Problem 26 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=2 x^{3}+x^{2}+3 x-8, \quad a=1\)

Step-by-Step Solution

Answer

The Taylor series is \(-2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3.\)

Step 1: Define the Taylor Series

The Taylor series of a function \( f(x) \) around \( a \) is given by the formula \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \] We'll calculate these terms for \( f(x) = 2x^3 + x^2 + 3x - 8 \) at \( a = 1 \).

Step 2: Calculate \( f(a) \)

Substitute \( x = 1 \) into the function \( f(x) = 2x^3 + x^2 + 3x - 8 \):\[ f(1) = 2(1)^3 + (1)^2 + 3(1) - 8 = 2 + 1 + 3 - 8 = -2. \]

Step 3: Find the First Derivative \( f'(x) \) and \( f'(a) \)

Calculate the first derivative: \( f'(x) = 6x^2 + 2x + 3 \). Substituting \( x = 1 \), we find \( f'(1) = 6(1)^2 + 2(1) + 3 = 6 + 2 + 3 = 11 \).

Step 4: Find the Second Derivative \( f''(x) \) and \( f''(a) \)

Calculate the second derivative: \( f''(x) = 12x + 2 \). Substituting \( x = 1 \) gives \( f''(1) = 12(1) + 2 = 14 \).

Step 5: Find the Third Derivative \( f'''(x) \) and \( f'''(a) \)

Calculate the third derivative: \( f'''(x) = 12 \). This is a constant function, so \( f'''(1) = 12 \).

Step 6: Assemble the Taylor Series

Using the Taylor series formula, substitute the derivatives into their respective terms: \[\begin{align*}f(x) &= f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 \ &= -2 + 11(x-1) + \frac{14}{2}(x-1)^2 + \frac{12}{6}(x-1)^3 \ &= -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3.\end{align*}\]

Key Concepts

First Derivative

Understanding the first derivative, denoted as \( f'(x) \), is crucial when working on Taylor series expansion. The first derivative represents how the function \( f(x) \) changes as \( x \) changes, basically indicating the slope of the function. This is important for interpreting the initial linear behavior of the function near the point \( a \).

In the given exercise, the original function is \( f(x) = 2x^3 + x^2 + 3x - 8 \). To find the first derivative, we differentiate each term separately:

• The derivative of \( 2x^3 \) is \( 6x^2 \) because 3 times 2 is 6 and we subtract 1 from the exponent 3.
• The derivative of \( x^2 \) is \( 2x \) for similar reasons. The exponent 2 comes down then we reduce the power by 1.
• The derivative of \( 3x \) is 3, as we multiply the coefficient by the exponent, which is 1, resulting in just the coefficient.
• The derivative of a constant, \(-8\), is 0, since constants do not change.

Hence, the first derivative is \( f'(x) = 6x^2 + 2x + 3 \). When we substitute \( x = 1 \) in \( f'(x) \), it results in 11, which helps us in forming the linear term in the Taylor series around that point.

Second Derivative

The second derivative of a function, \( f''(x) \), reveals how the rate of change itself is changing. In simpler terms, it tells us if the slope of \( f(x) \) is increasing or decreasing. This is helpful to assess the function's concavity and is crucial for forming the quadratic term in the Taylor series.

For the function \( f(x) = 2x^3 + x^2 + 3x - 8 \), we find the second derivative by differentiating \( f'(x) = 6x^2 + 2x + 3 \). Calculate it as follows:

• The derivative of \( 6x^2 \) is \( 12x \), where we bring down the 2 and multiply it by 6.
• The derivative of \( 2x \) is 2.
• The derivative of a constant, 3, is 0.

So, the second derivative becomes \( f''(x) = 12x + 2 \). Substituting \( x = 1 \) into \( f''(x) \), we get 14. This result is used to evaluate the behavior of the quadratic term \( \frac{f''(1)}{2!}(x-1)^2 \) in the Taylor series expansion.

Third Derivative

The third derivative, denoted as \( f'''(x) \), represents the rate of change of the second derivative. It is useful for identifying the "jerk" in the context of applied physics or how the concavity is changing, contributing to the cubic term in a Taylor series.

In our given function, \( f(x) = 2x^3 + x^2 + 3x - 8 \), the third derivative is found by taking the derivative of the second derivative \( f''(x) = 12x + 2 \).

• The derivative of \( 12x \) is 12, as we multiply 12 by 1.
• The derivative of a constant, 2, is 0.

Thus, the third derivative \( f'''(x) \) is a constant 12. Notably, evaluating \( f'''(1) \) still gives 12, indicating that this term remains constant at this value. In the Taylor series, it determines the coefficient for the cubic term \( \frac{f'''(1)}{3!}(x-1)^3 \), which helps to see how bending of the curve is shaped.