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Power Series for sin x and cos x

Suppose we are given a function f(x) and a point c, and we wish to represent f(x) as the sum of a power series in x - c. This will be possible for some functions (the analytic functions), but not for all. Theorem 1 shows that if there is such a power series it is the Taylor series for f(x). Thus we use the following steps to represent f(x) as a power series.

Step 1 Compute all the derivatives f(n)(c), n = 0,1, 2,.... If these derivatives do not all exist, f(x) is not the sum of a power series in powers of x - c.

Step 2 Write down the Taylor series of f(x) at x = c and find its radius of convergence r.

Step 3 If possible, show that f(x) is equal to the sum of its Taylor series for c - r < x < c + r.

We shall now use Steps 1-3 to obtain the power series for sin x, cos x, and (1 + x)p.

THE POWER SERIES FOR sin x

Step 1

This step was carried out in the preceding section. The values of f(n)(0) for n = 0, 1, 2, ... are

0, 1, 0, -1, 0, 1, 0, -1,....

Step 2

The MacLaurin series for sin x is

09_infinite_series-763.gif

Let 09_infinite_series-764.gif. We use the Ratio Test,

09_infinite_series-765.gif

Therefore the series converges for all x and has radius of convergence x.

Step 3

We use MacLaurin's Formula,

09_infinite_series-766.gif

09_infinite_series-767.gif

Let us show that the remainders approach zero. We have

09_infinite_series-768.gif

Since the even terms are zero, R2n-1(x) = R2n(x). Therefore limn→∞ Rn(x) = 0.

Conclusion: Since the remainders approach zero, the MacLaurin polynomials approach sin x. So for all x,

09_infinite_series-769.gif

THE POWER SERIES FOR cos x

This power series can be found by the same method as was used for sin x. However, it is simpler to differentiate the power series for sin x.

09_infinite_series-770.gif


Last Update: 2006-11-08