Other thread reached the bump limit
SQT Stupid Question Thread
Cameron Sullivan
Grayson Stewart
Fucking expected value.
As far as I understand it, its the sum of each possible value of x times their respective probabilities, right?
So, is the expected value actually the mean value we can expect for any random iteration of whatever probabilistic process we're studying? Say we're calculating it based off a few previous rolls of a die. Is the expected value we get the most likely result of a subsequent roll based on the data we gathered?
Gabriel Brooks
When finding the derivative of 1/3x-1 or any 1/x-k problem, I get different answers when using the power rule, and then using the long method. Why is this?
For this example when using the power rule I get -1/(3x-1)^2, but doing the long method I get -3/(3x-1)^2. Why am I getting different results here?
Jayden Wood
The expected value is more like the mean, and has nothing to do with the most likely result of your next roll. Notably, the expected value usually is not even a possible outcome. For example, the expected value is 3.5 on the roll of a dice
Dominic Jenkins
What is the strongest set theory you can set up,
so that there being a surjection from N into every set is consistent
I.e.
[math] \forall X.\ \exists (f: {\mathbb N} \to X). \ \forall (x \in X).\ \exists (n \in {\mathbb N}). \ \, f(n) = x [/math]
The question might turn into the question which of the stronger axioms you need to drop
to not make the powerset of the naturals inherently uncountable.
What axioms rules out the Cantor argument being usable?
Aaron Cruz
You're not applying the power rule properly and fully. I guess what you consider the power rule is only valid for any a*x^k. Your problem is of the form f(x)^k, and as such, you need to apply the chain rule: k * f(x)^(k-1) * (d f(x) / d x).
Sorry, too lazy for markup.
Daniel Fisher
Ah thanks, I'm only on my first year of calculus and haven't gotten to the chain rule yet.
Oliver Butler
So if it is more like the mean, then whats the point? Why not just calculate the mean?
Christopher Perez
Expected value and mean are literally the same thing.
Mason Allen
[eqn]\left(\frac{1}{f(x)}\right)'=-\frac{f'(x)}{f(x)^2}[/eqn]
[eqn]\left(\frac{1}{3x-1}\right)'=-\frac{3}{(3x-1)^2}[/eqn]
------------------------------------------------------------------
[eqn](f(x)^n)'=nf(x)^{n-1}\cdot f'(x)[/eqn]
[eqn]((3x-1)^{-1})'=-(3x-1)^{-2}\cdot3=-\frac{3}{(3x-1)^2}[/eqn]