#1 Beckhorn | #2 Benson | #3 Chan | #4 Chang | #5 | #6 Cohen |
#7 | #8 Greenbaum | #9 | #10 Hedberg | #11 | #12 |
#13 | #14 Oleynick | #15 | #16 | #17 |
1.Prove that 2n-3 <= 2^(n-2) for all n >= 5, n in N.
Base Case:
Induction Step: Sent by Graham Beckhorn on Tue, 25 Feb 2003 21:57:57 |
2. Prove using the definition of limit that
lim((1+2n^2)/(3+4n^2))=1/2.
According to definition 3.1.3. A sequence is said to converge to x if
for every eps > 0, there exists k in N s.t. for all n>=k, the terms xn
satisfy |x_n-x| Sent by Michael Benson on Sun, 2 Mar 2003 21:32:19 |
3.
Suppose a,b are in R and suppose that for every epsilon>0
we have a <= b+ epsilon. a) Prove that a <= b. b) Show by example that it does not follow that a < b.
Since we want to prove that a <= b, we will try to contradict using
a>b. Sent by Man Wai Chan on Tue, 4 Mar 2003 22:17:52 |
4.If A and B are any nonempty sets of numbers which
have the property that a <= b for all a in A and all b in B,
prove: a) sup A <= b for all b in B. b) sup A <= inf B. Proof of a): Let z be any upper bound of A and let u be sup A. From the definition of Supremum, u <= z. Also by the definition of Supremum, for given b in B, a <= b for all a in A, so b is an upper bound of the set A for all b in B. Thus by the Completeness Axiom, sup A exists. This means u <= b which means to sup A <= b, for all b in B. Proof of b): Since sup A <= b for all b in B (from above), sup A is a lower bound of set B. By the Completeness Axiom, inf B must exist. All lower bounds of B are less than or equal to inf B. This means sup A, which is one of the lower bounds, is also less than or equals to inf B. Thus, sup A <= inf B. Sent by "Soyun Chang on Tue, 4 Mar 2003 17:50:12 |
6. Prove that if lim(x_n) = x and if x >,
then there is a positive integer M such that x_n > 0 for all n
gt;=M.
Since lim(x_n) = x for all epsilon > 0, there exists a K(epsilon) that belongs to N such that if n >= K(epsilon), then |x_n-x| epsilon. Since x > 0, let x = epsilon and let M = K(epsilon). Then for n >= M, |x_n-x|<x, which implies -x < x_n -x < x. Adding x to both sides gives 0 < x_n. Sent by Adam Cohen on Sat, 1 Mar 2003 03:28:25 |
8.
Suppose (a_n) and (b_n) are any sequences of real
numbers with the following properties: (a_n) converges, and lim(a_n) is 0. (b_n) is bounded above by 3 and below -3: so |b_n|<=3 for all n in N. Prove that the sequence (c_n) with c_n defined by c_n := a_n · b_n converges, and that lim(c_n) = 0.
We know that (a_n) converges to 0 so 0 <= abs(a_n*b_n) =
abs(a_n)abs(b_n).
We also know that abs(b_n) <= 3 and since abs(a_n) is nonnegative
then abs(a_n)abs(b_n)<= 3abs(a_n)
hence: 0 <= abs(a_n)abs(b_n)<= 3 abs(a_n). Sent by Jackie Greenbaum on Tue, 4 Mar 2003 20:20:36 Further comment by the management Since |&nbps;|"a number"|-0 |=|"a number"-0| it seems apparent that a sequence (w_n) converges to 0 if and only if the sequence (|w_n|) converges to 0. This result is being used implicitly in the solution above, I believe. Another different solution to the problem can be written ("... not that there's anything wrong with" the one that is given). Since (a_n) converges to 0, given blah>0 there is K(blah) in N so that if n>=K(blah), then |a_n-0|&*lt;0. But we need to show that if epsilon>0, there is some element of N so that when n is bigger than it, |c_n-0|<epsilon. Take K(epsilon/3) for that element. Then when n>=K(epsilon/3), |c_n-0|=|a_n·b_n|<(epsilon/3)·3=epsilon, so we are done. |
10. Let (x_n) be an infinite sequence. Assume lim(x_n)=L. Let M
be a positive integer and suppose S={x_n:n>=M}. Prove that inf
S<=L.
Let's prove this via contradiction; assume that inf S > L. Firstly, inf S exists because S in non-empty (x_M is a member of S) and all convergent sequences are bounded, so S is bounded. Because (inf S - L) > 0, let's take epsilon = (inf S - L). Then for n >= K(inf S - L), we have |x_n - L| < (inf S - L). Unrolling this inequality yields: -(inf S - L) < x_n - L < (inf S - L). Now if we add L to all sides we get (2L - inf S) < x_n < inf S. But it is not the case that x_n < inf S: contradiction! Therefore, inf S <= L. Sent by Joe Hedberg on Sat, 1 Mar 2003 13:45:51 |
14. Let a, b, c, d be real numbers such that a < b and c < d.
Prove the inequality bd + ac > ad + bc.
If a < b and c < d, then by the definition of greater than, b - a is an element of P, the set of positive numbers, and d - c is an element of P. By the order properties, the product of two positive numbers is also positive so (b - a)(d - c) is also an element of P. Multiplying out those terms gives bd - bc - ad + ac as an element of P. By the commutative property, bd - bc - ad + ac = bd + ac - ad - bc and by the distributive and associative properties, bd + ac - ad - bc = (bd + ac) - (ad + bc), which is an element of P. If (bd + ac) - (ad + bc) is an element of P then by the definition of greater than, bd + ac > ad + bc. Sent by John Oleynick on Fri, 28 Feb 2003 16:23:51 |