What is the Taylor series generated by g(x) = x^3-2x+2 at a=5?
as always, it is
g(5)/0! + g'(5)(x-5)/1! + g"(5)(x-5)^2/2! + g"'(5)(x-5)^3/3! + ...
117 + 73(x-5) + 15(x-5)^2 + (x-5)^3 + ...
but all the higher derivatives are zero, so that's all you get.
f(x) = x³ -2x+2----> f(5) = 117
f'(x) = 3x² -2-------> f'(5) =73
f''(x) = 6x --------------> f''(5) = 30
f'''(x) = 6 --------------------> f'''(1) = 6
f(x) = 117/0! + 73/1! (x−5) + 30/2! (x−5)² + 6/3! (x−5)³
f(x) = 117+ 73(x−5) + 15(x−5)² + (x−5)³
f(x) = 117 + 73x−365 + 15x²−150x +375 + x³ -15x²+75x-125
f(x) = x³ -2x +2 which is the same as the original function!
To find the Taylor series of a function at a specific point, first we need to calculate the derivatives of the function. The n-th derivative will be used to determine the coefficient of the n-th term in the Taylor series.
Let's start by calculating the derivatives of the function g(x) = x^3 - 2x + 2:
g'(x) = 3x^2 - 2
g''(x) = 6x
g'''(x) = 6
Now, let's evaluate these derivatives at the given point a = 5:
g(5) = 5^3 - 2(5) + 2 = 123
g'(5) = 3(5)^2 - 2 = 73
g''(5) = 6(5) = 30
g'''(5) = 6
Using these values, we can write the Taylor series for g(x) centered at a = 5. The general formula for a Taylor series is:
f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ...
So, plugging in the calculated values, we have:
g(x) = 123 + 73(x - 5) + (30/2!)(x - 5)^2 + (6/3!)(x - 5)^3 + ...
Simplifying, we get:
g(x) = 123 + 73(x - 5) + 15(x - 5)^2 + (2/3)(x - 5)^3 + ...
This is the Taylor series generated by g(x) = x^3 - 2x + 2 at a = 5.