Question

Use the remainder term to estimate the maximum error in the following approximations on the given interval. Error bounds are not unique. $$\sin x \approx x-x^{3} / 6 ;[-\pi / 4, \pi / 4]$$

Short answer

The maximum error in the approximation is approximately 0.000982.

Step-by-step solution

Recall Taylor's Inequality

Taylor's Inequality states that, if $$|f^{(n + 1)} (x)| \leq M$$ for all x in the interval [a, b], then the error E(x) of the nth degree Taylor polynomial approximation of the function f(x) satisfies: $$|E(x)| \leq \frac{M |x - a|^{n + 1}}{(n + 1)!}$$ For our problem, the nth degree Taylor polynomial approximation is $$P_n(x) = x - \frac{x^3}{6}$$ for the function $$f(x) = \sin x$$ and we are working with the interval $$[-\pi/4, \pi/4]$$. Now we just need to find the maximum value of the (n + 1)th derivative to plug into the inequality.

Find the Fourth Derivative of sin(x)

Since the given polynomial is of degree 3, we need to find the maximum value of the fourth derivative of $$\sin x$$ in the given interval which will be used in Taylor's Inequality. 1st Derivative: $$f'(x) = \cos x$$ 2nd Derivative: $$f''(x) = -\sin x$$ 3rd Derivative: $$f^{(3)}(x) = -\cos x$$ 4th Derivative: $$f^{(4)}(x) = \sin x$$

Find the Maximum Value of the Fourth Derivative

We now need to find the maximum value of the fourth derivative of $$\sin x$$ on the interval $$[-\pi/4, \pi/4]$$. This can be done by finding the absolute maximum of $$|\sin x|$$ within the given interval. Since $$-1 \leq \sin x \leq 1$$, the maximum absolute value of $$\sin x$$ in this interval is 1.

Apply Taylor's Inequality

Now that we have all the necessary information, let's apply Taylor's Inequality to estimate the maximum error: $$|E(x)| \leq \frac{1 * |x - 0|^4}{(3+1)!} = \frac{|x|^4}{24}$$ Since the interval is symmetric about zero and E(x) is an even function, we only need to find the maximum error for the positive half of the interval, which will also be the maximum error for the entire interval $$[-\pi/4, \pi/4]$$. Now we have to find the maximum value of $$\frac{|x|^4}{24}$$ in the interval $$[0, \pi/4]$$.

Find the Maximum Error

We need to find the maximum value of $$\frac{x^4}{24}$$ for x in the interval $$[0, \pi/4]$$. We can achieve this by finding the derivative of the function and analyzing its critical points, but as the function is increasing in this interval, the maximum value occurs at the endpoint $$x = \pi/4$$. Plugging this x value into the error function, we get: $$|E(\pi/4)| \leq \frac{(\pi/4)^4}{24} \approx 0.000982$$ Therefore, the maximum error in the approximation $$\sin x \approx x - \frac{x^3}{6}$$ for $$x \in [-\pi/4, \pi/4]$$ is approximately 0.000982.

Key concepts

Error Bound

When approximating functions using Taylor series, it's important to understand the concept of an error bound. This tells us how far off our approximation might be from the true value.
In the context of Taylor series, the error bound is often discussed using Taylor's Inequality. This inequality helps us estimate the maximum error in our approximation. For a Taylor polynomial of degree \(n\), if we can determine an upper bound \(M\) for the \((n+1)\)th derivative of the function over the interval of interest, then the error bound is given by:

• \(|E(x)| \leq \frac{M |x - a|^{n+1}}{(n+1)!}\)

This formula essentially tells us that the error depends on how big \(M\) is, the distance \(|x - a|\) raised to the power of \(n+1\), and the factorial \((n+1)!\).
For our specific problem with \(\sin(x)\), the error bound shows us the maximum error, which was calculated as approximately 0.000982 when \(x\) is in \([-\pi/4, \pi/4]\). This means that our simple approximation \(\sin x \approx x - \frac{x^3}{6}\) is extremely close to the actual sine function in this range.

Derivative

A derivative represents how a function changes as its input changes. It tells us the rate of change or slope of the function. Derivatives are fundamental in calculus and essential for Taylor series approximations.
In our task of approximating \(\sin(x)\) by a Taylor polynomial, understanding derivatives is crucial, as every term in the Taylor series involves derivatives of the function. To estimate the error accurately, one needs to examine derivatives up to the \((n+1)\)th degree.
For \(\sin(x)\), the derivatives cycle every four terms, which is quite predictable:

• The 1st derivative: \(f'(x) = \cos(x)\)
• The 2nd derivative: \(f''(x) = -\sin(x)\)
• The 3rd derivative: \(f^{(3)}(x) = -\cos(x)\)
• The 4th derivative: \(f^{(4)}(x) = \sin(x)\)

We needed to find the 4th derivative in this exercise to determine the error bound, because we are using a third-degree polynomial in our approximation. Knowing the periodic nature of \(\sin(x)\), we realize that its 4th derivative, \(\sin(x)\), is maximized at 1 in the interval \([-\pi/4, \pi/4]\).

Remainder Term

The remainder term in the Taylor polynomial indicates how far the polynomial approximation diverges from the actual function. It's what remains when you use the polynomial to approximate a function. This term becomes very important when trying to understand how accurate your approximation is.
For a Taylor polynomial, the remainder term after \(n\) terms is linked closely with the error of the approximation. In our exercise, the remainder term is related to the function's derivatives beyond the ones used directly in the polynomial.
Mathematically, for a function \(f(x)\), the remainder term \(R_n(x)\) accounts for the difference:

• \(R_n(x) = \frac{f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}\) for some \(c\) between \(a\) and \(x\)

This term informs us how much of the actual function \(f(x)\) is "left over" after our polynomial fit. For \(\sin x\), being able to calculate the remainder gives us a handle on estimating its value's accuracy when approximated by \(x - \frac{x^3}{6}\). Recognizing these remainder impacts helps make sense of why our error bound approximation is reliable.

Trigonometric Function

Trigonometric functions like \(\sin(x)\) play a critical role in various fields, including science and engineering. These functions relate angles to ratios of sides in right-angled triangles and appear frequently in wave and oscillation studies.
The sine function, \(\sin(x)\), is periodic and oscillates between -1 and 1. Its behavior is well understood under transformations like derivatives and integrals, supporting its analysis through Taylor series expansion.
Why use trigonometric functions in Taylor series? First, they're infinitely differentiable, which makes them excellent candidates for series expansion. Plus, their periodic properties mean their derivatives cycle predictably. This can simplify calculations in repeated function analysis.
In our example, approximating \(\sin x\) as \(x - \frac{x^3}{6}\) is effective in minimizing complexity while maintaining approximation precision within the specified range. Studying functions such as these can provide insights into oscillatory behaviors in natural systems.