178 CHAPTER 8. POWER SERIES
All that is required is to multiplyex︷ ︸︸ ︷
1+ x+x2
2!+
x3
3!· · ·
sinx︷ ︸︸ ︷x− x3
3!+
x5
5!+ · · ·
From the above theorem the result should be
x+ x2 +
(− 1
3!+
12!
)x3 + · · ·
= x+ x2 +13
x3 + · · ·
You can continue this way and get the following to a few more terms.
x+ x2 +13
x3− 130
x5− 190
x6− 1630
x7 + · · ·
I don’t see a pattern in these coefficients but I can go on generating them as long as I want.(In practice this tends to not be very long.) I also know the resulting power series willconverge for all x because both the series for ex and the one for sinx converge for all x.
Example 8.10.3 Find the Taylor series for tanx centered at x = 0.
Lets suppose it has a Taylor series a0 +a1x+a2x2 + · · · . Then
(a0 +a1x+a2x2 + · · ·
)cosx︷ ︸︸ ︷
1− x2
2+
x4
4!+ · · ·
=
(x− x3
3!+
x5
5!+ · · ·
).
Using the above, a0 = 0,a1x = x so
a1 = 1,(
0(−12
)+a2
)x2 = 0
so a2 = 0. (a3−
a1
2
)x3 =
−13!
x3
so a3− 12 =− 1
6 so a3 =13 . Clearly one can continue in this manner. Thus the first several
terms of the power series for tan are
tanx = x+13
x3 + · · · .
You can go on calculating these terms and find the next two yielding
tanx = x+13
x3 +2
15x5 +
17315
x7 + · · ·
This is a very significant technique because, as you see, there does not appear to be a verysimple pattern for the coefficients of the power series for tanx. Of course there are some