Taylor series is an example of infinite series derived from the expansion of `f(x)` about a single point. It is represented by infinite sum of ` f^n(x)` centered at ` x=c` . The general formula for Taylor series is:
`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`
or
`f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...`
To apply the definition of Taylor series for the given function `f(x) = 1/(1-x)` centered at c=2, we list `f^n(x)` using the Power rule for differentiation: `d/(dx) u^n= n *u^(n-1) *(du)/(dx) ` and basic differentiation property: `d/(dx) c* f(x)= c * d/(dx) f(x)` .
`f(x)= 1/(1-x)`
Let `u=1-x` then `(du)/(dx)= -1` .
The derivative of f(x) will be:
`d/(dx) (1/(1-x)) =d/(dx) (1-x)^(-1)`
`= (-1)*(1-x)^(-1-1)*(-1)`
`=(1-x)^-2 or 1/(1-x)^2`
Then, we list the derivatives of `f(x)` as:
`f'(x) = d/(dx) (1/(1-x))`
`=(1-x)^-2 or 1/(1-x)^2`
`f^2(x)= d/(dx) (1-x)^(-2)`
`=-2*((1-x)^(-2-1))*(-1)`
`=2(1-x)^(-3) or 2/(1-x)^3`
`f^3(x)= d/(dx) 2(1-x)^(-3)`
`=2*d/(dx) (1-x)^(-3)`
`=2* (-3*(1-x)^(-3-1))*(-1)`
`=6(1-x)^(-4) or 6/(1-x)^4`
`f^4(x)= d/(dx)6(1-x)^(-4)`
`=6*d/(dx) (1-x)^(-4)`
`=6* (-4*(1-x)^(-4-1))*(-1)`
`=24(1-x)^(-5) or 24/(1-x)^5`
Plug-in `x=2` , we get:
`f(2)=1/(1-2)`
`=1/(-1)`
`=-1`
`f'(2)=1/(1-2)^2`
`=1/(-1)^2`
`=1/1`
`=1`
`f^2(2)=2/(1-2)^3`
` =2/(-1)^3`
`=2/(-1)`
`=-2`
`f^3(2)=6/(1-2)^4`
`=6/(-1)^4`
`=6/1`
`=6`
`f^4(2)=24/(1-2)^5`
`=24/(-1)^5`
`=24/(-1)`
`=-24`
Plug-in the values on the formula for Taylor series, we get:
`1/(1-x) =sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n`
` =f(2)+f'(2)(x-2) +(f^2(2))/(2!)(x-2)^2 +(f^3(2))/(3!)(x-2)^3 +(f^4(2))/(4!)(x-2)^4 +...`
`=-1+1*(x-2) + (-2)/(2!)(x-2)^2 +6/(3!)(x-2)^3 + (-24)/(4!)(x-2)^4 +...`
`=-1+(x-2) -2/2(x-2)^2 +6/6(x-2)^3 -24/24(x-2)^4 +... `
`=-1+(x-2) -(x-2)^2 + (x-2)^3 -(x-2)^4 +...`
The Taylor series for the given function `f(x)=1/(1-x)` centered at `c=2` will be:
`1/(1-x)=-1+(x-2) -(x-2)^2 + (x-2)^3 -(x-2)^4 +...`
or
`1/(1-x) = sum_(n=0)^oo (-1)^(n+1)(x-2)^n`
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