Problem 27 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=x^{4}+x^{2}+1, \quad a=-2\)

Step-by-Step Solution

Answer

The Taylor series is \( 21 - 36(x+2) + 25(x+2)^2 - 8(x+2)^3 + (x+2)^4 \).

Step 1: Calculate the Function Value at a

Evaluate the function at the given point, which is when \( x = -2 \).The function is \( f(x) = x^4 + x^2 + 1 \).Substituting \( x = -2 \), we have:\[f(-2) = (-2)^4 + (-2)^2 + 1 = 16 + 4 + 1 = 21\]Thus, \( f(-2) = 21 \).

Step 2: Find the First Derivative and Evaluate at a

First, find the first derivative of the function:\(f'(x) = 4x^3 + 2x\)Evaluate this derivative at \( x = -2 \):\[f'(-2) = 4(-2)^3 + 2(-2) = 4(-8) - 4 = -32 - 4 = -36\]So, \( f'(-2) = -36 \).

Step 3: Find the Second Derivative and Evaluate at a

Next, find the second derivative of the function:\(f''(x) = 12x^2 + 2\)Evaluate this derivative at \( x = -2 \):\[f''(-2) = 12(-2)^2 + 2 = 12(4) + 2 = 48 + 2 = 50\]So, \( f''(-2) = 50 \).

Step 4: Find the Third Derivative and Evaluate at a

Now, compute the third derivative:\(f'''(x) = 24x\)Evaluate this derivative at \( x = -2 \):\[f'''(-2) = 24(-2) = -48\]So, \( f'''(-2) = -48 \).

Step 5: Find the Fourth Derivative and Evaluate at a

Calculate the fourth derivative:\(f''''(x) = 24\)Since this is a constant, \( f''''(-2) = 24 \) for any \( x \).

Step 6: Construct the Taylor Series at a

The Taylor series for \( f(x) \) about \( x = a \) is given by:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]For this particular function, up to the fourth derivative (since the fifth and higher derivatives are zero), we have:\[f(x) \approx 21 - 36(x + 2) + \frac{50}{2}(x + 2)^2 - \frac{48}{6}(x + 2)^3 + \frac{24}{24}(x + 2)^4\]Simplify the coefficients:\[21 - 36(x+2) + 25(x+2)^2 - 8(x+2)^3 + (x+2)^4\]

Step 7: Write the Final Taylor Series

The Taylor series of the function \( f(x) = x^4 + x^2 + 1 \) around \( x = -2 \) is:\[21 - 36(x+2) + 25(x+2)^2 - 8(x+2)^3 + (x+2)^4\] This expression can be used to approximate \( f(x) \) near \( x = -2 \).

Key Concepts

Derivatives

The concept of derivatives is fundamental to understanding how functions change. A derivative provides the rate of change of a function with respect to a variable. In the context of Taylor series, derivatives are used to capture the behavior of a function near a certain point.
For our exercise, the function is given as:
\[f(x) = x^4 + x^2 + 1\]
We computed four derivatives, starting with the first one:

• First derivative: \(f'(x) = 4x^3 + 2x\)
• Second derivative: \(f''(x) = 12x^2 + 2\)
• Third derivative: \(f'''(x) = 24x\)
• Fourth derivative: \(f''''(x) = 24\)

Derivatives of higher order (beyond the fourth in this case) are zero, making them irrelevant for this particular task of finding the Taylor series to the fourth degree.

Polynomials

Polynomials are expressions composed of variables and coefficients, adding powers of a variable. They have the form:
\[a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\]
In the given exercise, the function \(f(x) = x^4 + x^2 + 1\) is a polynomial of degree 4. It includes terms with non-negative integer exponents.
Polynomials are smooth and continuous, properties that make them ideal for approximation methods like the Taylor series. Understanding the structure of polynomials helps in calculating the series as each term's derivative adds more detail about the function's behavior around the point of interest.
Thus, converting derivatives back into polynomial form is critical in representing those behaviors correctly in the Taylor series.

Mathematical Approximations

Mathematical approximations are strategies to find simpler, accurate representations of more complex functions. The Taylor series offers such approximations by using polynomial expressions to approximate a function close to a specified point.
The Taylor series for a function \(f(x)\) near \(x = a\) is given by:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]
Terms in the Taylor Series:

• The zero-order term gives the value of the function at \(x = a\).
• Higher-order terms incorporate derivatives, which refine the approximation.

In our exercise, up to fourth-order derivatives were used:\[21 - 36(x+2) + 25(x+2)^2 - 8(x+2)^3 + (x+2)^4\]
This is a powerful tool allowing us to approximate the function \(f(x)\) effectively around \(x = -2\). The more derivatives we include, the better the approximation. However, practically there's a limit on how many terms you might include for convenience and precision balance.

Function Evaluation

Evaluating a function means substituting a specific value into a function and calculating the result. This is essential when forming Taylor series, as each derivative needs to be evaluated at the point \(x = a\) to develop the series.
In our task, function evaluation focused on these aspects:

• Calculating \(f(x)\) at \(x = -2\), yielded \(f(-2) = 21\).
• Similarly, each derivative was evaluated: \(f'(-2) = -36\), \(f''(-2) = 50\), \(f'''(-2) = -48\), and \(f''''(-2) = 24\)

Evaluating these derivatives at the point specified gives the coefficients you'll use in the polynomial form of the Taylor series. These steps ensure that the series captures the original function's characteristics accurately around the specific point \(x = -2\).
This evaluation process underlines the precision of Taylor expansions and ensures reliable approximations with the provided function.