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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