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I have the following problem:

How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

My attempt: \begin{align*} \exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!} \end{align*} Now split the sum in 4 sums because $i^n$ has period 4, more precisely $i^{4n}=1$, $i^{4n+1}=i$, $i^{4n+2}=-1$ and $i^{4n+3}=-i$ for $n\geq 0$. \begin{align*} \sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\ &=\sum_{n=0}^{\infty}\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)+i\sum_{n=0}^{\infty}\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right) \end{align*} Then $4n$ and $4n+2$ are nonnegative and even numbers and $4n+1$ and $4n+3$ are odd. How can I use this to convert the two last expressions into \begin{align*} \cos x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m}}{(2m)!}\\ \sin x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m+1}}{(2m+1)!} \end{align*} I don't want to see proofs like \begin{align*} \exp(ix)&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}\ldots\\ &=\Big(1-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ldots\Big)+i\Big(x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\ldots\Big)\\ &=\cos x+i\sin x \end{align*}

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You split the sum like that:

\begin{align*} \exp(ix) = \sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\ &=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} + i\left( \sum_{n=0}^{\infty}\frac{x^{4n+1}}{(4n+1)!} - \sum_{n=0}^{\infty}\frac{x^{4n+3}}{(4n+3)!}\right) \end{align*} Now we join the sums as follows: \begin{align*} \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} &= \sum_{n=0}^{\infty} (-1)^{2n}\frac{x^{4n}}{(4n)!} + \sum_{n=0}^{\infty}(-1)^{2 n+ 1}\frac{x^{4n+2}}{(4n+2)!} \\ &= \sum_{n=0}^{\infty} (-1)^{2n}\frac{x^{2(2n)}}{(2(2n))!} + \sum_{n=0}^{\infty}(-1)^{2n+ 1}\frac{x^{2(2n+1)}}{(2(2n+1))!} \\ &= \sum_{m \in \mathbb N_0 \text{ even}} (-1)^{m}\frac{x^{2m}}{(2m)!} + \sum_{m \in \mathbb N_0 \text{ odd}} (-1)^{m}\frac{x^{2m}}{(2m)!} \\ &= \sum_{m=0}^{\infty} (-1)^{m}\frac{x^{2m}}{(2m)!} = \cos(x). \end{align*} The same trick works for $\sin(x)$, so you get $$ \exp(ix) = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} + i\left( \sum_{n=0}^{\infty}\frac{x^{4n+1}}{(4n+1)!} - \sum_{n=0}^{\infty}\frac{x^{4n+3}}{(4n+3)!}\right) = \cos(x) + i \sin(x).$$ I hope that helps you :)

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We obtain

\begin{align*} \cos(x)&=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\ &=\sum_{{n=0}\atop{n \text{ even}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}+\sum_{{n=0}\atop{n \text{ odd}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\ &=\sum_{{n=0}}^{\infty} \frac{x^{4n}}{(4n)!}-\sum_{{n=0}}^{\infty} \frac{x^{4n+2}}{(4n+2)!}\\ &=\sum_{{n=0}}^{\infty}\left( \frac{x^{4n}}{(4n)!}- \frac{x^{4n+2}}{(4n+2)!}\right) \end{align*}

and similarly for $\sin(x)$.

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Your approach assumes that you know the Taylor series for $\cos x, \sin x$. It is possible to prove directly that $$\exp(ix) =\cos x+i\sin x$$ by using the derivatives of circular functions.


I will provide an approach in the hope that you may like it. Also note that this is not exactly the answer for your query regarding the series transformation.

Consider the function $f(x) =\exp(ix) $ where $x$ is a real variable. Using binomial theorem for positive integral index and the rule for multiplication of infinite series it is easy to prove that $$\exp(z+w) =\exp(z) \exp(w) $$ and hence $$f'(x) =\lim_{h\to 0}\frac{\exp(ix+ih)-\exp(ix)}{h}=\exp(ix)\lim_{h\to 0}\frac{\exp(ih)-1}{h}$$ Next we can note that $$\left|\frac{\exp(ih) - 1}{h}-i\right|=\left|\frac{i^{2}h}{2!}+\frac{i^{3}h^{2}}{3!}+\cdots\right|\leq \frac{|h|} {2!}+\frac{|h|^{2}}{3!}+\cdots$$ and the sum on the right in above equation does not exceed $$\frac{|h|} {2}+\frac{|h|^{2}}{2^{2}}+\cdots=\frac{|h|}{2-|h|}$$ provided that $|h|<2$. Thus if $|h|<2$ then $$0\leq\left|\frac{\exp(ih)-1}{h}-i\right|\leq \frac{|h|} {2-|h|}$$ and applying Squeeze Theorem we can see that $$\lim_{h\to 0}\frac{\exp(ih)-1}{h}=i$$ and therefore we have $f'(x) =if(x) $.

If $g(x) =f(x) (\cos x-i\sin x) $ then \begin{align} g'(x) &=f'(x) (\cos x-i\sin x) - f(x)(\sin x+i\cos x)\notag \\ &=if(x) (\cos x-i\sin x) - f(x) (\sin x+i\cos x)\notag \\ &=0\notag \end{align} and hence $g(x) $ is constant. We have $g(x) =g(0)=1$ and hence $$\exp(ix) =f(x) =\frac{1}{\cos x-i\sin x} =\cos x+i\sin x$$


Note: Although the Mean Value Theorem does not hold for complex valued functions of a real variable, one of its consequences namely the implication "derivative vanishes everywhere $\Rightarrow $ function is constant everywhere" does hold for complex valued functions of a real variable and this fact has been used here for the function $g(x) $. The proof for the complex valued function is given by treating real and imaginary parts separately.

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  • $\begingroup$ @ParamandhSingh: OP is explicitly asking for a specific series transformation. $\endgroup$ – Markus Scheuer Jul 15 '17 at 16:09
  • $\begingroup$ @MarkusScheuer: I don't think I can really add anything more than your or user "Yaddle"'s answer (both upvoted) as far as the series approach is concerned. Therefore I have tried to give another approach. I will edit to highlight that this is not exactly an answer to the question. $\endgroup$ – Paramanand Singh Jul 15 '17 at 16:12
  • $\begingroup$ No problem at all. I do appreciate many of your answers. I just wanted to indicate OPs focus in case you are not aware of it. $\endgroup$ – Markus Scheuer Jul 15 '17 at 16:19

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