Recent content by J$C

What they probably meant is since T(0)=0, if you can quickly see that 0 is not in the range then certainly it's not going to be a linear transformation because T(0)=something else. If T(0)=a and a,b are nonzero then b*T(0)=b*a, but if T is linear then b*T(0)=T(b*0)=T(0)=a \neqa*b. Or a more...

You have the right derivative, and the right critical values. To tell whether the function is at a max or a min at these points, look to see if the derivative is going from positive to negative (max) or negative to positive (min) at your critical points. Conveniently enough cos(x)*e^sin(x)...

Well it looks like Mark beat me to it. What he said!

Remember that when the volume is changing, the height and width are also changing. The size of the trough will be the same, but since the amount of liquid is changing the height of the liquid in the trough is a function, and so is the width. If you take some liquid out of the trough the width...

Looks good.

The key is that for matrices A and A', T(A+A') will have ones in the upper left and lower right entries by the nature of the transformation. T(A), and T(A') will also have that. So look at these entries and compare them on T(A+A') and T(A)+T(A').

Direct Proof: Assume A U B=A as you have. Then to show a set B is a subset of a set A the standard technique is to let x be in B, then show it is also in A. Notice if x is in B then it is clearly also in A U B and the conclusion follows from your initial assumption. Converse: If B is a...

Well the question asks if it is a linear transformation. Try to find a counter-example and you will see why you can't prove it. Notice the transformation always turns (1,1) and (2,2) into 1.

When I see a graph of a Fourier Transform, or something in the frequency domain, say band-limited from -\Omega to \Omega, I'm confused to what the interpretation is of the negative frequencies. Physically it would seem as though something considered in cycles/second for example, should be...

You're right, I was stuck in a continuous mindset I suppose. Thanks, good discussion.

So a function such as 1/sqrt(x) is considered a member of L1[0,1] even though the function is not defined at 0? And yes, Maze's infinite domain solution is quite slick.

Maze, I just looked at your example 1/sqrt(x) from 0 to 1. This of course makes sense. So to clarify my last post then, here we have an example on a finite interval in L1 but not in L2. I need to add the condition that my professor was referring to a closed interval i.e. L1[a,b]. Perhaps...

Very interesting discussion. I assure you it is not a homework question, although it did stem from a lecture. The professor seemed to imply it was a simple concept, which kept me from asking a question (It was also a bit off topic for that particular lecture). Of course with the...

So f(x)=1/x would fall into the L2 but not L1 category. I'm interested to see if there are functions in L1 but not in L2, or if L1 is a subset of L2.

It is clear that there are functions in L2 that are not in L1, but what about the other way? And what effect does considering L2(R) versus L2([a,b]) have? Thanks.