Taylor Series

f(x)=c0+c1(xa)+c2(xa)2+c3(xa)3+c4(xa)4+f(a)=c0+0+0+0+0+c0=f(a)\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}

f(1)(x)=1c1+2c2(xa)+3c3(xa)2+4c4(xa)3+f(1)(a)=1c1+0+0+0+c1=f(1)(a)1\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}

f(2)(x)=(21)c2+(32)c3(xa)+(43)c4(xa)2+f(2)(a)=(21)c2+0+0+c2=f(2)(a)21\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}

f(3)(x)=(321)c3+(432)c4(xa)+f(3)(a)=(321)c3+0+c3=f(3)(a)321\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}

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

f(x)=n=0cn(xa)nf(x)=n=0f(n)(a)n!(xa)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
4 months, 2 weeks 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 cotx\displaystyle \cot x centered at x=0\displaystyle x=0

Aaghaz Mahajan - 4 months, 2 weeks ago

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

Jacob Dreiling - 3 months ago

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you couldn't, but you could find the Taylor series of xcot(x)x\cot(x) centered at 00, and play around with that to get a series for cot(x)\cot(x)

Aareyan Manzoor - 2 months, 2 weeks ago

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

Aaghaz Mahajan - 2 months, 1 week ago

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