(I assume, I never actually proved that statement :)
If the sum can be infinite, then it is certain, obviously
You might try putting together two odd functions whose absolute values are periodic.
@Ethan Actually, it is even easier than what I said: $$f_1(x)=\begin{cases} 0\text{ if }x<0\\ \sin x\text{ if }x\ge 0\end{cases}$$ and $$f_2(x)=\begin{cases}\sin x\text{ if } x<0\\ 0\text{ if } x\ge 0\end{cases}$$
@Ethan But if the summands are periodic, I would suspect they all have the period of their sum or better.
and to all the C/C++ master race arians: have a look what a lot of supposed C/C++ numerics libraries use - oh, is it automatically translated Fortran 77 code (netlib/BLAS/LAPPACK) - i'm in tears
for example this guy maths.ox.ac.uk/contact/details/trefethen is MathWorks gifting him expensive beach-resort trips every second week, so that he is a total matlab shill the rest of the time?
suppose i have an ODE f(y'',y',y,x) = 0 . Does the domain of x where f is defined have anything to do with the domain in which a soluction y = f(x) satisfies the equation?
suppose i have an ODE f(y'',y',y,x) = 0 . Does the domain of x where f is defined have anything to do with the domain in which a soluction y = f(x) satisfies the equation? Knowing a priori the domain of x where f(y'',y',y,x) is defined, do i know in advance something about the domain of the solution y = f(x) that satisfies it
@PeterSheldrick What the hell? Java was till very recently (1-2 years ago) was really slow as compared to C++. Also, parallelism support is even now very lagging in Java. C++ use FORTRAN only because people who wrote FORTRAN optimized it to the last details and FORTRAN works with C++ well. Its more of already invested effort in FORTRAN rather than something else. (Almost no one uses the translated libraries because they do not have a lot of functions). Still, there are a lot of new libraries that have come up in C++ which do not have any LAPACK/BLAS Baggage.
suppose i have an homogeneous function f(x,y) , how is the fact that f(mx,my) = m^n f(x,y) identical to the fact that f(x,y) can be regarded as a function of y/x ?
@Lord_Farin Hmm. Given a complex power series $A(z)$ as above and a real power series of nonnegative coefficients $p_n$ we write $$A(z) \ll P(z)$$ if $|a_n|\leq p_n$ for each $n$.
I am asked to show that if $A(z)\ll P(z)$ and $A^*(z)\ll P^*(z)$ then $A+A^*\ll P+P^*$ and $AA^*\ll PP^*$ but I don't know what the star means.
n choose k is the number of images for an injective function [k]->[n], which is less than the number of injective functions [k]->[n], which is less than the number of functions [k]->[n]
@PeterTamaroff that one I am actually not qualified to tell if it's a real question. I wouldn't be surprised if it did make sense, given there are many vague and mysterious connections between zeta and physics
but you said you installed it? where did you put it? do you know what a bookmark is? what internet browser are you using? does it have a bookmarks bar?
Write $A(z)\ll P(z)$ whenever $A(z)=a_0+a_1z+\cdots$ is a series of complex coefficients and $P(z)=p_0+p_1z+\cdots$ is a sequence of real nonnegative coefficients and $|a_n|\leq p_n$
I see two strategies to attempt first on the agenda: multiply out as $z(1-z)f'(z)=(1+z)f(z)$, then compare coefficients of both sides, or consider the logarithmic derivative of $f(z)$ (it is of the form $1+O(z)$) multiply by $z$ then compare directly to $(1+z)/(1-z)$
@blob you can scroll up the page to where it was linked to you earlier
In the sentence "Every subgroup of an abelian group is normal and gives rise to a quotient group", what does "gives rise to a quotient group" mean? I have an idea of what a quotient group is, but I suppose I'm more confused with the "gives rise to".
@AlanH Given a group $G$, a congruence on $G$ is an equivalence relation $\equiv$ such that $a\equiv b$ and $a'\equiv b'$ imply $aa'\equiv bb'$.
In particular, given a group $G$, there is a one to one correspondence between the possible congruences on it and its normal subgroups.
The correspondence is as follows: If $N$ is a normal subgroup, we define $a\equiv b\mod N$ if $a^{-1}b\in N$. And if $\equiv$ is a congruence in $G$, then the class $\hat 1=N$ of the identity is a normal subgroup, and all other classes are $\hat a=aN=Na$, cosets of the class $\hat 1=N$.
I'm reading the help center of stackexchange, I'm wondering about this line:
"You should only ask practical, answerable questions based on actual problems that you face. "
I'm not sure this really applies to MSE, as it seems like there many questions asked because of curiousity, instead of an actual problem that is being faced.
Yeah, I think this sentence is as vague as it can get: "You should only ask practical, answerable questions based on actual problems that you face. "
What is meant with practical ? Mathematics is in essence not practical, but theoretical. Some questions aren't know to be answerable.. (in mathematics) .. :p
@AlanH if the equivalence relation has absolutely nothing to do with the group structure then what's the point of putting it on a group?
"things invariant under group actions" is a ubiquitous thing-to-check throughout mathematics
invariant under left actions gives you a left coset space, invariant under right actions gives you a right coset space, invariant under both gives a left or equivalently right coset space where the subgroup is normal
Usually people say: "Hi!" we reply: "Hi", and then the guy asks: "Is it true that the conformity of bananolitic limits is dense on the set of real numbers?!"
