i. (2nd Principle of Induction): Suppose that a1 = 2 and a2 = 4 and for n > 2, an = 5an-1 – 6an-2. Prove that for all n e N, an = 2". (This is easy. Show precisely where you need the 2nd Principle.)
Exercise 1 Consider utility maximization problem: Kuhn Tucker Theorem max U (x1, x2) = x1+x2 21,22 subject to Tị 2 0, r2 > 0, p1x1+p2r2 < I, where p1, p2 and I are positive constants. Exercise 1 Consider utility maximization problem: Kuhn Tucker Theorem max U (x1, x2) = x1+x2 21,22 subject to Tị 2 0, r2 > 0, p1x1+p2r2 < I, where p1, p2 and I are positive constants.
Let s = {k=1CkXAz be a simple function, where {A1, A2, ... , An} are disjoint. Prove that for every p>0, |CK|PXAR
3. (14 pts.) Let the sequence an be defined by ao = -2, a1 = 38 and an = 2an-1 + 15an-2 for all integers n > 2. Prove that for every integer n > 0, an = 4(5") + 2(-3)n+1.
Let a1 = 3 and an+1 = a + 5 2an for all n > 1 Prove that (an)nen converges and find limn7oo an:
5: Question 3. - 10.0 pts possible Four charges are placed at the corners of a square, where q is positive (q > 0). There is no charge in the center of the square. Q1 = -9 Q2 = +9 ° 1. U = 2 v7 keq? 02.0 = 2 ke q² a © 3. U = 4 va keq? a ° 4. U = vz kieg? a Q4 = -2 Q3 = + O5. U = 2 ke q²...
Proposition 7.27. Suppose fn: G + C is continuous, for n > 1, (fn) converges uniformly to f :G+C, and y C G is a piecewise smooth path. Then lim n-00 $. fn = $. . 7.23. Let fn(x) = n2x e-nx. (a) Show that limn400 fn(x) = 0 for all x > 0. (Hint: Treat x = O as a special case; for x > 0 you can use L'Hôspital's rule (Theorem A.11) — but remember that n is...
Extra Credit Question:[4+4=8 pts) If E [exp(aX)] exists for a given constant a, then show that for to (a) exp(-at)P(x >t) <E (exp(aX)], if a > 0. (b) exp(-at)P(X <t) <E (exp(aX)], if a < 0.
2 Double summation Let a1, A2, A3, ... be a sequence of real numbers, and let n > 1 be an integer. Which of the following are always equal? пі пп nn nn « ££« Žia. E£« L« i=1 j=1 i=1 j=1 i= 1 i=1 j=1 j=1 i=j
2) (3 pts) Use mathematical induction to show that when n is an exact power of 2, the solution of the recurrence 2, ifn=2 T(n) =127G)+n, ifn=2.for k > 1 ISI(72) = n lg n.