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Problem 6. -x² we get, justification) to get o(-1)"x²n Observe that for x = 1 we obtain Calculating 7: From the geometric series on , if |x| < 1, replacing a for 12, if x < 1. It is tempting to integrate each side of that equation (but this needs a formal x x²n+1 2n + 1 Σ(-1)". n=0 ∞ Σ (-1)" n=0 1 = tan x. 1 2011 - 2n + 1 πT This formula is usually attributed to G. Leibnitz, but also to I. Newton and to J. Gregory. Nevertheless, much earlier in the region of Kerala in southwest India, it was already known by Nilakantha and Madhave around the XV century. This series converges very slowly. Compute the first 5 partial sums of it.

Problem 6. -x² we get, justification) to get o(-1)"x²n Observe that for x = 1 we obtain Calculating 7: From the geometric series on , if |x| < 1, replacing a for 12, if x < 1. It is tempting to integrate each side of that equation (but this needs a formal x x²n+1 2n + 1 Σ(-1)". n=0 ∞ Σ (-1)" n=0 1 = tan x. 1 2011 - 2n + 1 πT This formula is usually attributed to G. Leibnitz, but also to I. Newton and to J. Gregory. Nevertheless, much earlier in the region of Kerala in southwest India, it was already known by Nilakantha and Madhave around the XV century. This series converges very slowly. Compute the first 5 partial sums of it.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 6. -x² we get, n=0 justification) to get o(-1)x²n Observe that for x = 1 we obtain Calculating : From the geometric series Σ0x² xn = 1¹, if |x| < 1, replacing a for 1+2, if |x| < 1. It is tempting to integrate each side of that equation (but this needs a formal ∞ x²n+1 2n + 1 Σ(-1)". n=0 -1 = tan X. 1 2n + 1 Σ(-1). = ㅠ n=0 This formula is usually attributed to G. Leibnitz, but also to I. Newton and to J. Gregory. Nevertheless, much earlier in the region of Kerala in southwest India, it was already known by Nilakantha and Madhave around the XV century. This series converges very slowly. Compute the first 5 partial sums of it.
Transcribed Image Text:Problem 6. -x² we get, n=0 justification) to get o(-1)x²n Observe that for x = 1 we obtain Calculating : From the geometric series Σ0x² xn = 1¹, if |x| < 1, replacing a for 1+2, if |x| < 1. It is tempting to integrate each side of that equation (but this needs a formal ∞ x²n+1 2n + 1 Σ(-1)". n=0 -1 = tan X. 1 2n + 1 Σ(-1). = ㅠ n=0 This formula is usually attributed to G. Leibnitz, but also to I. Newton and to J. Gregory. Nevertheless, much earlier in the region of Kerala in southwest India, it was already known by Nilakantha and Madhave around the XV century. This series converges very slowly. Compute the first 5 partial sums of it.
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