Johannes Kloos

4,206 reputation

624

bio	website	mpi-sws.org
	location	Kaiserslautern, Germany
	age	31
visits	member for	1 year, 2 months
	seen	11 hours ago
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PhD student at MPI-SWS.

	4,206 reputation	bio	website	mpi-sws.org	visits	member for	1 year, 2 months
624 badges		location	Kaiserslautern, Germany		seen	11 hours ago

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May 15	comment	Complex Analyisis: Exponential Function Note that $e^{a+ib} = e^a (\cos b + i\sin b)$,hence $\|e^{a+ib}\| = e^a$. Do you know that $e^\cdot: \mathbb{R} \to (0, \infty)$?
May 15	comment	Complex Analyisis: Exponential Function As said in my earlier comment, you will only be able to show that $e^\cdot: D \to \mathbb C \setminus \{ 0 \}$ is bijective, since $0$ is not in the image.
May 15	comment	Complex Analyisis: Exponential Function In your notation, does one-to-one mean injective or bijective? In the latter case, this is not true: 0 is not in the image of the complex exponential function.
Apr 24	comment	how discrete mathematics is related to computerscience Just as an example: Graph Theory, as an area of discrete mathematics, is used all over the place. Since you mention complexity, it is interesting to note that many graph-theoretic problems are natural examples of NP-complete problems.
Apr 5	comment	Riemann-Stieltjes Integral is Everywhere Zero Hint: Use the fact that $f$ is not just integrable, but continuous. In particular, if there is some $x$ such that $f(x) > 0$, there is also an interval $I$ with $x \in I$ such that $f(y) > 0$ for all $y \in I$ (similar for $f(x) < 0$).
Mar 18	comment	What is true for a ring with exactly two right ideals As an example for a ring that satisfies II, but not I, consider the quaternions. For a finite ring that satisfies the property, take any finite field (e.g., $\mathbb Z/2Z$. As for proving II, I guess it should not be too hard to adapt the standard proof that a commutative ring with exactly two ideals is a field.
Mar 16	comment	Regular Languages Algorithm? You're almost there. Do you know the algorithm for computing the product of two automata?
Mar 12	comment	Monomial ordering deglex It is, in fact, that easy.
Mar 2	comment	Solving Differential equation with laplace transformation Hello and welcome to math.stackexchange.com. Since you are new here, I'd like to point out two things: 1. People don't like being addressed with imperatives here; please reformulate your question into an actual question format. 2. It's considered polite to describe not just the problem you are trying to solve, but also what you have tried and where you are stuck. This is particularly true if the question you ask stems from a homework problem (in which case it is also recommended to use the "homework" tag). About the question: Do you have a table of standard Laplace transforms available?
Mar 2	comment	Not a Zero Divisor While it is true that $x-x=0$, this is not what was asked here. Also, note that we are talking about general rings and not numbers: The notion $b>c$ might not even make sense in the given ring. For example, in $\mathbb Z/2\mathbb Z$, there is no ordering compatible with addition.
Mar 2	comment	Is there a notation for $f(x,y) - f(y,x)$? @vrich: And what does that have to do with the question?
Feb 28	comment	What is this physics simulation formula doing? It reminds me of a numerical ODE solver. It's similar to, but not quite the same as, a fourth-order Runge-Kutta method.
Feb 26	comment	Maximal ideals in polynomial rings Actually, this requires that the isomorphism is a $K$-algebra isomorphism (see Hurkyl's answer). I assume that this is the case here.
Feb 26	comment	Maximal ideals in polynomial rings Yes; I'll edit the answer accordingly.
Feb 24	comment	$\mid$ in simply typed lambda calculus If I interpret your question correctly, the $\mid$ simply means an alternative in the syntax: an expression $e$ is either a variable ($x$), a $\lambda$ term ($\lambda x: \tau. e$), a function application ($ee$) or a constant ($c$). See en.wikipedia.org/wiki/Backus%E2%80%93Naur_Form.
Feb 23	comment	solving $\sqrt{3-\sqrt{3+x}}=x$. They probably fall outside the interval, or are complex.
Feb 23	comment	solving $\sqrt{3-\sqrt{3+x}}=x$. The function $\sqrt{3-\sqrt{3+x}}-x$ is strictly decreasing on $[0,\sqrt{3}]$, so there's at most one zero (this is easy to check by differentiation). Since all solutions must lie in this interval, there can be at most one solution. I think that solving this equation without going to a fourth-order polynomial is impossible, but I'll try to check this now.
Feb 23	comment	solving $\sqrt{3-\sqrt{3+x}}=x$. What's the problem with guessing here, though? Anyway, it's easy to check that $x \in [0, \sqrt{3}]$; I'll try to derive further constraints from that.
Feb 23	comment	What graph is this? Isn't this a cartesian product of cycles?
Feb 20	comment	Polynomial root finding The "small enough" thing you can do as follows: As soon as you have seperated the zeroes into intervals (use Sturm's Theorem), it is guaranteed that there is a sign change inside every interval (because there are no repeated zeroes). You can then use, e.g., bisection to find a sub-interval in which Newton's method converges quadratically.