Question

Assume that the Maclaurin series converges to the function. Show that \(1-\cos x=x^{2} / 2\) with an error less than 0.003 for \(|x|<\frac{1}{2}\).

Short answer

The error is less than \[0.003\] for \[ |x| < \frac{1}{2} \], so \[ 1 - \cos x \approx \frac{x^2}{2} \].

Step-by-step solution

Recall the Maclaurin Series for Cosine

The Maclaurin series for \(\cos x\) is given by \[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]

Derive the Expression for 1 - Cos x

Subtract the Maclaurin series for \(\backslash\backslash\text{cos } x\) from 1:\[ 1 - \cos x = 1 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots \right) = \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720} - \cdots \]

Estimate the Error

Consider the first term not included in the approximation \[ \frac{x^4}{24} \]. For \[ |x| < \frac{1}{2} \], \(\backslash x \backslash\backslash \) is at most \(\backslash \backslash 0.5 \backslash\backslash \). Calculate the magnitude of this term as follows:\[ \left| \frac{(0.5)^4}{24} \right| = \frac{0.0625}{24} = 0.0026 \]

Verify the Error

The remaining terms of the series contribute an error less than \[ 0.003 \] when \[ |x| < \frac{1}{2} \]. Thus, \[ 1 - \cos x \approx \frac{x^{2}}{2} \] with the required accuracy.

Key concepts

Cosine Function

The cosine function, denoted as \( \cos x \), is a crucial trigonometric function used in mathematics to describe the relationship between the angles and sides of a right-angled triangle.

It is a periodic function, meaning it repeats its values in regular intervals or periods. The value of \( \cos x \) oscillates between -1 and 1.

Here's a quick look at the fundamental properties of the cosine function:

• The cosine of 0 degrees (or 0 radians) is 1, i.e., \( \cos(0) = 1 \).
• It has a period of \( 2\backslashpi \), meaning \(\cos(x + 2\backslashpi) = \cos(x) \)
• The cosine function is an even function, which means it satisfies the equation \(\cos(-x) = \cos(x) \).

Understanding how to expand \(\cos x \) using a series helps in calculating its values for small angles accurately. The Maclaurin series is one such method.

Error Approximation

In mathematical analysis, error approximation is a critical concept. It helps us understand how far an approximated value is from the true value.

When you use series expansions like the Taylor or Maclaurin series to approximate functions, the difference between the actual function value and the series value is the error.

Here’s a closer look at the error when approximating functions using series:

• The error term is often denoted by R_n (remainder term) in the series.
• For the Maclaurin series of \(\cos x\), the error after including the first two terms is \( \float{x^4}{24} \), which is small for small values of \( x \).
• In our example, with \( \frac{1}{2} \) as the upper limit for \( |x| \, the error \left| \frac{(0.5)^4}{24} \right| = 0.0026 \) is less than the desired threshold of 0.003.

This concept ensures that series approximations are practical for real-world applications.

Taylor Series

The Taylor series is a powerful tool in mathematics for approximating functions. A Taylor series expands a function into an infinite sum of terms calculated from its derivatives at a single point. When the point is zero, it’s called a Maclaurin series.

The general formula for the Taylor series of a function \( f(x) \) around the point \( a \) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] For the Maclaurin series (where \( a = 0 \)), it simplifies to:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] For the cosine function, we use its Maclaurin series expansion:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots \] By understanding and applying the Taylor series, you can approximate complex functions easily. This method provides insights into how functions behave near specific points, thereby aiding in error minimization and efficient computation.