(A.17) cos(x) 2 + sin(x) 2 = 1. Let us also recall basic properties of complex numbers at this point: For every complex

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answerhappygod
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(A.17) cos(x) 2 + sin(x) 2 = 1. Let us also recall basic properties of complex numbers at this point: For every complex

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(A.17) cos(x) 2 + sin(x) 2 = 1. Let us also recall basic
properties of complex numbers at this point: For every complex
number z ∈ C there exist a, b ∈ R, r ≥ 0 and φ ∈ [0, 2π) such
that
A 17 Cos X 2 Sin X 2 1 Let Us Also Recall Basic Properties Of Complex Numbers At This Point For Every Complex 1
A 17 Cos X 2 Sin X 2 1 Let Us Also Recall Basic Properties Of Complex Numbers At This Point For Every Complex 1 (192.43 KiB) Viewed 36 times
3. Taylor's theorem THEOREM A.17. Let I be an interval and let f e C'n+1(I), i.e all derivatives of f up to order n +1 are continuous in I. Fix a E I. Then for all x E I. f(k)(a) (A.34) f(α) = Σ -(x – a)* + Rn(x, a) k! n k=0 where = a R.(5,0) = 1 * (1974)" pella+1)(e)dt fa n! (x – a)n+1 n! (A.35) (1 – s)" f(n+1)(a + s(x – a))ds PROOF. We first observe that the second version and the first version of the remain- der term are equivalent by changing variables (via the substitution t = a + s(x – a), dt = (x – a)ds; note that t ranges from a to x as s ranges from 0 to 1). For n=0 the formula reads = (A.36) f(x) = f(a) + = + [*f'(t)dt which just follows from the fundamental theorem of calculus. We also find that by integration by parts for f e C(n+2)(I) ( * -t) not f(n+1)(0) C -(x – t)n++ f(n+2)(t)dt [ (– t)" g(n+)(E) dt : - | n+1 2 n+1 (x – a)n+1 n +1 - f(n+1) (a) + /* (* — 4)+1 -f(n+2)(t)dt. n+1 which shows 1 (A.37) Rn(x, a) = (x – a)n+1 - f(n+1)(a) + Rn+1(x, a) n+1
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