# Taylor Series

\begin{aligned} f\left( x \right) & = { c }_{ 0 }+{ c }_{ 1 }{ (x-a) }+{ c }_{ 2 }{ (x-a) }^{ 2 }+{ c }_{ 3 }{ (x-a) }^{ 3 }+{ c }_{ 4 }{ (x-a) }^{ 4 }+\cdots \\ f\left( a \right) & = { c }_{ 0 }+0+0+0+0+\cdots \\ \Rightarrow { c }_{ 0 } & = f\left( a \right) \end{aligned}

\begin{aligned} f^{ (1) }\left( x \right) & = 1{ c }_{ 1 }+2{ c }_{ 2 }{ (x-a) }+3{ c }_{ 3 }{ (x-a) }^{ 2 }+4{ c }_{ 4 }{ (x-a) }^{ 3 }+\cdots \\ f^{ (1) }\left( a \right) & = 1{ c }_{ 1 }+0+0+0+\cdots \\ \Rightarrow { c }_{ 1 } & = \frac { f^{ (1) }\left( a \right) }{ 1 } \end{aligned}

\begin{aligned} f^{ (2) }\left( x \right) & = (2\cdot 1){ c }_{ 2 }+(3\cdot 2){ c }_{ 3 }{ (x-a) }+(4\cdot 3){ c }_{ 4 }{ (x-a) }^{ 2 }+\cdots \\ f^{ (2) }\left( a \right) & = (2\cdot 1){ c }_{ 2 }+0+0+\cdots \\ \Rightarrow { c }_{ 2 } & = \frac { f^{ (2) }\left( a \right) }{ 2\cdot 1 } \end{aligned}

\begin{aligned} f^{ (3) }\left( x \right) & = (3\cdot 2\cdot 1){ c }_{ 3 }+(4\cdot 3\cdot 2){ c }_{ 4 }{ (x-a) }+\cdots \\ f^{ (3) }\left( a \right) & = (3\cdot 2\cdot 1){ c }_{ 3 }+0+\cdots \\ \Rightarrow { c }_{ 3 } & = \frac { f^{ (3) }\left( a \right) }{ 3\cdot 2\cdot 1 } \end{aligned}

$\Longrightarrow { c }_{ n }=\frac { f^{ (n) }\left( a \right) }{ n! }$

\begin{aligned} f\left( x \right) & = \displaystyle \sum _{ n=0 }^{ \infty }{ { c }_{ n }{ (x-a) }^{ n } } \\ \quad & \Longrightarrow \boxed { f\left( x \right) =\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { f^{ (n) }\left( a \right) }{ n! } { (x-a) }^{ n } } } \end{aligned} Note by Gordon Chan
8 months ago

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@Gordon Chan If you have just learnt about this, then here's an interesting question.......How would you find the Taylor expansion of $\displaystyle \cot x$ centered at $\displaystyle x=0$

- 7 months, 4 weeks ago

Cotangent has an asymptote at x = 0 so I don't think that would work

- 6 months, 2 weeks ago

you couldn't, but you could find the Taylor series of $x\cot(x)$ centered at $0$, and play around with that to get a series for $\cot(x)$

- 5 months, 4 weeks ago

Yeah......that was exactly what I had in mind......

- 5 months, 3 weeks ago

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- 2 months, 3 weeks ago