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comment Prove that $\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)$
both sides are multiplicative functions. So it is enough to check the equality for prime powers
Jul
1
comment Resources about infinite primes of form $n^2 + 1$
Look at this post math.stackexchange.com/questions/44126/… The best partial progress for the polynomial values that are prime (in two dimensions though) is the work of Iwaniec and Friedlander en.wikipedia.org/wiki/Friedlander–Iwaniec_theorem
Jul
1
comment Resources about infinite primes of form $n^2 + 1$
the problem is open. You want resources that contain partial progress?
Jun
28
answered How prove this Stronger AM-GM inequality $\frac{n^2-1}{6}\min_{1\le i<j\le n}\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^2\le A_{n}-G_{n}$
Jun
27
revised Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?
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Jun
27
revised Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?
added 217 characters in body
Jun
27
revised Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?
added 300 characters in body
Jun
27
revised Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?
added 300 characters in body
Jun
27
answered Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$?
Jun
25
reviewed Approve suggested edit on Integrate $\tan^a(u)$ from 0 to $\pi/2$
Jun
25
comment Is there a sharper bound than exponential for $\sum_{k\ge0}\frac{m!(k+n-m)!}{(k+n)!}\frac{s^k}{k!}$?
my suggestion was basically the same to what Greg Martin mentioned: the Binomial coefficient essentially behaves in the same way as the expression I gave.
Jun
25
answered Prove that there exists $\xi \in (a,b)$, such that $f(\xi)+f'(\xi)=\xi + 1$
Jun
25
comment Prove that there exists $\xi \in (a,b)$, such that $f(\xi)+f'(\xi)=\xi + 1$
you might want to add some condition for the derivative... Say that it is continuous. As to the problem, if such point \ksi does not exist then we have strict inequality between two sides. You can then deduce the inequality for $f...$
Jun
25
comment Is there a sharper bound than exponential for $\sum_{k\ge0}\frac{m!(k+n-m)!}{(k+n)!}\frac{s^k}{k!}$?
If you want to keep everything precise, then the best way to estimate binomial coefficient is to apply Stirling formula. The cheap way to improve your bound is to put smth like $\frac{m!e^s}{ P(s)}$ where $P(s)$ is certain polynomial.
Jun
25
comment Limit of the sequence $ x_{n+1}=x_n+\sqrt{a+x_n^2}$
You are welcome!
Jun
25
answered Limit of the sequence $ x_{n+1}=x_n+\sqrt{a+x_n^2}$
Jun
25
comment Is there a sharper bound than exponential for $\sum_{k\ge0}\frac{m!(k+n-m)!}{(k+n)!}\frac{s^k}{k!}$?
That depends, what you are after. If you fix $m,n$ and think of $s$ being large, then can be more precise. The binomial coefficient is roughly $k^{m}$ times a constant for large $k...$
Jun
22
answered $f'$ is bounded and isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\lim_{x\to y}f'$ does not exist
Apr
3
reviewed Reject suggested edit on Do sections defined in different patches give the same element in an associated bundle?