To find the fifth degree Taylor polynomial approximation of the function f(x) = ex centered at a = 0, we first need to compute the derivatives of the function at the point a = 0.
The function f(x) = ex has the property that all its derivatives are also equal to ex. Therefore:
- f(0) = e0 = 1
- f'(0) = e0 = 1
- f”(0) = e0 = 1
- f”'(0) = e0 = 1
- f^{(4)}(0) = e0 = 1
- f^{(5)}(0) = e0 = 1
Using the formula for the Taylor polynomial, we have:
Tn(x) = f(a) + f'(a)(x – a) + rac{f”(a)}{2!}(x – a)^{2} + rac{f”'(a)}{3!}(x – a)^{3} + rac{f^{(4)}(a)}{4!}(x – a)^{4} + rac{f^{(5)}(a)}{5!}(x – a)^{5}
Substituting a = 0 and the values we computed:
T5(x) = 1 + 1(x – 0) + rac{1}{2!}(x – 0)^{2} + rac{1}{3!}(x – 0)^{3} + rac{1}{4!}(x – 0)^{4} + rac{1}{5!}(x – 0)^{5}
This simplifies to:
T5(x) = 1 + x + rac{x^{2}}{2} + rac{x^{3}}{6} + rac{x^{4}}{24} + rac{x^{5}}{120}
Therefore, the fifth degree Taylor polynomial approximation of f(x) = ex centered at a = 0 is:
T5(x) = 1 + x + rac{x^{2}}{2} + rac{x^{3}}{6} + rac{x^{4}}{24} + rac{x^{5}}{120}