Problem 24 - Calculus

Question

Find a simplified formula for \(P_{5}(x),\) the fifth-degree Taylor polynomial approximating \(f\) near \(x=0\). Let \(f(0)=-1\) and, for \(n > 0, f^{(n)}(0)=-(-2)^{n}\)

Step-by-Step Solution

Answer

\( P_5(x) = -1 + 2x - 2x^2 + \frac{4}{3}x^3 - \frac{2}{3}x^4 + \frac{4}{15}x^5 \)

Step 1: Understanding Taylor Polynomials

A Taylor polynomial of degree \( n \) for a function \( f(x) \) centered at \( x = a \) is given by: \[ P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]. Since we're finding the Taylor polynomial around \( x = 0 \), our \( a = 0 \) and the formula simplifies in terms of \( f(0), f'(0), \ldots, f^{(5)}(0) \).

Step 2: Calculating the Necessary Derivatives

We need \( f(0), f'(0), f''(0), f'''(0), f^{(4)}(0), \) and \( f^{(5)}(0) \). According to the problem:- \( f(0) = -1 \)- \( f^{(n)}(0) = -(-2)^{n} \) for \( n > 0 \). Let's compute the values: - \( f'(0) = -(-2)^1 = 2 \) - \( f''(0) = -(-2)^2 = -4 \) - \( f'''(0) = -(-2)^3 = 8 \) - \( f^{(4)}(0) = -(-2)^4 = -16 \) - \( f^{(5)}(0) = -(-2)^5 = 32 \).

Step 3: Constructing the Taylor Polynomial

Using the values calculated for the derivatives, we construct \( P_5(x) \):\[P_5(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5\]Substituting the derivative values:\[P_5(x) = -1 + \frac{2}{1!}x - \frac{4}{2!}x^2 + \frac{8}{3!}x^3 - \frac{16}{4!}x^4 + \frac{32}{5!}x^5\]

Step 4: Simplifying the Expression

Simplify each term:- \( \frac{2}{1!} = 2 \)- \( \frac{4}{2!} = 2 \)- \( \frac{8}{3!} = \frac{8}{6} = \frac{4}{3} \)- \( \frac{16}{4!} = \frac{16}{24} = \frac{2}{3} \)- \( \frac{32}{5!} = \frac{32}{120} = \frac{4}{15} \)Thus, the simplified polynomial is:\[P_5(x) = -1 + 2x - 2x^2 + \frac{4}{3}x^3 - \frac{2}{3}x^4 + \frac{4}{15}x^5\]

Step 5: Verifying the Simplified Formula

Double-check each calculation to ensure correctness in simplification:- Calculations for derivatives were verified with the formula \( f^{(n)}(0) = -(-2)^{n} \).- Simplifications of terms were done radially by dividing correctly and confirmed.Now, our final 5th-degree Taylor polynomial is:\[P_5(x) = -1 + 2x - 2x^2 + \frac{4}{3}x^3 - \frac{2}{3}x^4 + \frac{4}{15}x^5\]

Step 6: Conclusion

After verifying calculations and substitutions, the Taylor polynomial \( P_5(x) \) is accurately represented as:\[P_5(x) = -1 + 2x - 2x^2 + \frac{4}{3}x^3 - \frac{2}{3}x^4 + \frac{4}{15}x^5\]

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand how things change. It is fundamentally about two concepts: differentiation and integration. Differentiation involves finding the rate at which something changes, while integration finds the total accumulation of a quantity.
Understanding these concepts in calculus is crucial for solving many real-life problems, such as finding the speed of a car at any point in time or the amount of fuel consumed over a journey.

• Limits: In calculus, limits help us get closer and closer to a particular value. They are essential for defining derivatives and integrals.
• Derivatives: These are used to determine the rate of change for a function. For example, how fast a car accelerates.
• Integrals: These are used to find areas under curves or the total accumulation of quantities over a region.

Calculus provides the foundation for understanding more advanced mathematical concepts daily, including the idea behind Taylor Polynomials.

Taylor Series

The Taylor Series is a mathematical tool used to approximate functions. It represents a function as an infinite sum of terms derived from its derivatives at a single point. This can be particularly helpful for calculations when dealing with complex functions that are difficult to evaluate directly.
The Taylor Series takes the form:
\[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]
where \(a\) is the point around which the series is centered. By choosing a suitable point and using a finite number of terms, we get what's called a Taylor Polynomial.

• Convergence: Not all Taylor series converge to the actual function value. It's important to consider the interval in which they are valid.
• Application: Taylor series are widely used in science and engineering, especially for approximating functions that are difficult to solve analytically.

The Taylor Polynomial is specifically useful for providing polynomial approximations of functions, which are easier to handle computationally.

Polynomial Approximation

Polynomial approximation involves estimating a function with a polynomial. This approach simplifies complex operations because polynomials are easy to handle, differentiate, and integrate. The goal of this approximation is to represent functions that are complicated, such as trigonometric or exponential functions, with easier-to-manage polynomial forms.
In mathematical notation, a function \(f(x)\) is approximated around a point \(x=a\) by a polynomial \(P_n(x)\) of degree \(n\). This is precisely what a Taylor polynomial does. By using a few initial terms of the Taylor series, we achieve a polynomial approximation, which is especially beneficial for calculations without computing infinite series.

• Utility: Polynomial approximations are useful in computations for computers because they reduce complex formulas to simpler polynomial operations.
• Limitations: While polynomial approximations are powerful, they may lose accuracy outside specific domains or intervals of convergence.

With polynomial approximation, we balance simplicity and precision, making it a reliable method in applied mathematics.

Function Derivatives

Function derivatives are a cornerstone of calculus and are essential in understanding how a function behaves with respect to changes in its input. Derivatives tell us the slope of the function at any given point, offering valuable insights into the function's kneeling behavior.
To compute the derivatives, you follow a straightforward rule of finding the rate of change. Each derivative provides information on turning points and inflection points in a function:

• First Derivative: This tells you the slope, or how steep the graph is at that point.
• Second Derivative: This indicates the concavity of the function, helping to determine whether the function is curving upwards or downwards.

In the context of Taylor Polynomials, derivatives are used to find coefficients for the polynomial terms of an approximated function. For example, with \( f^{(n)}(0) = -(-2)^n \), we compute various orders of derivatives at zero to build our Taylor polynomial. By understanding derivatives, you gain more insight into not just the function's current state but also its future direction and tendency.