Problem 7 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=\sin x, \quad a=\pi / 4\)

Step-by-Step Solution

Answer

Taylor polynomials: - Order 0: \( P_0(x) = \frac{\sqrt{2}}{2} \) - Order 1: \( P_1(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) \) - Order 2: \( P_2(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 \) - Order 3: \( P_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 \)

Step 1: Find the Function and Its Derivatives

Start with the function \( f(x) = \sin x \). To find the Taylor polynomials, we need the derivatives of \( f(x) \) evaluated at \( x=a \) which is \( a = \frac{\pi}{4} \). Calculate the first few derivatives: - \( f(x) = \sin x \) - \( f'(x) = \cos x \) - \( f''(x) = -\sin x \) - \( f'''(x) = -\cos x \)

Step 2: Evaluate the Derivatives at \(a = \frac{\pi}{4}\)

Substitute \( x = \frac{\pi}{4} \) into each derivative: - \( f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)- \( f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)- \( f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \)- \( f'''\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \)

Step 3: Taylor Polynomial of Order 0

The 0th-order Taylor polynomial, \( P_0(x) \), is simply the function value at \( a \):\[ P_0(x) = \frac{\sqrt{2}}{2} \]

Step 4: Taylor Polynomial of Order 1

Include the first derivative in the polynomial:\[ P_1(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) \]

Step 5: Taylor Polynomial of Order 2

Add the second derivative term:\[ P_2(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 \]

Step 6: Taylor Polynomial of Order 3

Include the third derivative term:\[ P_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 \]

Key Concepts

Derivatives

When working with Taylor polynomials, one of the first steps is to calculate the derivatives of the function in question. A derivative provides the rate at which a function changes. In other words, it tells us how a function behaves as its input changes.

To find a Taylor polynomial for a given function like \( f(x) = \sin x \), the derivatives of that function are needed at a specific point, here given as \( a = \frac{\pi}{4} \).
Examples of derivatives include:

• First derivative \( f'(x) = \cos x \): This derivative shows the instantaneous rate of change, or slope, of \( \sin x \) at any point \( x \).
• Second derivative \( f''(x) = -\sin x \): Represents the rate of change of the rate of change of \( \sin x \). It helps understand the concavity of the function.
• Third derivative \( f'''(x) = -\cos x \): Continues to describe the behavior of \( \sin x \) as one analyzes changes further.

Each successive derivative gives deeper insight into the function's behavior, which is crucial for accurately constructing Taylor polynomials.

Function Evaluation

Once derivatives are calculated, the next step in constructing Taylor polynomials involves evaluating these derivatives at the given point \( a = \frac{\pi}{4} \). Functional evaluation is necessary to anchor our Taylor polynomial accurately on the specific point of interest.

At \( x = \frac{\pi}{4} \), each derivative of the function \( \sin x \) is substituted:

• \( f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \): The value of the function itself.
• \( f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \): The slope of \( \sin x \) at \( \frac{\pi}{4} \).
• \( f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \): Concavity measurement at the same point.
• \( f'''\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \): Further insights on how fast the concavity is changing.

These evaluations are crucial for establishing points in the approximation of the Taylor polynomial.

sin x

The function \( \sin x \) is periodic and oscillates between -1 and 1. It is one of the fundamental trigonometric functions commonly encountered in mathematics.

In the Taylor series context, \( \sin x \) can be approximated around a point using its derivatives. This approximation becomes more accurate as more terms are added.

• The angle \( x = \frac{\pi}{4} \) results in some recognizable values like \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
• These values allow easier calculation and provide a point of reference around which the function behaves predictably.

Understanding how \( \sin x \) works, especially at special points like \( \frac{\pi}{4} \), helps in creating effective Taylor polynomials.

cos x

The function \( \cos x \) completes the trigonometric duo alongside \( \sin x \). It, too, is periodic with a similar oscillation pattern, albeit shifted from \( \sin x \).

In generating Taylor polynomials, \( \cos x \) plays a vital role as it represents the first derivative of \( \sin x \). Each derivative sequence will cycle through \( \sin x \) and \( \cos x \) and their negatives:

• \( \cos x \) gives the slope of \( \sin x \) — crucial for Taylor polynomial terms.
• It has its own unique point values with \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), similar to \( \sin x \) at \( \frac{\pi}{4} \).

Grasping \( \cos x \) allows deeper insight into creating accurate approximations through its interaction with \( \sin x \).