Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 Re: Help Me ! » Fascinating limit » 2014-01-24 09:59:47
Yes.
Is there a way to prove that that sequence is always lesser than 3 without e?
#2 Help Me ! » Fascinating limit » 2014-01-23 06:34:23
- Bezoux
- Replies: 3
Hi, everyone.
I had an interesting problem in class today. Find
Now, it's fairly obvious that the answer is 0, the question is how to prove it.
I had an immediate suggestion, obviously if
,
then the answer is 0. So, we had to prove that.
By definition,
Since both and M are positive, I can raise them to the degree of n, which is when I get , which is obviously correct for n>= a certain n0.
Now, my teacher told me this was wrong and the class ended before he could elaborate on why, and the test is next class, which worries me a bit. I do a lot of my proofs this way.
What she said is that we need to prove , after which we can easily use the squeeze theorem to solve the problem. Unfortunately, I've had no success in proving that statement.
Can you point out where I was wrong in my solution, or help me prove the teacher's statement? I keep getting stuck at , but the teacher doesn't want us to use e (or infinite geometric series, which would make the proof fairly straightforward by using the binomial formula).
Thank you in advance, I know I'm asking a lot.
#3 Help Me ! » A very peculiar surjection » 2013-12-18 05:07:52
- Bezoux
- Replies: 2
Hey everyone.
I have a very strange question that I know is true, but seems to be bugging me.
I define so that it doesn't include 0.
This function is obviously surjective, as in for any y from the codomain there is an x in the domain such that f(x)=y.
Is there a good way to prove it?
#4 Re: Help Me ! » polynomial » 2013-11-21 00:26:03
Here's my, in my opinion, a fairly elegant solution 
.
There might be a silly arithmetic mistake in there, but I think that was the idea.
#5 Re: Help Me ! » Polynomials and matrixes » 2013-11-20 10:56:40
Essentially, my first idea was to assume the opposite, i.e. that there is a polynomial such that p(A)=B, for example (I'd do the p(B)=A one separately).
Then,
I found a very useful link on finding the nth degree of a matrix (I can't link it unfortunately, but it's the first result that comes up when you google "finding the nth power of a matrix").
Still, something makes me think there's a much more elegant solution that I'm not seeing.
#6 Help Me ! » Polynomials and matrixes » 2013-11-17 10:50:07
- Bezoux
- Replies: 10
Hi everyone,
I've encountered a problem while studying matrices.
A={{2,1},{3,-1}}, B={{4,-2},{3,-1}}
Prove that there does not exist a polynomial with real coefficients such that p(A)=B or p(B)=A.
I've read up on eigenvalues, eigenvectors, characteristic polynomials and diagonalization, but nothing seems to be making sense, as the whole thing gets way too complicated for a high school problem.
I think there's a way to do this without using any of the aforementioned.
Can you help me out, please?
Pages: 1