# Euler Identity Derivation-Is this allowed?

I was reading about how Euler derived his famous identity, $e^{i{\pi}}$. It said that it was discovered when Euler took the Taylor Expansion for $e^x$, and he multiplied the $x$ by $i$, and it gave him the formula: $$e^{ix}=1+ix-\frac {x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}...$$ He then separated the series by imaginary and real parts, and found $$e^{ix}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-...)+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)$$ He then recognized the two series as taylor series for sine and cosine, $$e^{ix}=\cos x+{i}\sin x$$

Now, my question is, why was he able to split up the infinite series like that? From my understanding, if you change the order of terms in an infinite series you'll change the value of the sum, so why was he able to?

• The rule is that you are able to rearrange the terms in a sequence if the sequence is absolutely convergent. That is if $|x_n|$ converges then you can rearrange the terms without changing the value of the limit. So in this case you can expand those terms since the series is absolutely convergent. Feb 9, 2017 at 4:58
• Actually that statement is slightly (it can be easily generalized to other spaces) weaker than you need because you're considering complex numbers. The general statement applies to norms not just absolute values. The norm of complex numbers is called the complex modulus and $|a + bi| = \sqrt{a^2 + b^2}$. Feb 9, 2017 at 5:04
• May be this will make sense to you. If $i$ is multiplied by a real number, then it is called as a complex number (loosely speaking). Infact, a complex number has two parts, which are referred as the in-phase and quadrature term. Example, a complex number $Z = A + iB$, then in-phase is $A$ and quadrature part is $B$. Say $Z = 1 + i1 \neq i2 \neq 2$. So, you can not just add arithmetically $A + iB$ because imagine complex-number as a two-dimensional vector in real domain. Does that make sense to you? Apr 27, 2018 at 15:31

This is completely fine. Suppose I had the series $S$ such that $S=1+2+3+4+5+6+...$
If I split this series into series $T=1+3+5+...$ and $U=2+4+6+...$, then $T+U=S$ clearly holds. The same thing has been done in your example.
• That's not true for divergent series like that. In fact, if were true, then: $U-T=1+1+1+1+1+1+1+...$, $$T=1+3+5+...=1+(1+2)+(1+4)+(1+6)...=(1+1+1+1+1+1+...)+(2+4+6+8+....)=(U-T)+U=2U-T$$ but then $U=T$ and then $S=2U$ implies $S=\frac{1}{2}$. I think Euler didn't know about divergent series but he was very lucky because exponential series is absolutely convergent. Apr 27, 2018 at 15:31
• Doesn't $S=2U$ imply $U=\frac{S}{2}$ which we would expect considering $U=T$ and $U+T=S$? Apr 27, 2018 at 23:13