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Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.

Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0. 35. v aniwollo 36. sin x?
Transcribed Image Text:Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0. 35. v aniwollo 36. sin x?
Table 11.5 1 1 + x + x² +· . .+ x* + • Σ Ex, for x < 1 1 k=0 00 1 = 1 – x + x² – . ..+ (-1)*x* + . (-1)**, for |x| < 1 for x< 1 1 + x k=0 X et = 1 + x + 2! + k! for x < ∞ k!' k=0 (-1)* x2k+1 +.. Σ (-1)k x2k+1 (2k + 1)! sin x = x for x < ∞ 3! 5! (2k + 1)! k=0 (-1)* x* (2k)! (-1)* x Σ x² cos x = 1 for x < 0 2! 4! (2k)! k=0 (-1)*+1 xk (-1)*+ly* In (1 + x) = x Σ for -1 < x< 1 %3D 2 k k k=1 x² -In (1 – x) = x+ x3 th k for -1 < x < 1 %3D k ' k=1 tan x = x - 3 (-1) x2k+1 (-1)* x²k+1 +. .. = for x < 1 5 2k + 1 2k + 1 k=0 x2k+1 00 x2k+1 sinh x = x +t 3! +...+ 5! for x < ∞ (2k + 1)! k=0(2k + 1)! cosh x = 1 + 2! %3D +. (2k)! for x < ∞ 4! k=0 (2k)!' (1 + x)* = . r*, for x<1 and k (P) = P(p - 1)(p - 2) · (p – k + 1) | %3D k 0 = 1 k! | ल
Transcribed Image Text:Table 11.5 1 1 + x + x² +· . .+ x* + • Σ Ex, for x < 1 1 k=0 00 1 = 1 – x + x² – . ..+ (-1)*x* + . (-1)**, for |x| < 1 for x< 1 1 + x k=0 X et = 1 + x + 2! + k! for x < ∞ k!' k=0 (-1)* x2k+1 +.. Σ (-1)k x2k+1 (2k + 1)! sin x = x for x < ∞ 3! 5! (2k + 1)! k=0 (-1)* x* (2k)! (-1)* x Σ x² cos x = 1 for x < 0 2! 4! (2k)! k=0 (-1)*+1 xk (-1)*+ly* In (1 + x) = x Σ for -1 < x< 1 %3D 2 k k k=1 x² -In (1 – x) = x+ x3 th k for -1 < x < 1 %3D k ' k=1 tan x = x - 3 (-1) x2k+1 (-1)* x²k+1 +. .. = for x < 1 5 2k + 1 2k + 1 k=0 x2k+1 00 x2k+1 sinh x = x +t 3! +...+ 5! for x < ∞ (2k + 1)! k=0(2k + 1)! cosh x = 1 + 2! %3D +. (2k)! for x < ∞ 4! k=0 (2k)!' (1 + x)* = . r*, for x<1 and k (P) = P(p - 1)(p - 2) · (p – k + 1) | %3D k 0 = 1 k! | ल
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