Consider the power series expansion

Find the coefficient in this expansion.

First, we recall the identity for the sine of a sum,

So we have

Then, we know the expansions of and are

Therefore, we have

(Where denotes the greatest integer less than or equal to .) So, the 98th coefficient is given by

Only cos(2x) plays a role for x^98, so it can be solved even simpler than presented here.

I presume this exercise is much simpler to do by direct differentiation of sin(2x + pi/4) and induction. We find the derivative f”(0), and by induction the derivative f^{98}(0). This is the expansion of sin(2x + ..) around 0. Since the series converges everywhere, we simply use the derivative to find the coefficient a_98.

Could you please elaborate on the last step of the demonstration please? I don’t get it.

The last step before the calculation of (the one where we combine the even and odd terms) or the actual computation of ?

Assuming it is the combining the even and odd terms. We start with

For each in the sum we get two terms, one with even powers and one with odd powers, and they are either both positive or both negative. So, we’re actually getting every term , but instead of alternating with each term, it is two positive terms and then two negative terms. That’s what the does. Looking at the sum we combine them into

So, I think it does work as claimed.