mth.122.handout.21

# X 5 5 cos x 1 x 2 2 x 4 4 try using long divsion to

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+ x 5 5! - · · · , cos x = 1 - x 2 2! + x 4 4! - · · · . Try, using long divsion, to find the power series for tangent. Answer: This is torture! But you might be able to see a pattern if you start the division. sin x cos x = x - x 3 3! + x 5 5! - · · · 1 - x 2 2! + x 4 4! - · · · = x + x 3 3 + 2 x 5 15 + 17 x 7 315 + 62 x 9 2835 + · · · Yikes, if there’s a pattern, I don’t see it! Anyway, you can always look them up. tan x = X n =1 B 2 n ( - 4) n (1 - 4 n ) (2 n )! x 2 n - 1 The interval of convergence for this series is ( - π/ 2 , π/ 2). The B 2 n are Bernoulli numbers. The main point I want to make here is that power series in general can be a pain to generate, but if you know how to use a computer you be able to it quickly! 5. Use the method of fitting higher degree polynomials (see problem on e x ) to find the power series for f ( x ) = ln (1 - x ). Answer: We are given f ( x ) = ln (1 - x ) and suppose we’re looking for a polynomial of degree six that fits this, so let P 6 ( x ) = A + Bx + Cx 2 + Dx 3 + Ex 4 + Fx 5 + Gx 6 , and the derivatives of both f and P 6 f ( x ) = ln (1 - x ) f 0 ( x ) = - (1 - x ) - 1 f 00 ( x ) = - (1 - x ) - 2 f 000 ( x ) = - 2 (1 - x ) - 3 f (4) ( x ) = - 6 (1 - x ) - 4 f (5) ( x ) = - 24 (1 - x ) - 5 f (6) ( x ) = - 120 (1 - x ) - 6 P 6 ( x ) = A + Bx + Cx 2 + Dx 3 + Ex 4 + Fx 5 + Gx 6 P 0 6 ( x ) = B + 2 Cx + 3 Dx 2 + 4 Ex 3 + 5 Fx 4 + 6 Gx 5 P 00 6 ( x ) = 2 C + 6 Dx + 12 Ex 2 + 20 Fx 3 + 30 Gx 4 P 000 6 ( x ) = 6 D + 24 Ex + 60 Fx 2 + 120 Gx 3 P (4) 6 ( x ) = 24 E + 120 Fx + 360 Gx 2 P (5) 6 ( x ) = 120 F + 720 Gx P (6) 6 ( x ) = 720 G 12

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Now, using x = 0 we can determine the constants A , B , C , D , E , F , and G . P 6 (0) = f (0) = 0 = A P 0 6 (0) = f 0 (0) = - 1 = B P 00 6 (0) = f 00 (0) = - 1 = 2 C P 000 6 (0) = f 000 (0) = - 2 = 6 D P (4) 6 (0) = f (4) (0) = - 6 = 24 E P (5) 6 (0) = f (5) (0) = - 24 = 120 F P (6) 6 (0) = f (6) (0) = - 120 = 720 G We’re getting a fairly nice pattern, and I’m going to make a guess 4 here and say that ln (1 - x ) = - x + x 2 2 + x 3 3 + x 4 4 + · · · = - X n =1 x n n . The interval of convergence is [ - 1 , 1). Let’s look at the graph of f ( x ) and - 20 n =1 x n n . -1 0 1 -1 1 Figure 5: Partial graphs of f ( x ) [black], and the first twenty terms of its power series [red]. Here we have (seen before) that ln 2 = 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 5 + · · · 4 We’ll discuss the actual method soon. 13
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