Chapter 10: Problem 71
By multiplying the Taylor series for \(e^{x}\) and \(\sin x,\) find the terms through \(x^{5}\) of the Taylor series for \(e^{x} \sin x .\) This series is the imaginary part of the series for \begin{equation}e^{x} \cdot e^{i x}=e^{(1+i) x}.\end{equation} Use this fact to check your answer. For what values of \(x\) should the series for \(e^{x} \sin x\) converge?
Short Answer
Step by step solution
Write the Taylor Series for \(e^x\)
Write the Taylor Series for \(\sin x\)
Multiply the Two Series
Calculate Terms up to \(x^5\)
Relate to \(e^{(1+i)x}\) and Check Answer
Determine the Convergence of the Series
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series of \(e^x\)
- Exponential growth means each term eventually outpaces the one before it, contributing to \(e^x\) approaching its true value as more terms are included.
Taylor Series of \(\sin x\)
- Odd symmetry around the origin implies sine mirrors itself across the axis, hence the presence of only odd powered terms.
- The factorial in each denominator slows the growth of the terms, reflecting the bounded nature of \(\sin x\) between -1 and 1.
Understanding Convergence
- The Taylor series for \(e^x\) is convergent for all real \(x\), thanks to polynomial growth countered by factorial growth in denominators.
- Similarly, \(\sin x\) converges for all real \(x\) due to the bounded nature of sine and its alternating series which stabilizes the terms.
Exploring the Imaginary Part
- The real part corresponds to \(e^x \cos x\), capturing the oscillations in the real plane.
- While the imaginary part is \(e^x \sin x\), aligning perfectly with our earlier derived series for the imaginary component.
Multiplication of Series
- Each product's term is formed by pairing terms from both series, such that their power sums to a target degree, like \(x^3\) or \(x^5\).
- This is akin to polynomial multiplication but extends infinitely, making it crucial to keep track of similar powers as the series progress.