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To find the Taylor polynomial for a function, you need to understand derivatives. Derivatives represent the rate at which a function changes. They're essential for constructing a Taylor polynomial, as each derivative contributes a term to the polynomial.
For the function \( f(x) = x^4 \), the first four derivatives are crucial. Let's break them down:

• First derivative (\( f'(x) \)): This measures the slope or rate of change of the function at any point. For \( f(x) = x^4 \), the first derivative is \( f'(x) = 4x^3 \).
• Second derivative (\( f''(x) \)): It tells us how the rate of change itself is changing. Here, it is \( f''(x) = 12x^2 \).
• Third derivative (\( f'''(x) \)): It can be thought of as the rate at which the second derivative changes. This result is \( f'''(x) = 24x \).
• Fourth derivative (\( f^{(4)}(x) \)): It shows the rate of change of the third derivative. The outcome is a constant \( f^{(4)}(x) = 24 \).

Interestingly, because \( x^4 \) is a polynomial of degree 4, all higher-order derivatives (5th, 6th, etc.) are zero. Understanding these derivatives helps build the Taylor polynomial.

The degree of a polynomial is a fundamental concept in understanding Taylor polynomials. It's the highest power of the variable in a polynomial expression. For example, in the function \( f(x) = x^4 \), the highest power is 4. Thus, it is a degree 4 polynomial. Why is knowing the polynomial degree important?

• It tells us how many derivatives we need to check when crafting a Taylor polynomial. For \( x^4 \), it’s up to the 4th derivative.
• Polynomials of degree \( n \) can be perfectly represented by a Taylor polynomial of order \( n \).

Understanding the polynomial degree is vital for ensuring your Taylor polynomial accurately reflects the original function, especially since each term's importance diminishes as the order exceeds the original polynomial's degree.

Constructing a Taylor polynomial involves several calculus steps that ensure the polynomial closely approximates the original function. Let’s look at this process using our example function, \( f(x) = x^4 \), centered at \( x = 2 \):Firstly, evaluate the derivatives at the base point (\( x = 2 \)). This gives us specific values:

• \( f(2) = 16 \)
• \( f'(2) = 32 \)
• \( f''(2) = 48 \)
• \( f'''(2) = 48 \)
• \( f^{(4)}(2) = 24 \)

The calculus formula for a Taylor polynomial is:\[T_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]Plugging in the relevant values for each term and simplifying, we achieve:\[T_4(x) = 16 + 32(x - 2) + 24(x-2)^2 + 8(x-2)^3 + 1(x-2)^4\]In this context, understanding each calculus step is essential for confirming that the Taylor polynomial mirrors the original function. Finally, note that once expanded, \( T_4(x) \) should perfectly match the original polynomial, \( f(x) = x^4 \), proving its exact representation.