Question

a. Find the nth-order Taylor polynomials for the given function centered at the given point a, for \(n=0,1,\) and 2 b. Graph the Taylor polynomials and the function. $$f(x)=\sin x, a=\pi / 4$$

Short answer

Question: Find the nth-order Taylor polynomials for the function \(f(x) = \sin x\), centered at \(a = \frac{\pi}{4}\) for \(n = 0,\) 1, and 2. Then, graph the Taylor polynomials and the function. Answer: 0th-order Taylor polynomial: \(P_0(x) = \frac{1}{\sqrt{2}}\) 1st-order Taylor polynomial: \(P_1(x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right)\) 2nd-order Taylor polynomial: \(P_2(x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right) - \frac{1}{2}\left(x-\frac{\pi}{4}\right)^2\) To graph the Taylor polynomials and the function, use graphing software like Desmos or GeoGebra to enter the function \(f(x) = \sin x\) and the Taylor polynomials \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\) that we found. Compare how well these polynomials approximate the original function around \(a = \frac{\pi}{4}\).

Step-by-step solution

1. Compute the derivatives

We will compute the derivatives of \(f(x) = \sin x\) up to order 2 and evaluate them at \(a = \frac{\pi}{4}\). First derivative: \(f'(x) = \cos x\) $$f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$ Second derivative: \(f''(x) = -\sin x\) $$f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$$

2. Construct the Taylor polynomials

To construct the nth-order Taylor polynomial, we will use the general formula for the Taylor series centered at \(a\): $$P_n(x)= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n$$ For our function and given values of \(n\), the Taylor polynomials will be: 0th-order Taylor polynomial: \(P_0(x) = f\left(\frac{\pi}{4}\right)\) $$P_0(x) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$ 1st-order Taylor polynomial: \(P_1(x) = P_0(x) + f'\left(\frac{\pi}{4}\right)(x-\frac{\pi}{4})\) $$P_1(x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right)$$ 2nd-order Taylor polynomial: \(P_2(x) = P_1(x) + \frac{f''\left(\frac{\pi}{4}\right)}{2!}(x-\frac{\pi}{4})^2\) $$P_2(x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right) - \frac{1}{2}\left(x-\frac{\pi}{4}\right)^2$$

3. Graph the Taylor polynomials and the function

To graph the Taylor polynomials and the function, you can use graphing software like Desmos or GeoGebra. First, enter the function \(f(x) = \sin x\), then enter the Taylor polynomials \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\) that we just found. You can compare how well these polynomials approximate the original function around \(a = \frac{\pi}{4}\). Note: In this text-based format, we can't directly include graphs. You need to plot them separately using graphing software.

Key concepts

Taylor Series

A Taylor Series is a powerful tool in mathematics. It's used to approximate complicated functions with polynomials. These series are constructed from the function's derivatives at a single point, known as the center of the series. For example, in our case scenario with the function \(f(x) = \sin x\), the Taylor series is centered at \(a = \frac{\pi}{4}\). This means we're using derivatives of \(\sin x\) at \(\frac{\pi}{4}\) to create our polynomial approximations.
A Taylor polynomial of order \(n\) is a finite sum of terms up to \(\frac{f^{(n)}(a)}{n!}(x-a)^n\). These terms give us successively better approximations of the function. The more terms we include, the closer the Taylor polynomial's graph resembles the function's graph around the center point. Hence, each added term enhances the approximation.
In practice, you'll often use a Taylor polynomial rather than an infinite Taylor series because it's a simpler representation while still providing valuable approximation insights.

Derivative Calculation

Derivatives are fundamental when constructing Taylor Series. They measure how a function changes as its input changes, and this information feeds directly into the computation of Taylor polynomials. Each derivative corresponds to a coefficient in the polynomial, scaled appropriately by factorials.
In our solution, we found the first and second derivatives of the function \(f(x) = \sin x\). Here's how we did it:

• First derivative: \(f'(x) = \cos x\). At \(x = \frac{\pi}{4}\), \(f'(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\).
• Second derivative: \(f''(x) = -\sin x\). At \(x = \frac{\pi}{4}\), \(f''(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\).

Taking these derivatives correctly is crucial for creating an accurate Taylor polynomial. Missteps in derivative calculations can lead to erroneous polynomial approximations, so be precise and systematic.

Graphing Functions

Graphing shows us how well our Taylor polynomials approximate the original function. By plotting both the function and its Taylor polynomials, you can visualize how similar their paths are.With graphing software like Desmos or GeoGebra, you start by entering the function \(f(x) = \sin x\). Next, input each Taylor polynomial:

• 0th-order to 2nd-order - \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\).

By comparing these graphs, you can see how each Taylor polynomial increasingly matches the function's curve around \(x = \frac{\pi}{4}\). This visual representation strengthens your understanding of the approximation process.
Seeing the graphs together allows you to notice how the higher-order polynomials (with more terms) yield better approximations over a broader range around the center point. This method is a practical way to grasp the concept and utility of Taylor Series in approximating complex functions.