Question

Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at \(0 .\) Estimates are not unique. $$\sin 0.3 ; n=4$$

Short answer

The \(4th\)-order Taylor polynomial approximation of \(\sin(0.3)\) is approximately \(0.295\), and the estimated absolute error of this approximation is less than or equal to \(0.00000648\).

Step-by-step solution

Find the Taylor polynomial for \(\sin(x)\) centered at 0

To find the \(4th\)-order Taylor polynomial, we need to determine the first four derivatives of \(\sin(x)\) at \(x=0\). Note that the derivatives of \(\sin(x)\) are cyclic, with the \(0th\), \(2nd\), and \(4th\) derivatives being \(\sin(x)\), and the \(1st\), \(3rd\), and \(5th\) derivatives being \(\cos(x)\). Evaluate the derivatives at \(x=0\): For \(k=0\): \(f^{(0)}(x) = \sin(0) = 0\) For \(k=1\): \(f^{(1)}(x) = \cos(0) = 1\) For \(k=2\): \(f^{(2)}(x) = -\sin(0) = 0\) For \(k=3\): \(f^{(3)}(x) = -\cos(0) = -1\) For \(k=4\): \(f^{(4)}(x) = \sin(0) = 0\) Now, substitute these values into the Taylor polynomial formula: $$P_4(x) = 0+\frac{1}{1!}x^1 + \frac{0}{2!}x^2 - \frac{1}{3!}x^3 + \frac{0}{4!}x^4 = x - \frac{1}{6}x^3$$

Calculate the approximation of \(\sin(0.3)\) using the Taylor polynomial

Now, we can approximate \(\sin(0.3)\) using our \(4th\)-order Taylor polynomial. Substitute \(x=0.3\) into \(P_4(x)\): $$P_4(0.3) = (0.3) - \frac{1}{6}(0.3^3) = 0.3 - \frac{1}{6}(0.027) \approx 0.295$$

Estimate the absolute error using the remainder term

The absolute error can be estimated using the remainder term \(R_4(x)\). According to the remainder term bound, for \(f(x) = \sin(x)\) and \(n=4\): $$|R_4(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$$ In this case, \(M\) is the maximum value of \(|f^{(5)}(x)|\) on the interval \([0, 0.3]\). Since \(f^{(5)}(x) = \cos(x)\), \(0\leq\cos(x)\leq1\) for all \(x\) on the interval, so \(M=1\). Substitute the given values into the equation: $$|R_4(0.3)| \leq \frac{1}{5!}(0.3)^5 \approx 0.00000648$$ So, the absolute error in approximating \(\sin(0.3)\) with the \(4th\)-order Taylor polynomial centered at \(0\) is estimated to be less than or equal to \(0.00000648\).

Key concepts

Taylor polynomial

The Taylor polynomial is a useful tool for approximating complex functions with simpler polynomials. It is derived from the function's derivatives at a single point, typically denoted as the center of the approximation. In our case, we are focusing on the Taylor polynomial for the \(\sin(x)\) function centered at 0. This is called a Maclaurin series, a specific type of Taylor series centered at zero.

The formula for the nth-order Taylor polynomial is \[P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]For the \(\sin(x)\) function centered at 0, we calculate each derivative of the function up to the fourth order, as given in the original exercise. Plugging these values into the Taylor polynomial formula yields a polynomial that approximates \(\sin(x)\) for values near zero.

• 0th derivative \(\rightarrow\) \(\sin(x)\)
• 1st derivative \(\rightarrow\) \(\cos(x)\)
• Process repeats using cyclical derivatives

By substituting these into the polynomial series formula, we get \(P_4(x) = x - \frac{1}{6}x^3\).This fourth-order polynomial provides a good approximation of \(\sin(x)\) around \(\x = 0\).

remainder term

When approximating a function with a Taylor polynomial, it is crucial to understand the remainder term, as it provides insight into the potential error in the approximation. The remainder term, often denoted as \(R_n(x)\), represents the difference between the actual function value and the polynomial approximation.

For the Taylor series of \(\sin(x)\) centered at 0 with a fourth-order polynomial, the remainder term is essential for estimating the accuracy of our approximation. The formula used to evaluate the remainder term is:\[ |R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1} \]Here, \(M\) represents the maximum value of the \((n+1)\)th derivative of the function over the interval we are considering. In our exercise, this corresponds to the maximum value of \(|f^{(5)}(x)|\), which equals 1 because \(\cos(x)\) is bounded between -1 and 1.

The remainder term allows us to estimate how close the Taylor polynomial \(P_4(x)\) is to \(\sin(x)\) when \(x = 0.3\). It gives a bound on the absolute error in our approximation, ensuring we understand the amount of error that might be present in the calculation.

error estimation

Error estimation is a fundamental aspect of working with Taylor polynomials because it helps assess the accuracy of our approximations. By determining how closely a Taylor polynomial approximates a function, we can understand its reliability.

In the context of the \(\sin(x)\) Taylor polynomial centered at 0, we used the remainder term to estimate the error. The absolute error tells us how much the approximate value \(P_4(0.3)\) might differ from true \(\sin(0.3)\). This was computed to be less than \(0.00000648\) using the formula for the remainder term:

• Set M = 1, based on \(cos(0.3)\) peak value in interval
• Calculate the bound using the fifth derivative constraint

This way, the error estimation assures us that the approximation obtained using \(P_4(x)\) is extremely close to the actual value, making it a reliable approximation within our specified range.

differentiation

Differentiation is the mathematical process we use to find derivatives, which are the building blocks of the Taylor polynomial. Derivatives represent the rate of change of a function and play a crucial role in constructing the polynomial.

For the \(\sin(x)\) function, knowing the pattern of its derivatives at zero helps in forming the Taylor polynomial efficiently. The \(\sin(x)\) and \(\cos(x)\) functions have cyclical derivatives, which make the process of differentiation straightforward for this particular exercise.
In our example:

• The first derivative of \(\sin(x)\) is \(\cos(x)\)
• The second derivative is \(-\sin(x)\)
• The third derivative regains a similar form: \(-\cos(x)\)
• Further differentiation continues this cycle

The critical cyclical nature means that each successive derivative aligns with either \(\sin(x)\) or \(\cos(x)\), making it predictable. Knowing this cycle helps efficiently derive terms for the Taylor polynomial, simplifying both the calculation and approximation process.