Question

5 f(x) = xeˣ, a = 01 + x⁵ - 10. Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of a.

A) (f(x) = (1 + x⁵ - 10)e¹⁺ˣ⁵⁻¹⁰
B) (f(x) = xeˣ + x²eˣ + x³eˣ + x⁴eˣ)
C) (f(x) = xeˣ + (x⁵ - 10)eˣ⁵⁻¹⁰
D) (f(x) = xeˣ + xeˣ⁵-¹⁰ + x²eˣ⁵⁻¹⁰ + x³eˣ⁵-¹⁰

Answer 1

Using the Taylor series formula, the first four nonzero terms for the function f(x) = xe^x when expanded around a = 0 are x, x^2, x^3, and x^4, which corresponds to option B of the provided choices.

Step-by-step explanation:

To find the first four nonzero terms of the Taylor series for f(x) = xex centered at a = 0, we need to use the formula for the Taylor series expansion of a function about a point a. The formula is f(a) + f'(a)(x-a) + f''(a)(x-a)2/2! + f'''(a)(x-a)3/3! + ..., where f', f'', and so on, are the successive derivatives of f evaluated at a.

First, let's find the first four derivatives of f(x):

• f'(x) = ex + xex
• f''(x) = 2ex + xex
• f'''(x) = 3ex + xex
• f''''(x) = 4ex + xex

Now we evaluate these derivatives at a = 0:

• f(0) = 0
• f'(0) = e0 = 1
• f''(0) = 2e0 = 2
• f'''(0) = 3e0 = 3
• f''''(0) = 4e0 = 4

Substituting these into the Taylor series formula, we get the first four nonzero terms:

• f(0) = 0 (this term is zero, so it does not contribute to the nonzero terms)
• f'(0)(x-0) = x
• f''(0)(x-0)2/2! = x2
• f'''(0)(x-0)3/3! = x3
• f''''(0)(x-0)4/4! = x4

So the first four nonzero terms of the series are:

f(x) = x + x2 + x3 + x4

This matches option B.