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Chapter 6: Problem 9

If \(g(x):=\sin x\), show that the remainder term in Taylor's Theorem converges to zero as \(n \rightarrow \infty\) for each fixed \(x_{0}\) and \(x\).

Short Answer

Expert verified
The remainder term in Taylor's theorem for the sine function does indeed converge to 0 as \(n\) approaches infinity for any fixed \(x_0\) and \(x\). This is due to the specific properties of the sine function and the fact that the factorial growth in the denominator of the remainder term outweighs any polynomial growth in the numerator, causing the entire term to shrink towards 0.

Step by step solution

01

Recall Taylor's Theorem and the Remainder

Taylor's theorem states that any function can be locally approximated by a polynomial, with the error of this approximation given by the remainder term. For a function \(g(x)\), the Taylor series is given by \(\sum_{n=0}^{\infty} \frac{g^{(n)}(x_0)}{n!} (x-x_{0})^{n}\), where \(g^{(n)}(x_0)\) denotes the nth derivative of \(g(x)\) evaluated at \(x_0\). The remainder \(R_n(x)\) after \(n\) terms in this series is \(R_n(x)=g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(x_0)}{k!} (x-x_{0})^{k}\).
02

Properties of Sine Function

Notice the function given \(g(x)=\sin(x)\) is periodic with period \(2\pi\), and its maximum and minimum values are 1 and -1 respectively, no matter how many times it is differentiated. This means that the value of the \(n\)th derivative of \(g(x)\) at any point \(x_0\) is always between -1 and 1.
03

Evaluate The Remainder

Given \(R_n(x)=g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(x_{0})}{k!} (x-x_{0})^{k}\), we can bound \(R_n(x)\) by taking the absolute value and applying the triangle inequality: \(|R_n(x)| \leq |g(x)|+\left|\sum_{k=0}^{n} \frac{g^{(k)}(x_{0})}{k!} (x-x_{0})^{k}\right|\). Given that \(|g^{(k)}(x_{0})|\leq 1\) for all \(k\) and \(x_0\), we have \(|R_n(x)| \leq 1+\left|\sum_{k=0}^{n} \frac{1}{k!} |x-x_{0}|^{k}\right|\). As \(n \rightarrow \infty\), the expression \(\frac{1}{k!} |x-x_{0}|^{k}\) converges to 0 since the factorial growth in the denominator outweighs any polynomial growth in the numerator. Therefore \(|R_n(x)| \rightarrow 0\) as \(n \rightarrow \infty\).
04

Conclusion

This shows that the remainder term in Taylor's theorem for the sine function converges to zero as \(n\) approaches infinity for any fixed \(x_{0}\) and \(x\). This is a restatement of the fact that the Taylor series for the sine function converges to the sine function itself as \(n \rightarrow \infty\).

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