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

The Taylor polynomials at \(a = \frac{\pi}{4}\) are: - Order 0: \(T_0(x) = \frac{\sqrt{2}}{2}\) - Order 1: \(T_1(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4})\) - Order 2: \(T_2(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2\) - Order 3: \(T_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} (x - \frac{\pi}{4})^3\)

Step 1: Evaluate the Function at the Given Point

First, determine the value of the function at the point \(a = \frac{\pi}{4}\). For \(f(x) = \sin x\), we have:\[f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.\]

Step 2: Find the First Derivative and Evaluate at the Point

Compute the first derivative of the function and evaluate it at \(a = \frac{\pi}{4}\). The derivative is:\[f'(x) = \cos x.\]Therefore:\[f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.\]

Step 3: Find the Second Derivative and Evaluate at the Point

Compute the second derivative of the function and evaluate it at \(a = \frac{\pi}{4}\). The second derivative is:\[f''(x) = -\sin x.\]Thus, we have:\[f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}.\]

Step 4: Find the Third Derivative and Evaluate at the Point

Compute the third derivative and evaluate it at \(a = \frac{\pi}{4}\). The third derivative is:\[f'''(x) = -\cos x.\]Therefore:\[f'''\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}.\]

Step 5: Construct the Taylor Polynomial of Order 0

The Taylor polynomial of order 0 is simply the function evaluated at the point \(a = \frac{\pi}{4}\):\[T_0(x) = f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.\]

Step 6: Construct the Taylor Polynomial of Order 1

The Taylor polynomial of order 1 includes the function value and the first derivative term:\[T_1(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}).\]

Step 7: Construct the Taylor Polynomial of Order 2

The Taylor polynomial of order 2 includes up to the second derivative:\[T_2(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2.\]

Step 8: Construct the Taylor Polynomial of Order 3

The Taylor polynomial of order 3 includes up to the third derivative term:\[T_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} (x - \frac{\pi}{4})^3.\]

Key Concepts

Taylor polynomial

Taylor polynomials are approximations of functions near a certain point. These approximations are made using derivatives. Regardless of the function, the goal is to create a polynomial expression that closely represents the function's behavior around a particular point. This point is often denoted as \(a\). The more derivatives we use, the more accurate the approximation becomes. For a Taylor polynomial of order \(n\), we consider derivatives up to the \(n\)-th order.

The process starts by evaluating the function at a specific point. For example, with \(f(x) = \sin x\) at \(a = \pi/4\), the zero-order polynomial is simply \(T_0(x) = f(a)\). As we increase the order of the polynomial, we add terms involving the function's derivatives. This is why the polynomial expands into an expression like \(T_n(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + ... + c_n(x-a)^n\), where each \(c_i\) is related to derivatives of the function at \(a\).

derivatives

Derivatives are central to Taylor polynomials. They provide information on how a function changes at a certain point. The first derivative, often noted as \(f'(x)\), gives us the slope of the tangent line to the function at a point. It tells us how much the function is increasing or decreasing. Evaluating it at \(a = \pi/4\) provides one of the coefficients of the polynomial, helping determine the linear component.

Each successive derivative adds more information. The second derivative, \(f''(x)\), reveals the curvature of the function. It's essentially a way to see how the slope itself changes. When constructing a Taylor polynomial, each derivative lets us add a term that makes our polynomial fit the function’s shape more closely. For \(f(x) = \sin x\), derivatives evaluated at \(a = \pi/4\) are used step-by-step to construct polynomial terms, up to the desired accuracy level.

order of polynomials

The order of a Taylor polynomial determines how many terms are included in the polynomial from our expansion process. A higher order typically means a more accurate approximation. Starting with order zero, we just look at the function’s value at point \(a\). For a first-order Taylor polynomial, we include a term based on the first derivative, resulting in a linear approximation.

As we increase to second-order, we incorporate the second derivative, allowing us to capture the function's curvature. Third-order polynomials consider up to the third derivative, adding further precision. Essentially, the order tells us how 'in-depth' our polynomial reaches down the derivative chain. With more terms, the polynomial becomes more adaptable to replicating the behavior of the original function in the vicinity of \(a\). This makes Taylor polynomials powerful tools for approximation in mathematical analysis.