Some Series Stuff (Part I)

Okay, this is something of an experiment.  I have translated a LaTeX file into HTML using tth (link on the official math links page on the UD site).  This should look okay if you are using a mozilla or firefox browser.  Otherwise you are taking your chances.

The idea of this post is to compute Taylor series for sines and cosines without using Calculus.

The rest of the post is after the jump.  (Click on the post title to get to the whole article and comments.)

Infinite Polynomials for Sine and Cosine

We seek to derive power series for the functions sin(x) and cos(x) using only basic algebra and trigonometric facts.

Trigonometry: Two identities are required.

sin2(x) + cos2(x) = 1             (1) 2 sin(x) cos(x) = sin(2x)             (2)

We shall also use the symmetries of the functions, namely that sin(x) is odd and that cos(x) is even. We thus will write the functions in the following forms:

sin(x) = a1 x + a3 x3 + …+ a2k+1 x2k+1 + …             (3) cos(x) = a0 + a2 x2 + …+ a2kx2k + …             (4)

Our goal then is to discover the coefficients a₀, a₁, …. (Reminder: These power series expansions are predicated on the assumption that the angle x is measured in radians.)

It is a fairly easy fact that a₀ = 1. It is less easy to see why a₁ = 1. However, with these two values, we are in a position to calculate the remaining coefficients.

For instance, only keeping track of terms up to second order, by plugging in our power series expansions (3, 4) into our first trigonometric identity (1), we obtain

1 = (x + …)2 + (1 + a2 x2 + …)2 = (x2 + …) + (1 + 2a2 x2 + …) = 1 + (2a2 + 1)x2 + …

The coefficients of the different powers of x on both sides of the equation must match. We see that the constant terms do match, and that for the same to be true of the coefficients of x², we require that 0 = 2a₂ + 1, and thus that a₂ = −1/2.

Similarly, plugging in the power series expansions into the second trigonometric identity (2), and keeping track of terms up to third order, we obtain

2 (x + a3 x3 + …) (1 − x2/2 + …) = (2x + a3(2x)3 +…) 2x + (2a3 − 1) x3 + … = 2x + 8 a3 x3 + …

Again, the coefficients of each power of x of both sides of the equation must match. We see that they do for the first powers, already, and by comparing third powers, we require 2a₃ − 1 = 8 a₃, hence a₃ = −1/6.

We may continue in this way, alternating between the two trigonometric identities and keeping track of ever higher orders as we go, to obtain the coefficients in sequence.

It is not too difficult to show that for even values of k ≥ 2 we have

ak = − 1/2( [(k)/2]−1 ∑ j=1 a2j ak−2j + [(k)/2]−1 ∑ j=0 a2j+1 ak−2j−1)             (5)

and for odd values of k ≥ 3 we have

ak = 1 2k−1 − 1 [(k−1)/2] ∑ j=1 a2j ak−2j.             (6)

Exercises

1. Prove the two trigonometric identities in equations (1) and (2).
   
   
   
   
2. Prove that a function expressible as a power series will be an odd function if and only if all even powers have coefficients equal to zero.
   
   Similarly, functions expressible as a power series are even if and only if the coefficients for the odd powers are equal to zero.
   
   
   
   
3. How do we know that a₀ = 1? How do we know that a₁ = 1?
   
   
   
   
4. Prove the recursion formulas (6, 5) for the coefficients of the odd and even powers.
   
   
   
   
5. From the pattern to the coefficients revealed (hopefully!) by the first several terms, discover and prove a simple formula for the coefficients.
   
   
   
   
6. Compute approximations to the quantities sin(1) and cos(1) using the first several terms of the power series for each function. Note: 1 radian is slightly less than π/3, hence corresponds to an angle only slightly less than 60^°.
   
   
   
   
7. Other trigonometric identities could be used instead of the ones given. Have fun exploring other paths to these answers. Some may even work out more simply!
   
   
   
   

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File translated from
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by
T_TH,
version 3.86.
On 18 Sep 2009, 11:47.