Meital

Has anyone studied Differential Geometry from Do Carmo's book, Riemannian Geometry? I am trying to work problem 4 page 57, but I don't even know how to start. I will scan the problem and post it laer.

ok i figured it out!

ODE/Sturm-Liouville problem. consider the Sturm-Liouville problem: y" + [ lambda p (t) - q(t)] y= 0 in (0,1), alpha y(0) + beta y'(0) = 0 gamma y(1) + delta y'(1) = 0, Where alpha, beta, gamma, delta are real constants, and p: [0,1] -> R and q:[0,1] -> R are continuous functions with p(t) > 0. (a) Suppose alpha*beta doesn't equal 0. Show that if f_n(t) and g_n(t) are eigenfunctions associated with a given eigenvalue lambda_n of the Sturm-Liouville problem, then f_n(t) = c g_n(t), t belongs to [0,1], for some constant c in R. (b) Can one remove the restriction alpha*beta doesn't equal to 0 in part (a)? [ Explain and justify your answer.]

Prove or disprove: If the boundary of set omega in R^d has an outer measure zero, then omega is Lebesgue measurable. I was trying to come up with a counter example, but I couldn't. Then I tried to prove it, yet I was not able to do so. Please help me

If A is Lebesgue measurable and x is an element of Rn, then the translation of A by x, defined by A + x = {a + x : a ∈ A}, is also Lebesgue measurable and has the same measure as A.

Can you guys tell me if my answer is correct? Determine if the following functions satisfy local or uniform Lipschitz condition. 1). te^y I found d/dy (te^y) = te^y, and this is not bounded above for any value of y, so this made me conclude that it has locally Lipschitz condition since the Lipschitz constant here changes as the reagion changes? Am I right? I used the equation | f(t,y_1) - f(t, y_2) | = d/dy(f(t,y)) + | y_1 - y_2| 2). y t^2/ (1 + y^2) I used the same approach here and d/dy (f ) = t^2 - 2y + y^2/ ( 1 + y^2)^2, which is clearly could be bounded above by a constant but this constant changes as the reagion changes so it is local lipschitz.

I am reading ODE ( Ordinary DE) notes, and there is a statement that says " A function that is not uniform, even a continuous function that is not uniform, cannot have a lipschitz constant. As an example is the function 1/x on the open interval (0,1). I want to see if I understood this correctly. I will assume that we can find a number k >= 0 such that | f(x) - f(y) | =< K*|x - y| we have x and y in (0,1) so 0 < x < 1 0 < y < 1 .....(1) Now, | 1/x - 1/y | =< k*|x - y| Find common denominator | y - x|/|xy| =< k*|x-y| | y - x | = |x - y| then we cancel the term from both sides of inequality, so we get 1/|xy| =< k ...($) k >= 0, and from (1) 1/xy > 1 so is 1/|xy| > 1 but then the equation ( $ ) becomes 1 < 1/|xy| =< k , but k >= 0 so this is contradiction? I am not sure, can someone tell me why 1/x doesn't have a lipschitz constant? I mean since we are in (0,1) then there is no upper bound for 1/|xy| so we can't have an upper bound for it (lipschitz constant)

I figured it out..I don't know why I thought it was difficult! Thanks though

Can someone help me with the following proof: Suppose f ,g: X -> [- infinity, + infinity] are measurable. Prove that the sets { x : f(x) < g(x) } , {x: f(x) = g(x) } are measurable.

ok i figured it out

I am trying to find the set A such that For r > 0 let A ={w, w = exp (1/z) where 0<|z|<r}.

what is the radius of convergence of the sum (n=0 to infinity) of z^(n!) ?

ok I got it..no need for any replies..thanks though

Can someone find an example of a sequence such that it doesn't have a limit, but it has a lim sup? and find the lim sup? I thought about the sequence of all rational numbers in the interval [0,1], but not sure if that's a correct example.

I see, so we can take the colletion: R ( real numbers), open set (0,2), and the empty set. This is topology since it satisfies all 3 axioms of topology, but not sigma-algebra because we don't have the comp of (0,2) in the collection.

Can someone give me an example of a topology, which is not sigma algebra?

Stereographic projection on complex plane -------------------------------------------------------------------------------- Let V be a circle lying in S. Then there is a unique plane P in R^3 such that p /\ S = V ( /\ = intersection). Recall from analytuc geomerty that P = { (x_1,x_2,x_3) : x_1 b_1 + x_2 b_2 + x_3 b_3 = L, where L is a real number}. Where ( b_1,b_2,b_3) is a vector orthogonal to P . It can be assumed that (b_1)^2 + (b_2)^2 + (b_3)^2=1. Use this information to show that if V contains the point N then its seteographic projection on the complex plane is a straight line. Otherwise, V projects onto a circle in complex plane. N = (0,0,1) the north pole on S Please any hints will be appreciated!!

Actuallty this Q is from a book called " functions of one complex variable I" 2nd edition.

cis k = cos k + i sin k

Can someone help me with this problem. I need any hints to help me get started. I am assuming limit points of cisx are -1- i and 1+i I am not sure if this is correct. And if it, I don't know how to show that it is dense, I know the def of a dense subset, but I don't know how to apply it here. Please help me. Show that {cis K : K is a non-negative integer) is dense in T = { z in C ( Complex space) : |z|= 1}. For which values of theta is {cis ( k*theta) : k is non-negative integer} dense in T?

By using the def of e^x ( series function) and considering the sequence of partial sums, how can you prove that f(x) = e^x is differentiable on R and that (e^x)' = e^x?

Basically it is the same Q I have on page 153 of James R.Munkres book, Topology 2nd edition. My question is on Theorem 24.1 in case 1, how does it follow that there exists some interval of the form (d,c] in B_o, just from knowing that B_o is open in [a,b]?

A is a non-empty subset of [a,b], where [a,b] is a subset of an ordered, linear continuum space. Now that A is non-empty, let's assume b and x are 2 elements of A. Now knowing that x is in A, and that A is open, can we say that there exists y < x such that (y,x] is in the subset A? If so, how to show that. This claim is part of a proof I am doing to prove that [a,b] is connected. So I assumed that 2 non-empty disjoint open sets of [a,b], call them A and B, cover [a,b], ( as you see I am doing proof by contradiction), and the claim stated above ( my question) is where I am stuck at..so can you help me please?

I have a general question on properties of intervals from a topological point of view( intervals in any ordered , linear continuum space, not only reals), so if we have a non-empty open set A in [a,b], and let b be in A for example, and let x < b , x in A. I need to prove that there exists an interval (y,x] in A. I believe that the fact that A is open and the betweenness of A has to do something with that proof. Again, A is any subset of [a,b], not necessary a convex subset. Will someone help me with this, please? Thanks in advance.

Abd-Al-Karim's answer makes more sense to me..any more puzzles redalert?