2. Find the converse and the contrapositive of the following statemnents:(a) If m divides 2+j², then m divides i and n divides j (Here, m, i and j are firednumbers.)(b) If f(r) = S(y), then r = (Here the numbers r, y and the function f are fired.)(c) If the sequence (a,) of real mumbers is convergent, then it's bounded. (Here (a,)is fized, )(d) The set A is a subset of the set B if any element a in A is an element of B.(Here,A and B are fized.)

Answered: Image /qna-images/answer/e9b6cdcb-7ec9-40a9-b5ce-30942fc3d2c8.jpg

Answered: 2. Find the converse and the…

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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(Babylonian Method). All numerical answers should be rounded to 6-digit floating-point numbers. (0) If A > 0 is a positive number, then given a real number xo, the sequence (xn) defined recursively by where n > 0 is a natural number, converges to √A, that is, provided that the initial term xo is chosen 'not too badly'. In practice, calculation of terms xn is carried out till this or that stopping criterion is triggered. (i) Let (xn) be the sequence defined recursively by (ii) Generate the terms and x0 = 94 (as it could be guessed from part (0), the sequence (xn) converges to √946). x() = Xn+1= = = (x++) ₁ (n ≥ 0), x1 = x2 = of the sequence (xn) till you find the term x N satisfying the condition x3 = fl(xN) = fl(xN-I (our stopping criterion), where fl(z) denotes the result of rounding of a real number z to a 6-digit floating-point number. Along the way, once the k-th term xk is generated, enter in the corresponding input field the result fl(xk) of rounding of xk to a 6-digit…
For the recursively defined function T(n)= ST(1/2) + 3 Which of the following best reflects the two statements: A) T (n) =_-_ (nlgn) B) T(n) = O(2.25) A. Only A is TRUE B. Only B is TRUE C. Both A and B are TRUE D. Both A and B are FALSE
3. Compute ao, a1, a2, a3, a4 the of the following sequences. (a) ao = 2, an = 3an-1 for n > 1. (b) ao = -5, a, = an-1 +5 for n > 1. (c) ao = 0, an = an-1 +n for n > 1.
4. Prove the following statements using an inductive argument: _n(n+1)(2n+1) 6 72 a. Σiª: = i=1 12 b. If a = 2a- and b = 1-(1-a)², show that a = b for all natural numbers n. Assume that a is a fixed constant. c. Let a be the sequence defined by a₁ =1, a₁ = 8, and a₁ = a₁-11 +2a2 for n ≥ 3. Use an inductive argument to prove that if b₁ =3×2"-¹ +2(−1)", the a = b for all PL 72 neN
5. Let 03 = √5+4√√5 +4√5, a₁ = √5, 9₂ = √5 +4√5, (a) Find the recursion formula (b) Assuming the sequence converges find its limit ⠀
7. Prove that there is no sequence of real numbers whose set of accumulation points is exactly Q.
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Suppose we have a recursive sequence ƒ1, ƒ2, ƒ3, . . .. For the purposes of this problem, it does not matter exactly how the fi are defined, only that they are recursively defined. = i=1 For integer n ≥ 1, let P(n) be the predicate that Σï-1 fi = 2fn+2 − 3. Don't worry about whether this predicate "makes sense"; we haven't defined the fi so you won't be able to "make sense" of the P(n). It's not important for this problem. Consider a proof by induction that Vn ≥ 1: P(n). Suppose we've gotten to the inductive step, and suppose that the first steps of the inductive step are k+1 k+1 Σ1 fi = f1 + f2 + Σh fi = f₁ + f2 + Σkt k+1, i=3 (fi−1 + fi−2) = f₁ + f2 + ¹ fi- 1 + k fi 2 f1 k+1 k+1 k = f1 + f2 + Σh_2 fi + [h=1² fi i=2 = f2 + Σ²²₁ fi+fi i=1 i=1 True or false: Based on the information given, we will need strong induction for this proof. True False
(e) Find a closed formula for the sequence (Cn)neN defined as follows: 1 if n = 1, Cn = 1 if n = 2, 8. Cn-1 · (Cn-2)6 if n ≥ 3. Explain your answer. Hint: While the recurrence for (cn) provided here is not given in a form discussed earlier in the lectures, one can transform the problem into an equivalent one with a non-homogeneous linear recurrence. Try to find such transformation and then solve the obtained non-homogeneous recurrence.

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