10.6.6 Convergence of the Taylor Series
           for Sine and Cosine

The sine and cosine Taylor series have terms of alternating sign for every value of the input variable, but we can not apply the Alternating Series Test to show that these series always converge unless we know that the terms of these series decrease to zero in size. Now we can show this — and, therefore, show that these series do indeed converge for every real input. As with the exponential and logarithmic series, we postpone until the next section the question of whether the series actually converge to "the right stuff," namely, values of the sine and cosine functions.

Here again is the series for the sine function:

sin ⁢     x = x - 1 6 ⁢ x 3 + 1 120 ⁢   x 5 - ⋅⋅⋅ + ( - 1 ) k ( 2 k + 1 ) ! ⁢   x 2 k + 1 + ⋅⋅⋅ .

Note first that the series has the form of a power series: We may consider the even-numbered powers all to have 0 as coefficient. Having made that observation, we now ignore all those even-power terms because they don't fit our pattern for terms of the series. Thus a convenient numbering is the one indicated in the notation above: Start with `k=0`, and only count odd-power terms.

Checkpoint 4

In order to apply the Alternating Series Test, we have to decide whether the terms of the series eventually get small. The question is not trivial for large values of `x` — check the sizes of the first few terms with `x=10`. In fact, this was the same problem we had with the exponential series: The size of the `k`th term is `text[|]xtext[|]^mtext[/]m!`, where `m=2k+1`. But we saw in Activity 3 that the limiting value of such terms, as `m` (or `k`) becomes large, is zero — no matter how big `x` is.

The Alternating Series Test requires more — not just that the terms approach zero, but that they do so in a disciplined way, with each term larger (in absolute value) than the one following it. Is that eventually true? Let's compare the term numbered `k` with the term numbered `k+1`. The absolute values of these terms are, respectively,

a k = 1 ( 2 k + 1 ) ! ⁢   | x | 2 k + 1    and    a k + 1 = 1 ( 2 k + 3 ) ! ⁢   | x | 2 k + 3 .

The ratio of `a_(k+1)` to `a_k` is

a k + 1 a k = ( 2 k + 1 ) ! ( 2 k + 3 ) ! ⁢   | x | 2 k + 3 | x | 2 k + 1   = | x | 2 ( 2 k + 3 ) ( 2 k + 2 )   .

Now, for any fixed `x`, we can make the expression on the right as small as we like by taking `k` sufficiently large. In particular, we can make it less than `1` by choosing `k` to be, say, bigger than `text[|] x text[|]`. (Half that big would be big enough because of the factors of `2` in the denominator.) Thus, eventually, we have `a_(k+1)` smaller than `a_k` for all the remaining terms of the series. Now we have satisfied all three conditions of the Alternating Series Test, so we can conclude that

1. the Taylor series for `sin x` converges for every real number `x`, and
2. the error after summing `n` terms (the ones numbered from `0` to `n-1`) is no bigger than

a n = 1 ( 2 n + 1 ) ! ⁢   | x | 2 n + 1 .

Checkpoint 5

The cosine series is treated in exactly the same way as the sine series. We leave the details to you in Exercise 2.