Hello,I need some advice on a problem.
Let f,g:R\rightarrow R (where R denotes the real numbers) be two continuous functions, assume that f(x) < g(x) \forall x \neq 0 ,
and f(0) = g(0).Define A = \left\{(x,y)\neq (0,0): y< f(x),x \in R\right\}
B = \left\{(x,y)\neq (0,0): y> g(x),x \in...
for (i) let d = d(X,Y), and O = \bigcup_{x\in X}(x - d,x+d) so O\bigcap Y = empty set, and X \subseteq O then since O is measurable we get:
m(X\cup Y) = m((X\cup Y)\cap O) + m((X\cup Y)\cap O^{c}) = m(X) + m(Y).
forgot to mention the definition of rectifiable here: a (bounded) set S is rectifiable if
\int_{S} 1 exists. (so it has volume.)update: PROBLEM HAS BEEN SOLVED.
sorry that's only true if the E_{i} where increasing, but
lim_{i\rightarrow\infty}\left(\sum^{\infty}_{j=i}m(X_{j})\right) =lim_{i\rightarrow\infty}m(X_{i}) .
recall that if an infinite series converges you can make the remainder sum arbitrarily small.
let g:[a,b] -> R be a function that is continuous almost everywhere. assume that g(x) > 0 on [a,b]. Show that the set
S = { (x,y): 0 <= y <= g(x) , a <= x <= b} is rectifiable.
One way to attack it, is to show that S is bounded and boundary of S has measure zero. the problem I am having is...
I get that T(\dot{x}) = \frac{1}{2}m \dot{x}^{2}
U(x) = mgx ,
\frac{\partial L}{\partial x}= - m g
\frac{d}{dt} \frac{\partial L}{\partial \dot{x}}= m\ddot{x}
Lagranges equation implies m\ddot{x} + mg = 0
but I'm a little confused, if the chain is of infinite length, would it then...
are you referring to this equation:
\frac{\partial L}{\partial x_{i}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x_{i}}} = 0, i=1,2,3
where L = T - U is the lagrange function,
how can I use this to model my problem?
(ps. I am apologize if my physics is wrong, unfortunately I am a math...
Hello all, I'm having trouble with the following problem:
Pb: A chain with constant density and infinite length is slipping down from the table without friction. Determine the position of the tip of the chain at time t.
I know there are a few ways to approaching this problem, namely from...
Im having trouble showing that given a sequence of uniformly bounded C^1([a,b],R) functions,
the derivative sequence is uniformly bounded.
Any suggestions are helpfull
Define f: R^{2} \rightarrow R , by f(x,y) = \int^{sin(x sin(y sin z))}_{a} g(s) ds
where g:R -> R is continuous. Find the gradient of f.
I tried using the FTC, and differentiating under the integral, but did not get anywhere,
thanks for any suggestions.
Sorry for the discrepancy, the problem is to show that the supremum of |f(x)| over [0,1] is less than or equal to the integral from 0 to 1 of |f ' (x)|, where f ' (x) is the derivative of f.
I'm having trouble with this inequality:
let f be (real valued) continuously differentiable on [0,1] with f(0)=0, prove that
sup_{x\in[0,1]} \left|f(x)\right| \leq \int^{1}_{0}\left|f\acute{}(x)\right| dx
Thanks for any help.