Counterpoint: ##\mathbb{Q}\subset \mathbb{R}##, is a 1 dimensional vector space over itself but it's not a subspace of the one dimensional real vector space ##\mathbb{R}##. I would argue the addition is the same :)
You can kind of just compute this exactly. For any possible choice of x, you know what fraction of the time your sampling will return a 0 instead of a 1. Then you can compute things like what value of x makes it so you would only see at least as extreme a result as you got 5% of the time (in...
I suspect it's a typo and the 3 in the denominator is supposed to be gone.
Edit: actually I take it back. Let's say ##M_k>N_k##. Then where ##M_k## is realized, ##f_N## is within ##\epsilon/(3(b-a))## of ##M_k## and is guaranteed to not be larger than ##N_k##, showing the inequality in your...
The Landau case, ##\frac{\partial L}{\partial v^2}## is the same thing, you could let ##x_1=v^2## and you're just doing the chain rule. ##\frac{\partial L}{\partial v^2}=\frac{\partial L}{\partial x_1}## the ##v^2## in the denominator is the author's way of just making sure you don't compute...
Yeah, this is very abused notation. I think it helps to separate the variables a bit. Let ##f=f(x_1)##, and let ##x_1=y+\alpha \eta##. Then it's really ##\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial \alpha}##. Unfortunately people don't want to need 2n variables for an n...
The simple example you are looking at is just the chain rule. ##\frac{\partial f}{\partial \alpha} =\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}## (where in this case the partial derivatives are full derivatives)
I think the simplest thing to consider is just like, consider the curve sketched out by##(t,t^2)## compared to ##(t^3,t^6)##.
Both of them traverse the parabola ##y-x^2=0## but the behavior of a particle traveling according to each of the two functions of t is pretty different. For example...
If you subtract ##A\cap C## from the left side of everything, and ##B\cap D## on the right side, then you have reduced to the case of no intersection as long as you can show the following result:
If ##B\subset A##, ##D\subset C##, ##D\approx B##, ##A\approx C##, then ##A-B \approx C- D##
A and B are matrices you know, and ##\,omega## is a number that you know, so you can plug them all in and you get like like ##Cx=0## for some matrix ##C## that you know. You might at first guess that gives you two equations and two unknown (##x## is a vector of 2 dimensions, the 0 on the right...
I find it a bit weird to suggest someone compute the derivative of cos(x) as a substep in their struggle to compute the derivative of cos(x). I don't think the suggestion is really that appropriate - either we know the derivative of cos(x) is sin(x) and we should use that fact directly, or we...
I think if a Hilbert space has a countable Schauder basis, then it has a countable Hilbert basis. You can basically just apply Grahm Schmidt on the Schauder basis and you get a set with the same span, and everything is orthonormal.
I suspect the same is true (cardinality of Schauder and...