that gives an easy proof of the divergence of the harmonic sum. pretty sure i've seen it before, but i don't remember the proof off the top of my head (aka it's a fine problem for me to get distracted by :p)
Theorem: Let $U$ be a bounded , open subset of $\mathbb{R}^n$ , and suppose $\partial{U}$ is $C^1$. Assume $1 \leq p<n$, and $u \in W^{1,p}(U)$. Then $u \in L^{p^{\ast}}(U)$ , with the estimate $||u||_{L^{p^{\ast}}}(U) \leq C ||u||_{W^{1,p}(U)}$, the constant $C$ depending only on $p,n$, and $U$....
Go ahead with the chapter 8 stuff. I maintain from earlier that if you're not being challenged you should supplement with Hirsch, doing most of the exercises (there are many and they are difficult).
I am a bit confused as how the expected value of a random variable differs from the the random variable itself when considering indicator functions. Say there is a geometric distribution. X means the number of the failures before success, E(X) means the expected number of failures.
Now I want to compute the Expected value given p= prob. of success and q otherwise.
Let c=E(X);
Now I do start-step analysis. c=0.p+(1+c)q
In this step, I don't understand the coefficient of q. In (1+c), 1 makes sense but why c?
Oh, so embedding manifolds in Euclidean space is much easier than Whitney. It's just gluing the graphs of the coordinate charts by a partition of unity. I should have realized that way earlier.
Sorry for asking again, does my question make sense. When computing the expected value: it is the kp^kq^k. So the coefficient is the value of the random variable. But in my example it is the expected value at the next step which confuses me.
You should post it as a question in MSE instead of asking multiple times as it is less likely that you'd get an answer and more likely that people will be annoyed by getting spammed by the same question getting asked so many times, even though you do mention that you are sorry for it.
I don't think any of us who are in the chat right now are capable of answering the stuff you are asking.
@AbhishekBhatia You need to take a step back and reevaluate what a random variable is.
That is where your confusion lies.
In your probability space, you have a set of elementary events $\Omega$. A random variable is a function $X : \Omega \to \mathbb R$. Why do we need this function?
One reason we need this function is to be able to describe measured relationships between events.
If you flip five coins, you can have two events $HHHTT$ and $HTHTT$. What does it mean for $HHHTT>HTHTT$? Absolutely nothing.
Now, if we construct the random variable $\text{Number of Heads} : \Omega \to \mathbb R$, we can do something with it.
$\text{Number of Heads}(HHHTT) = 3$ $\text{Number of Heads}(HTHTT) = 2$
@AbhishekBhatia Now, Random Variables are functions defined over a probability space. The expectation function is a function from Random Variables to numbers. Do you see how these are two completely different mathematical objects?
