Discover P5(x) and 4th Order Taylor Series of Sin(x) and xSin(2x)

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Homework Help Overview

The discussion revolves around finding the 5th order Taylor series of sin(x) and subsequently the 4th order Taylor series for x sin(2x) about x = 0. Participants are exploring the relationship between these two series and the necessity of calculating the higher-order series to derive the lower-order one.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning why the 5th order Taylor series of sin(x) is required to find the 4th order series for x sin(2x). Some suggest that it may not be necessary to compute the higher-order series at all.

Discussion Status

The discussion is active, with participants providing differing viewpoints on the necessity of finding the 5th order polynomial. Some participants have offered insights into the implications of the term "hence" and its mathematical context, suggesting that the original poster's assumption may not hold.

Contextual Notes

There is an underlying assumption regarding the relationship between the orders of the Taylor series for sin(x) and x sin(2x), which is being critically examined. The discussion also reflects on the implications of polynomial multiplication in relation to Taylor series.

Ry122
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Find P5(x), the 5th order Taylor series, of sin (x) about x = 0. Hence find the 4th
order Taylor series for x sin (2x) about x = 0.

In this question why is it required to find the 5th order taylor series of sin(x) to find the 4th order taylor series of xsin(2x)?
 
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After you have P5(x), it's a simple matter to find the power series for xsin(2x). "Hence" in this problem just means "after that," I believe.
 
Mark44, I believe Ry122's question was specifically why it would be necessary to find the 5th order polynomial of one in order to find the 4th order polynomial of the other. And, in fact, it is not necessary. Multiplying an nth order polynomial by x gives, of course, an n+1 order polynomial.

It would be sufficient to find the 3rd order Taylor's polynomial for sin(x) in order to find the 4th order Taylor's polynomial for x sin(2x). It would have made sense if you were looking for the 4th order Taylor's polynomial of sin(2x)/x.
 
My answer was a little oblique. I don't think it is necessary to find the 5th order Taylor polynomial. My comment about "hence" was intended to convey my belief that this word was not used in it usual mathematics sense of "it therefore follow that..."
 

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