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Chapter 37: Problem 973

Expand the function: \(\cos \mathrm{x}\), in powers of \(\mathrm{x}-\mathrm{a}\), where \(\mathrm{a}=-(\pi / 4)\), and determine the interval of convergence.

Short Answer

Expert verified
The Taylor series expansion of the cosine function at \(x=-\frac{\pi}{4}\) is given by: \[\cos x = \sum_{n=0}^{\infty}{\frac{((-1)^{n/2})/\sqrt{2}}{n!}(x+\frac{\pi}{4})^n}, n \text{ even},\] \[\hspace{45pt} + \sum_{n=0}^{\infty}{\frac{((-1)^{(n+1)/2})/\sqrt{2}}{n!}(x+\frac{\pi}{4})^n}, n \text{ odd}\] The interval of convergence is \((-\infty, \infty)\).

Step by step solution

01

Determine the Taylor series expansion of the cosine function at x=a

The Taylor series expansion of a function f(x) at the point x=a is given by: \[f(x) = \sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n}\] For the cosine function, we have: \[f(x)=\cos x\] For any even value of n (0, 2, 4, ...), the derivative of cos(x) is either cos(x) or -cos(x), and for any odd value of n (1, 3, 5, ...), the derivative of cos(x) is either sin(x) or -sin(x). Now, our point of expansion is \(a=-\frac{\pi}{4}\). Let's differentiate \(\cos x\) until we can see a pattern and find the general expression for its nth derivative evaluated at \(x=a\).
02

Derivatives of cos(x) at x=a

For \(n=0\), \(f^{(0)}(x)=\cos x\). So, \[f^{(0)}(-\frac{\pi}{4}) = \cos(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\] For \(n=1\), \(f^{(1)}(x)=-\sin x\). So, \[f^{(1)}(-\frac{\pi}{4}) = -\sin(-\frac{\pi}{4}) =-\frac{1}{\sqrt{2}}\] For \(n=2\), \(f^{(2)}(x)=-\cos x\). So, \[f^{(2)}(-\frac{\pi}{4}) =-\cos(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\] For \(n=3\), \(f^{(3)}(x)=\sin x\). So, \[f^{(3)}(-\frac{\pi}{4}) = \sin(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\] From these results, we can see a pattern: 1. If n is even, \(f^{(n)}(-\frac{\pi}{4}) = \frac{(-1)^{n/2}}{\sqrt{2}}\) 2. If n is odd, \(f^{(n)}(-\frac{\pi}{4}) = \frac{(-1)^{(n+1)/2}}{\sqrt{2}}\)
03

Expression for the Taylor series expansion of cos(x) at x=a

Using the pattern found in Step 2, we can now write the Taylor series expansion of the cos(x) function at x=a: \[\cos x = \sum_{n=0}^{\infty}{\frac{f^{(n)}(-\frac{\pi}{4})}{n!}(x+\frac{\pi}{4})^n}\] \[\cos x = \sum_{n=0}^{\infty}{\frac{((-1)^{n/2})/\sqrt{2}}{n!}(x+\frac{\pi}{4})^n}, n \text{ even},\] \[\hspace{45pt} + \sum_{n=0}^{\infty}{\frac{((-1)^{(n+1)/2})/\sqrt{2}}{n!}(x+\frac{\pi}{4})^n}, n \text{ odd}\]
04

Interval of convergence

The interval of convergence for the Taylor series of the cosine function is the same for any value of a, and it converges to cos(x) for all x in the real line. Thus, convergence is guaranteed for any value of x: \[\text{Interval of Convergence} = (- \infty, \infty)\] So, we have found the Taylor series expansion of the cosine function at \(x=-\frac{\pi}{4}\) and determined its interval of convergence.

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Most popular questions from this chapter

Using the sine series for \(\mathrm{f}(\mathrm{x})=\mathrm{x}, 0<\mathrm{x}<\pi\), and the mean square value, obtain the relation: \(\left(\pi^{2} / 6\right)=1+\left(1 / 2^{2}\right)+\left(1 / 3^{2}\right)+\left(1 / 4^{2}\right)+\ldots \ldots\)

Expand \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}},-\pi<\mathrm{x}<\pi\), in a Fourier series,

Find all values of \(\mathrm{x}\) for which the series: \(1+(1 / 2) \mathrm{x}+[(1 \cdot 3) /(2 \cdot 4)] \mathrm{x}^{2}+[(1 \cdot 3 \cdot 5) /(2 \cdot 4 \cdot 6)] \mathrm{x}^{3}\) \(+\ldots \ldots\) converges.

Test the alternating series: \([(1+\sqrt{2}) / 2]-[(1+\sqrt{3}) / 4]+[(1+\sqrt{4}) / 6]-[(1+\sqrt{5}) / 8]+\ldots\) for convergence.

Establish the convergence or divergence of the series: \([1 /(1+\sqrt{1})]+[1 /(1+\sqrt{2})]+[1 /(1+\sqrt{3})]+[(1 /(1+\sqrt{4})]+\ldots \ldots\)
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