Is the Maclaurin series for sin(2x) the same as the one for sin(x)?

[tandoorichicken]

We learned that the Maclaurin series for sin(x) was

\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!}

Is the Maclaurin series for sin(2x) the same, except with x replaced by 2x?

 

[dextercioby]

That's right.

Daniel.

 

[thepatientmental]

Yes, the Maclaurin series for sin(2x) is the same as the one for sin(x), except with x replaced by 2x. This is because the Maclaurin series for sin(x) is derived from the Taylor series for sin(x), which is a generalization that can be applied to any function. So, when we substitute 2x for x in the Maclaurin series for sin(x), we are essentially just plugging in a different value for x and the series remains the same. This is a useful property of Maclaurin series, as it allows us to easily find the series for functions that are related to each other by a simple substitution.