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a. Use the given Taylor polynomial \(p_{2}\) to approximate the given quantity. b. Compute the absolute error in the approximation assuming the exact value is given by a calculator. Approximate \(\sqrt{1.05}\) using \(f(x)=\sqrt{1+x}\) and \(p_{2}(x)=1+x / 2-x^{2} / 8\)

Short Answer

Expert verified
Question: Use the Taylor polynomial \(p_{2}(x)=1+\frac{x}{2}-\frac{x^{2}}{8}\) to approximate \(\sqrt{1.05}\) and compute the absolute error in the approximation by using the exact value given from a calculator. Answer: Step 1: Substitute the appropriate value of x into the Taylor polynomial. \(p_2(0.05) = 1+\frac{0.05}{2}-\frac{{(0.05)}^{2}}{8} \approx 1.0246875\) Step 2: Calculate the exact value of \(\sqrt{1.05}\) using a calculator. Exact value: \(\sqrt{1.05} = 1.0246950766\) Step 3: Compute the absolute error between the approximation and the exact value. Absolute Error = \(|f(x)-p_2(x)|=|1.0246950766 - 1.0246875| \approx 0.0000075766\) The approximate value of \(\sqrt{1.05}\) using the Taylor polynomial is approximately 1.0246875 and the absolute error in the approximation is about 0.0000075766.

Step by step solution

01

Substitute the appropriate value of x into the Taylor polynomial

To approximate \(\sqrt{1.05}\), we need to substitute \(x = 0.05\) in the Taylor polynomial \(p_2(x)\). \(p_2(0.05) = 1+\frac{0.05}{2}-\frac{{(0.05)}^{2}}{8}\) Now, calculate the numerical value of this expression.
02

Calculate the exact value of \(\sqrt{1.05}\) using a calculator

Using a calculator, compute the exact value of \(\sqrt{1.05}\).
03

Compute the absolute error between the approximation and the exact value

The absolute error can be computed as the absolute difference between the approximate and the exact value. Absolute Error = \(|f(x)-p_2(x)|\) Now, calculate the absolute error using the values computed in Steps 1 and 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Approximation
Understanding approximation is key when dealing with Taylor polynomials. An approximation finds a simple form of a complex function. For functions like square roots or trigonometric functions, they can be approximated using polynomials for ease of calculation. A Taylor polynomial is a sum of derivatives of a function at a particular point. It's expanded in such a way that it gets closer and closer to the function's actual curve. This is particularly useful when direct computation is hard or impossible due to lack of calculator access. The polynomial provides a very close estimate that directly facilitates calculations on hand. By using the Taylor polynomial to approximate \(\sqrt{1.05}\), we are simplifying the process and getting a practical answer without complicated tools.
Ultimately, polynomials offer an effective means to estimate and approximate values.
Absolute Error
Absolute error helps us understand how accurate our approximation is. It defines the difference between the approximate value and the exact value. When we calculate an approximation using a polynomial and compare it to the exact computation, the absolute error tells us how off the estimate might be. To find it, subtract the approximate value obtained from the Taylor polynomial from the exact value acquired using a calculator. Next, take the absolute value of the result since we are interested in the magnitude of this error, not its direction. By calculating the absolute error:
  • It shows us how close the approximate value is.
  • It helps make informed decisions, especially in engineering or scientific computations where precision is necessary.
Understanding absolute error is essential for evaluating and refining the methods we use for approximation, ensuring our tools are reliable.
Exact Value
The exact value is the precise calculation of a mathematical expression. In our example, the exact value of \(\sqrt{1.05}\) is computed using a calculator, which provides a high-accuracy result. Exact values are critical as they serve as benchmarks to measure accuracy and precision against approximated or estimated results.
In scientific and engineering fields, exact values are sought after for their precision, though they can often be cumbersome to compute manually without technological aid. Therefore, while Taylor polynomials give us useful approximations for the sake of computational ease, having the exact value allows us to gauge the degree of error introduced by our approximation.
This balance between simplicity and precision is key to effective problem-solving in quantitative disciplines, guiding how we assess polynomial approximations.
Polynomial Substitution
Polynomial substitution is a process where a specific value is placed into the polynomial expression to simplify computation. In our Taylor polynomial, we substitute \(x = 0.05\) to estimate \(\sqrt{1.05}\), effectively tailoring the polynomial to our specific needs.
This act of substitution transforms a generic polynomial expression into a specific number, allowing quick computation without solving complex equations every time. By substituting the value into Taylor's polynomial:
  • We derive a functional form that's ready for a simple arithmetic calculation.
  • It's a vital step in acquiring an approximation, as it personalizes the polynomial's adaptability to any function we wish to explore around a certain point.
Thus, polynomial substitution not only enhances computational efficiency but also expands the polynomial's utility to solve various mathematical problems as per requirement.

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Most popular questions from this chapter

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty} 2^{k} x^{2 k+1}$$

Find the function represented by the following series and find the interval of convergence of the series. $$\sum_{k=1}^{\infty} \frac{(x-2)^{k}}{3^{2 k}}$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\frac{1}{x^{4}+2 x^{2}+1}$$

Teams \(A\) and \(B\) go into sudden death overtime after playing to a tie. The teams alternate possession of the ball and the first team to score wins. Each team has a \(\frac{1}{6}\) chance of scoring when it has the ball, with Team \(\mathrm{A}\) having the ball first. a. The probability that Team A ultimately wins is \(\sum_{k=0}^{\infty} \frac{1}{6}\left(\frac{5}{6}\right)^{2 k}\) Evaluate this series. b. The expected number of rounds (possessions by either team) required for the overtime to end is \(\frac{1}{6} \sum_{k=1}^{\infty} k\left(\frac{5}{6}\right)^{k-1} .\) Evaluate this series.

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-3 x}$$

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