0)!}=\frac{(i\theta)^{0}}{0!}$$ $$n=0: \frac{(i\theta)^{2\cdot0+1}}{(2\cdot0+1)!}=\frac{(i\theta)^{1}}{1!}$$ $$n=1: \frac{(i\theta)^{2\cdot1}}{(2\cdot1)!}=\frac{(i\theta)^{2}}{2!}$$ $$n=1: \frac{(i\theta)^{2\cdot1+1}}{(2\cdot1+1)!}=\frac{(i\theta)^{3}}{3!}$$

Notice the pattern? $$\frac{(i\theta)^{n}}{n!}$$ The first term ($\cos$) provides the even terms in the second sequence, and the second term ($\sin$) provides the odd terms in the second sequence. Since we are summing over an infinite sequence of integers, the upper bound doesn't have to change.

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  • $\begingroup$ I've changed multiple instances of $a*b$ to $a\cdot b$. The asterisk for ordinary multiplication is a workaround for occasions when one is limited to the characters on the keyboard and must reserve the lower-case x for things other than using it as a substitute for $\times$. I also changed $sin$ and $cos$ to $\sin$ and $\cos$, coded as \sin and \cos. That not only prevents italicization, but also provides proper spacing in things like $a\cos b$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 28 '14 at 1:57
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$$ \begin{array}{c|c|c|c} n & & 2n & 2n+1 \\ \hline 0 & & 0 & 1 \\ 1 & & 2 & 3 \\ 2 & & 4 & 5 \\ 3 & & 6 & 7 \\ 4 & & 7 & 8 \\ 5 & & 8 & 9 \\ 6 & & 12 & 13 \\ 7 & & 14 & 15 \\ \vdots & & \vdots & \vdots \end{array} $$ As $n$ goes through the list $0,1,2,3,4,\ldots$, the two later columns together also go through the list $0,1,2,3,4,\ldots$, each column going through half of it. Therefore $$ \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty (a_{2n} + a_{2n+1}). $$

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Just expand $$\sum\limits_{n=0}^\infty\left(\frac{(i\theta)^{2n}}{(2n)!}+\frac{(i\theta)^{2n+1}}{(2n+1)!}\right) =\left(\frac{1}{0!}+\frac{i\theta}{1!}\right)+\left(\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}\right)+\left(\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}\right)+\dotsb .$$ Now assuming all the nice convergence properties we can rewrite this (without the parentheses) as $$\text{RHS} =\sum\limits_{n=0}^\infty{\frac{(i\theta)^n}{n!}}.$$

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If you write out the first few terms on each side, you will see that the left side groups pairs of terms from the right side.

The idea is: $$(a_0 + a_1) + (a_2 + a_3) + \cdots = a_0+a_1+a_2+a_3 + \cdots$$

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