Chapter 9: Problem 43
Use the Remainder Estimation Theorem to find an interval containing \(x=0\) over which \(f(x)\) can be approximated by \(p(x)\) to three decimal-place accuracy throughout the interval. Check your answer by graphing \(|f(x)-p(x)|\) over the interval you obtained. $$f(x)=\sin x ; p(x)=x-\frac{x^{3}}{3 !}$$
Short Answer
Expert verified
The interval is approximately \([-0.182, 0.182]\).
Step by step solution
01
Understand the Problem
We need to use the Remainder Estimation Theorem to find an interval around \(x=0\) where the approximation \(p(x) = x - \frac{x^3}{3!}\) is within three decimal places of accuracy for the function \(f(x) = \sin x\).
02
Identify Derivatives and Remainder Term
The Taylor series expansion for \(\sin x\) at \(x=0\) is \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\). We need to consider the remainder term \(R_n(x)\), where \(n=3\) since \(p(x)\) includes terms up to \(x^3\). Step to calculate remainder: \[|R_n(x)| = \left|\frac{f^{(n+1)}(c)\cdot x^{n+1}}{(n+1)!}\right|\] where \(f^{(4)}(c)\) for some \(c\) in the interval is the derivative \(\cos c\).
03
Calculate Maximum Value of the Fourth Derivative
The fourth derivative \(f^{(4)}(x)\) is \(\cos x\), which is bounded by 1 for all \(x\). So, the maximum value is \(M = 1\).
04
Apply Remainder Estimation Theorem
We want \[|R_3(x)| = \left|\frac{\cos(c) \cdot x^4}{4!}\right| \leq \left|\frac{x^4}{24}\right| < 0.0005\]This condition guarantees three-decimal-place accuracy. Solve \(\frac{x^4}{24} < 0.0005\) by multiplying through by 24: \[x^4 < 0.012\].
05
Solve for Interval
Solve the inequality \(x^4 < 0.012\) by taking the fourth root. Let \(x = \pm\sqrt[4]{0.012}\):\[x \approx \pm 0.182242\]Thus, the interval is \(-0.182, 0.182\) to three decimal places.
06
Verification by Graphing
Plot \(|f(x)-p(x)|\) over the interval \([-0.182, 0.182]\). Check the maximum deviation from zero; it should be less than 0.0005, confirming three-decimal accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
The Taylor series is a powerful tool in calculus used for approximating complex functions with polynomials. A Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. For the function \(\sin x\), its Taylor series expansion at the point \(x=0\) is:
- \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\)
Approximation Accuracy
Approximation accuracy is essential when using polynomials to represent more complex functions. It refers to how closely the polynomial approximation approaches the true value of the function within a chosen interval. The Remainder Estimation Theorem helps determine the error bound, or the difference between the actual function \(f(x)\)and its approximation \(p(x)\). For instance, in using the Taylor polynomial \(p(x) = x - \frac{x^3}{3!}\) to approximate \(\sin x\), the remainder term \(|R_n(x)|\) is given by:
- \(\left|\frac{f^{(n+1)}(c)\cdot x^{n+1}}{(n+1)!}\right|\)
Interval Estimation
Interval estimation is pivotal when applying Taylor series for approximation. It identifies the range around a point, often \(x=0\), within which the approximation remains accurate to a specific degree. For our problem, to ensure three-decimal-place accuracy for \(\sin x\) using \(p(x)\), we derived the condition\(\frac{x^4}{24} < 0.0005\). Solving this inequality involves evaluating the maximum error where it occurs and simplifying it to find \(x^4 < 0.012\). By extracting the fourth root of 0.012, we determined the interval
- \([-0.182, 0.182]\)