Recent content by MatthewD

This is the second part...

but there's no p in the second part of the problem...?

The entire expression, or just the integral part? Could I do: lim_{x\rightarrow b^-} \frac{1}{g(x)} \leq lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a F(t)g'(t)dt \leq lim_{x\rightarrow b^-} \int^x_a M g'(t)dt where the right hand side becomes =lim_{x\rightarrow b^-}...

i see it now, using parts helped a lot, thanks i'm suppose to follow it up with this addition: assume further that \int^b_a f(t)dt converges and show lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x f(t)g(t)dt = 0 i used parts again and came down to the fact: lim_{x \rightarrow b^-}...

When I tried parts, i let u= g(t) and dv = f(t)dt so du = g'(t)dt and v = F(t) and I got \displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} (g(x)F(x) -g(a)F(a)- \int^x_a F(t)g'(t)dt)} \displaystyle{= lim_{...

How would I show an integral is bounded? If I know obviously \int_a^x f(t)dt is bounded, how would I know anything about the g function. The information given basically denies that its bounded right? Since the limit is infinity? I'm so lost!

Homework Statement Suppose f is continuous and F(x)=\int_a^x f(t)dt bounded on [a,b). Given g>0, g'\geq 0 and g' locally integrable on [a,b) and lim_{ x\rightarrow b^-} g(x) = infinity. prove for p>1 \displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}...

Do I have to use Cayley-Hamilton? Could I use the fact that A would be orthogonally equivalent to a diagonal matrix by definition of symmetric, so for some orthogonal matrix Q and diagonal matrix D: A=Q*DQ then det(A)=det(Q*DQ)=det(D) D is diagonal=>det(D)=product of diagonal entries... but...

Is there anyway to show that for a symmetric or normal matrix A, that det(A) = \prod \lambda_i without using Jordan blocks? I want to show this result using maybe unitary equivalence and other similar matrices... any ideas? It's obviously easy with JCF...

Which is clearly a contradiction! I realized my mistake: at the end I kept using Rolle's to say there exists a point c where f'(c)=0 instead of looking at the new function. It seems so easy now... I'm sorry! Thank you for your time, I really appreciate it!

So this is how I've been starting, as you said, and I think I'm confusing myself... can't I only apply Rolle's Theorem if f(x1) = f(x2)? I have: suppose f'(x) not= 1 on [0,1] and suppose there exists 2 pts, x1 and x2 in [0,1] such that f(x1) = x1 f(x2) = x2 and let some function g be...

Thank you so much---it worked out better for me because I had to prove to myself that the case worked, which helped me understand so much better. Thank you!:smile:

I thought that 1/(x-1) was unbounded, and therefore not integrable? Am I mistaken? I'm still trying to understand this whole chapter... Would it work if I had f(x): = 0 g(x): = 1, x is rational g(x): = 0, x is irrational Because then f is constant ->integrable and fg=0 is...

Homework Statement I need to find 2 functions on [0,1]: one that is integrable, one that isn't, such that their product is integrable. I'd like to use functions that are non-constant, if possible Homework Equations The Attempt at a Solution I was thinking of using g(x) :=...

Homework Statement Let the function: f : I→ I be continuous on I and differentiable on the open set I for I := [0,1] Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t Homework Equations I know...