Problem: Expand sin(2x + 1) around x = 0. THE WRONG WAY: We know that

sin z = z − \frac{z^{3}}{6} + \frac{z^{5}}{120} + ⋯,

so it seems natural to substitute 2x + 1 for z:

sin(2x + 1) = (2x + 1) − \frac{(2x + 1)^{3}}{6} + \frac{(2x + 1)^{5}}{120} + · · · .

Next, trying to combine terms, we multiply everything out and get

1 + 2x − \frac{1}{6}(1 + 6x + 12x^{2} + 8x^{3}) + \frac{1}{120}(1 + 10x + ⋯) + ⋯

= (1 − \frac{1}{6} + \frac{1}{120} + \underbrace{⋯}_{?} ) + x(2 − 1 + \frac{1}{12} + ⋯) + · · · .

This is a disaster! Even the very first term (the constant term) depends on the terms of arbitrarily high order in the Taylor series of the sine. The point is that this method of finding the Taylor series of the composite function won’t work unless the expression substituted for z approaches 0 as x → 0.

A RIGHT WAY: Note that 2x + 1 approaches 1, not 0, as x approaches 0. Therefore, we should be using the Taylor expansion of sin z around z = 1:

sin z = sin(1) + cos(1) (z − 1) − \frac{1}{2} sin(1) (z − 1)^{2} + · · · .

Then sin(2x + 1) can be successfully approximated for x near 0 by substituting 2x + 1 for z — that is, 2x for z − 1.

sin(2x + 1) = sin(1) + 2 cos(1) x − 2 sin(1) x^{2} + · · · .

(Of course, we can’t find exact numerical values of sin(1) and cos(1), but that’s life; they are the correct coefficients in this series.)