For reflexive, show that (m, n) ~ (m, n) is true. For symmetric, show that if (m, n) ~ (p, q) then (p, q) ~ (m, n). Just check these explicitly to see if they work out.
Is it basically by definition? Since f is continuous, f-1(N) is a neighborhood of x for every neighborhood N of f(x). Maybe I don't understand it, heh.
Ok, if I use that fact, then if f(x) != g(x), there exist subsets U, V disjoint and open with f(x) in U and g(x) in V. Since f is continuous, f-1(U) is open in X (because U is open in Y). Similarly, g-1(V) is open in X (because V is open in Y). Thus {x in X | f(x) != g(x)} is open, and it's...
Homework Statement
Let X be a topological space, Y a Hausdorff space, and let f:X -> Y and g:X -> Y be continuous. Show that {x \in X : f(x) = g(x)} is closed. Hence if f(x) = g(x) for all x in a dense subset of X, then f = g.
Homework Equations
Y is Hausdorff => for every x, y in Y with...
Homework Statement
Let f be of class C1 on [a, b], with f(a) = f(b) = 0. Show that \int_a^b xf(x)f'(x)dx = -1/2 \int_a^b [f(x)]^2 dx.
Homework Equations
If F is an antiderivative of f, then \int_a^b f(t)dt = F(b) - F(a)
The Attempt at a Solution
I'm just really not sure how to begin this...
Ok, I have another Riemann sum, but this time, it actually says what the integral evaluates to. The question is the following:
Show that the limit as n -> infinity \sum_{k=1}^nn/(n2+k2) = pi/4.
So I'm thinking that f(x) = sqrt(1-x2) on [0, 1]. But I'm having a hard time figuring out what to...
As usual, I'm just not seeing it. :redface: I guess the k^3 should be a dead giveaway, but I'm not sure what to do about the n^4. Is the limit = \intx^3 from a to b?
By using the definition of a Riemann sum, I get: \sum_{k=1}^n \xi3(xk-xk-1). But what are the xks?
Homework Statement
Find the limit, as n -> infinity, of \sum_{k=1}^nk3/n4
Homework Equations
Riemann sum: S(f, \pi, \sigma) = \sum_{k=1}^nf(\xi)(xk - xk-1)
The Attempt at a Solution
My guess is that I should try to put this sum in terms of a Riemann sum, and then taking n -> infinity will...
Homework Statement
Let p > 1, and put q = p/(p-1), so 1/p + 1/q = 1. Show that for any x > 0, y > 0, we have
xy <= xp/p + yq/q, and find the case where equality holds.
Homework Equations
The Attempt at a Solution
This is in the differentiation chapter of my analysis book (Browder)...
How do I know that AC > AD from that? If that's true, then I have a contradiction, since AB is supposed to be the longest side, but AC is longer than it.
Homework Statement
Let ABC be a triangle, and suppose that AB is the largest side. Prove that the perpendicular from C to the line AB crosses at some point D with A*D*B.
Homework Equations
The Attempt at a Solution
I know that since AB is the largest side, that angle ACB is the largest...
OK, let's see if this works.
Define an = (x2 + n)/n2. lim(an) = 0, so the series \sum(-1)nan converges. By lim(an) = 0, there exists N such that n > N implies an < \epsilon.
Then |\sum_{n=0}^\infty(-1)nan - \sum_{n=0}^N(-1)nan| < aN+1 < \epsilon. So the series converges uniformly.
Is...
I'm not sure why what I did was incorrect. I'm not really sure what you're getting at here. Sorry. :redface: What do you mean by "estimate the error you made"?
Thanks for your response, cellotim.
How about if I use the fact that (g_n) converges uniformly to g on I = (a, b) iff lim(sup|g(x) - g_n(x)|) = 0, where x is in I.
So sup|g(x) - g_n(x)| = sup(x2 + n)/n2 = (b2 + n)/n2. So lim(sup{(b2 + n)/n2}) = 0, and the series is uniformly convergent for...
Homework Statement
Show that the series \sum(-1)n(x2+n)/n2 is uniformly convergent on every bounded interval in R, but is not absolutely convergent for any x.
Homework Equations
Weierstrass M-Test
The Attempt at a Solution
Take g_n(x) = (-1)n(x2+n)/n2. Then |g_n(x)| = (x2+n)/n2. To be...
Homework Statement
1. Suppose A * B * C and A * C * D.
a) Prove that no two of A, B, C, D are all equal.
b)Prove that A, B, C, D are all on one line.
2. Suppose that A, B, C are points not all on one line. Prove that AB and BC have no points in common except B.
Homework Equations...
For induction, you have to prove the base case. Then you assume your induction hypothesis, which in this case is 2n >= n2. After that you want to prove that it is true for n + 1, i.e. that 2n+1 >= (n+1)2. You will use the induction hypothesis in the proof (the assumption that 2n >= n2).
