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NAME: SID: Midterm 2 Problem 1. (14 points) i) (2 points) State the second fundamental theorem of calculus. ii) (4 points) Using the second fundamental theorem of calculus compute Z a(x) d f (t)dt. dx b(x) Prove your claim. iii) (2 points) Write down the definition of pn (x), the Taylor polynomial of f at x0 (assume f has enough derivatives). iv) (2 points) Write down the main statement of the Lagrange remainder theorem. v) (2 points) Write down the formula for βk for β ∈ R and k ≥ 1. What is the definition β0 ? vi) (2 points) Write down Newton’s binomial expansion, that is the the series expansion for (1 + x)β for β ∈ R and |x| < 1. Rx Solution. i) If f : [a, b] → R is continuous, then F (x) = differentiable at all x ∈ (a, b). Moreover, Z x d F 0 (x) = f (t) dt = f (x). dx a a f (t) dt is ii) d dx Z a(x) b(x) d f (t) dt = dx "Z d = dx "Z a(x) f (t) dt + 0 # 0 Z f (t) dt b(x) a(x) f (t) dt − 0 # b(x) Z f (t) dt 0 = a0 (x)f (a(x)) − b0 (x)f (b(x)). iii) pn (x) = n X f (k) (x0 )(x − x0 )k k! k=0 . iv) Let f (x) and pn (x) be functions as in part iii). For each x, there exists c (depending on x, x0 and n) between x and x0 such that f (x) = pn (x) + f (n+1) (c)(x − x0 )n+1 (n + 1)! 1 2 v) β (β)(β − 1) . . . (β − k + 1) = , k k! β =1 0 vi) β (1 + x) = ∞ X β k k=0 xk if x ∈ (−1, 1). Problem 2. (12 points) Let f : R → R be a continuous function and g : R → R be a differentiable function. a) (4 points) Compute Z x d g(x)f (t)dt. dx 1 b) (8 points) Define x Z (x − t)f (t)dt. G(x) = 1 Prove that G00 (x) = f (x). Solution. a) Note that Z x x Z g(x)f (t)dt = g(x) f (t)dt 1 1 Using the product rule and the fundamental theorem of calculus Z x Z x d 0 g(x)f (t)dt = g (x) f (t)dt + g(x)f (x). dx 1 1 b) We write Z x Z f (t)dt − G(x) = x 1 x tf (t)dt. 1 and use part a) to compute Z x Z 0 G = f (t)dt + xf (x) − xf (x) = 1 x f (t)dt. 1 Next, it follows G00 (x) = f (x). Problem 3. (10 points) Use the Lagrange remainder theorem to prove that: 1 x 2x2 x 1+ − < (1 + x) 5 < 1 + , ∀x > 0 5 25 5 Solution. From the Lagrange remainder theorem, it follows 1 f (x) = (1 + x) 5 = 1 + x f 00 (c) 2 + x , 5 2! 0 < c < x. 3 Since f 00 (c) = 1 5 9 9 4 · (− 45 )(1 + c)− 5 = − 25 (1 + c)− 5 and 0 < c < x we have 2x2 f 00 (c) 2 < x < 0. 25 2! from which the claimed inequality follows. − 4 Problem 4. (14 points) Let f : R → R which has derivatives of all orders. Assume that f 00 (x) = 2f 0 (x), ∀x ∈ R. a) Find a recursive formula for the coefficients of the Taylor series of f at x0 = 0. b) Prove that f equals its Taylor series at x0 = 0 for all x ∈ R. c) Consider now the Taylor series of f at x0 6= 0. Does f equal its Taylor series for all x ∈ R? Solution. a) From f 00 (x) = 2f 0 (x), ∀x ∈ R it is easy to see (by induction) that, for n ≥ 2, f (n) (x) = 2f (n−1) (x), ∀x ∈ R and f (n) (x) = 2n−1 f 0 (x), ∀x ∈ R. We thus have, for n ≥ 2, f (n) (0) 2f (n−1) (0) 2(n − 1)!an−1 2an−1 = = = . n! n! n! n b) Fix r ∈ R and denote I = [−r, r]. Since f (n) (x) = 2n−1 f 0 (x), ∀x ∈ R we obtain an = |f (n) (x)| ≤ 2n−1 |f 0 (x)| ≤ 2n−1 sup |f 0 (x)| ≤ 2n−1 M x∈I |f 0 (x)|. Using a theorem in the book we conclude that where M = supx∈I f equals its Taylor series on [−r, r] and since r is arbitrary, we obtain the result on R. c) The analysis above carries on for any other point x0 ∈ R, simply replace I = [x0 − r, x0 + r].