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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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- Thread starter Ry122
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- #1

- 565

- 2

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)?

- #2

Mark44

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- #3

HallsofIvy

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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.

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Mark44

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