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Dec
28
comment What is tangent to a curve or function?
after all if tangency required that it not intersect at any point then all you'd be left with are the convex/concave functions. In fact you can't even restrict non-intersection to a local neighborhood, e.g. $y=x^3$ at $(0,0)$.
Dec
28
comment What is tangent to a curve or function?
I don't know what texts you're reading but the definition you give is non-standard as far as I'm concerned.
Dec
28
comment A bijective linear map with no inverse, how is it possible?
that's not what onto means, onto means that it hits every point in the co-domain, which it clearly doesn't.
Dec
28
comment A bijective linear map with no inverse, how is it possible?
bijective requires that a map be one-to-one and onto, your matrix $A$ is not onto since it doesn't send anything to points off the x-y plane. You can of course restrict the co-domain of an injective map to its range and turn it into a bijective mapping, which in the case of $A$ would be the identity function.
Dec
28
comment What is tangent to a curve or function?
it just means it doesn't cross at the point at which it is tangent. It may or may not cross at other points.
Dec
27
comment Find the condition that fourth degree equation $x^4+rx+s=0$ will have no real roots.
do you know calculus?
Dec
20
comment Understanding the random variable definition of Markov chains
If I understand you correctly you're talking about just looking at a particular path of the chain so the probability of the path would just be the product of each transition probability for that path. If there are $n$ states and you want to look at paths of length $N$ then I assume the (inherited) state space would be $\{1,...,n\}^N$ and the random variables would be $N$-variate, but I'll say no more because this is getting beyond my depth now.
Dec
19
comment Understanding the random variable definition of Markov chains
you can look at $(P^n)_{ij}$ for the probability that a path that starts in state $i$ will end in state $j$ after $n$ steps. If you want a state space on the set of all paths then that is probably going to be something like $\prod_{i=1}^{\infty}\Omega_{i}$. But a particular path in that state space won't be of the form $s_0,s_1,...$ it will be a sequence of states $s_{0j_0},s_{1j_1},...$
Dec
19
comment Understanding the random variable definition of Markov chains
Well when looking at it in its transition matrix form I would say viewing it as a sequence of pmf's is how it's viewed since that's what the transition matrix does, it takes one pmf to another pmf.
Dec
19
comment Understanding the random variable definition of Markov chains
@joshphysics, it's probably called the state space because you can think of it as the state space of the identity random variable $Y$ with the probability measure inherited via the original random variable $X$. Or you could just call $Y$ $X$ and forget about the original $X$ defined on $\Omega$
Dec
19
revised Understanding the random variable definition of Markov chains
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Dec
19
revised Understanding the random variable definition of Markov chains
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Dec
19
answered Understanding the random variable definition of Markov chains
Dec
17
awarded  Benefactor
Dec
17
comment Do the singular matrices form a topological manifold
thanks for expanding on Orangeskid's answer, if I could give you both the bounty I would. I don't fully follow what you wrote but I can sort of 'see' it, I'll probably return to your answer and try to fully understand it once I have some more geometry and topology under my belt.
Dec
17
accepted Do the singular matrices form a topological manifold
Dec
16
revised Do the singular matrices form a topological manifold
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Dec
16
comment Optimization problem involving rectangles [Calculus 1]
$x=-\frac{2}{5}y + 1020$
Dec
15
comment Uniform convergence - fancy way of saying sequence of functions is Cauchy?
Any convergent sequence (of points or functions, and for any choice of metric) is automatically Cauchy, however the reverse, that any Cauchy sequence converges, is only true if your space is complete.
Dec
15
comment Uniform convergence - fancy way of saying sequence of functions is Cauchy?
As user160738 said, uniform convergence is really just convergence in your function space when you give it the $L^{\infty}$ norm. I think the point is that while looking at pointwise convergence is sometimes useful, it unfortunately doesn't correspond to convergence in a normed function space.