📂Calculus

Exponential, Sine, and Cosine Functions' Taylor Series Expansion

Theorem¹

$$ \begin{equation} { { e ^ x } }=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ n } }{ n! } } \end{equation} $$

$$ \begin{equation} \sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } } \end{equation} $$

$$ \begin{equation} \cos x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } } \end{equation} $$

Explanation

The Maclaurin series for the exponential function, sine function, and cosine function can be easily obtained without the use of difficult techniques. Combining these three results in the famous Euler’s formula. One tip is to remember that sine contains only odd-degree terms while cosine contains only even-degree terms.

Proof

$(1)$

Since

${ \left( { { e ^ x } } \right) ^{ (n) } }={ { e ^ x } }$

$$ { { e ^ x } }=\frac { { x } ^{ 0 } }{ 0! } { e }^{ 0 } +\frac { { x } ^{ 1 } }{ 1! } { e }^{ 0 } +\frac { { x } ^{ 2 } }{ 2! } { e }^{ 0 } + \cdots =\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ n } }{ n! } } $$

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$(2)$

Regarding

$k=0,1,2, \cdots$

$$ { (\sin x) } ^{ (n) }= \begin{cases} \cos x & , n=4k+1 \\ \pm \sin x & , n=2k \\ -\cos x & , n=4k+3 \end{cases} $$

Therefore

$$ { (\sin 0) } ^{ (n) }=\begin{cases} 1 & , n=4k+1 \\ 0 & , n=2k \\ -1 & , n=4k+3 \end{cases} $$

And when represented as a series

$$ \begin{align*} \sin x =& \frac { { x } ^{ 0 } }{ 0! }0+\left( \frac { { x } ^{ 1 } }{ 1! }1+\frac { { x } ^{ 2 } }{ 2! }0+\frac { { x } ^{ 3 } }{ 3! }(-1)+\frac { { x } ^{ 4 } }{ 4! }0 \right) + \cdots \\ =& \frac { x }{ 1! }-\frac { { x } ^{ 3 } }{ 3! }+\frac { { x } ^{ 5 } }{ 5! }-\frac { { x } ^{ 7 } }{ 7! }+ \cdots \\ =& \sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } } \end{align*} $$

To summarize, we obtain $\displaystyle \sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } }$.

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$(3)$

Regarding

$k=0,1,2, \cdots$

$$ { (\cos x) }^{ (n) } = \begin{cases} \cos x & , n=4k \\ \pm \sin x & , n\neq 2k \\ -\cos x & , n=4k+2 \end{cases} $$

Therefore

$$ { (\cos 0) } ^{ (n) } = \begin{cases} 1 & , n=4k \\ 0 & , n\neq 2k \\ -1 & , n=4k+2 \end{cases} $$

And when represented as a series

$$ \begin{align*} \cos x =& \left( \frac { { x } ^{ 0 } }{ 0! }1+\frac { { x } ^{ 1 } }{ 1! }0+\frac { { x } ^{ 2 } }{ 2! }(-1)+\frac { { x } ^{ 3 } }{ 3! }0 \right) +\frac { { x } ^{ 4 } }{ 4! }1+ \cdots \\ =& \frac { 1 }{ 0! }-\frac { { x } ^{ 2 } }{ 2! }+\frac { { x } ^{ 4 } }{ 4! }-\frac { { x } ^{ 6 } }{ 6! }+ \cdots \\ =& \sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } } \end{align*} $$

To summarize, we obtain $\displaystyle \cos x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } }$.

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