Infimums and supremums..?
Draw a picture. It'll help you to see what's going on. Remember that sin(θ) is the y-coordinate of a point on the unit circle rotated counterclockwise by θ from (1,0)). The closer the point is to the top of the circle, i.e. to (0,1), the greater the sin of its angle, etc. When n = 1, sin(pi/n + pi/4) = -sqrt(2)/2, so -sqrt(2)/2 is in A. Hence inf(A) ≤ -sqrt(2)/2. To show that actually inf(A) = -sqrt(2)/2, we need to show that for all n ∈ ℤ+ we have sin(pi/n + pi/4) ≥ -sqrt(2)/2. We can do that by showing that for all n ∈ ℤ+ we have 5pi/4 ≥ pi/n + pi/4 ≥ -pi/4, i.e. pi ≥ pi/n ≥ -pi/2, i.e. 1 ≥ 1/n ≥ -1/2, which is clearly the case. To find sup(A) is slightly trickier. Notice that the sequence pi/n + pi/4 (n=1,2,3, ...) starts with 5pi/4 and gets less each time, tending to pi/4 from above. However, sin is maximal at pi/2 so we need to find when pi/n + pi/4 "passes" pi/2. In other words, if we can find the least n for which pi/n + pi/4 ≤ pi/2, then either pi/n + pi/4 or pi/(n-1) + pi/4 will be closest pi/2, with the corresponding sin being a maximal element of A. So let's inspect the first few values of pi/n + pi/4: n || pi/n + pi/4 ========== 1 || 5pi/4 2 || 3pi/4 3 || 7pi/12 4 || pi/2 Bingo! You can't get any closer to pi/2 than pi/2. Hence sup(A) = sin(pi/4 + pi/4) = sin(pi/2) = 1. To sum up, inf(A) = -sqrt(2)/2 and sup(A) = 1.
What is the supremum and infimum of 1/2^n?
What is the supremum and infimum of 1/2^n?Assume this is the sequence{[math]a_n[/math]} = {[math]\frac{1}{2^n}[/math]}.If the index [math]n[/math] starts at 0, then the supremum, or least upper bound (LUB), is [math]1[/math].If the index [math]n[/math] starts at 1, the supremum is [math]\frac{1}{2}\text{.}[/math]In either case, the infimum, or the greatest lower bound (GLB), is 0.
What is infimum and supremum?
It is easy to understand this by an example.Consider the set of positive real numbers [math]R^{+*}[/math].(i.e real numbers greater than zero). What is the minimum of this set? Impossible to determine, because any given element in [math]R^{+*}[/math], could be divided into a smaller number, which still lies in [math]R^{+*}[/math](For example, given 1, we can find 0.5).However, we can determine the greatest lower bound for this set i.e a bound below which a positive real number does not exist - the number zero. This number does not not belong to the set of positive real numbers, but a superset of it - the set of real numbers. This number zero is called the infimum of the set of positive real numbers.It is important to understand the thinking here. We can’t find a minimum/lower bound within the set of positive real numbers. Hence we choose/create a superset of it(the set of real numbers) and then find a lower bound within it. This lower bound is not unique (as any number below zero will also form a lower bound). Hence we choose the greatest lower bound.Now consider the set of negative real numbers [math]R^{-*}[/math](i.e real numbers less than zero). We can’t determine the maximum of this set, because for any given element in [math]R^{-*}[/math], we can find a greater number belonging to this set.(For example, given -1, we can find -0.5.).So we choose a superset of this set - the set of real numbers. Within this set, we can find an upper bound, beyond which, a negative real number does not exist - the number zero. This upper bound, is not unique, as any number greater then zero will also form an upper bound. Hence we choose the lowest upper bound. This lowest upper bound is called the supremum of the set of negative real numbers.For the idea of an infimum or supremum to work, it is necessary, that the superset have some property of an order, without which it would be difficult to compare its elements.
Supremum and infimum of empty set?
Usually negative infinity is defined as the supremum and positive infinity as the infimum.consider the supremum, it is the least upper bound. Since the set is empty, any number is an upper bound and negative infinity is the least.
Find supremum and infimum of sets?
1) π/2, noting that arctan x < π/2 for all x, and lim(x→∞) arctan x = π/2. 2) -∞, since we can make x arbitrarily small (negative). 3) Since f(x) = e^x is increasing for all x, and the domain of x in the set is x = 0, the infimum is e^0 = 1, and the supremum is ∞. I hope this helps!
Supremum and infimum, how to show this?
Yes, intuitively clear. It takes practice to be able to pick out just how to approach a formal proof. Mainly you just state what's given, apply definitions, and put down the implications. Eventually you get exactly what you need. This one's not too bad. 1) Assume sup A > inf B Then inf B is not an upper bound for A, and therefore there is some a in A such that sup A > a > inf B Since a is greater than inf B, a is not a lower bound for B, so there must be some b in B such that a > b >= inf B But every a in A is <= every b in B, so that is a contradiction. Therefore sup A cannot be > inf B. 2) Suppose sup A = inf B and eps > 0 Let d = eps/2 Since sup A is the least upper bound on A, there must be some a > supA - d. Similarly, since inf B is the greatest lower bound on B, there must be some b in B such that b < inf B + d (Inequality 1) You also have a > sup A - d Multiply that by -1 to get -a < -sup A + d (Inequality 2) Add Inequalitites 1 and 2 and you have b - a < -supA + inf B + 2d = 0 + 2d = 2d = e That proves one direction. Now assume for any eps > 0, there exists a and b such that b-a < eps. We know that b >= inf B a <= sup A so -a >= - sup A So eps > b-a >= inf B - sup A Since the difference is < eps for any eps > 0, the difference must be <= 0. It can't be < 0 -- that was proven in the first part of the question. Therefore the difference must = 0, and inf B = sup A.
Prove the supremum and infimum of this series?
Call "X" the infinite expression. Then because it's infinite in length of defining it, you can substitute it in another expression of itself. I.e.: X equals sqrt(2+sqrt(2+sqrt(...))) equals sqrt(2+X) You see what I did with that last step? The infinite expression shows up in the sub-expression if you look at it right. X=sqrt(2+X) X^2=2+X X^2 - X - 2 = 0 (X-2)(X+1) = 0 X= 2 or -1 -1 is a false solution introduced by the squaring step. So yes, your supremum is 2.
Infimum and Supremum question?
For a) You've assumed that A is a bounded set of R containing at least two points. The proposition in part (a) follows from that and from the definition of infimum and supremum. Namely, A is bounded (has an infimum and supremum) and the infimum and supremum are disparate (since A contains at least two points). For b) The proposition in part (b) follows from the properties of subsets and from the definition of infimum and supremum. Suppose B is a nonempty subset of A. An infimum is less than or equal to all the elements of a subset, so of course infB
How do you define infimum and supremum to a layman?
There may be many elements of T which are [math]\leq[/math] all the elements of S. Then the infimum of S is the greatest of those elements.For example, consider the set S = {3, 36, 239, 869} as a subset of the natural numbers. What natural numbers are less than or equal to all elements of S? The answer is 0, 1, 2 and 3. So what is the infimum of S? It is the greatest of those numbers, so it's 3.Notice that the infimum doesn't necessarily have to be in the subset. For example, take S to be the open interval (0,1) as a subset of the real numbers. Then the greatest real number which is less than or equal to all elements of S is 0, so that's the infimum.