Exponential, Sine, and Cosine Functions' Taylor Series Expansion
📂Calculus Exponential, Sine, and Cosine Functions' Taylor Series Expansion Theorem e x = ∑ n = 0 ∞ x n n !
\begin{equation}
{ { e ^ x } }=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ n } }{ n! } }
\end{equation}
e x = n = 0 ∑ ∞ n ! x n
sin x = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! ( − 1 ) n
\begin{equation}
\sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } }
\end{equation}
sin x = n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 ( − 1 ) n
cos x = ∑ n = 0 ∞ x 2 n ( 2 n ) ! ( − 1 ) n
\begin{equation}
\cos x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } }
\end{equation}
cos x = n = 0 ∑ ∞ ( 2 n )! x 2 n ( − 1 ) n
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 ) (1) ( 1 ) Since
( e x ) ( n ) = e x { \left( { { e ^ x } } \right) ^{ (n) } }={ { e ^ x } } ( e x ) ( n ) = e x
e x = x 0 0 ! e 0 + x 1 1 ! e 0 + x 2 2 ! e 0 + ⋯ = ∑ n = 0 ∞ x n n !
{ { 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! } }
e x = 0 ! x 0 e 0 + 1 ! x 1 e 0 + 2 ! x 2 e 0 + ⋯ = n = 0 ∑ ∞ n ! x n
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( 2 ) (2) ( 2 ) Regarding
k = 0 , 1 , 2 , ⋯ k=0,1,2, \cdots k = 0 , 1 , 2 , ⋯
( sin x ) ( n ) = { cos x , n = 4 k + 1 ± sin x , n = 2 k − cos x , n = 4 k + 3
{ (\sin x) } ^{ (n) }= \begin{cases} \cos x & , n=4k+1
\\ \pm \sin x & , n=2k
\\ -\cos x & , n=4k+3 \end{cases}
( sin x ) ( n ) = ⎩ ⎨ ⎧ cos x ± sin x − cos x , n = 4 k + 1 , n = 2 k , n = 4 k + 3
Therefore
( sin 0 ) ( n ) = { 1 , n = 4 k + 1 0 , n = 2 k − 1 , n = 4 k + 3
{ (\sin 0) } ^{ (n) }=\begin{cases} 1 & , n=4k+1
\\ 0 & , n=2k
\\ -1 & , n=4k+3 \end{cases}
( sin 0 ) ( n ) = ⎩ ⎨ ⎧ 1 0 − 1 , n = 4 k + 1 , n = 2 k , n = 4 k + 3
And when represented as a series
sin x = x 0 0 ! 0 + ( x 1 1 ! 1 + x 2 2 ! 0 + x 3 3 ! ( − 1 ) + x 4 4 ! 0 ) + ⋯ = x 1 ! − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! ( − 1 ) n
\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*}
sin x = = = 0 ! x 0 0 + ( 1 ! x 1 1 + 2 ! x 2 0 + 3 ! x 3 ( − 1 ) + 4 ! x 4 0 ) + ⋯ 1 ! x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ⋯ n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 ( − 1 ) n
To summarize, we obtain sin x = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! ( − 1 ) n \displaystyle \sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } } sin x = n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 ( − 1 ) n
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( 3 ) (3) ( 3 ) Regarding
k = 0 , 1 , 2 , ⋯ k=0,1,2, \cdots k = 0 , 1 , 2 , ⋯
( cos x ) ( n ) = { cos x , n = 4 k ± sin x , n ≠ 2 k − cos x , n = 4 k + 2
{ (\cos x) }^{ (n) } = \begin{cases} \cos x & , n=4k
\\ \pm \sin x & , n\neq 2k
\\ -\cos x & , n=4k+2 \end{cases}
( cos x ) ( n ) = ⎩ ⎨ ⎧ cos x ± sin x − cos x , n = 4 k , n = 2 k , n = 4 k + 2
Therefore
( cos 0 ) ( n ) = { 1 , n = 4 k 0 , n ≠ 2 k − 1 , n = 4 k + 2
{ (\cos 0) } ^{ (n) } = \begin{cases} 1 & , n=4k
\\ 0 & , n\neq 2k
\\ -1 & , n=4k+2 \end{cases}
( cos 0 ) ( n ) = ⎩ ⎨ ⎧ 1 0 − 1 , n = 4 k , n = 2 k , n = 4 k + 2
And when represented as a series
cos x = ( x 0 0 ! 1 + x 1 1 ! 0 + x 2 2 ! ( − 1 ) + x 3 3 ! 0 ) + x 4 4 ! 1 + ⋯ = 1 0 ! − x 2 2 ! + x 4 4 ! − x 6 6 ! + ⋯ = ∑ n = 0 ∞ x 2 n ( 2 n ) ! ( − 1 ) n
\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*}
cos x = = = ( 0 ! x 0 1 + 1 ! x 1 0 + 2 ! x 2 ( − 1 ) + 3 ! x 3 0 ) + 4 ! x 4 1 + ⋯ 0 ! 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + ⋯ n = 0 ∑ ∞ ( 2 n )! x 2 n ( − 1 ) n
To summarize, we obtain cos x = ∑ n = 0 ∞ x 2 n ( 2 n ) ! ( − 1 ) n \displaystyle \cos x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n } }{ (2n)! }{ { (-1) }^{ n } } } cos x = n = 0 ∑ ∞ ( 2 n )! x 2 n ( − 1 ) n
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