Recent content by eeMath

I see... I was confusing continuity as lining up everything perfectly. Even if the terms overlap, the bumps will just look a little wobbly over the overlapping interval. One last thing, for L1[0,infinity) do the centers always need same distance from each other, or can the distance get larger...

But if I pick some sequence for ck that goes to infinity, how will the bumps line up to create a continuous function? I may be totally missing something here - could you please give me an example of a function for ck that goes to infinity and how the whole function would look? At this point I am...

So I figured out a function that starts with an area of 2 under the first bump and each successive area is cut in half... f(t) = \sum^{\infty}_{k=0} abs{cos(2^{k}(t - \frac{3(2^{k-1})\pi }{2^{k+1}}} where \frac{3(2^k)\pi - 4 \pi}{2^{k+1}} \leq t \leq \frac{3(2^k)\pi - 2 \pi}{2^{k+1}} The...

Both astrology and astronomy are very interesting fields of study. If you can carry out an experiment based on measurable factors and come to logical conclusions based on that experiment, its a science.

In the title.

Totally; I think I was momentarily confusing the value of the function (will always fluctuate between 1 and 0) and the area under the curve. So I guess the final thing I don't know how to prove is that the integral for the entire absolute value of the function is less than infinity so that the...

Thanks, this should help, but I meant how do I show that that the limit as x-> doesn't exist?

The function f=abs(cos x) satisfies f(0)=1 and f(pi/2)=0. The area under the curve is 2 for every bump (-pi/2 to pi/2). If you take f(2x), the area under the curve is 1 for the bump (-pi/4 to pi/4). I see how the area and interval are cut by the value of c, so if the c increases each bump has...

Thanks for the reply... With abs(sin(e^x)) I have a function that does that, but I am getting a limit of 0 according to wolfram alpha... am I doing something wrong?

Thanks, that sounds more like it. I have been trying some functions like sin(e^x) which is continuous and has no limit, but I can't prove it is in L1. Is it in L1? Do I need to change the function in some way to get it into L1?

My professor said clearly that you can do it with a continuous function. With a discontinuous function, you could put in pieces with measure zero that would mess up the limit, while still being in L1... Any more ideas?

Homework Statement I am trying to come up with a continuous function in L1[0,infinity) that doesn't converge to 0 as the function goes out to infinity. Homework Equations I am trying to show an example of an f in L1[0,infinity) (i.e. ∫abs(f) < infinity) where the limit as the function...