Taylor Series

f(x)=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+c4(x−a)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(x−a)+3c3(x−a)2+4c4(x−a)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)=(2⋅1)c2+(3⋅2)c3(x−a)+(4⋅3)c4(x−a)2+⋯f(2)(a)=(2⋅1)c2+0+0+⋯⇒c2=f(2)(a)2⋅1\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)=(3⋅2⋅1)c3+(4⋅3⋅2)c4(x−a)+⋯f(3)(a)=(3⋅2⋅1)c3+0+⋯⇒c3=f(3)(a)3⋅2⋅1\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=0∞cn(x−a)n⟹f(x)=∑n=0∞f(n)(a)n!(x−a)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}

#Calculus

Note by Gandoff Tan
2 years ago

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Comments

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

Aaghaz Mahajan - 2 years ago

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

Jacob Dreiling - 1 year, 11 months ago

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 - 1 year, 10 months ago

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

Aaghaz Mahajan - 1 year, 10 months ago
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