C 3 x x 3 3 x 5 5 x 7 7 arctan x c 3 0 3 3 5 5 7 7

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C 3 + x - x 3 3 + x 5 5 - x 7 7 + · · · = arctan x C 3 + 0 - 0 3 3 + 0 5 5 - 0 7 7 + · · · = arctan 0 C 3 = 0 So now we can say that arctan x = x - x 3 3 + x 5 5 - x 7 7 + · · · = X n =0 ( - 1) n x 2 n +1 2 n + 1 which converges for [ - 1 , 1]. Here’s the graph of arctangent and the first ten terms (degree 19) of its power series. -1 0 1 -1 1 Figure 4: Partial graphs of arctan x [black], and the first ten terms of its power series [red]. Here you might want to note 3 that arctan 1 = π 4 = 1 - 1 3 + 1 5 - 1 7 + · · · . 3 Leibniz’s famous formula for finding π . You may want to compare this one with Srinivasa Ramanujan’s formula 1 π = 2 2 9801 X n =0 (4 n )! (1103 + 26390 n ) ( n !) 4 396 4 n . Although Leibniz’s famous formula looks easier, it’s converges much slower than Srinivasa Ramanujan’s formula. You might want to try computing three terms of each series to see the difference. If you’re going to go mad and compute π as an obsession, I strongly suggest you use Srinivasa Ramanujan’s formula. The Japanese use a variation of Ramanujan’s power series, and they, among others, are obsessed with finding digits of π . I believe the Japanese are well beyond 1,241,100,000,000 decimal digits. That’s scary. 9
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3 Examples 1. Use long division to find the power series for f ( x ) = 1 1 - x 4 , and indicate the radius of convergence. 2. Use what you already know about the power series for 1 1 - x = 1 + x + x 2 + x 3 + · · · , to verify the power series obtain above. 3. Find the power series for 1 1 + x 7 and then use this power series to find a power series for Z 1 1 + x 7 d x. Also, state the interval of convergence for each series. 4. We know that tan x = sin x cos x , and we know that sin x = x - x 3 3! + x 5 5! - · · · , cos x = 1 - x 2 2! + x 4 4! - · · · . Try, using long divsion, to find the power series for tangent. 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 ). 4 Answers 1. Use long division to find the power series for f ( x ) = 1 1 - x 4 , and indicate the radius of convergence. Answer: The division is quite simple, but we’ll do this in class. f ( x ) = 1 1 - x 4 = 1 + x 4 + x 8 + x 12 + · · · = X n =0 x 4 n And the interval of convergence is ( - 1 , 1). 10
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2. Use what you already know about the power series for 1 1 - x = 1 + x + x 2 + x 3 + · · · , to verify the power series obtain above. Answer: First, just rewrite 1 1 - x 4 = 1 1 - ( x 4 ) , and then just plug it in! 1 1 - x = 1 + x + x 2 + x 3 + · · · 1 1 - ( x 4 ) = 1 + x 4 + x 8 + x 12 + · · · 3. Find the power series for 1 1 + x 7 and then use this power series to find a power series for Z 1 1 + x 7 d x. Also, state the interval of convergence for each series. Answer: First, just rewrite 1 1 + x 7 = 1 1 - ( - x 7 ) , and then just plug it in! 1 1 - x = 1 + x + x 2 + x 3 + · · · 1 1 - ( - x 7 ) = 1 - x 7 + x 14 - x 21 + · · · = X n =0 ( - 1) n x 7 n This series has an interval of convergence ( - 1 , 1). Using this power series to integrate: Z 1 1 + x 7 d x = Z ( 1 - x 7 + x 14 - x 21 + · · · ) d x = C + x - x 8 8 + x 15 15 - x 22 22 + · · · = C + X n =0 ( - 1) n x 7 n +1 7 n + 1 and this series has an interval of convergence ( - 1 , 1) 11
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4. We know that tan x = sin x cos x , and we know that sin x = x - x 3 3!
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