Recent content by bolzano

Hi, thanks for the replies. However is this something trivial or is it hard to prove (that thery're equal a.e.)?

Hi, I was wondering whether if ∫f×g dμ=∫h×g dμ for all integrable functions g implies that f = h? Thanks

But are you assuming that ##\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)##? Oh wait I think I understood. I'll let you know later thanks a lot.

This is the last part to a Fourier Series problem. After using Parseval's Identity the following series emerges: \frac{1}{1^{4}}+\frac{1}{3^{4}} +\frac{1}{5^{4}} +\frac{1}{7^{4}}+\cdots = \frac{\pi^{4}}{96} We are then asked to show: \frac{1}{1^{4}}+\frac{1}{2^{4}} +\frac{1}{3^{4}}...

I think it is true, since the shape is a sphere radius one centered at the origin (the infinite cylinder equation would be x^{2}+y^{2}=1). Think of projecting the sphere onto the xy-plane. What would it's shadow look like?

So (\sqrt[n]{a_{n}})=2r,r,\sqrt[3]{2}r,r,\sqrt[5]{2}r,... which converges to r right? :)

Hi i have to show that the series 1+2r+r2+2r3+r4+2r5+... converges for r=\frac{2}{3} and diverges for r=\frac{4}{3} using the nth root test. The sequence \sqrt[n]{a_{n}}comes a bit complicated so i was wondering if I can remove the 1st term a1=1 and show that 2r+r2+2r3+r4+2r5+... converges...

Thanks a lot! :)

Hi tiny-tim :) Do i draw a line between A and C and use it as diameter? :)

Can someone please explain how i can find a point X on the tangent line such that AX and XC are perpendicular (A is the centre)? Thanks : ) Here's a diagram: http://img856.imageshack.us/i/62852104.png/

Yes precisely, the sequence I'm considering obviously isn't Cauchy, otherwise the space won't be complete. In fact the theorem tells us that the sequence can't be Cauchy since it converges to a function "outside" the space of Cont. Functions. However i really felt that this sequence was...

Hi Landau, i know that the sequence converges pointwise and the limit function is discontinuous on [0,1]. However in the metric space we're treating the functions as points; have You looked at the metric we're using on [0,1]? It's d(x,y) = max|x(t)-y(t)| for all t in [0,1]. The convergence...

Well intuitively it seems so, but later on i tried checking this out as follows: i) compute the derivative of x^(n-1) - x^n. ii)Equating this to zero will give the point x in [0,1] at which this is highest. iii)Substituting this value back into x^(n-1) - x^n will give the disatance...

Hi I'm using Kreyszig's Introductory Functional Analysis with Applications and he proves that the set of continuous functions on an interval [a,b] under the metric d(x,y) = max|x(t)-y(t)| is complete. Standard proof nothing hard over there. But isn't the sequence of x^n s on [0,1] a Cauchy...