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'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...
To evaluate the given function f(x) =root(3)(x) , we may express it in terms of fractional exponent using the radical property: root(n)(x)= x^(1/n) . The function becomes:
f(x) = (x)^(1/3) .
Apply the definition of the Taylor series by listing the f^n(x) up to n=3 .
We determine each derivative using Power Rule for differentiation: d/(dx) x^n = n*x^(n-1) .
f(x) = (x)^(1/3)
f'(x) = 1/3 * x^(1/3-1)
= 1/3x^(-2/3) or1/(3x^(2/3) )
f^2(x) = d/(dx) (1/3x^(-2/3))
= 1/3 * d/(dx) (x^(-2/3))
= 1/3*(-2/3x^(-2/3-1))
= -2/9 x^(-5/3) or -2/(9x^(5/3))
f^3(x) = d/(dx) (-2/9 x^(-5/3))
= -2/9*d/(dx) (x^(-5/3))
= -2/9*(-5/3x^(-5/3-1))
=10/27 x^(-8/3) or 10/(27x^(8/3))
Plug-in x=8, we get:
f(8) = (8)^(1/3)
= 2
f'(8)=1/(3*8^(2/3) )
=1/(3*4)
=1/12
f^2(8)=-2/(9*8^(5/3))
=-2/(9*32)
= -2/288
=-1/144
f^3(8)=10/(27*8^(8/3))
=10/(27*256)
= 10/6912
=5/3456
Applying the formula for Taylor series centered at c=8 , we get:
sum_(n=0)^3 (f^n(8))/(n!)(x-8)^n
=f(8) + f'(8) (x-8)+ (f'(8))/(2!) (x-8)^2+ (f'(8))/(3!) (x-8)^3
=2+ (1/12) (x-8)+ (-1/144)/(2!) (x-8)^2+ (5/3456)/(3!) (x-8)^3
=2+ 1/12 (x-8)-1/(144*2) (x-8)^2+ 5/(3456*6) (x-8)^3
=2+ 1/12 (x-8)-1/288 (x-8)^2+ 5/20736 (x-8)^3
The Taylor polynomial of degree n=3 for the given function f(x)=root(3)(x) centered at c=8 will be:
P(x)=2+ 1/12 (x-8)-1/288 (x-8)^2+ 5/20736 (x-8)^3
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