Homework Statement
Prove that if fn -> f uniformly on a set S, and if gn -> g uniformly on S, then fn + gn -> f + g uniformly on S.
Homework Equations
The Attempt at a Solution
fn -> f uniformly means that |fn(x) - f(x)| < \epsilon/2 for n > N_1.
gn -> g uniformly means that |gn(x) -...
If you look at the graph (the image has finally been approved), you will see that the origin and also the two end points of the S all lie on the line y = x, so there are actually 3 points that the line y = x intersects. The end points of that graph should really be labeled, though.
The number of points on the line y = x is infinite, but the number of points where the line y = x intersects your "S" is not. Draw the line y = x as well as the "S," and the points where the lines cross will be the intersection points.
For example, in...
The line y = x consists of all the points where *drumroll...* y = x. For example, (0, 0), (1, 1), (2.439, 2.439), etc. Plot two of these points on the graph (maybe (0, 0) and (1, 1) ). The straight line that connects these points is y = x. It extends out infinitely past both points.
If a1 >= a2 >= ... >= an >= ... >= 0 and lim(an) = 0, then the alternating series \sum(-1)nan converges. But, like I said before, do I know that a1 >= a2 >= ... >= an because the series converges at R?
I don't know that one. But the comparison test in my book says the following:
Let \suman be a series where an >=0 for all n.
(i) If \suman converges and |bn| <= an for all n, then \sumbn converges.
If I let an = anRn, this is >=0 for all n. And if I let bn = an(-R)n, then I have |bn| <=...
Homework Statement
Suppose that \sumanxn has finite radius of convergence R and that an >= 0 for all n. Show that if the series converges at R, then it also converges at -R.
Homework Equations
The Attempt at a Solution
Since the series converges at R, then I know that \sumanRn = M...
Ok, for the kernel, since M(T, B) is diagonal, M(T, B)*(x) = 0 = M(T, B)k*(x), where (x) is a 1 x n vector consisting of all 0s. So Ker(T) = Ker(Tk).
For the image, since M(T, B) and M(Tk, B) are both diagonal matrices, when multiplied by a vector, they will produce another vector
\left(...
It's not that I don't want to; I just didn't think of it! I'm not sure if I know what you're saying.
So if T can be diagonalized, then M(T, B) is a diagonal matrix for some basis B. If M(T, B) = 0, then M(T, B)k = 0. But if M(T, B)k = 0, this doesn't necessarily mean that M(T, B) = 0. This...
Let Tk(v) = Tk-1(T(v)) = 0. By the induction hypothesis, since Tk-1(T(v)) = 0, then T(T(v)) = T2(v) = 0. I'm not seeing where the normality of T and the adjoint will come into play here.
Homework Statement
Prove that if T in L(V) is normal, then Ker(Tk) = Ker(T) and Im(Tk) = Im(T) for every positive integer k.
Homework Equations
The Attempt at a Solution
Since T is normal, I know that TT* = T*T, and also that ||Tv|| = ||T*v|| and <Tv, Tv> = <T*v, T*v>.
Ker(T) is the...
Thanks for the help! Let's see if I got it...
limsup(a_n + b_n) = sup(A + B) = lim(a_n + b_n) = lim(a_n) + lim(b_n). Since (a_n), (b_n) are <= sup(A), sup(B), respectively, then so are lim(a_n) and lim(b_n). So we have limsup(a_n + b_n) <= sup(A) + sup(B). But we have sup(A) = sup(a_n) =...
Homework Statement
Show that limsup(s_n + t_n) <= limsup(s_n) + limsup(t_n) for bounded sequences (s_n) and (t_n).
Homework Equations
The Attempt at a Solution
My book gives a hint that says to first show that sup{s_n + t_n : n > N} <= sup{s_n : n > N} + sup{t_n : n > N}. I'm not...
You're right, it is a doozy! This series is known as a harmonic series, and according to http://plus.maths.org/issue12/features/harmonic/index.html" [Broken], "there is no simple formula, akin to the formulae for the sums of arithmetic and geometric series, for the sum."
Cool. Thanks for the confirmation. Now I'm trying to figure out why I can say that even though my matrix equals its conjugate transpose, it is not a contradiction. I'm thinking it has to do with how sparse the matrix is, specifically on the diagonal, but I'm not sure. Anyone? Thanks!
Can I just do this:
An operator is self-adjoint iff <Tv, w> = <v, Tw> for all v, w. So using the defined inner product, we have <T(a_0 + a_1*x + a_2*x2), b_0 + b_1*x + b_2*x2> = (1/2)*a_1*b_0 + (1/3)a_1*b_1 + (1/4)*a_1*b_2.
But we also have <a_0 + a_1*x + a_2*x2, T(b_0 + b_1*x + b_2*x2)> =...
:rolleyes: I wasn't paying attention to the problem and was calculating the inner-product wrong. It's obviously defined to be that integral between 0 and 1, not the dot product as I was doing before.