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. $$\sqrt[3]{126}$$

Short answer

The approximation of the cube root of 126 using a Taylor polynomial of degree 3 is approximately 5.02776, and the absolute error of this approximation is approximately 0.00029.

Step-by-step solution

Identify the function and the point of approximation

We are given the function $$f(x) = \sqrt[3]{x}$$ and we need to approximate $$f(126)$$, or in other words, the cube root of 126. We'll pick a point close to 126 where the cube root can be easily calculated, and that point is 125 since $$\sqrt[3]{125} = 5$$.

Find the Taylor polynomial of degree 3

To find the Taylor polynomial of degree 3 centered at 125, we need to compute the first three derivatives of the function $$f(x)$$ and their respective values at the point 125. The first few derivatives are as follows: $$f'(x) = \frac{1}{3}x^{-\frac{2}{3}}, f''(x) = -\frac{2}{9}x^{-\frac{5}{3}}, f'''(x) = \frac{10}{27}x^{-\frac{8}{3}}$$ Now find the values of these derivatives at the point $$x=125$$: $$f'(125) = \frac{1}{3}(125)^{-\frac{2}{3}} = \frac{1}{15}, f''(125) = -\frac{2}{9}(125)^{-\frac{5}{3}} = -\frac{2}{225}, f'''(125) = \frac{10}{27}(125)^{-\frac{8}{3}} = \frac{2}{3375}$$ Now, we can construct the Taylor polynomial of degree 3: $$P_3(x) = f(125) + f'(125)(x-125) + \frac{f''(125)(x-125)^2}{2!} + \frac{f'''(125)(x-125)^3}{3!}$$

Calculate the approximation

Now, we can use the Taylor polynomial to approximate $$f(126)$$. Plug in $$x=126$$ into the polynomial $$P_3(x)$$: $$P_3(126) = 5 + \frac{1}{15}(126-125) - \frac{1}{225}(126-125)^2 + \frac{1}{50625}(126-125)^3$$ After calculating, we get: $$P_3(126) \approx 5.02776$$

Calculate the exact value and absolute error

Using a calculator, we find the exact value of the cube root of 126: $$\sqrt[3]{126} \approx 5.02747$$ Now, we can compute the absolute error: $$\text{Absolute Error} = |P_3(126) - \sqrt[3]{126}| \approx |5.02776 - 5.02747| \approx 0.00029$$

Conclusion

The approximation of the cube root of 126 using a Taylor polynomial of degree 3 is approximately $$5.02776$$, with an absolute error of approximately $$0.00029$$.

Key concepts

Approximation

Let's start with the idea of approximation, which is about finding a value that is close to the exact one but generally easier to work with. It's like an educated guess that gets you close enough for practical purposes. In mathematics, especially when dealing with complex functions, we often use Taylor polynomials to approximate these functions.
Taylor polynomials are mathematical tools that express a function as a sum of terms calculated from the values and derivatives of that function at a certain point. They allow us to approximate functions when computing them exactly is difficult or impossible.
For example, when we approximate the cube root of 126 using a Taylor polynomial centered around 125, we find a formula that requires simpler calculations. This method is particularly helpful when we need a fast estimate without the use of complex tools like computers or calculators.

Derivative

Derivatives are at the core of calculus and understanding how functions change. You could think of a derivative as a measure of how a function's output changes in response to a change in its input; it's like the slope of a graph at any given point.
In the context of Taylor polynomials, derivatives help us build the approximation. For the function \(f(x) = \sqrt[3]{x}\), each derivative represents how \(f(x)\) changes as \(x\) changes. These calculations are foundational when deriving a Taylor polynomial.

• The first derivative, \(f'(x)\), shows the rate of change of the function itself.
• The second derivative, \(f''(x)\), indicates the rate at which the first derivative is changing - essentially showing the rate of acceleration.
• The third derivative, \(f'''(x)\), informs us on how the second derivative is changing, giving us deeper insight into the function's curvature.

By calculating these derivatives at a point, like at 125 in our example, we can incrementally build terms of the Taylor polynomial, each contributing to a more accurate approximation.

Absolute Error

Absolute error informs us how far our approximation is from the exact value. It's an important concept because it quantifies the difference and helps us understand the accuracy of our approximation method.
In this exercise, after constructing the third-degree Taylor polynomial \(P_3(x)\) to approximate the cube root of 126, we found that our approximation was \(5.02776\). By using a calculator, we determined the true cube root to be \(5.02747\). The absolute error is then calculated as:\[\text{Absolute Error} = |P_3(126) - \sqrt[3]{126}| \approx |5.02776 - 5.02747| = 0.00029\]The value of 0.00029 tells us how close our Taylor polynomial approximation is to the actual cube root. The small absolute error in this example indicates that our approximation is quite precise.

Cube Root

Understanding cube roots is key to solving the given problem. A cube root of a number \(x\) is a value \(y\) such that \(y^3 = x\). In other words, it is the number that, when multiplied by itself twice, gives the original number.
For instance, the cube root of 125 is 5, because \(5 \,\times\, 5 \,\times\, 5 = 125\). When dealing with numbers like 126, which aren't perfect cubes, finding a precise cube root becomes complex.
In this exercise, approximating \(\sqrt[3]{126}\) using the Taylor polynomial helped bypass the need for complex deriving. Such approximations are useful in a wide range of fields, from physics to engineering, where exact values might not always be necessary or possible. This makes the understanding of cube roots and their approximations essential in both academic and practical applications.