Problem 7 - Active Calculus

Question

The examples we have considered in this section have all been for Taylor polynomials and series centered at 0 , but Taylor polynomials and series can be centered at any value of \(a\). We look at examples of such Taylor polynomials in this exercise. a. Let \(f(x)=\sin (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=\frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(x=\frac{\pi}{2}\). Why should you have expected the result? b. Let \(f(x)=\ln (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=1\). Then find the Taylor series for \(f(x)\) centered at \(x=1\).

Step-by-Step Solution

Answer

For\(\sin(x)\)centered at \(\frac{\pi}{2}\):\(1 -\frac{(x - \frac{\pi}{2})^2}{2} +\frac{(x - \frac{\pi}{2})^4}{24}\). For \(\ln(x)\)centered at 1: \( (x-1) -\frac{(x-1)^2}{2} +\frac{(x-1)^3}{3} -\frac{(x-1)^4}{4}\)

Step 1: Define the function and derivatives

For part (a), let the function be defined as: \[ f(x) = \sin(x) \]. The derivatives up to the fourth order are: - First derivative: \[ f'(x) = \cos(x) \] - Second derivative: \[ f''(x) = -\sin(x) \] - Third derivative: \[ f'''(x) = -\cos(x) \] - Fourth derivative: \[ f^{(4)}(x) = \sin(x) \].

Step 2: Evaluate the derivatives at the center point

The center point for the Taylor series is at \( x = \frac{\pi}{2} \). Evaluating the derivatives at this point, we get: - \[ f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \] - \[ f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \] - \[ f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1 \] - \[ f'''(\frac{\pi}{2}) = -\cos(\frac{\pi}{2}) = 0 \] - \[ f^{(4)}(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \].

Step 3: Construct the Taylor polynomial up to the fourth order for \(f\)

Using the evaluated derivatives, the Taylor polynomial up to the fourth order centered at \( x = \frac{\pi}{2} \) is: \[ P_4(x) = f(\frac{\pi}{2}) + f'(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})}{1!} + f''(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^2}{2!} + f'''(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^3}{3!} + f^{(4)}(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^4}{4!} \]. Substituting the values, we obtain: \[ P_4(x) = 1 + 0 \cdot (x - \frac{\pi}{2}) - \frac{(x - \frac{\pi}{2})^2}{2} + 0 \cdot \frac{(x - \frac{\pi}{2})^3}{6} + \frac{(x - \frac{\pi}{2})^4}{24} \]. Simplified, this gives: \[ P_4(x) = 1 - \frac{(x - \frac{\pi}{2})^2}{2} + \frac{(x - \frac{\pi}{2})^4}{24} \].

Step 4: Identify the Taylor series for \(f(x) = \sin(x)\) centered at \( x = \frac{\pi}{2} \)

The pattern found in the polynomial suggests a general Taylor series of: \[ \sum_{n=0}^{\infty} (-1)^n \frac{(x - \frac{\pi}{2})^{2n}}{(2n)!} \]. This follows because all odd-power terms are zero and the even-power terms alternate in sign and are divided by the factorial of the order.

Step 5: Define the function and derivatives for part (b)

For part (b), let the function be defined as: \[ f(x) = \ln(x) \]. The derivatives up to the fourth order are: - First derivative: \[ f'(x) = \frac{1}{x} \]. - Second derivative: \[ f''(x) = -\frac{1}{x^2} \]. - Third derivative: \[ f'''(x) = 2\frac{1}{x^3} \]. - Fourth derivative: \[ f^{(4)}(x) = -6\frac{1}{x^4} \].

Step 6: Evaluate the derivatives at the center point for part (b)

The center point for the Taylor series is at \( x = 1 \). Evaluating the derivatives at this point, we get: - \[ f(1) = \ln(1) = 0 \]. - \[ f'(1) = \frac{1}{1} = 1 \]. - \[ f''(1) = -\frac{1}{1^2} = -1 \]. - \[ f'''(1) = 2\frac{1}{1^3} = 2 \]. - \[ f^{(4)}(1) = -6\frac{1}{1^4} = -6 \].

Step 7: Construct the Taylor polynomial up to the fourth order for \(f\)

Using the evaluated derivatives, the Taylor polynomial up to the fourth order centered at \( x = 1 \) is: \[ P_4(x) = f(1) + f'(1) \frac{(x - 1)}{1!} + f''(1) \frac{(x - 1)^2}{2!} + f'''(1) \frac{(x - 1)^3}{3!} + f^{(4)}(1) \frac{(x - 1)^4}{4!} \]. Substituting the values, we obtain: \[ P_4(x) = 0 + 1 \cdot (x - 1) - \frac{(x - 1)^2}{2} + 2 \cdot \frac{(x - 1)^3}{6} - 6 \cdot \frac{(x - 1)^4}{24} \]. Simplified, this gives: \[ P_4(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} \].

Step 8: Identify the Taylor series for \(f(x) = \ln(x)\) centered at \( x = 1 \)

The pattern found in the polynomial suggests a general Taylor series of: \[ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 1)^n}{n} \].

Key Concepts

Taylor polynomials

Taylor polynomials are approximations of functions using polynomials. They are particularly useful for estimating the value of a function near a specific point. A Taylor polynomial of order n centered at a point a for a function f(x) can be expressed as:











































































































































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The Taylor polynomial helps give a good local approximation for functions where calculating the exact value may be complicated.

Taylor series

A Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. While a Taylor polynomial is a finite approximation, the Taylor series extends this to an infinite number of terms for potentially better accuracy. For a function f(x) centered at point a, the Taylor series is:































































The main advantage of the Taylor series is its ability to represent complex functions with simpler polynomial expressions. This makes it very useful in various fields like physics and engineering.

• Approximation Accuracy: The more terms you add from the series, the closer you get to the actual function.
• Summing Up: Taylor series often sum up to give a good representation of a wide range of functions.

Function derivatives

Derivatives measure how a function changes as its input changes. They are an essential part of calculus. Derivatives provide the rates at which functions increase or decrease. The nth-order derivative provides information on the nth rate of change. For a function f(x), the first derivative, denoted as f'(x), represents its rate of change. Higher-order derivatives like the second (f''(x)), third (f'''(x)), and so on provide deeper insights.



















































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Derivatives are essential in constructing Taylor polynomials and series. They form the coefficients of the terms in these polynomials and series, giving an accurate representation of the original function.

• First Derivative: Tells us about the slope or the rate of change.
• Second Derivative: Provides information on the concavity and acceleration of the function.
• Higher-order Derivatives: Offer deeper insights into the behavior of functions beyond mere slope or acceleration.