@JM Sorry if I'm boring you again.. If, in applying partial fraction, I have a quadratic numerator and a cubic denominator, I'd have to reiterate the partial fraction?
Related result is given here, they assume continuity of $f$.
Theorem 9.1.3: Let $f\colon [a,b]\to\mathbb R$ be a continuous function. If some of Dini derived number is non-negative (non-positive) for each $x\in(a,b)$, then $f$ is non-decreasing (non-increasing) on $[a,b]$.
This results is obtained as a corollary of the following (Theorem 9.1.2): Let $f$ be a continuous function on $(a,b)$. Then $$\sup_{x\in(a,b)} D^+ f(x)=\sup_{x\in(a,b)} D_+ f(x)=\sup_{x\in(a,b)} D^- f(x)=\sup_{x\in(a,b)} D_- f(x)=\sup_{\substack{x_1\ne x_2\\x_{1,2}\in(a,b)}} \frac{f(x_2)-f(x_1)}{x_2-x_1}.$$
I've copied it from a Slovak texbook I have here.
My translation, so ignore grammatical mistakes.
But I guess it would be possible to find another reference in English, if you have to.
I was thinking about how to explain semigroups to people.
Then I figured out it is an excellent method to justify the use of infinite dimensional spaces. I mean, our space is finite dimensional and then we use these seemingly esoteric things.
@MartinSleziak I do understand now. So basically you just keep stacking languages ad infinitum. Just like you would stack finite-dimensional spaces to get an infinite-dimensional one! 8-).
Is this an appropriate way of using the big-oh notation? 'all elements are of the form $e^{\lambda t} \, \mathcal{O}(t^k)$ for some eigenvalue $\lambda$'
I mean, is that a good way to say: all elements are some exponential of t times a polynomial of t of order k
@harmen: But actually that only means that the elements are bounded above by $Ce^{\lambda t}P(t^k)$ for some polynomial $P$ and sufficiently large $t$. If the elements are actually equal to $e^{\lambda t}P(t^k)$, then that is a stronger statement that you should make.
I have a question, and I don't want to post it in math.SE because the answer is probably very readily available.
It is in Linear Algebra
In my book, one theorem says that if we have a finite-dimensional linear space, then any linearly independent set with the same cardinality as a basis for this set will be a basis.
Can this theorem be generalized for infinitely-dimensional linear spaces?
Hi all, I was wondering if someone may be able to help me with a few questions regarding a Markov chain? I need clarification on what my initial state distribution is; and how to determine the number of steps that I need to take
Well, if you can separate the equation, they are not so heavily intertwined right? You can just write stuff as product of two functions each depending on some other variable.
So, what you actually do want to know are other properties of the solution like boundedness, maximal regularity and so on.
And you don't need a solution for that, and very often the solution you get in that basis it not so insightful and you can derive the similar properties directly from the equation.
@Eugene No? I mean that if you can solve it this way the equation is pretty useless or you will write a program to do it anyway. Takes like forever to evaluate an infinite series.
How to arrive at the following from given $ x = r\sin \theta \cos \phi, y = r\sin \theta \sin \phi, z=r\cos\theta $
$$ \begin{bmatrix}
A_x\\
A_y\\
A_z
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \cos \theta \cos \phi & -\sin\phi\\
\sin \theta \sin \phi & \cos \th...
