hopefully peter will figure that out too. myself I'm tempted to start doing hw de-tagging to help drive him off the first active page, except that puts other questions in the crossfire.
probably the best thing we could do right now is try to give an actual answer to the question that was posted. as it is that person's just caught in the middle.
@IanMateus There is also a Samta Claus 3. I have notified a Community Manager. I have to go out for a while, but let me know if that account posts spam, too.
glad to hear that. it is almost impossible to improve a question with a stubborn OP with nonstandard terms and author wanting examples for everything.
@blue, so, have you thought of a proof without Hilbert's irreducibility theorem? I wrote in the comments that I suspect one proof goes by noting that $x^n - x - 1$ has galois group the whole symmetry group.
Btw i'll share that IMO problem in case you haven't seen it yet : Prove that you can choose $2k$ numbers from the set $\{1, 2, 3, …, 3k−1\}$ in such a way that the chosen set contains no averages of any two of its elements.
It might even possible (?) to choose $(3^k-1)/2$ of those elements (except for k=1)
Urm i'm seeing some false problem here, where am I wrong ? (typing the problem below)
sorry my connection went down
I saw $\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} = \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(\left\{ u \right\}\right) \: \frac{\mathrm{d} u}{u(u-1)} $
But if i take $u=1/x$ then for me $du=d(1/x)=-\frac{dx}{x^2}$
$$\lim_{n\to\infty}\frac{\displaystyle \binom{4n}{2n}}{\displaystyle 4^n \binom{2n}{n}}$$ No need for Stirling, no need for Gamma function or other special functions.
No need for central binomial coefficient approximation.
hello, I was helping my sister with her algebra class, and the professor in the conference said that 0^0 = undefined, I learned in college many years ago that the convention is 0^0 = 1, I would like tell the professor that he has to update his knowledge, but where can I find a valid reference to prove the convention, Is there any some of official guide or standard that I can show to the professor
Now I got a doubt (-3)^2 + 0^0 where 0^0 is undefined, how can I handle that sum, for example in programming undefined throw an error undefined is <> 0
I'm still having trouble with $\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(\1/x}\right) \frac{ \mathrm{d}x}{1-x} \\ =\int_{k}^{k+1} f \left(u\right) \: \frac{\mathrm{d} u}{u(u-1)}$ :/
I believe we do integration by substitutuin 'reversed' but
It was $\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(1/x\right) \frac{ \mathrm{d}x}{1-x} =\int_{k}^{k+1} f \left(u \right) \: \frac{\mathrm{d} u}{u(u-1)}$
If i have $\phi(x)=1/x$ then $\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(1/x\right) \frac{ \mathrm{d}x}{1-x} =\int_{\phi(k+1)}^{\phi(k)} f \left(\phi(x)\right) \frac{ \mathrm{d}x}{1-x}$
@TedShifrin not lately. I'm trying to relax a bit before the year starts.
:-)
@AlexanderGruber Actually have you thought about staying away from the gym completely, giving your entire body a rest; until your neck starts feeling better?
I have found that injuries start to heal faster that way.
@Chris'ssis If it's that easy, can you make a one line solution ?
@Chris'ssis I gtg now, if you can find an (easy) way to prove the result it would be great (it's from sos440's blog sos440.tistory.com/189). If you don't have the time it's ok, but if you do find something for me ping me here so that i get notified on MSE's banner :)
I have no reason to believe that this indicates some deep connection between the divisor function and the sigma function. What you are saying is essentially that $\sigma_{-1}(n)$ and $H_n$ "looks alike" (one is the divisor sum of reciprocal of integers upto $n$ and one is just the usual sum)
@robjohn It does, I think. If $k$ doesn't divide $n$, the term in the parenthesis tends to $\infty$ for $s\to 0$. The reciprocal thus tends to $0$ then.
In fact, we can start out with $$\lim_{n\to\infty} \frac{(2n+1)(2n+3)\cdots (4n-1)(4n+1)}{(2n)(2n+2)(2n+4)\cdots (4n)}$$ and to compute this limit we may square it and then use the inequality $n(n+2)<(n+1)^2$
@Chris'ssis The way I did the other problem was by writing $$\frac{\left(\frac{n+2}{3}\right) \left(\frac{n+2}{3}+1\right) \left(\frac{n+2}{3}+2\right) \dots\left(\frac{n+2}{3}+n-1\right)} {\left(\frac{n+1}{3}\right) \left(\frac{n+1}{3}+1\right) \left(\frac{n+1}{3}+2\right) \dots\left(\frac{n+1}{3}+n-1\right)}$$
noting that by increasing $n$ by $3$, the denominator is increased by a factor of $4$
@robjohn actually, that technique is pretty old and it was developed many years ago by a Romanian mathematician. There is a generalization related to these limits.
