Problem 7 - A guide to MATLAB: For Beginners and Experienced Users

Question

Find the Taylor polynomial of the indicated degree \(n\) at the indicated point \(c\) for the following functions: (a) \(f(x)=e^{x}, n=6, c=0\) (b) \(f(x)=\sin x, n=4\) and \(6, c=0\) (c) \(f(x)=\sin x, n=5, c=2\) (d) \(f(x)=\tan x, n=6, c=0\) (e) \(f(x)=\ln x, n=4, c=1\)

Step-by-Step Solution

Answer

(a) Taylor polynomial is \( T_6(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \), (b) For \(n=4, T_4(x)=x-\frac{x^3}{3!}\), \(n=6, T_6(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}\), (c) Compute using derivatives at \(2\), (d) \( T(x) = x + \frac{x^3}{3} + \ldots \) (e) \( T_4(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} \).

Step 1: Understanding the Problem

We need to find the Taylor polynomial for different functions at specified degrees and centers. Problems are broken down into separate parts for various functions and conditions: exponential, trigonometric, tangent, and logarithmic.

Step 1: Exponential Function Taylor Series

The Taylor series for the function \(f(x) = e^x\) at \(c = 0\) is given by \( T_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!} \). For \(n = 6\), the Taylor polynomial is \( T_6(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \).

Step 2: Sine Function Taylor Series at Different Degrees

The Taylor series for \(f(x) = \sin x\) at \(c = 0\) is \( T_n(x) = \sum_{k=0}^{n} (-1)^k \frac{x^{2k+1}}{(2k+1)!} \). For \(n = 4\), it is \( x - \frac{x^3}{3!} \). For \(n = 6\), it is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} \).

Step 3: Sine Function Taylor Series at Non-zero Center

To find the Taylor polynomial of \(f(x) = \sin x\) for \(n = 5\) at \(c = 2\), calculate the derivatives of \(\sin x\) and evaluate them at \(x = 2\). First few derivatives are: \(f'(x) = \cos x\), \(f''(x) = -\sin x\), \(f'''(x) = -\cos x\), etc. Plug the derivatives into the Taylor series formula.

Step 4: Tangent Function Taylor Series

The Taylor series for \(f(x) = \tan x\) at \(c = 0\) involves computing derivatives manually as formulas are cumbersome. Derive the Taylor polynomial by calculating the necessary derivatives up to \(x^6\) and evaluating them at 0. This generally results in a series starting with \(x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots\) where complex higher-order derivatives are involved.

Step 5: Logarithmic Function Taylor Series

For \(f(x) = \ln x\) at \(c = 1\), the Taylor series is given by \(\ln(1 + (x-1)) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} \) for \(n = 4\). Extend derivatives up to fourth order using this centered at 1.

Key Concepts

Exponential Function

The exponential function, denoted as \( f(x) = e^x \), is one of the fundamental mathematical functions with numerous applications. In the context of Taylor polynomials, expanding this function at the center \( c = 0 \) makes use of the formula:

\[T_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots + \frac{x^n}{n!}\]
For a 6th-degree polynomial, or \( n = 6 \), the series covers terms up to \( x^6 \). This series is essentially a polynomial representation of \( e^x \) providing an approximation that becomes more accurate as more terms are included.

• For \( n = 6 \), the Taylor polynomial is:
  \( T_6(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} \)

Understanding how to compute each term helps in recognizing patterns and calculating derivatives more easily.

Sine Function

The sine function, written as \( f(x) = \sin x \), is a periodic trigonometric function. For the Taylor series at \( c = 0 \), the formula is slightly more complex as it involved alternating series:

\[T_n(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\]
For different degrees:

• At \( n = 4 \): The approximation is \( x - \frac{x^3}{6} \).
• At \( n = 6 \): It is extended to \( x - \frac{x^3}{6} + \frac{x^5}{120} \).

The alternating pattern (positive and negative) arises from the nature of sine function's derivatives, which also alternate in sign. When centering not at zero, like \( n = 5 \) at \( c = 2 \), calculations involve evaluating various derivatives of sine at \( x = 2 \). This can become quite intricate due to the repetitive derivative of sine and cosine functions.

Tangent Function

The tangent function \( f(x) = \tan x \) has a Taylor series which is more complex to derive. It is not as straightforward as other trigonometric functions due to its repetitive increase and asymptotic behavior near its vertical asymptotes. At \( c = 0 \), we need to manually compute derivatives up to the 6th degree. The series approximates as:

• \( T(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots \)

Calculating the tangent Taylor polynomial involves finding higher-order derivatives of \( \tan x \), which can be quite complicated. These derivatives include combinations of \( \sec^2 x \) and \( \sec^4 x \), which are derived from the primary definition \( \frac{d}{dx}(\tan x) = \sec^2 x \).

Logarithmic Function

The natural logarithm function, \( f(x) = \ln x \), is another essential function in mathematics. Its Taylor series when centered at \( c = 1 \) provides a polynomial approximation:

\[T(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4}\]

• This series is valid for \( \ln(1 + (x-1)) \), which simplifies the function to be centered around 1.
• It also represents how the logarithm behaves when \( x \) is close to 1, providing useful approximations in many practical applications.

Although logarithmic functions are continuous and smooth, Taylor polynomials offer a way to simplify calculations over short ranges.