Calculus and the Euler Formula
Theorem
- Euler’s Formula:
$$ { e }^{ ix }= \cos x + i \sin x $$
- Euler’s Identity:
$$ { e }^{ i\pi }+1=0 $$
Explanation
Euler’s Formula is in itself so peculiar that even Euler did not know where it might be used, but nowadays, it is utilized in so many fields that it is difficult to summarize its usefulness. It is even more astonishing when considering it was discovered at a time when imaginary numbers were still not well accepted in academia. Its derivation can be simply done through the Taylor series expansions of the exponential function, sine function, and cosine function.
$$ { { e ^ x } }=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ n } }{ n! } } $$
$$ \sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } } $$
$$ \cos x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } } $$
Derivation (Euler’s Formula)
$$ \begin{align*} { e }^{ ix } =& \sum _{ n=0 }^{ \infty }{ \frac { { (ix) } ^{ n } }{ n! } } \\ =&\frac { { (ix) } ^{ 0 } }{ 0! }+\frac { { (ix) } ^{ 1 } }{ 1! }+\frac { { (ix) } ^{ 2 } }{ 2! }+\frac { { (ix) } ^{ 3 } }{ 3! }+\frac { { (ix) } ^{ 4 } }{ 4! }+ \cdots \\ =&\frac { 1 }{ 0! }+\frac { ix }{ 1! }-\frac { { x }^{ 2 } }{ 2! }-\frac { i { x }^{ 3 } }{ 3! }+\frac { { x } ^{ 4 } }{ 4! }+ \cdots \\ =& \left( \frac { 1 }{ 0! } - \frac { { x } ^{ 2 } }{ 2! }+\frac { { x } ^{ 4 } }{ 4! }-\frac { { x } ^{ 6 } }{ 6! }+\cdots \right) + i\left( \frac { x }{ 1! } - \frac { { x } ^{ 3 } }{ 3! }+\frac { { x } ^{ 5 } }{ 5! }-\frac { { x } ^{ 7 } }{ 7! }+\cdots \right) \\ =& \cos x + i \sin x \end{align*} $$
Therefore,
$$ { e }^{ ix }= \cos x + i \sin x $$
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Especially, by substituting $x=\pi$, we get what is called ’the most beautiful equation in the world,’ the Euler’s identity. Moreover, by manipulating the Euler’s identity, we can also find the value of the power of the imaginary unit $i$, that is, ▷eq4◀. Surprisingly, this value is a real number, and the proof goes as follows.
Proof
$$ \begin{align*} && { e }^{ i\pi }+1 =& 0 \\ \implies && { e }^{ i\pi }=&-1 \\ \implies && { e }^{ \frac { i\pi }{ 2 } } =& \sqrt { -1 } \\ \implies && { \left( { e } ^{ \frac { i\pi }{ 2 } } \right) }^{ i } =& { \sqrt { -1 } }^{ i } \\ \implies && { e }^{ \frac { i\pi }{ 2 }i } =& { i } ^{ i } \\ \implies && { i }^{ i } =& { e }^{ -\frac { \pi }{ 2 } } \\ \implies && { i }^{ i } =& \frac { 1 }{ \sqrt { { e }^{ \pi } } } \end{align*} $$
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