Question

In Exercises \(35-38,\) use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. $$ \arcsin (0.4) \approx 0.4+\frac{(0.4)^{3}}{2 \cdot 3} $$

Short answer

The upper bound for the error calculated using Taylor's theorem is 0. The actual error is approximately 0.001816487.

Step-by-step solution

Apply Taylor's Theorem to Find the Error Bound

Apply Taylor's Theorem formula for error bound. Start by finding the 3rd derivative of \(arcsin(x)\) at \(x = 0\). This will give the maximum value of \(M\), because on the interval from 0 to 0.4, the 3rd derivative of \(arcsin(x)\) is increasing. The 3rd derivative of \(arcsin(x)\) at \(x=0\) is 0, thus \(M = 0\). Hence, the upper bound of the error \(R_2\) is 0 in this case.

Calculate the Exact Value of arcsin(0.4)

Obtain the exact value of \(arcsin(0.4)\) using a calculator, which is approximately 0.411516846.

Calculate Exact Value of the Approximation

Calculate the approximate value of \(arcsin(0.4)\) using the Taylor series approximation provided: \(0.4+\frac{(0.4)^{3}}{2 \cdot 3}\), which gives approximately 0.413333333.

Calculate Actual Value of Error

Calculate the exact error by subtracting the exact value from the approximated value. Since the approximated value is larger, the error is positive: \(0.413333333 - 0.411516846 = 0.001816487.\)

Key concepts

Understanding Taylor's Theorem

Taylor's Theorem is a foundational principle used for approximating the values of functions that may otherwise be difficult to calculate. At its core, it expresses a function as an infinite sum of terms, calculated from its derivatives at a single point. This sum is called the Taylor series.

When we represent a function as a Taylor series, we're creating a polynomial that closely resembles the function within a certain interval around the point of expansion. The more terms we include, the more accurate our approximation will be. However, we usually truncate the series after a few terms for simplicity, which introduces an error - the difference between the true value of the function and the value given by the finite Taylor series.

To quantify this error, we use the 'remainder' or 'error bound' of Taylor's Theorem. This error bound gives us a way to measure the accuracy of the approximation. Specifically, it tells us how far off our Taylor series approximation might be from the function's true value, under certain conditions.

The Error Bound in Approximations

The error bound is an essential concept when working with Taylor series. It provides an upper limit on how much the Taylor series of a function, truncated at a finite number of terms, can differ from the actual value of the function.

To find this error bound, we often look at the derivative that comes after the last term used in the approximation. For instance, if we truncate the series after the second term, we look at the third derivative. The maximum value this derivative takes on in the interval of interest sets the value of 'M' in Taylor's Theorem, contributing to the upper bound of the remainder, or error.

Calculating Error Bound

In practice, this means we need to assess the behavior of the derivatives of our function within the range we are interested in. For the given exercise, the third derivative of \(arcsin(x)\) determines our error bound. An important point to remember is that if the maximum value of this derivative (M) turns out to be 0, as in the example, our upper bound is also zero - indicating that the approximation will be very accurate within the specified range.

Arcsin Approximation Using Taylor Series

The Taylor series can be particularly useful when approximating trigonometric functions, such as the arcsine function. The arcsine function, \(arcsin(x)\), can be difficult to compute exactly for non-special angles. Thus, a Taylor series approach offers an efficient way to approximate its value.

The approximation provided in the exercise uses the first two non-zero terms of the Taylor series for \(arcsin(x)\) expanded at \(x=0\). When \(x=0.4\), these two terms create the polynomial \(0.4 + \frac{(0.4)^3}{2 \cdot 3}\), which represents the arcsine function in the vicinity of \(x=0\).

Even with just two terms used, the Taylor series can provide a close approximation, as evidenced by the small error value reported in the solution steps. This highlights the power of the Taylor series: even a truncated series can yield a practical estimation of complex functions, simplifying calculations in various fields such as physics, engineering, and finance.