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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...

Got it. Thanks!

Thanks statdad, but I've already figured that one out with Dick's help. See the post right before yours for the one I'm now having trouble with.

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...

Cool. Thanks!

Ok, so if I choose \xi = k/n = xk, then I get \sum(k/n)3(k/n - (k-1)/n) = \sum(k3/n3)(1/n) = \sumk3/n4. So the limit of this sum is \int_{0}^1x^3 ?

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?

You should be calculating \int2sqrt(x) - x from 0 to 4.

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...

Use the product rule with -2x and sin(x^2). You'll have to use the chain rule again when you calculate the derivative of sin(x^2).

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)...

Duh! :rolleyes: Thanks a lot for your help. :smile:

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...

I think it should be g(x) = f(-x + pi). There must have been a typo in the book. g(x) = pi - (-x + pi) = pi + x - pi = x.

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...

It doesn't span R3, but it does span R2, which you can think of as a plane in R3.

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?

Ok! :smile: Thanks once again for your help.

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| <=...

Thanks! So since \sumanRn converges, and an(-R)n <= anRn for all n, then \suman(-R)n converges.

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...

That's ok! Thanks for your help. Does anybody else see a strategy for proving 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."

Oh, sorry. I read the original post too quickly.

Maybe I'm misunderstanding your question, but wouldn't it just be \sum_{n=1}^\infty 1/n ?

I see now. Thanks for the clarification. :smile:

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!

ln(e^x) = x. You can take the natural log of both sides of an equation.

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.