Problem 48 - Calculus Early Transcendentals

Question

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given by a calculator. $$\tanh (0.5)$$

Step-by-Step Solution

Answer

The Taylor polynomial of degree 3 for the function tanh(x) at x=0 is P_3(x) = x - (1/3)x^3. Using this polynomial, we approximated tanh(0.5) as 0.45833. The actual value of tanh(0.5) is approximately 0.46212, resulting in an absolute error of about 0.00379 in the approximation.

Step 1: Find the Maclaurin series of \(\tanh(x)\)

To find the Maclaurin series of \(\tanh(x)\), we need to compute its derivatives at \(x=0\). Recall that \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), where \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). Compute the derivatives of \(\tanh(x)\): 1. First derivative: \(\frac{d}{dx}\tanh(x)=\frac{1}{\cosh^2(x)}\) 2. Second derivative: \(\frac{d^2}{dx^2}\tanh(x)=-2\times\frac{\sinh(x)}{\cosh^3(x)}\) 3. Third derivative: \(\frac{d^3}{dx^3}\tanh(x)=6\times\frac{\sinh^2(x)}{\cosh^4(x)}-2\times\frac{1}{\cosh^2(x)}\) Now we need to compute these derivatives at \(x=0\): 1. \(\tanh(0)=\frac{\sinh(0)}{\cosh(0)}=0\) 2. \(\frac{d}{dx}\tanh(0)=\frac{1}{\cosh^2(0)}=\frac{1}{1}=1\) 3. \(\frac{d^2}{dx^2}\tanh(0)=-2\times\frac{\sinh(0)}{\cosh^3(0)}=0\) 4. \(\frac{d^3}{dx^3}\tanh(0)=6\times\frac{\sinh^2(0)}{\cosh^4(0)}-2\times\frac{1}{\cosh^2(0)}=-2\)

Step 2: Truncate the Maclaurin series to a Taylor polynomial of degree 3

Using the Maclaurin series, we can write the Taylor polynomial of degree 3 as: $$P_3(x) = \tanh(0) + \frac{d}{dx}\tanh(0) \cdot x + \frac{1}{2!}\frac{d^2}{dx^2}\tanh(0) \cdot x^2 + \frac{1}{3!}\frac{d^3}{dx^3}\tanh(0) \cdot x^3$$ Substitute the values obtained in step 1: $$P_3(x) = 0 + 1 \cdot x + 0 + \frac{-2}{6} \cdot x^3 = x-\frac{1}{3}x^3$$

Step 3: Approximate \(\tanh(0.5)\) using the Taylor polynomial

Now we will approximate \(\tanh(0.5)\) using \(P_3(x)\): $$P_3(0.5) = 0.5 - \frac{1}{3}(0.5)^3 = 0.5 - \frac{1}{24} = \frac{11}{24} \approx 0.45833$$

Step 4: Compute the absolute error in the approximation

Assuming that the exact value is given by a calculator, we can now find the absolute error: Exact value: \(\tanh(0.5) \approx 0.46212\) Approximate value (using Taylor polynomial): \(P_3(0.5) \approx 0.45833\) Absolute error: \(|0.46212 - 0.45833| \approx 0.00379\) Thus, the absolute error in this approximation is about \(0.00379\).

Key Concepts

Maclaurin Series

The Maclaurin series is a special case of the Taylor series that expands a function around zero. It's quite useful when dealing with functions that are difficult to evaluate directly. In the case of \(\tanh(x)\), we break it down into its hyperbolic sine and cosine components: \(\sinh(x)\) and \(\cosh(x)\). This helps us find the derivatives necessary for forming the series.
For \(\tanh(x)\), the derivatives at \(x=0\) were computed as follows:

• First derivative: \(1\)
• Second derivative: \(0\)
• Third derivative: \(-2\)

With these, we construct the Taylor polynomial of degree 3, a truncated form of the series that provides an approximation of the function up to the third order term. The polynomial we've used is \(x - \frac{1}{3}x^3\).

Approximation

In many mathematical contexts, exact solutions are difficult to identify or not possible to use easily. That's where approximation comes in. Using Taylor polynomials, we can approximate values of functions around a point, in this case, zero (due to the Maclaurin series).
For \(\tanh(0.5)\), we used the polynomial \(P_3(x) = x - \frac{1}{3}x^3\) to find an approximate value. By substituting \(x=0.5\) into this polynomial, we calculated:
\[P_3(0.5) = 0.5 - \frac{1}{3}(0.5)^3 = \frac{11}{24} \approx 0.45833\]This value is a close approximation, demonstrating that such polynomial expansions can effectively estimate function values near the point of expansion.

Absolute Error

Absolute error provides insight into the accuracy of an approximation compared to the exact value. It is calculated as the absolute difference between the approximate and exact values. In mathematical terms, if \(A\) is an approximate value and \(E\) is the exact value, the absolute error is given by \(|E - A|\).
For our exercise, the exact value for \(\tanh(0.5)\) was found to be approximately 0.46212 using a calculator. Our approximation using the Taylor polynomial was 0.45833. Thus, the absolute error can be calculated as:
\[|0.46212 - 0.45833| \approx 0.00379\]
This relatively small error shows that our approximation was quite close to the exact value.

Hyperbolic Functions

Hyperbolic functions, such as \(\sinh(x)\) and \(\cosh(x)\), are analogous to trigonometric functions but for a hyperbola. They have essential applications in various fields including engineering, physics, and mathematics. The function \(\tanh(x)\), which stands for hyperbolic tangent, is a key component that appears in many contexts, just like its trigonometric counterpart \(\tan(x)\).
Understanding hyperbolic functions involves knowing their definition:

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

The relationship \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\) helps in developing the Maclaurin series, as seen in the exercise we explored.
These functions increase exponentially as their argument moves away from zero, making them similar to exponentials yet retaining properties akin to sine and cosine.