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bio website math.ntnu.no/~hanche
location Trondheim, Norway
age
visits member for 2 years, 11 months
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Dec
5
comment free abelian group
So what did you try? Please show your work and tell us where you got stuck.
Dec
5
comment N-th derivative of a expotential derivative
I suspect it is rather tricky to come up with a good formula for that. What one usually needs to know (for the most common use of this function) is that it has the form $p_n(1/x)e^{-1/x^2}$, where $p_n$ is a polynomial – so that the limit as $x\to0$ is zero. This is easy to prove by induction, along with a recursion formula for $p_n$.
Dec
5
comment Polynomial growth, using the Cauchy Integral Formula,
If $f$ is not assumed to be entire, then any rational function is a counterexample. So that really is an essential assumption.
Dec
5
revised Polynomial growth, using the Cauchy Integral Formula,
Much better this way. Also address the question more carefully.
Dec
5
answered Polynomial growth, using the Cauchy Integral Formula,
Dec
4
answered Prove $gcd(a,b)=gcd(a,2a+b)$
Dec
4
comment Prove $gcd(a,b)=gcd(a,2a+b)$
$b=(2a+b)-2a$ …
Dec
3
revised what can be said about the solution of the following differential equation
Fixed a sign mistake
Dec
3
answered what can be said about the solution of the following differential equation
Dec
3
comment what can be said about the solution of the following differential equation
I lost you at $y(x)\le f(x)$. How did that come about?
Dec
3
comment Is the union of nowhere dense sets which are separated by disjoint open sets nowhere dense?
@AsafKaragila Wonderful! Thanks.
Dec
3
comment What tools should be used to prove that a real function is one-to-one and onto?
Not sure how to answer this one. Tools? This is done with bare hands, no tools necessary. Except perhaps some elementary algebra, it's just applying the definitions of $f$, one-to-one, and onto. And knowing that $y=f(x)$ is equivalent to $x=f^{-1}(y)$.
Dec
3
comment Bounded polynomials are Lipschitz
If you restrict a polynomial to a bounded interval, however, then it is Lipschitz, because the derivative is also a polynomial, and hence bounded on any bounded interval.
Dec
3
answered Existence of integer $n > 2$ such that for any abelian group $G$ , $G_n:=\{e\} \cup \{a \in G :o(a)=n \} $ is a subgroup of $G$
Dec
3
comment Is $P(\bigcup_{k=1}^\infty A_k)=\lim_{n\rightarrow\infty}P(\bigcup_{k=1}^n A_k)$ equivalent to countable additivity?
… I forgot to mention $P(A)\ge0$ …
Dec
3
comment Is $P(\bigcup_{k=1}^\infty A_k)=\lim_{n\rightarrow\infty}P(\bigcup_{k=1}^n A_k)$ equivalent to countable additivity?
They are equivalent, if you assume finite additivity. Otherwise, most likely not. To elaborate: You mention the “third axiom”, but there are more than one way to write up the axioms. I am used to having just the one you gave together with $P(\emptyset)=0$ and $P(\Omega)=1$.
Dec
2
comment Why doesnt the sum of $\tan(1/n)$ converge?
To be pedantic, your inequality is nonsense, as it claims $\infty>\infty$. If you replace it by $\ge$, all is well.
Dec
2
comment If $e^{f(z)}$ is analytic then $f(z)$ is analytic
Hint: Any continuous branch of the logarithm function is analytic.
Dec
2
answered Is the union of nowhere dense sets which are separated by disjoint open sets nowhere dense?
Dec
2
comment Is the union of nowhere dense sets which are separated by disjoint open sets nowhere dense?
Do you have any thoughts? What have you tried?